Class 8 Maths - Practical Geometry

Construct a rectangle $ABCD$ given
$AB=6\ cm, BC=5\ cm$

We know that in rectangle measure of each angle is

$90^o$ and opposite sides are equal.

$\Rightarrow$ Draw a line segment

$AB=6\,cm$ .

$\Rightarrow$ Construct angle of

$90^o$ at points

$A$ and

$B$

$\Rightarrow$ Cut off

$AD=5\,cm$ and

$BC=5\,cm$

$\Rightarrow$ Join

$CD$ .

1254659_1162483_ans_ce42015623444986a7bc07d84ea7299f.jpeg

Construct the quadrilaterals with the measurements given:
Square $LUDO$ with $LU=4.5cm$ .

Steps of construction :

Step 1. Draw a line segment

$LU= 4.5$ cm horizontally.

Step 2. Construct

$\angle LUX=90^\circ$ at point

$U$ .

Step 3. On the line

$UX$ , make

$UD=4.5$ cm.

Step 4. With

$L$ as centre, draw an arc of radius

$=4.5$ cm.

Step 5. With

$D$ as centre, draw an arc of radius

$=4.5$ cm to cut the previous arc at point

$O$ .

Step 6. Join

$LO$ and

$DO$ .

Hence,

$LUDO$ square has been constructed.

1690524_568576_ans_8a4da5d8c1b34146b0d9838be88ce6c2.jpg

Construct a quadrilateral $PQRS$ where $PQ =4\ cm,QR=6\ cm,RS=5\ cm. PS=5.5\ cm$ and $PR =7\ cm$ .

1. Draw a line segment

$PR = 7\ cm$ .

2. With

$4\ cm$ as radius, draw an arc from

$P$ .

3. Similarly, with radius as

$6$ cm from

$R$ , cut the arc just drawn.

4. They will intersect at

$Q$ . Join

$PQ$ and

$QR$ .

5. With

$5.5\ cm$ as radius, draw an arc from

$P$ . Similarly, with radius as

$5\ cm$ from

$R$ , cut the arc just drawn.

6. They will intersect at

$S$ . Join

$PS$ and

$RS$ .

$PQRS$ is the required quadrilateral.

Construct a rectangle ABCD, when:

Its sides are 6.0 cm and 7.2 cm.

Draw the base

$AB = 7.2 cm$

Construct right angles at both A and B and draw vertical lines along it.

Using a compass, cut arc of length

$6 cm$ . Mark the points of intersection as C and D.

Join ABCD. It is the required rectangle.

Construct a quadrilateral $\text{ABCD}$ , when: $\text{AB} = 3.2\ \text{cm}, \text{BC} = 5.2\ \text{cm}, \text{CD} = 6.2\ \text{cm}, \text{DA} = 4.2\ \text{cm}$ and $\text{BD} = 5.2\ \text{cm}$ .

Steps of Construction :

$\Rightarrow$ Draw a line segment

$\text{AB = 3.2 cm}$ .

$\Rightarrow$ With A as a center and radius

$\text{AD = 4.2 cm}$ draw an arc.

$\Rightarrow$ Then with B as a center and

$\text{BD = 5.2 cm}$ radius draw an arc intersecting the point at D.

$\Rightarrow$ Again B as a center and radius

$\text{BC = 5.2 cm}$ draw an arc.

$\Rightarrow$ Then with D as a center and

$\text{CD=6.2 cm}$ radius, draw an arc intersecting the point at C.

$\Rightarrow$ Join

$\text{BC, CD, DA,}$ and

$\text{BD}$ .

Thus,

$\text{ABCD}$ is the required quadrilateral.

Construct a rhombus ABCD, given it's diagonal AC = 6.3 cm and diagonal BD = 5.8 cm.

We know that, diagonals of a rhombus are perpendicular bisectors of each other.
Using the same property, follow the below steps to construct a rhombus ABCD:

1. Draw $AC = 6.3 \ cm$

2. Draw perpendicular bisector XY of AC intersecting AC at O.

3. Taking O as centre and $\dfrac {1}{2} \times 5.8 = 2.9 \ cm$ as radius,
draw two arcs intersecting OX and OY at points B and D respectively.

4. Join AB, BC, CD and AD.

The figure ABCD is the required rhombus.

Construct a square ABCD, given it's one diagonal = 4.5 cm

A square has diagonals perpendicular to each other and are equal

1. Draw a rough diagram and mark the given points

2. Draw a line segments $DO=4.5 cm$

3. Construct a perpendicular bisector XY of DO.XY intersect DO at P and DP

$=PO=2.25 cm$

4.With D as centre and radius $2.25 cm$ draw arcs cutting XY at G and L

.

5.Join DL,DG,OL,OG.

6.GOLD is the required square

By constructing a rectangle $WXYZ$ of base = $4.8 \, cm$ and height $= 5\ cm$ , find the square root of its perimeter (where perimeter is in cm) upto the nearest integer.

Fist we draw a line YZ=4.8 cm

Make angle 90 at point Y ,And draw a line cut at point Y, YX =5 cm by compass

After that with compass make cut from point X ,4.8 cm and cut at point Z , 5 cm both cuts meeting at point W

Join W from X and Z we get rectangle XYZW

The perimeter of rectangle =

$2(4.8+5)=2\times 9.8=19.6 cm$

Then perimeter of square =19.6 cm sq cm

Then square root of perimeter of square =

$\sqrt{19.6}=4.42\approx 4$

erimeter is equal to $24\ cm.$

Let the length of the side of the square is equal to

$a.$

Then,

$Perimeter=4a$

$24=4a$

$a=6\ cm$

Steps of construction:

First draw a line segment $BC=6\ cm.$
Draw right angle at point $B$ and $C$ and extend it as $BX$ and $CY.$
Make an arc at a distance $6\ cm$ from $B$ intersecting $BX$ at $A.$
Make an arc at a distance $6\ cm$ from $C$ intersecting $CY$ at $D.$
Joint $A$ and $D.$

$ABCD$ is the required square.

Construct a quadrilateral $MORE$ where $MO=6\ cm,OR=4.5\ cm ,\angle M =60^o, \angle O=105^o$ and $\angle R=105^o$ .

Steps of construction :

$\Rightarrow$ Draw line segment MO = 6cm.

$\Rightarrow$ At M and O, construct

$\angle$ 60

$^\circ$ and

$\angle$ 105

$^\circ$ , respectively.

$\Rightarrow$ Cut off OR = 4.5cm.

$\Rightarrow$ Construct an angle of 105

$^\circ$ at R and let it meet ME at E.

$\Rightarrow$ Join RE.

Thus, MORE is required quadrilateral.

Construct a square $ABCD$ , given it's one side = 4.5 cm.

Draw the base $AB = 4.5 cm$ using a ruler.
Construct right angles at both points $A$ and $B$ and draw vertical lines along it.
Using compass, mark an arc of length $4.5 cm$ on both the vertical lines from points
$A$ and $B$ respectively.
Mark the points as $D$ and $C$ respectively.
Join $ABCD$ which is the required square.

Construct a quadrilateral $ABCD,$ given that $BC=4.5\ cm ,AD=5.5\ cm$ and $CD=5\ cm$ , the diagonal $AC=5.5\ cm$ and the other diagonal $BD=7\ cm$ .

Steps of construction are given below :

$\Rightarrow$ Draw line segment BC of length 4.5cm.

$\Rightarrow$ With C as a center and radius CD = 5cm draw an arc.

$\Rightarrow$ With B as a center and radius BD = 7cm draw an arc intersecting the previous arc at D.

$\Rightarrow$ Again with C as the center and 5.5cm as a radius draw an arc.

$\Rightarrow$ Then with D as the center and 5.5cm as radius, draw an arc intersecting the point at A.

$\Rightarrow$ Join AB, AD, CD, BD, and AC.

Thus, ABCD is a required quadrilateral.

Draw a square WXYZ with side $5.2$ cm.

As we know that square has all side equal and each angle is

$90^0$

$\underline step:1$ draw a line

$WX$ of 5.2cm

$\underline step:2$ make angle of

$90^0$ from point

$W$ and

$X$

$\underline step:3$ cut an arc of

$5.2$ cm from point

$W$ and

$X$ and give the cut point

$Y$ and

$Z$ .

$\underline step:4$ meet the point

$YZ$ .

Construct a quadrilateral $NILE$ such that $NI=2.5\ cm , IL=3.5\ cm , NL= 4\ cm , IE=5\ cm , EL=3\ cm.$

$\Rightarrow$ Draw a line segment NI = 2.5cm.

$\Rightarrow$ With N as a center and radius NL = 4cm, draw an arc.

$\Rightarrow$ With I as a center and radius IL = 3.5cm, draw an arc intersecting the previous arc at point L.

$\Rightarrow$ Again take I as a center with radius IE = 5cm, draw an arc.

$\Rightarrow$ With L as a center and radius LE = 3cm, draw an arc intersecting the previous arc at point E.

$\Rightarrow$ Join NL, IL, IE, LE and NE.

Construct a quadrilateral $LIFT$ , when:
$LI=4\ cm, IF=3\ cm ,TL =2.5\ cm ,LF=4.5\ cm ,IT=4\ cm$ .

$\Rightarrow$ Draw a line segment LI = 4cm.

$\Rightarrow$ With L as a center and radius LF = 4.5cm draw an arc.

$\Rightarrow$ With I as a center and radius IF = 3cm draw an arc intersecting the previous arc at point F.

$\Rightarrow$ Again take L as a center with radius LT = 2.5cm, draw an arc.

$\Rightarrow$ With I as a center and radius IT = 4cm, draw an arc, intersecting the previous arc at point T.

$\Rightarrow$ Join LF, IF, LT, IT and FT.

Thus, LIFT is a required quadrilateral.

Construct a quadrilateral $PQRS$ such that $PQ=3.5\ cm ,QR=4.5\ cm, RS=4\ cm, \angle Q=110^o$ and $\angle R =80^o$ .

$\Rightarrow$ Draw a line segment QR = 4.5cm.

$\Rightarrow$ Construct angles 110

$^\circ$ and 80

$^\circ$ at points Q and R.

$\Rightarrow$ Cut off PQ = 3.5cm.

$\Rightarrow$ Cut off RS = 4cm.

$\Rightarrow$ Join PS.

Thus, PQRS is a required quadrilateral.

Construct a quadrilateral $JUMP$ , when: $JU=3.5\ cm, UM =4\ cm, MP=5\ cm, PJ=4.5\ cm, PU=6.5\ cm$

First draw a rough sketch

Steps of construction:

$1$ . Draw diagonal PU of length

$6.5$ cm.

$2$ . Taking

$4.5$ cm as radius & P as center, draw an arc on top of PU.

$3$ . Taking

$3.5$ cm as radius & U as center, draw an arc on top of PU.

$4$ . Taking

$5$ cm as radius & P as center, draw an arc on bottom of PU.

$5$ . Taking

$4$ cm as radius & U as center, draw an arc on bottom of PU.

$6$ . Let J be the point where arcs intersect on both of PU & M be the point where arcs intersect below PU.

$7$ . Join JP, JU, MP, MU & label the sides.

1009820_425706_ans_6c208829490541b1a80dc8d0c7fc344c.jpg

Construct a quadrilateral $ABCD$ such that $AB=3\ cm , BC=4\ cm ,$ $\angle A=90^O$ , $\angle B=65^o$ and $\angle C=110^o$ .

$\Rightarrow$ Draw a line segment BC = 4cm.

$\Rightarrow$ At B and C, construct

$\angle$ 65

$^\circ$ and

$\angle$ 110

$^\circ$ , respectively.

$\Rightarrow$ Cut off BA =3cm.

$\Rightarrow$ Construct an angle of 90

$^\circ$ at A and let it meet CD at D.

Thus, ABcD is required quadrilateral.

Construct a quadrilateral $ABCD$ with $AB=6\ cm, BC=3\ cm, CD=8\ cm,$ $\displaystyle \angle B={ 60 }^{ \circ },\angle C={ 95 }^{ \circ }.$

1. Draw a line segment AB
2. Draw a line such that angle =

$\displaystyle \angle B = 60˚$
3. With that point as a centre and radius

$=3\ cm$ , draw an arc to cut that line at third side.
4. Draw next line which makes an angle

$\displaystyle \angle C$ with the previous line segment.
5. Mark last point on that line segment.
6. Join them, you will get the required quadrilateral.

Construct a quadrilateral $ABCD$ , with $BC=5$ cm, $CD=9$ cm, $DA=6$ cm, $\displaystyle \angle C={ 60 }^{ \circ },\angle D={ 75 }^{ \circ }$

1. First draw a line segment

$CD = 9$ cm.

2. Make

$\angle C= 60^{0}$ at point

$C$ and

$\angle D= 75^{0}$ at point

$D$ .

3. Open compass

$5$ cm and mark curve at angle line from point

$C$ this cut at point

$B$ .

3. After open compass

$6$ cm and mark curve at angle line from point

$D$ this cut at point

$A$ .

5. Join

$A$ and

$B$

Hence,

$ABCD$ is required quadrilateral.

Construct a rectangle $ABCD$ with $AB = 8\ \mathrm{ cm}$ and $BC = 6\ \mathrm{cm}$ .

We have been given the sides of rectangle which are

$AB = 8 \ \mathrm{cm}$ and

$BC = 6\ \mathrm{cm}$ .

We know that opposite sides of rectangle are equal.

$\implies AB = DC =8\ \mathrm{cm}$ and

$BC = AD = 6\ \mathrm{cm}$

$\textbf{Steps of construction :}$

$1.$ Draw a line segment

$AB = 8\ \mathrm{cm}$

$2.$ Since, we know all angles of rectangle are

$90^\circ$ then construct an angle of

$90^\circ$ at

$A$ and

$B$ .

$3.$ Cut off

$AD$ and

$BC$ at

$6\ \mathrm{cm}.$

$4.$ Join

$CD$ .

Thus, we get the required rectangle

$ABCD$ .

Construct a quadrilateral $PQRS$ , with $PQ=3$ cm, $QR=9$ cm, $RS=5$ cm, $\displaystyle \angle Q={ 45 }^{ \circ },\angle R={ 60 }^{ \circ }$

First draw a line segment

$QR = 9$ cm.

Make

$\angle Q= 45^{0}$ at point Q and

$\angle C= 60^{0}$ at point

$R$ .

Open compass

$3$ cm and mark curve at angle line from point

$Q$ this cut at point

$P$ .

After open compass

$5$ cm and mark curve at angle line from point

$R$ this cut at point

$S$ .

Join

$P$ and

$S$ .

Hence,

$PQRS$ is required quadrilateral.

Construct a quadrilateral $ABCD$ in which $AB=6$ cm, $BC=8$ cm, $CD=5$ cm, $AD=4$ cm and diagonal $AC=4.5$ cm

$\Rightarrow$ First draw a line segment $BC=8$ cm.

$\Rightarrow$ Open compass

$4.5$ cm and mark curve at point

$C$ .

$\Rightarrow$ And open compass

$6$ cm and mark curve at point

$B$

$\Rightarrow$ These both curve cut at point

$A$ .

$\Rightarrow$ Join

$A$ to

$B$ and

$A$ to

$C$ .

$\Rightarrow$ Mark an other curve

$5$ cm from point

$C$

$\Rightarrow$ And

$A$ curve

$4$ cm draw from point

$A$ .

$\Rightarrow$ These both curve cut each other at point

$D$ .

$\Rightarrow$ Join

$D$ to

$A$ and

$D$ to

$C$

$\Rightarrow$ Hence,

$ABCD$ is the required quadrilateral.

Construct a quadrilateral $ABCD$ with $AB=6\ cm, BC=8\ cm,$ $\displaystyle \angle A={ 75 }^{ \circ },\angle B={ 90 }^{ \circ },\angle C={ 45 }^{ \circ }$ .

$\Rightarrow$ Draw a line segment AB = 6cm

$\Rightarrow$ Construct angles of 75

$^\circ$ and 90

$^\circ$ at points A and B.

$\Rightarrow$ Cut off BC = 8cm.

$\Rightarrow$ Construct angle of 45

$^\circ$ at point C in such way that it will cuts the arm of angle 75

$^\circ$ at point D.

$\Rightarrow$ Join AD.

Thus, ABCD is a required quadrilateral.

Construct a quadrilateral $PQRS$ with $QR=8$ cm, $RS=4$ cm, $SP=9$ cm, $PQ=5.5$ cm and $QS=6$ cm.

First draw a line segment

$QR=8$ cm.

Open compass

$4$ cm and mark curve at point

$R$

And Open compass

$6$ cm and mark curve at point

$Q$

These both curve cut at point

$S$ , then join

$S$ to

$Q$ and

$S$ to

$R$ .

Mark an other curve

$5.5$ cm from point

$Q$

And a curve

$9$ cm draw from point

$S$

These both curve cut each other at point

$P$ , then join

$P$ to

$S$ and

$R$ to

$S$ .

Hence,

$PQRS$ is the required quadrilateral.

Construct a quadrilateral $ABCD$ with $AB=6\ cm, BC=4\ cm, CD=7\ cm, AC=8\ cm$ and $BD=7\ cm.$

$\Rightarrow$ Draw a line segment

$AB = 6cm$ .

$\Rightarrow$ With

$B$ as a center and radius

$BC = 4cm$ , draw an arc.

$\Rightarrow$ With

$A$ as a center and radius

$AC = 8cm$ , draw an arc intersecting the previous arc at point

$C$ .

$\Rightarrow$ Again take

$B$ as a center and radius

$BD = 7cm$ , draw an arc.

$\Rightarrow$ With

$C$ as a center and radius

$CD = 7cm$ , draw an arc intersecting the previous arc at point $$D$4.

$\Rightarrow$ Join

$AC, BC, BD, CD and AD$ .

Thus,

$ABCD$ is a required quadrilateral.

Construct a quadrilateral PQRS with PQ=9cm, QR=5cm, RS=8cm, PR=10cm, QS=13cm.

1. Draw

$QR$ of length

$5\ cm$ .

2. Cut an arc of

$13\ cm$ from

$Q$ and

$8\ cm$ from

$R$ . Their point of intersection is

$S$ .

3. Cut an arc of

$9\ cm$ from

$Q$ and

$10\ cm$ from

$R$ . Their point of intersection is

$P$ .

4. Join

$PQ, PS, SR, QR$ .

$PQRS$ is the required quadrilateral.

Construct a quadrilateral $PQRS$ with $PQ=7$ cm, $QR=5$ cm, $RS=3$ cm, $ST=8$ cm and $PR=5$ cm

First draw a line segment

$QR=5$ cm.

Open compass

$3$ cm and mark curve from point

$R$

And Open compass

$8$ cm and mark curve from point

$Q$

These both curve cut at point

$S$ , then join

$S$ to

$Q$ and

$S$ to

$R$ .

Mark an other curve

$5$ cm from point

$Q$ and a curve

$9$ cm draw from point

$R$

These both curve cut each other at point

$P$ , then join

$P$ to

$S$ and

$P$ to

$Q$ .

Hence,

$PQRS$ is the required quadrilateral.

Construct a quadrilateral $ABCD$ in which $AB=5\ cm, BC=6\ cm, CD=4\ cm, AD=3\ cm$ and diagonal $AC=5\ cm.$

$\Rightarrow$ Draw a line segment AB = 5cm.

$\Rightarrow$ With A as a center and radius AC = 5cm, draw an arc.

$\Rightarrow$ With B as a center and radius BC = 6cm, draw an arc, intersecting the previous arc at point C.

$\Rightarrow$ With A as a center and radius AD = 3cm, draw an arc.

$\Rightarrow$ Take C as a center with radius CD = 4cm draw an arc intersecting the previous arc at point D.

$\Rightarrow$ Join AC, BC, AD and CD.

Thus, ABCD is a required quadrilateral.

Construct a quadrilateral $ABCD$ with $AB=3$ cm, $BC=7$ cm, $CD=9$ cm, $CA=11$ cm, $BD=13$ cm.

First draw a line segment

$BC =7$ cm.

Open compass

$9$ cm and mark curve from point

$C$ .

Open compass

$13$ cm and mark curve from point

$B$ .

These curve cut each other at point

$D$ , then join

$B$ to

$D$ and

$C$ to

$D$

Open compass

$11$ cm and mark curve from point

$C$ .

Open compass

$3$ cm and mark curve from point

$B$ .

These curve cut each other at point

$A$ , then join

$A$ to

$B$ and

$A$ to

$D$

Hence,

$ABCD$ is the required quadrilateral.

Construct a quadrilateral $ABCD$ with $AB=\ 9cm$ , $BC=6\ cm$ , $CD=7\ cm$ , $AC=9\ cm$ and $BD=11\ cm$ .

1. Draw a line segment BC

$=6\,\, cm$ .

2. Open compass of length

$7$ cm and mark curve from point C.

3. Open compass of length

$11\,\, cm$ and mark curve from point B.

4. These curve cut each other at point D.

5. Join B to D and C to D

6. Open compass of length

$9\,\, cm$ and mark curve from point C.

7. Open compass of length

$9 \,\,cm$ and mark curve from point B.

8. These curve cut each other at point A

9. Join A to B and A to D

Hence ABCD is the required quadrilateral.

Construct a quadrilateral PQRS with PQ=7cm, QR=8cm, RS=6cm, PR=11cm, QS=12cm.

1. Draw QR of length 8cm.

2. Cut an arc of 12 cm from Q and 6 cm from R. Their point of intersection is S.

3. Cut an arc of 7 cm from Q and 11cm from R. Their point of intersection is P.

4. Join PQ, PS, SR, QR.

PQRS is the required quadrilateral.

Construct the following quadrilaterals.
(i) Quadrilateral $ABCD$
$AB = 4.5 cm$
$BC = 5.5 cm$
$CD = 4 cm$
$AD = 6 cm$
$AC = 7 cm$

(ii) Quadrilateral $JUMP$
$JU = 3.5 cm$
$UM = 4 cm$
$MP = 5 cm$
$PJ = 4.5 cm$
$PU = 6.5 cm$

(iii) Parallelogram $MORE$
$OR = 6 cm$
$RE = 4.5 cm$
$EO = 7.5 cm$

(iv) Rhombus $BEST$
$BE = 4.5 cm$
$ET = 6 cm$

$\textbf {Step 1:}$

We need to construct the Quadrilateralsa according to the Data given to us.

$\textbf {Step 2:}$

We will use the following Steps of construction:

$(1)$

$1.$ Draw

$AD$ of length

$6\ cm$ .

$2.$ Cut an arc of

$7 \ cm$ from

$A$ and

$4\ cm$ from

$D$ .

Their point of intersection is

$C$ .

$3.$ Cut an arc of

$4,5\ cm$ from

$A$ and

$5.5\ cm$ from

$C$ .

Their point of intersection is

$B$ .

$4.$ Join all the points.

$ABCD$ is the required quadrilateral.

$(2)$

$1.$ Draw

$UM$ of length

$4\ cm$ .

$2.$ Cut an arc of

$6.5\ cm$ from

$U$ and

$5\ cm$ from

$M$ .

Their point of intersection is

$P$ .

$3.$ Cut an arc of

$3.5\ cm$ from

$U$ and

$4.5\ cm$ from

$P$ .

Their point of intersection is

$J$ .

$4.$ . Join all the points.

$JUMP$ is the required quadrilateral.

$3)$

$1.$ Draw

$OR$ of length

$6\ cm$ .

$2.$ Cut an arc of

$7.5\ cm$ from

$O$ and

$4.5\ cm$ from

$R$ . Their

point of intersection is

$E$ .

$3.$ Cut an arc of

$4.5\ cm$ from

$O$ and

$6\ cm$ from

$E$ . Their point of intersection is

$M$ .

$4.$ Join all the points.

$MORE$ is the required parallelogram.

$4)$

$1.$ Draw

$ET$ of length

$6\ cm$ .

$2.$ Draw its perpendicular bisector.

$3.$ Cut an arc of

$4.5\ cm$ from

$E$ on the perpendicular bisector both

above and below

$ET$ , the point of intersections are

$B$ and

$S$ .

$4.$ Join all the points.

$BEST$ is the required rhombus.

2133603_463393_ans_a016421e343f446dbd0fffc25cee7949.png

Construct the following quadrilaterals
(i) Quadrilateral $MORE$
$MO = 6\ cm$
$OR = 4.5\ cm$
$\angle M = 60^{\circ}$
$\angle O = 105^{\circ}$
$\angle R = 105^{\circ}$

(ii) Quadrilateral $PLAN$
$PL = 4\ cm$
$LA = 6.5\ cm$
$\angle P = 90^{\circ}$
$\angle A = 110^{\circ}$
$\angle N = 85^{\circ}$

(iii) Parallelogram $HEAR$
$HE = 5\ cm$
$EA = 6\ cm$
$\angle R = 85^{\circ}$

(iv) Rectangle $OKAY$
$OK = 7\ cm$
$KA = 5\ cm$

$\textbf{Step 1:}$

We need to make the constructions according to the information given.

$\textbf{Step 2:}$

We will follow the following steps of construction.

$1)$

$1.$ Draw

$MO$ of length

$6$ cm.

$2.$ Draw an angle of

$105$ degree on

$O$ and cut an arc of

$4.5$ cm on it. The point of intersection is

$R$ .

$3.$ Draw an angle of

$105$ degree on

$R$ and draw an angle of

$60$ degree on

$M$ . Their point of intersection is

$E$ .

$4.$ Join all the points.

$MORE$ is the required quadrilateral.

$2)$

$1.$ Draw

$PL$ of length

$4$ cm.

$2.$ Calculate the value of

$\angle L$ , which comes out to be

$75^{o}$ .Draw an angle of

$75^{o}$ on

$L$ and cut an arc of

$6.5$ cm on it. The point of intersection is

$A$ .

$3.$ Draw an angle of

$110^{o}$ on

$A$ and draw an angle of

$90^{o}$ on

$P$ . Their point of intersection is

$N$ .

$4.$ Join all the points.

$PLAN$ is the required quadrilateral.

$3)$

$1.$ Draw

$EA$ of length

$6$ cm.

$2.$ Draw an angle of

$85^{o}$ on

$E$ and cut an arc of

$5$ cm on it. The point of intersection is

$H$ . (

$\angle R= \angle E = 85^o$ and

$\angle H = 95^o$ )

$3.$ Draw an angle of

$95^{o}$ on

$H$ and cut an arc of

$6$ cm. Their point of intersection is

$R$ .

$4.$ Join all the points.

$HEAR$ is the required parallelogram.

$4)$

$1.$ Draw

$OK$ of length

$7$ cm.

$2.$ Draw an angle of

$90^{o}$ on

$K$ and cut an arc of

$5$ cm on it. The point of intersection is

$A$ .

$3.$ Draw an angle of

$90^{o}$ on

$A$ and cut an arc of

$7$ cm from

$A$ . The point of intersection is

$Y$ .

$4.$ Join all the points.

$OKAY$ is the required rectangle.

2133611_463395_ans_78f2e9aa2b1f4367b84eb7c2aff87dc9.png

Construct a quadrilateral $ABCD$ with $AB=5$ cm, $BC=9$ cm, $\displaystyle \angle A={ 30 }^{ \circ },\angle B={ 90 }^{ \circ },\angle C={ 15 }^{ \circ }$ .

First draw a line segment

$BC=9$ cm.

Make angle

$90^{0}$ at point

$B$ and angle

$15^{0}$ at point

$C$ .

Open compass

$5$ cm and mark curve at angle line from point

$B$

This cut point is

$A$ , then make angle

$30^{0}$ at point

$A$ .

This angle line cut the angle line at point point

$C$ at point

$D$ .

Join

$D$ to

$C$ .

Hence,

$ABCD$ is the required quadrilateral.

Construct a quadrilateral $PQRS$ with $PQ=8$ cm, $QR=6$ cm, $\displaystyle \angle P={ 60 }^{ \circ },\angle Q={ 45 }^{ \circ },\angle R={ 60 }^{ \circ }$ .

First draw a line segment

$QR=6$ cm.

Make angle

$45^{0}$ at point

$Q$ and angle

$60^{0}$ at point

$R$ .

Open compass

$8$ cm and mark curve at angle line from point

$Q$ .

This cut point is

$P$ .

Make angle

$60^{0}$ at point

$P$ , this angle line cut the angle line at point point

$S$ at point

$D$ .

Join

$S$ to

$R$ and hence

$PQRS$ is the required quadrilateral.

Construct a quadrilateral $MORE$ where $MO=6\ cm$ , $OR=4.5 \ cm$ , $\angle M =60^o$ , $\angle O=105^o$ and $\angle R=105^o$
Also, find the length of side EM (in cm)

$\Rightarrow$ Draw a line segment

$MO$ of

$6\,cm$ .

$\Rightarrow$ Take

$M$ as a center draw an angle of

$60^o$ .

$\Rightarrow$ Take point

$O$ as a center draw an angle of

$105^o$ .

$\Rightarrow$ Cut

$OR$ at

$4.5\,cm$

$\Rightarrow$ Take

$R$ point as a center draw an angle of

$105^o$ .

$\Rightarrow$ The arc of

$105^o$ angle and the arc of

$60^o$ angle intersect with each other at

$E.$

$\Rightarrow$ Hence,

$MORE$ is a required Quadrilateral.

$\Rightarrow$ Now, by using scale measure the length of side

$EM$

We get,

$EM=7.5\,cm$

1248554_528574_ans_0ec30d8b667f400ab29890a169be7e4a.jpeg

Construct a quadrilateral $PQRS$ with $PQ=7$ cm, $QR=6$ cm, $\displaystyle \angle P={ 45 }^{ \circ },\angle Q={ 60 }^{ \circ },\angle R={ 75 }^{ \circ }$ .

First draw a line segment

$QR=6$ cm.

Make angle

$60^{0}$ at point

$Q$ and angle

$75^{0}$ at point

$R$ .

Open compass

$7$ cm and mark curve at angle line from point

$Q$ .

This cut point is

$P$ .

Make angle

$45^{0}$ at point

$P$ , this angle line cut the angle line at point

$R$ at point

$S$ .

Join

$S$ to

$R$ and hence

$PQRS$ is the required quadrilateral.

Construct a quadrilateral $ABCD$ with $AB=6$ cm, $BC=9$ cm, $\displaystyle \angle A={ 75 }^{ \circ },\angle B={ 60 }^{ \circ },\angle C={ 45 }^{ \circ }$ .

First draw a line segment

$BC=9$ cm.

Make angle

$60^{0}$ at point B and angle

$45^{0}$ at point

$C$ .

Open compass

$6$ cm and mark curve at angle line from point

$B$ .

This cut point is

$A$ .

Make angle

$75^{0}$ at point

$A$ , this angle line cut the angle line at point point

$C$ at point

$D$ .

Join

$C$ to

$D$ and hence

$ABCD$ is the required quadrilateral.

Construct following quadrilaterals
(i) Quadrilateral $LIFT$
$LI = 4\ cm$
$IF = 3\ cm$
$TL = 2.5\ cm$
$LF = 4.5\ cm$
$IT = 4\ cm$

(ii) Quadrilateral $GOLD$
$OL = 7.5\ cm$
$GL = 6\ cm$
$GD = 6\ cm$
$LD = 5\ cm$
$OD = 10\ cm$

(iii) Rhombus $BEND$
$BN = 5.6\ cm$
$DE = 6.5\ cm$

$\textbf{Step 1:}$

We need to make the constructions according to the information given.

$\textbf{Step 2:}$

We will follow the following steps of construction.

$(i)$

$1.$ Draw

$LI$ of length

$4 cm$ .

$2.$ Cut an arc of

$2.5$ cm from

$L$ and

$4 cm$ from

$I$ . Their point of intersection is

$T$ .

$3.$ Cut an arc of

$4.5 cm$ from

$L$ and

$3 cm$ from

$I$ . Their point of intersection is

$F$ .

$4.$ Join all the points.

$LIFT$ is the required quadrilateral.

$(ii)$

$1.$ Draw

$OL$ of length

$7.5 cm$ .

$2.$ Cut an arc of

$10 cm$ from

$O$ and

$5 cm$ from

$L$ . Their point of intersection is

$D$ .

$3.$ Cut an arc of

$7.5 cm$ from

$L$ and

$6 cm$ from

$D$ . Their point of intersection is

$G$ .

$4.$ Join all the points.

$GOLD$ is the required quadrilateral.

$iii)$

$1.$ Draw

$DE$ of length

$6.5 cm$ .

$2.$ Draw a perpendicular bisector of

$DE$ . The point of intersection is

$O$ .

$3.$ Cut an arc of

$2.8 cm$ from

$O$ on the perpendicular bisector both above and below

$DE$ . The points of intersection are

$B$ and

$N$ .

$4.$ Join all the points

$BEND$ is the required rhombus.

2133609_463394_ans_60c9db2e1299493fb7dabebd8971c3c1.png

Construct a quadrilateral $ABCD$ with $AB=4.5cm, BC=5.5cm, CD=4cm, AD=6cm$ and diagonal $AC=7cm$ .

Steps for construction :

Step 1. Draw a line segment

$BC=5.5$ cm.

Step 2. With

$B$ as centre and radius

$=4.5$ cm, draw an arc.

Step 3. With

$C$ as centre and radius

$=7$ cm, draw an arc to cut the arc in previous step at point

$A$

Step 4. Join

$BA$ and

$CA$

Step 5. With

$C$ as centre and radius

$=4$ cm, draw an arc.

Step 3. With

$A$ as centre and radius

$=6$ cm, draw an arc to cut the arc in previous step at point

$D$

Step 4. Join

$AD$ and

$CD$

Hence, required quadrilateral

$ABCD$ is constructed.

1690560_568594_ans_451d59c560004bcfbb8aad06d4211541.png

Construct a square of side $5\ cm$ . Find its area also

Given :

$Side = 5\ cm$
Steps for construction
Step

$1$ : Draw a rough diagram and mark the given measurements.
Step

$2$ : Draw a line segment

$PQ = 5\ cm$ .
Step

$3$ : At

$P$ using a compass construct

$\vec {PX} \perp \vec {PQ}$ .
Step

$4$ : With

$P$ as centre, draw an arc of radius

$5\ cm$ cutting

$\vec {PX}$ at

$S$ .
Step

$5$ : With

$S$ as centre, draw an arc of radius

$5\ cm$ above the line segment

$\overline {PQ}$ .
Step

$6$ : With

$Q$ as centre and same radius, draw an arc, cutting the previous arc at

$R$ .
Step

$7$ : Join

$\overline {QR}$ and

$\overline {RS}$ .

$PQRS$ is the required square.

Construct rhombus $ABCD$ such that:
$AC = 7 cm$ , $BD = 4 cm$

Steps of construction:

1. Draw

$AC = 7$ cm.

2. Draw perpendicular bisector of

$AC$ to intersect at

$E$ .

3. Cut off

$ED = EB = 2$ cm.

4. Join

$AB, BC, CD$ and

$DA$ .

Hence,

$ABCD$ is the required rhombus is the above drawn figure.

Construct rhombus $ABCD$ such that diagonal $BD = 6.8 cm$ , $AC = 5.4 cm$ .

$AC = 5.4cm$

$BD = 6.8cm$

Steps of construction:

1. Draw

$BD = 6.8$ cm.

2. Draw

$PQ$ the perpendicular bisector of

$BD$ to intersect at

$E$ .

3. Cut off

$EA = EC = 2.7cm$ (half of the length of diagonal AC) on the perpendicular bisector

$PQ$ .

4. Join

$AB, BC, CD$ and

$DA$ .

Hence,

$ABCD$ is the required rhombus as shown in the above figure.

Construct square $ABCD$ which has side-length $2\ cm$ .

Steps of construction:

Draw $AB = 2\ cm.$
At $A$ , draw $AE$ perpendicular to $AB$ .
With $A$ as a center and $2\ cm$ as radius draw an arc to intersect $AE$ at $D$ .
With $D$ as a center and $2\ cm$ cm as radius draw an arc.
With $B$ as a center and $2$ cm as radius draw another arc to intersect at $C$ .
Join $BC$ and $DC$ .

Hence,

$ABCD$ is the required square in the above-drawn figure.

Construct a quadrilateral ABCD with AB=2.5 cm, BC=3.5 cm, AD=4.5 cm, $\angle A=90^0$ and $\angle B=120^0$ .

Draw $AB =2.5$ $cm$
Draw $90°$ at $A$
Mark $d$ such that $AD=4.5$ $cm$
Draw $120°$ at $B$
Mark $C$ such that $BC=3.5$ $cm$
Draw $CD$
$ABCD$ is the required quadrilateral

1226726_1228246_ans_f944b4450a924e74a0ab1eeff7dd9ec1.png

Construct the quadrilateral $PQRS$ with $PQ=3.5cm, QR=4cm, RS=5cm. PS=4.5cm$ and $QS=6.5cm$ .

Steps for construction :

Step 1. Draw a line segment

$PQ=3.5$ cm.

Step 2. With

$P$ as centre and radius

$=4.5$ cm, draw an arc.

Step 3. With

$Q$ as centre and radius

$=6.5$ cm, draw an arc to cut the arc in the previous step at point

$S$

Step 4. Join

$PS$ and

$QS$

Step 5. With

$S$ as centre and radius

$=5$ cm, draw an arc.

Step 3. With

$Q$ as centre and radius

$=4$ cm, draw an arc to cut the arc in the previous step at point

$R$

Step 4. Join

$SR$ and

$QR$

Hence, required quadrilateral

$PQRS$ is constructed.

1690569_568607_ans_39002c4761c7437c9651db38dd6c4785.jpg

Construct a rectangle $\text{ABCD}$ with $\text{AB=5 cm, BC=3.5 cm}$

Steps for construction :

Step 1. Draw a line segment

$AB=5\text{ cm}$ .

Step 2. Construct

$\angle ABC=90^\circ$ at points

$B$ .

Step 3. On

$AB$ , make

$BC=3\text{ cm}$ .

Step 4. With

$A$ as centre and radius

$=3.5 \text{ cm}$ , draw an arc.

Step 3. With

$C$ as centre and radius

$=5\text{ cm}$ , draw an arc to cut the arc in the previous step at point

$D$

Step 4. Join

$CD$ and

$AD$

Hence, the required rectangle

$ABCD$ is constructed.

1690628_568575_ans_9c57d27e927e439eba044a94eaad3f37.png

Construct a rhombus $PQRS$ with $PR = 8\ cm$ and $QS = 6\ cm$ and find its area.

Given:

$PR = 8\ cm$ and

$QS = 6\ cm$ .
Steps for construction
Step

$1$ : Draw a rough diagram and mark the given measurements.
Step

$2$ : Draw a line segment

$PR = 8\ cm$ .
Step

$3$ : Draw the perpendicular bisector

$\vec {XY}$ to

$\overline {PR}$ . Let it cut

$\overline {PR}$ at

$"O"$ .
Step

$4$ : With

$O$ as centre and

$3\ cm$ (half of

$QS$ ) as radius draw arcs on either side of

$'O'$ which cuts

$\vec {XY}$ at

$Q$ and

$S$ as shown in fig.
Step

$5$ : Join

$\overline {PQ}, \overline {QR}, \overline {RS}$ and

$\overline {SP}$ .

$PQRS$ is the required rhombus.
Step

$6$ : We know,

$PR = d_{1} = 8\ cm$ and

$QS = d_{2} = 6\ cm$ .
Calculation of area:
In the Rhombus

$PQRS$ ,

$d_{1} = 8\ cm$ and

$d_{2} = 6\ cm$ .
Area of the rhombus

$PQRS$

$= \dfrac {1}{2}\times d_{1}\times d_{2}$

$= \dfrac {1}{2} \times 8\times 6$

$= 24\ cm^{2}$ .

Construct a rectangle whose adjacent sides are of lengths $5$ $\text cm$ and $3.5$ $\text cm$ .

Steps of construction :

1. Take

$AB = 5 cm$ .

2. Draw

$AY \bot \ AB$

3. With

$A$ as centre and radius

$= 3.5 cm$ , describe an arc cutting

$AY$ at D.

4. With

$D$ as centre and radius

$5 cm$ , describe an arc with

$B$ as centre and radius

$3.5 cm$ , describe another arc intersecting the first arc at

$C$ .

5. Join

$BC$ and

$DC$ .

$ABCD$ is the required rectangle.

1808109_1054777_ans_4258e07a234045aa8b417a9edfebc447.png

Construct a rhombus whose diagonals are 6 cm and 4 cm

1432508_1017486_ans_7ded437809404703aac7ac8c46f63d8e.png

Construct the following quadrilaterals.
Quadrilateral MORE
$MO$ = $6\ cm$
$OR$ = $4.5\ cm$
$\angle M=60^o$
$\angle O=105^o$
$\angle R=105^o$

Here, we are given two adjacent sides and three angles.

Construct the quadrilateral using following steps of construction.

Step I: Draw side $MO$ of length $6\ cm$ .

Step II: Now, draw $60{}^\circ$ from point $M$ .

Step III: Now, draw $105{}^\circ$ from point $O$ .

Step IV: Take $4.5\ cm$ as radius and $O$ as the centre. Draw an arc. The arc intersects the line at point $R$ .

Step V: Again, draw $105{}^\circ$ from point $R$ .

Step VI: Mark point $E$ where the lines from point $M$ and point $R$ intersect. Now, label the sides of the quadrilateral.

Therefore,

$\Rightarrow MORE$ is the required quadrilateral as shown in the figure.
1027772_1042386_ans_63f99835f28c4096bdddf7db391230b0.PNG

Construct the following quadrilaterals.
Quadrilateral PLAN
$PL$ = $4cm$
$LA$ = $6.5cm$
$\angle P=90^o$
$\angle A=110^o$
$\angle N=85^o$

$\angle P+\angle L+\angle A+\angle N=360^o$

$\Rightarrow$

$90^o+\angle L+110^o+85^o=360^o$

$\Rightarrow$

$285^o+\angle L=360^o$

$\therefore$

$\angle L=75^o$

$\Rightarrow$ Draw line segment

$PL=4\,cm$

$\Rightarrow$ Now, take

$P$ as center and draw angle of

$90^o.$

$\Rightarrow$ Draw

$75^o$ angle at point

$L$ .

$\Rightarrow$ Cut off

$LA=6.5\,cm$

$\Rightarrow$ Now draw

$110^o$ from point

$A$ .

$\Rightarrow$ Mark point

$N$ where the lines from point

$P$ and point

$A$ intersect .

$\therefore$

$PLAN$ is required quadrilateral.

1250403_1042387_ans_75a37197394f4c789ec22a0b2a284977.jpeg

Construct a quadrilateral $ABCD$ in which $AB = 4.4$ cm, $BC = 4$ cm, $CD = 6.4$ cm, $DA = 3.8$ cm and $BD = 6.6$ cm.

1153675_1047800_ans_689f8d0a1d7140e787a99198fe0b8c3f.PNG

Construct $\square DEFG$ such that $l$ (DE) $=4.5$ cm, $l$ (EF) $=6.5$ cm, $l$ (DG) $=5.5$ cm, $l$ (DF) $=7.2$ cm, $l$ (EG) $=7.8$ cm.

$\Rightarrow$ Draw a line segment

$DE=4.5\,cm$

$\Rightarrow$ Take

$E$ as center with radius

$6.5\,cm$ and draw an arc.

$\Rightarrow$ Take

$D$ as center with radius

$7.2\,cm$ and draw an arc meeting previous arc at point

$F$

$\Rightarrow$ Take

$E$ as center with radius

$7.8\,cm$ and draw an arc.

$\Rightarrow$ Take

$D$ as center with radius

$5.5\,cm$ and draw an arc meeting previous arc at point

$G$

$\Rightarrow$ Join

$EF,DF,FG$ and

$DG$

$\therefore$

$DEFG$ is required quadrilateral.

1251755_1074225_ans_b9ae8235d7974180a5df67976ae4c3a3.jpeg

Construct a rhombus whose diagonals are $16$ cm and $12$ cm. What property of the diagonals is used?

$\Rightarrow$ Draw a line segment

$AC=16\,cm$

$\Rightarrow$ Draw a perpendicular bisector of

$AC$ , which intersect

$AC$ at point

$O.$

$\Rightarrow$ Take

$O$ as center with radius

$6\,cm$ and draw an arc above and below on perpendicular, We get point

$B$ and

$D$

$\Rightarrow$ Join

$AB,BC,CD$ and

$AD$

$\therefore$

$ABCD$ is required rhombus.

The properties which we have used are :

$\Rightarrow$ The intersection of the diagonals of a rhombus form

$90$ degree angles. This means that they are perpendicular.

$\Rightarrow$ The diagonals of a rhombus bisect each other. This means that they cut each other in half.

1253306_1084508_ans_989e59ff4b374b0693220de50fa51455.jpeg

Constant a quadrilateral $ABCD$ , where $AB=4\ cm,BC=5\ cm,CD=6.5\ cm$ and $\angle B=105^{o}$ and $\angle C=80^{o}$

Step (1):-

First Draw a line $BC=5\,cm$

Step (2):-

Construct an angle $\angle CBY={{105}^{o}}$

Step (3):-

With B as centre and radius $4$ cm, draw an arc to cut line by at point A.

Step (4):-

Again, construct an angle $\angle BCZ={{80}^{o}}$

Step (5):-

With C as centre and a radius $6.5\,cm$ draw an arc to cut $CZ$ at point $D$ .Join $AD$

Hence, $ABCD$ is the required quadrilateral.

1066303_1091020_ans_8e1e50abc7ee45ff99b57affc5cee514.png

Draw a quadrilateral LIFT, when LI = 4.0 cm, IF = 3.0 cm, TI = 4 cm, LF = 4.5 cm and LT = 2.5 cm.

Steps of construction:-

(a) Draw a line segment

$LI = 4 \; cm$ .

(b) Taking radius

$4.5 \; cm$ , draw an arc taking

$L$ as centre.

$3 \; cm$ taking

$I$ as centre which intersects the first arc at

$F$ .

(d) Join

$FI$ and

$FL$ .

(e) Draw another arc of radius

$2.5 \; cm$ taking

$L$ as centre and

$4 \; cm$ taking

$I$ as centre which intersect at

$T$ .

(f) Join

$TF, TL$ and

$TI$ .

Hence,

$LIFT$ is the required quadrilateral.

1134985_1153421_ans_aa0a7c0093324362ad355bd80dc7837d.png

Construct a quadrilateral $ABCD$ , where $AB=4\ cm,BC=5\ cm,CD=6.5\ cm$ and angle $B=75^{o}$ and angle $C=60^{o}$

$\Rightarrow$ Draw a line segment

$AB=4\,cm$

$\Rightarrow$ Take

$B$ as center and draw an angle of

$75^o.$

$\Rightarrow$ Cut off

$BC=5\,cm$

$\Rightarrow$ Take

$C$ as center and draw an angle of

$60^o.$

$\Rightarrow$ Cut off

$CD=6.5\,cm$

$\Rightarrow$ Join

$DA$

$\therefore$

$ABCD$ is a required quafrilateral.

1257860_1085489_ans_3ab48e4b74e746169069a9d1e894c368.jpeg

Quadrilateral GOLD
OL $=7.5$ cm
GL $=6$ cm
GD $=6$ cm
LD $=5$ cm
OD $=10$ cm.

$\Rightarrow$ Draw a line segment

$OL=7.5\,cm$

$\Rightarrow$ Take

$O$ as center with radius

$10\,cm$ and draw an arc.

$\Rightarrow$ Take

$L$ as center with radius

$5\,cm$ and draw an arc, which meets previous arc at point

$D$ .

$\Rightarrow$ Take

$L$ as center with radius

$6\,cm$ and draw an arc.

$\Rightarrow$ Take

$D$ as center with radius

$6\,cm$ and draw an arc, which meets previous arc at point

$G$ .

$\Rightarrow$ Now, join

$OD,LD,GD$ and

$GO$ .

1262158_1083002_ans_828a78dea74c4a9f909f8646cd5486bb.png

Construct a quadrilateral $ABCD$ with the following measures:
$l(AB)=6.4\ cm,\ l(BC)=4.8\ cm,\ \angle A=70^o,\ \angle B=50^o,\ \angle C=140^o$ .

Following are the construction steps:

(1). Draw a line segment

$AB=6.4\,cm$

(2). Take

$A$ as center and draw angle of

$70^o$ .

(3). Take

$B$ as center and draw angle of

$50^o$ .

(4). Take

$B$ as centre and cut an arc of

$4.8\ cm$ on the

$50^o$ angle line. It intersects at

$C$ . Hence,

$BC=4.8\,cm$ .

(5). Take

$C$ as center and draw angle angle of

$140^o$ , which meets

$\angle A$ at point

$D.$

Hence,

$ABCD$ is the required quadrilateral.

1252646_1085060_ans_75e2472c1755469c862aa5ac02267213.png

Construct a quadrilateral $ABCD$ in which $AB=6\ cm,CD=5\ cm,DA=3.5\ cm,\angle C=120^{o}$ and $\angle D=75^{o}$ .

$\Rightarrow$ Draw a line segment

$CD=5\,cm$

$\Rightarrow$ Draw an angle of

$120^o$ and

$75^o$ at points

$C$ and

$D$

$\Rightarrow$ Cut off

$DA=3.5\,cm$

$\Rightarrow$ Cut off

$AB=6\,cm$ and join

$AB$

$\therefore$

$ABCD$ is a required quadrilateral.

1270691_1219121_ans_ae77a0427e88468299c73b6095cfb9ad.png

Quadrilateral HELP with HE=6cm,EL=4.5cm, $\angle H=60^{0},\angle E=105^{0}$ and $\angle P=120^{0}$

We have,

$HE=6\,cm.$

$EL=4.5\,cm.$

$\angle H={{60}^{o}}$

$\angle E={{105}^{o}}$

$\angle P={{120}^{o}}$

We shall find the fourth angle first.

Sum of all the angles of a quadrilateral $={{360}^{o}}$

$\therefore \angle L={{360}^{o}}=\left( {{60}^{o}}+{{105}^{o}}+{{120}^{o}} \right)$

$\angle L={{750}^{o}}$

Step (1):-first draw base $HE=6\,cm.$

Step (2):- Construct $\angle H={{60}^{o}}$

Step (3):- Construct $\angle E={{105}^{o}}$

On this line of $\angle E$ cut an arc at $4.5$ cm.

Step(4):- Construct $\angle L={{75}^{o}}$

Complete figure.
1207936_1159860_ans_89ad4b3132bc41efaa2090b0e9052e9b.png

Construct a quadrilateral $ABCD$ in which $AB=4 \text{ cm}, BC=3.1 \text{ cm}, CD=3 \text{ cm}, AD=4.2 \text{ cm}$ and $AC=5 \text{ cm}$ . Measure the diagonal $BD$ .

$\Rightarrow$ Draw a line segment

$AB=4\,cm$

$\Rightarrow$ Take

$A$ as center with radius

$5\,cm$ and draw an arc.

$\Rightarrow$ Take

$B$ as center with radius

$3.1\,cm$ and draw an arc which cuts off previous arc at point

$C$

$\Rightarrow$ Take

$A$ as center with radius

$4.2\,cm$ and draw an arc.

$\Rightarrow$ Take

$C$ as center with radius

$3.\,cm$ and draw an arc which cuts off previous arc at point

$D$

$\Rightarrow$ Join

$BC,AC,CD,AD$ and

$BD$

$\Rightarrow$ Measure the length of

$BD$

$\Rightarrow$ We get,

$BD=5\,cm$

1271765_1169501_ans_830e6c296c934af1baab0d77604e2482.png

Construct a quadrilateral $READ$ , when $RE=3.51$ cm, $EA=4.3$ cm, $AD=4.8$ cm, $\angle E=105^{\circ}$ & $\angle R=55^{\circ}$ .

$\Rightarrow$ Draw a line segment

$RE=3.5\,cm$

$\Rightarrow$ Tale

$R$ and

$E$ as center and draw an arcs of angles of

$55^o$ and

$105^o.$

$\Rightarrow$ Cut off

$EA=4.3\,cm$

$\Rightarrow$ Cut off

$AD=4.8\,cm$

$\therefore$

$READ$ is a required quadrilateral.

1270506_1187858_ans_39482887658e4c9d97e696a733341074.png

Construct a quadrilateral $ABCD$ in which $AB = 4.2 cm, BC = 6 cm, CD = 5.2 cm, DA = 5 cm$ and $AC = 8 cm$ .

The given Quadrilateral

$ABCD$ can be drawn by following the steps.
Step 1: Draw a line

$AB = 4.2 cm$
Step 2: With

$A$ as the centre and radius

$= 8 cm$ draw an arc.
Step 3: with

$B$ as the centre and radius

$= 6 cm$ draw another arc cutting previous arc. Label this point as

$C$
Step 4: Join

$BC$
Step 5: With

$A$ as the centre and radius

$= 5 cm$ draw an arc.
Step 6: with

$C$ as the centre and radius

$= 5.2 cm$ draw another arc cutting previous arc. label this point as

$D$
Step 7: join

$CD$ and

$AD$ .

$\therefore$ The constructed quadrilateral looks like:
1607712_1218252_ans_688168d5954a405f8876cc5cfbbcd9f2.jpg

Construct a rectangle $ABCD$ given $AB=7\ cm, BC=5.5\ cm$ .

We know that in rectangle measure of each angle is

$90^o$ and opposite sides are equal.

$\Rightarrow$ Draw a line segment

$AB=7\,cm$ .

$\Rightarrow$ Construct the angle of

$90^o$ at points

$A$ and

$B$ with the help of protractor.

$\Rightarrow$ Cut off

$AD=5.5\,cm$ and

$BC=5.5\,cm$ .

$\Rightarrow$ Join

$CD$ .

1254698_1162487_ans_fd0aa52a97b941099ead77062a13f39b.png

Construct a quadrilateral MIND, where MI = 3 cm, IN = 5 cm, ND = 4 cm, DM = 5 cm and MN = 7 cm.

Following should be the steps of construction:

1. Draw a line segment

$MI=3\,cm$

2. Take

$M$ as center and radius

$7\,cm$ , draw an arc.

3. Take

$I$ as center and radius

$5\,cm$ , draw an arc which meets previous arc at point

$N$ .

4. Take

$M$ as center and radius

$5\,cm$ , draw an arc.

5. Take

$N$ as center and radius

$4\,cm$ , draw an arc which meets previous arc at point

$D$ .

6. Join

$IN,\ MN,\ ND$ and

$DM$ .

Hence,

$MIND$ is required quadrilateral.

1270474_1194819_ans_734730285cd744b4a4e36c57e626388a.jpeg

Construct a quadrilateral $ABCD$ in which $AB=3.6cm, BC=3.3cm, AD=2.7cm$ , diagonal $AC=4.6cm$ and diagonal $BD=4cm$

$\Rightarrow$ Draw a line segment

$AB=3.6\,cm$

$\Rightarrow$ Take

$A$ as center with radius

$4.6\,cm$ and draw an arc.

$\Rightarrow$ Take

$B$ as center with radius

$3.3\,cm$ and draw an arc which meets previous arc at point

$C$

$\Rightarrow$ Take

$A$ as center with radius

$2.7\,cm$ and draw an arc.

$\Rightarrow$ Take

$B$ as center with radius

$4\,cm$ and draw an arc which meets previous arc at point

$D$

$\Rightarrow$ Join

$BC,AC,DB,AD$ and

$CD$ .

$\therefore$

$ABCD$ is required quadrilateral.

1254726_1165363_ans_f89e8cb9fa5b4578961dc72a5a1af62b.jpeg

Construct a quadrilateral $PQRS$ with $PQ=6.2cm, QR=5.6cm, RS=6.2cm$ , $\angle{Q}={90}^{o}$ and $\angle {R}={120}^{o}$

1092303_1191389_ans_ec8e8c87b8734c97a34ff696f6179edc.png

Construct a quadrilateral $ABCD$ , when $AB = 3\ cm, CD = 3\ cm, DA = 7.5\ cm, AC = 8\ cm$ and $BD = 4\ cm$ .

If we consider a triangle

$ABD$ from the given data, then

$AB=3\ cm, BD=4\ cm, AD=7.5\ cm$

$AB+BD=3+4=7\ cm$

However, we know that the sum of the length of two sides of a triangles in always greater than the third side.

Therefore, construction is not possible from the given data.

Construct a quadrilateral $ABCD$ with $AB=5\;cm,BC=3.6\;cm,\angle ABC=80^{\circ},\angle DAB=60^{\circ}$ and $\angle ADC=120^{\circ}$ .

We know that sum of four angles of a quadrilateral is

$360^o.$

$\therefore$

$\angle A+\angle B+\angle C+\angle D=360^o.$

$\therefore$

$60^o+80^o+\angle C+\angle 120^o=360^o$

$\therefore$

$260^o+\angle C=360^o.$

$\therefore$

$\angle C=100^o.$

$\Rightarrow$ Draw a line segment

$AB=5\,cm$

$\Rightarrow$ Take

$A$ as a center and draw an angle of

$60^o$

$\Rightarrow$ Take

$B$ as a center and draw an angle of

$80^o$

$\Rightarrow$ Cut off

$BC=3.6\,cm$

$\Rightarrow$ Take

$C$ as a center and draw an angle of

$100^o$ , which cuts of

$\angle A$ at

$D.$

$\therefore$

$ABCD$ is a required quadrilateral.

1290001_1200057_ans_260c209033ce4ec6ba465b6b03afa228.jpeg

Construct a rectangle whose adjacent sides are $5.6\ cm$ and $4\ cm$

Construction:-

( 1). Draw a line segment $AB=5.6\,cm.$

( 2). Construct angle ${{90}^{o}}$ . At point $A$ and $B$ . ( Using compass)

( 3). Join $A$ to $D$ and Join $B$ to $C\ 4\ cm$ .

( 4). Set compass to $3.5\ cm$ . and draw an arc

( 5). Join $DC$ .

So,

$ABCD$ is required rectangle.
1237474_1276111_ans_72cfe0a6bcc649c9a23563138975b594.png

Construct a quadrilateral $PQRS$ with $PQ=4.7\ cm, QR=3.9\ cm, \angle QPS=75^{o}, \angle PQR=120^{o}$ and $\angle QRS=105^{o}$ . Measure $\angle PSR$ .

$\Rightarrow$ Draw a line segment

$PQ=4.7\,cm$

$\Rightarrow$ Take

$P$ and

$Q$ as a center and draw an angle of

$75^o$ and

$120^o.$

$\Rightarrow$ Cut off

$QR=3.9\,cm$

$\Rightarrow$ Take

$R$ as a center and draw an angle of

$105^o,$ which meet angle

$P$ at point

$S.$

$\Rightarrow$ Now, measure

$\angle PSR$

$\Rightarrow$ We get,

$\angle PSR=60^o.$

1318806_1301652_ans_6360b50eca5c43c7a1dead65ccfeda45.jpeg

Construct the following quadrilaterals :
Quadrilateral $RUNS$
$RU=4$ cm
$UN=5$ cm
$NS=4.2$ cm
$SR=5.8$ cm
$RN=7.2$ cm

$\Rightarrow$ Draw a line segment

$RU=4\,cm$

$\Rightarrow$ Take

$R$ as center, with radius

$7.2\,cm$ and draw an arc.

$\Rightarrow$ Take

$U$ as center, with radius

$5\,cm$ and draw an arc, which cuts off previous arc at

$N$

$\Rightarrow$ Take

$R$ as center, with radius

$5.8\,cm$ and draw an arc.

$\Rightarrow$ Take

$N$ as center, with radius

$4.2\,cm$ and draw an arc, which cuts off previous arc at

$S$

$\Rightarrow$ Join

$UN,NS,RS$ and

$RN$

$\therefore$

$RUNS$ is required quadrilateral.

1288424_1359328_ans_a7b45c72d53f4e8796579f182db56585.jpeg

Quadrilateral JUMP
JU=3.5 cm
UM=4 cm
MP=5 cm
PJ=4.5 cm
PU=6.5 cm

Draw a horizontal line

$UP=6.5cm$

From point U draw an arc with scale

$4cm$ above the line UP.

From point P draw an arc with scale

$5cm$ above the line UP.

Mark the point of intersection as M

From point U draw an arc with scale

$3.5cm$ below the line UP.

From point P draw an arc with scale

$4.5cm$ below the line UP.

Mark the point of intersection as J

Joint all the points.

1224973_1311999_ans_75139fa6f41c4688bc2eb33bef667869.jpg

Construct a quadrilateral $PQRS$ in which $QR = 7.5 cm, PR = PS = 6 cm, RS = 5 cm$ , and $QS = 10 cm$ .

The given Quadrilateral

$PQRS$ can be drawn by following the steps.
Step 1: Draw a line

$QR = 7.5 cm$
Step 2: Draw an arc of

$10 cm$ from

$Q$
Step 3: From

$R$ with

$5 cm$ as radius cut the arc drawn in step 2 and that point of intersection is

$S$ .
Step 4: Draw another arc from

$R$ with

$6 cm$ of radius
Step 5: From

$S$ take

$6 cm$ as radius and cut the arc drawn in step 4 and the point of intersection is

$P$ . now join

$PR$ and

$PS$ .
We get the required quadrilateral

$PQRS$

1617429_1498419_ans_e253d5d1606f41be8b92b0e4cd7b303f.jpg

Construct a quadrilateral $ABCD$ in which $\overline { AB } =\overline { BC } =\overline { CD } =\overline { AD } =5$ cm and diagonal $\overline { AC } =6.5$ cm.

Draw a line segment

$\overline{AB}=5\,cm$

Take

$A$ as center with radius

$6.5\,cm$ and draw an arc.

Take

$B$ as center with radius

$5\,cm$ and draw an arc, which cut off previous arc at point

$C$

Take

$A$ as center with radius

$5\,cm$ and draw an arc.

Take

$C$ as center with radius

$5\,cm$ and draw an arc, which cut off Now, join

$\overline{BC},\overline{CD},\overline{DA}$ and

$\overline{AC}$

$\therefore$

$ABCD$ is required quadrilateral.

1288021_1368351_ans_4478a32bf0184c138ea0ae909dbdf0d0.jpeg

Construct a quadrilateral $PQRS$ in which:
$PQ = 5.4 cm, QR = 4.6 cm, RS = 4.3 cm, SP = 3.5 cm$ , and Diagonal $PR = 4 cm$ .

The given Quadrilateral

$PQRS$ can be drawn by following the steps.
Step 1: Draw a line

$PR = 4 cm$
Step 2: with

$P$ as the centre draw an arc of

$5.4 cm$
Step 3: with

$4.6 cm$ as radius cut the arc drawn in step 2 from point R and that point of intersection is

$Q$ . join

$PQ$ and

$QR$ .
Step 4: Draw another arc from

$P$ with

$3.5 cm$ radius on opposite side of

$PR$ .
Step 5: From

$R$ take

$4.3 cm$ as radius and cut the arc drawn in step 4 and the point of intersection is

$S$ . now join

$PS$ and

$RS$ .
We get the required quadrilateral

$PQRS$

1617441_1507239_ans_e065cf27c03e48c3a21d02aa8077c083.jpg

Construct a quadrilateral $PQRS$ in which $PQ=3.8$ cm, $QR=6.8$ cm and three angles $\angle P=100^{o},\ \angle R=110^{o}$ and $\angle S=75^{o}$ are known.

We know that the sum of all four angles of a quadrilateral is

$360^\circ$

$\Rightarrow$

$\angle P+\angle Q+\angle R+\angle S=360^\circ$

$\Rightarrow$

$100^\circ+\angle Q+\angle 110^\circ+75^\circ=360^\circ$

$\Rightarrow$

$285^\circ+\angle Q=360^\circ$

$\therefore$

$\angle Q=75^\circ$

Steps to construct a quadrilateral when two adjacent sides and three angles are known:

Step 1: Draw a line segment

$PQ=3.8\ cm$

Step 2: Take

$Q$ as the center and draw an angle of

$75^\circ$

Step 3: Cut off

$QR=6.8\ cm$

Step 4: Take

$R$ as the center and draw an angle of

$110^o$

Step 5: Take

$P$ as the center and draw an angle of

$100^\circ,$ which cuts of

$\angle R$ at point

$S$ .

$\therefore$

$PQRS$ is required quadrilateral.

1288415_1356688_ans_660a02d30c164c0397ce106a153d9239.png

Construct a quadrilateral $ABCD$ with $AB=2.5\ cm,BC=3.5\ cm,CD=4.5\ cm,\angle A=90^{o}$ and $\angle B=120^{o}$ .

$1.$ Draw a line

$AB=2.5$ cm

$2.$ Construct an angle

${90}^{\circ}$ at

$A$ using protractor.

$3.$ Construct an angle

${120}^{\circ}$ at

$B$ using compass.

$4.$ Join

$BC$

$5.$ From

$C$ draw an arc of length

$4.5$ cm and name the point of intersection as

$D$

Result:

$ABCD$ is the required quadrilateral.

1299592_1362558_ans_a159b64c3e51459d9ea2e8a3d9ede42d.png

Construct a Rhombus $PQRS$ such that $PR= 8$ cm, $QS= 5.4$ cm.

Given :

$PR = 8$ cm and

$QS = 5.4$ cm
We know that,All sides of rhombus is equal and Diagonals bisect each other at right angles.
So,

$PA = AR = 4$ cm and

$QA=AS=2.7$ cm
To construct rhombus, we follow these steps—
Step

$1 )$ Draw a line

$QS = 5.4$ cm
Step

$2 )$ Now take radius of

$2.7$ cm and bisect the line

$QS$ . Midpoint of line

$QS$ is

$A$ .
Step

$3 )$ Construct a right angle on point

$A$ and extend into both directions. Now take radius of

$4$ cm and draw two arc on both side , by taking center

$A$ . we will get two points

$P$ and

$R$ .

Step

$4 )$ Now join

$PQ , PS , SR$ and

$QR$ .

Result:

$PQRS$ is the required rhombus.

A construct rectangle $FLAG$ with sides $F L = 6 \mathrm { cm }$ and $L A = 4.2 \mathrm { cm }$

We know that, in rectangle opposite sides are equal and each angle is

$90^o.$

$\therefore$

$FL=AG$ and

$LA=FG$

$\Rightarrow$ Draw a line segment

$FL=6\,cm$

$\Rightarrow$ Take

$F$ and

$L$ as center and draw an angle of

$90^o$ .

$\Rightarrow$ Cut off

$AL=4.2\,cm$ and

$FG=4.2\,cm$

$\Rightarrow$ Join

$AG$

$\therefore$

$FLAG$ is required quadrilateral.

1284097_1231672_ans_7c6a3e533b324de2b4fe338f279ae8d4.jpeg

Construct a quadrilateral $ABCD$ in which $BC = 7.5\ cm, AC = AD = 6\ cm, CD = 5\ cm$ and $BD = 10\ cm$ .

Steps of the constructions:

Step I:- Draw the

$AC=6\ cm$

Step II:- With

$A$ as the center and radius

$6\ cm$ and draw an arc.

Step III:- With

$C$ as the center and radius

$5\ cm$ and draw an arc to intersect the arc drawn step II at

$D$ .

Step IV:- With

$D$ as the center and radius

$10\ cm$ and draw an arc on the other side of the line segment

$AC$ .

Step V:- With

$C$ as the center and radius

$7.5\ cm$ and draw an arc to intersect the arc drawn in step IV at

$B$ .

Step VI:- Join

$BA, DA, BC$ , and

$CD$ to obtain the required quadrilateral.

1879368_1672857_ans_23b42be087c445e4bd3d19dace71a16d.jpg

Construct a quadrilateral $ABCD$ in which $AB = 3.5 cm, BC = 3.8 cm, CD = DA = 4.5 cm$ and Diagonal $BD = 5.6 cm$ .

The given Quadrilateral

$ABCD$ can be drawn by following the steps.
Step 1: Draw a line

$BD = 5.6 cm$
Step 2: with

$B$ as the centre draw an arc of

$3.5 cm$
Step 3: with

$4.5 cm$ as radius cut the arc drawn in step 2 and that point of intersection is

$A$ . join

$BA$ and

$DA$
Step 4: Draw another arc from

$B$ with

$3.8 cm$ radius on opposite side of

$BD$ .
Step 5: From

$D$ take

$4.5 cm$ as radius and cut the arc drawn in step 4 and the point of intersection is

$C$ . now join

$BC$ and

$CD$ .

$\therefore$ The constructed quadrilateral looks like:
1607725_1576168_ans_127e30967f9e4f35b5d5d1c790ef26d4.jpg

In the figure, $ABCD$ is a quadrilateral.
How many pairs of adjacent sides are there?

$A\text{ quadrilateral is defined as a closed two-dimensional shape which}$

$\text{has four straight sides}\text{.}$

$\text{Adjacent sides: two sides that meet at a point are called adjacent sides}$

$In\text{ the given figure }$

$\text{adjacent sides are :AB and BC, BC and CD,CD and DA,DA and AB}$

$Total\,adjacent\,side\,in\,given\,figure=4$

Construct a quadrilateral $ABCD$ given $AD = 5\ cm, AB = 5.5\ cm, BC = 2.5\ cm, AC = 7.1\ cm$ and $BD = 8\ cm$ .

Steps of the constructions:

Step I:- Draw the

$AB=5.5\ cm$

Step II:- With

$A$ as the center and radius

$7.1\ cm$ and draw an arc.

Step III:- With

$B$ as the center and radius

$2.5\ cm$ and draw an arc to intersect the arc drawn in step II at

$C$ .

Step IV:- With

$B$ as the center and radius

$8\ cm$ and draw an arc.

Step V:- With

$A$ as the center and radius

$5\ cm$ and draw an arc to intersect the arc drawn in step IV at

$D$ .

Step VI:- Join

$DA, DB, BC, AC$ and

$CD$ .

Then, the obtained the required quadrilateral

$ABCD$ .

1880505_1672860_ans_ad33c542ae3d42769ede6be9cfec1cb6.jpg

In the figure, $ABCD$ is a quadrilateral.
How many pairs of opposite sides are there?

$A\text{ quadrilateral is defined as a closed two-dimensional shape which}$

$\text{has four straight sides}\text{.}$

$\text{Opposite sides: side opposite to a side or side infront of the given side is called opposite sides}$

$In\text{ the given figure }$

$\text{Opposite sides are :AD and BC, AB and CD}$

$Total\,pair\,of\,opposite\,side\,in\,given\,figure=2$

Complete the following, so as to make a true statement:
A quadrilateral has ______ diagonals.

The line segment joining the opposite vertices of quadrilateral is the diagonal of the quadrilateral.

In the given figure, the two diagonals are AC and BD.

So there are 2 diagonals.

1508665_1672746_ans_607c78cf41574601a39e137d83d8701f.png

In the figure, $ABCD$ is a quadrilateral.
How many pairs of adjacent angles are there?

$A\text{ quadrilateral is defined as a closed two-dimensional shape which}$

$\text{has four straight sides}\text{.}$

$\text{Adjacent angles:Two angles are said to be adjacent if they have common side between them }\text{.}$

$In\text{ the given figure }$

$\,angle\,adjacent\,to:$

$\angle A\,are\,:\,\angle B\,and\,\angle D$

$\angle B\,are\,:\,\angle A\,and\,\angle C$

$\angle C\,are\,:\,\angle B\,and\,\angle D$

$\angle D\,are\,:\,\angle A\,and\,\angle C$

$Total\,pair\,of\,adjacen\operatorname{t}\,angles\,in\,given\,figure=4$

Construct a quadrilateral $ABCD$ given $AD = 3.5\ cm, BC = 2.5\ cm, CD = 4.1\ cm, AC = 7.3\ cm$ and $BD = 3.2\ cm$ .

Steps of the constructions:

Step I:- Draw the

$CD=4.1\ cm$

Step II:- With

$C$ as the center and radius

$7.3\ cm$ and draw an arc.

Step III:- With

$D$ as the center and radius

$3.5\ cm$ and draw an arc to intersect the arc drawn in step II at

$A$ .

Step IV:- With

$D$ as the center and radius

$3.2\ cm$ and draw an arc on the other side of

$AC$ .

Step V:- With

$C$ as the center and radius

$2.5\ cm$ and draw an arc to intersect the arc drawn in step IV at

$B$ .

Step VI:- Join

$BA, DA, BC, AC$ and

$BD$ .

Then, the obtained the required quadrilateral

$ABCD$ .

1881190_1672859_ans_1210999355324fb0b455291be8b00227.jpg

Construct a quadrilateral $ABCD$ , in which $AB = 6\ cm, BC = 4\ cm, CD = 4\ cm, \angle B = 95^{\circ}$ and $\angle C = 90^{\circ}$ .

Steps of the construction:

Step I:- Draw a line segment

$BC=4\ cm$

Step II:- Construct

$\angle ABC=95^0$ .

Step III:- With

$B$ as the center and radius

$6\ cm$ , cut off

$BA=6\ cm$ .

Step IV:- Construct

$\angle BCD=90^0$ .

Step V:- With

$C$ as the center and radius

$4\ cm$ , cut off

$CD=4\ cm$ .

Step VI:- Join

$AD$

Thus, the required quadrilateral

$ABCD$ is obtained.

1881226_1672868_ans_5a7e488b75d44661bc86be32799d0605.jpg

Construct a quadrilateral $PQRS$ , in which $\angle Q = 45^{\circ}, \angle R = 95^{\circ}; QR = 5\ cm, PQ = 9\ cm$ and $RS = 7\ cm$ .

Steps of the constructions:

Step I:- Draw the

$QR=5\ cm$

Step II:- Construct

$\angle PQR=45^0$ .

Step III:- With

$Q$ as the center and radius

$9\ cm$ , cut off

$QP=9\ cm$ .

Step IV:- Construct

$\angle QRS=95^0$ .

Step V:- With

$R$ as the center and radius

$7\ cm$ , cut off

$RS=7\ cm$ .

Step VI:- Join

$CD$

Since the line segment,

$PQ$ and

$RS$ intersect each other, the quadrilateral can not be constructed.

1881148_1672874_ans_746470a493b24a6399aa14bc737c7373.png

Describe completely the locus of points in each of the following cases:
point in a plane equidistant from a given line.

We know that

$AB$ is the given line and

$P$ is a point in the plane.
From the point

$P$ , construct a line CD and another line

$EF$ from

$P’$ parallel to

$AB$ .
Hence, CD and EF are the lines which are the locus of the point equidistant from AB.
1705721_1783533_ans_c2dffd8404c7477a8c55fb473bdb3afa.PNG

Construct a quadrilateral $ABCD$ , where $AB = 4.2\ cm, BC = 3.6\ cm, CD = 4.8\ cm, angle B = 30^{\circ}$ and $\angle C = 150^{\circ}$ .

Steps of the constructions:

Step I:- Draw the

$BC=3.6\ cm$

Step II:- Construct

$\angle ABC=30^0$ .

Step III:- With

$B$ as the center and radius

$4.2\ cm$ , cut off

$BA=4.2\ cm$ .

Step IV:- Construct

$\angle BCD=150^0$ .

Step V:- With

$C$ as the center and radius

$4.8\ cm$ , cut off

$CD=4.8\ cm$ .

Step VI:- Join

$AD$

Then, the obtained the required quadrilateral

$ABCD$ .

1880999_1672869_ans_5fa5e33473bc4e70881f71a14b69f0e9.jpg

Construct a quadrilateral $PQRS$ , in which $PQ = 3.5\ cm, QR = 2.5\ cm, RS = 4.1\ cm, \angle Q = 75^{\circ}$ and $\angle R = 120^{\circ}$ .

Steps of the constructions:

Step I:- Draw the line

$QR=2.5\ cm$

Step II:- Construct angle

$Q=75^0$ .

Step III:- Cut an arc of radius

$3.5\ cm$ with

$Q$ as the center to mark that point as

$P$ .

Step IV:- Construct angle

$R=120^0$ .

Step V:- Cut an arc of radius

$4.1\ cm$ with

$R$ as the center to mark that point as

$S$ .

Step VI:- Join

$QP, RS$ and

$PS$ .

Then, the obtained the required quadrilateral

$PQRS$ .

1881005_1672870_ans_5ba1e487d325410a9a6ff671089784b9.jpg

Construct a quadrilateral $BDEF$ , where $DE = 4.5\ cm, EF = 3.5\ cm, FB = 6.5\ cm, \angle F = 50^{\circ}$ and $\angle E = 100^{\circ}$ .

Steps of the constructions:

Step I:- Draw the

$EF=3.5\ cm$

Step II:- Construct

$\angle DEF=100^0$ .

Step III:- With

$E$ as the center and radius

$4.5\ cm$ , cut off

$DE=4.5\ cm$ .

Step IV:- Construct

$\angle EFB=50^0$ .

Step V:- With

$F$ as the center and radius

$6.5\ cm$ , cut off

$FB=6.5\ cm$ .

Step VI:- Join

$BD$

Then, the obtained the required quadrilateral

$ABCD$ .

1881205_1672876_ans_725fbba91aa341e4b5c5cfd80fb432d6.png

Construct a quadrilateral $ABCD$ in which $AB = BC = 3\ cm, AD = 5\ cm, \angle A = 90^{\circ}$ and $\angle B = 105^{\circ}$ .

Steps of the constructions:

Step I:- Draw the $AB=3\ cm$ .

Step II:- Construct $\angle DAB=90^0$ .

Step III:- With $A$ as the center and radius $5\ cm$ , cut off $AD=5\ cm$ .

Step IV:- Construct $\angle ABC=105^0$ .

Step V:- With $B$ as the center and radius $3\ cm$ , cut off $BC=3\ cm$ .

Step VI:- Join $$CD$.

Then, the obtained the required quadrilateral $ABCD$ .

1881140_1672875_ans_e13b305ee7714dc7968796879d472fce.png

Construct a quadrilateral $ABCD$ in which $BC = 4\ cm, CA = 5.6\ cm, AD = 4.5\ cm, CD = 5\ cm$ and $BD = 6.5\ cm$ .

Steps of the constructions:

Step I:- Draw the

$BC=4\ cm$

Step II:- With

$B$ as the center and radius

$6.5\ cm$ and draw an arc.

Step III:- With

$C$ as the center and radius

$5\ cm$ and draw an arc to intersect the arc drawn in step II at

$D$ .

Step IV:- With

$C$ as the center and radius

$5.6\ cm$ and draw an arc on the same side.

Step V:- With

$D$ as the center and radius

$4.5\ cm$ and draw an arc to intersect the arc drawn in step IV at

$A$ .

Step VI:- Join

$BA, DA, CD, AC$ and

$BD$ .

Then, the obtained the required quadrilateral

$ABCD$ .

1881196_1672861_ans_9890c4ef6ed248fbbe4ca0ec2c1788b7.jpg

Construct a quadrilateral $ABCD$ , in which $AD = 3.5\ cm, AB = 4.4\ cm, BC = 4.7\ cm, \angle A = 125^{\circ}$ and $\angle B = 120^{\circ}$ .

Steps of the constructions:

Step I:- Draw the

$AB=4.4\ cm$

Step II:- Construct

$\angle BAD=125^0$ .

Step III:- With

$A$ as the center and radius

$3.5\ cm$ , cut off

$AD=3.5\ cm$ .

Step IV:- Construct

$\angle ABC=120^0$ .

Step V:- With

$B$ as the center and radius

$4.7\ cm$ , cut off

$BC=4.7\ cm$ .

Step VI:- Join

$CD$

Then, the obtained the required quadrilateral

$ABCD$ .

1881009_1672873_ans_302cb4dffaf14ff2852c3eb8e309c36f.jpg

Construct a quadrilateral $ABCD$ given $BC = 6.6\ cm, CD = 4.4\ cm, AD = 5.6\ cm$ and $\angle D = 100^{\circ}$ and $\angle C = 95^{\circ}$ .

Steps of the constructions:

Step I:- Draw the

$DC=4.4\ cm$

Step II:- Construct

$\angle ADC=100^0$ .

Step III:- With

$D$ as the center and radius

$5.6\ cm$ , cut off

$DA=5.6\ cm$ .

Step IV:- Construct

$\angle BCD=95^0$ .

Step V:- With

$C$ as the center and radius

$6.6\ cm$ , cut off

$CB=6.6\ cm$ .

Step VI:- Join

$AB$

Then, the obtained the required quadrilateral

$ABCD$ .

1881006_1672872_ans_999318405a8c415d950508bc13adf3b7.jpg

Construct a quadrilateral $ABCD$ , where $\angle A = 65^{\circ}, \angle B = 105^{\circ}, \angle C = 75^{\circ}, BC = 5.7\ cm$ and $CD = 6.8\ cm$ .

1872291_1672881_ans_2f510a9794b249c788ee46d56faabd6d.jpg

Construct a quadrilateral $ABCD$ in which $AB = 3.6\ cm$ , $BC = 3.3\ cm$ , $AD = 2.7\ cm$ , diagonal $AC = 4.6\ cm$ and diagonal $BD = 4\ cm.$

The given Quadrilateral

$ABCD$ can be drawn by following the steps:-

Draw a line $AB = 3.6\ cm.$
Taking $4.6\ cm$ as radius draw an arc from $A.$
From $B$ taking the radius of $3.3\ cm$ and cut the arc drawn in step $2$ such that the point of intersection is $C.$ Join $BC.$
Draw another arc from $A$ with $2.7\ cm$ as radius.

Step 5: From

$B$ take

$4\ cm$ and cut the arc drawn in step 4 and the point of intersection is

$D.$ Join

$CD$ and

$AD.$

$\therefore$ We get the required quadrilateral

$ABCD$

1618815_1774747_ans_7c54cdde54274d9fb5475711d961a674.jpg

Construct a rectangle $ABCD$ whose adjacent sides are $11 cm$ and $8.5 cm$ .

The given rectangle

$ABCD$ can be drawn by following the steps.

Step

$1$ : Draw a line

$AB = 11 cm$
Step

$2$ : Draw

$BE$ perpendicular to

$AB$ , as we know that all angles of a rectangle are of

$90^o$
Step

$3$ : With

$8.5 cm$ from

$B$ draw an arc cutting

$BE$ at

$C$ .
Step

$4$ : With

$8.5\ cm$ as radius from

$A$ draw an arc
Step

$5$ : From

$C$ take

$11 cm$ of radius and cut the arc drawn in step

$4$ such that the point of intersection is

$D$ . Now join

$CD$ and

$AD$ .

$\therefore$ We get the required rectangle

1620792_1774803_ans_f9e0ecf5a4f44c6da2186e5584c7815c.jpg

Draw a rough sketch of a quadrilateral $PQRS$ . Draw its diagonals. Name them.

The meeting point

$O$ of the diagonals

$PR$ and

$QS$ of the quadrilateral

$PQRS$ is in the interior of the quadrilateral

$PQRS$ .

1720900_1787506_ans_dafaa0d4444a4f639122043ffe21149f.png

Without using set square or protractor, construct rhombus $ABCD$ with sides of length $4$ cm and diagonal $AC$ of length $5$ cm. Measure $ABC$ . Find the point $R$ on $AD$ such that $RB = RC$ . Measure the length of $AR.$

(i) Construct

$AB = 4$ cm.
(ii) Taking

$A$ as centre, construct an arc of radius

$5$ cm and with

$B$ as centre construct another arc of

$4$ cm radius which intersect each other at the point

$C$ .
(iii) Now join

$AC and BD.$
(iv) With

$A$ and

$C$ as centre, construct two arcs of 4 cm radius which intersect each other on

$D$ .
(v) Join

$AD$ and

$CD$ .
Hence,

$ABCD$ is the required rhombus and by measure

$\angle ABC= 78^{\circ }$
(vi) Construct perpendicular bisector of

$BC$ which intersects

$AD$ at the point

$R$ .
By measuring the length of

$AR = 1.2$ cm.
1708230_1783583_ans_b22dcf0a25dc4ae6aae0177f4a29234e.PNG

Draw and describe the locus in each of the following cases:
The locus of a point in rhombus $ABCD$ which is equidistant from $AB$ and $AD$ .

In a rhombus

$ABCD$ , join

$AC$ .

Here

$AC$ is the locus of a point in the rhombus which is equidistant from

$AB$ and

$AD$

We know that

Diagonal

$AC$ bisects

$∠A$ and

$∠C$

So any point on

$AC$ will be equidistant from

$AB$ and

$AD$ .

1705653_1783530_ans_59a75829e8f746b0a0f5d18f4663ff82.png

Construct a rhombus $PQRS$ whose diagonals $PR$ , $QS$ are $8$ cm and $6$ cm respectively. Find by construction a point $X$ equidistant from $PQ, PS$ and equidistant from $R, S$ . Measure $XR$ .

Take

$PR = 8$ cm and construct the perpendicular bisector of PR which intersects it at point

$O$ .
From the point

$O$ , cut off

$OS = OQ = 3 cm$ .
Now join

$PQ, QR, RS$ and

$SP$ .
Here

$PQRS$ is a rhombus with

$PR$ and

$QS$ as the diagonals.

$PR$ is the bisector of

$\angle SPQ$
Construct perpendicular bisector of SR which intersects PR at

$X$
Here

$X$ is equidistant from

$PQ$ and

$PS$ and also from

$S$ and

$R$ .
By measuring, length of

$XR = 3.2$ cm.
1706893_1783636_ans_1f7e5ba90ba84230a68caa22a6f601ca.PNG

Construct a quadrilateral $PQRS$ , in which $PQ = 4\ cm, QR = 5\ cm, \angle P = 50^{\circ}, \angle Q = 110^{\circ}$ and $\angle R = 70^{\circ}$ .

1872294_1672882_ans_15a99f769ffe40ae80c088da684b1524.jpg

Construct a square of side $4\ cm$ .

Step of Construction:

Construct a line

$PQ=4\ cm$

Draw

$\angle QPX=90^o$ and

$\angle PQY=90^o$

Cut the arc

$PS=4\ cm$ and

$PQ=4\ cm$ and join the line

$SR$

Therefore,

$PQRS$ is the required square.

1565123_1715824_ans_18304d714b8a4f01a9dbce57d767008c.png

Draw and describe the locus in each of the following cases:
The locus of a point in rhombus $ABCD$ which is equidistant from $AB$ and $AD$ .

In a rhombus

$ABCD$ , join

$AC$
Here

$AC$ is the diagonal of rhombus

$ABCD$
We know that

$AC$ bisects

$∠A$
Therefore, any point on

$AC$ is the locus which is equidistant from

$AB$ and

$AD$ .

1705628_1783529_ans_887226235fbe45efac2f474303ea7958.png

Fill in the blanks to make the statement true.
A quadrilateral can be constructed uniquely if its three sides and _________ angles are given.

A quadrilateral can be constructed uniquely if its three sides and the two included angles are given

Is it possible to construct quadrilateral $ABCD$ in which $AB= 3 \text{ cm}$ , $BC = 4 \text{ cm}$ , $CD = 5.4\text{ cm}$ , $DA = 5.9 \text{ cm}$ and diagonal $AC = 8\text{ cm}$ ? If not, why?

Given measures are

$AB = 3\text{ cm} , BC = 4\text{ cm}, CD = 5.4\text{ cm}, DA = 5.9\text{ cm}$ and

$AC = 8\text{ cm}$

Here, consider the

$\triangle ABC$ within the quadrilateral.

$AB + BC = 3 + 4 = 7\text{ cm}$ and

$AC = 8\text{ cm}$

i.e., the sum of two sides of a triangle is less than the third side, which is absurd.

Hence, we cannot construct such a quadrilateral.

Construct a quadrilateral NEWS in which NE $= 7 \,cm$ , EW $= 6 \,cm , \angle$ N $= 60^{\circ} , \angle$ E $= 110^{\circ}$ and $\angle$ S $= 85^{\circ}$ .

In quadrilateral

$NEWS$

$\angle S = 360 - (60 + 110 + 85) = 360^{\circ} - 255^{\circ} = 105^{\circ}$

Steps for construction :

Step 1. Draw

$NE = 7\,cm$

Step 2. Make

$\angle NEX = 110^{\circ}$

Step 3. With

$E$ as center and radius

$6\,cm$ , draw an arc, cutting

$EX$ at

$W$ .

Step 4. Make

$\angle EWY = 105^{\circ}$

Step 5. Make

$\angle ENZ = 60^{\circ}$ , so that

$NZ$ and

$WY$ intersect each other at point

$S$ .

Thus,

$NEWS$ is the required quadrilateral.

1687532_1793300_ans_47e343b86ab44fa3a3de84ae3b9630c9.png

Construct a square of side $3\,cm$ .

Steps of construction :

$1$ . Take AB =

$3\,cm$ .

$2$ . At A , draw AY

$\perp$ AB.

$3$ . With A as center and radius =

$3\,cm$ , describe an arc cutting AY at D.

$4$ . With B and D as centers and radii equal to

$3\,cm$ , draw arcs intersecting at C.

$5$ . Join BC and DC . ABCD is the required square .

1660275_1795040_ans_8b43b5342470420c867a64898ab57cb0.png

Construct a rhombus whose diagonals are $6.8\ \text{cm}$ and $5.2\ \text{cm}$ .

Steps of Construction:

(i) Construct $AC = 6.8\ \text{cm}$ .

(ii) Construct one bisector $PQ$ of $AC$ to meet it at the point $O$ .

(iii) From $POQ$ , cut off $and$ OD$$ such that

$OB = OD = \frac{1}{2} BD = \frac{1}{2} \times 5.2 = 2.6\ \text{cm}$

(iv) Now join $AB, BC, CD$ and $DA$ .

Therefore, $ABCD$ is the required rhombus.

1712650_1810233_ans_1084eb2e0443432a8782731058594e85.PNG

Construct a rectangle ABCD such that AB = 4.5 cm and BC = 3 cm. Draw its line (or lines) of symmetry.

Steps to construct:
Step 1 : Draw a line segment AB = 4.5 cm
Step 2 : Draw a line BQ from point B making an angle of

$90^{0}$
Step 3 : Taking B as centre and with radius = 3 cm, cut an arc on BQ, mark it as point C.
Step 4 : Now taking point C as centre and radius = 4.5 cm, draw an arc and with A as centre and radius = 3 cm, cutting the previous arc with marking that point as D.
Step 5 : Join CD and AD
Step 6 : ABCD is are required rectangle.
The lines of symmetry:
The lines of symmetry = 2, i.e. the lines joining the midpoints of opposite sides.
1764122_1809741_ans_542f85f7bee44db0b043770f98f94736.PNG

1764122_1809741_ans_542f85f7bee44db0b043770f98f94736.PNG

Construct each of the following and give justification :
A rhombus whose diagonals are $4\,cm$ and $6\,cm$ in lengths .

To construct: A rhombus ABCD.

Steps of construction :

$1$ . Take AC =

$6\,cm$ .

$2$ . Draw BD the right bisectors of AC.

$3$ . Cut off MB = MD =

$2\,cm$ .

$4$ . Join AB , BC , CD and DA .

Hence , ABCD is the required rhombus .

1660393_1795050_ans_433d0277b95f4be383307b5fa76e1bb4.png

Construct a quadrilateral PQRS where PQ = 4 cm, QR = 6 cm, P = $60\cdot$ Q = $90\cdot$ and R = $120^\cdot$

Steps of Construction:

(i) Construct a line segment QR = 6 cm.

(ii) At the point Q, construct a ray QX making an angle of $90^\cdot$ and cut off QP = 4 cm.

(iii) At the point P, construct a ray making an angle of $60^\cdot$ and at R, a ray making an angle $120^\cdot$ which meets each other at the point S.

Therefore, PQRS is the required quadrilateral.

1712165_1809572_ans_29772ce6354a4834ae1473e1a6222ff5.PNG

Construct a quadrilateral PQRS in which PQ = 3 cm, QR = 2.5 cm, PS = 3.5 cm, PR = 4 cm and QS = 5 cm.

Steps of Construction:

(i) Construct PQ = 3 cm.

(ii) Taking P as centre and 4 cm radius, construct an arc

Taking Q as centre and 2.5 cm radius, construct an arc which meets the previous arc at R

Now join PQ and QR

(iii) Taking P as centre and 3.5 cm radius, construct an arc

Taking Q as centre and 5 cm radius, construct an arc which meets the previous arc at S.

(iv) Join PS, QS and SR.

Therefore, PQRS is the required quadrilateral.

1712092_1809488_ans_5098eda9ae394a419430de34ac78835a.PNG

Construct a rectangle whose adjacent sides are 5.6 cm and 4 cm.

Steps of Construction:

(i) Construct $AB = 5.6 cm.$

(ii) At the point B, draw $∠ABP = 90^\cdot$

(iii) Cut off $BC = 4 cm$ from BP.

(iv) Taking C as centre and $5.6 cm$ radius, construct an arc.

(v) Taking A as centre and $4 cm$ radius, construct an arc which meets the previous arc at point D.

(vi) Now join AD and CD.

Therefore, ABCD is the required rectangle.

1712538_1810160_ans_83fb28a504a047eab6ca1a4a3e802395.PNG

Construct a quadrilateral ABCD in which AB = 3.5 cm, BC = 5 cm, CD = 5.6 cm, DA = 4 cm, BD = 5.4 cm.

Steps of Construction:

(i) Construct a line segment AB = 3.5 cm.

(ii) Taking A as centre and 4 cm radius construct and arc and with B as centre and 5.4 cm construct an arc which meets the previous arc at the point D.

Now join AD and BD.

(iii) Taking B as centre and 5 cm radius, construct an arc

Taking D as centre and 5.6 cm radius, construct an arc which meets the previous arc at the point C.

(iv) Join BC and CD.

Hence, ABCD is the required quadrilateral.

1712066_1809487_ans_88c50fb107d04f1190f6c0377b2f6c82.PNG

Construct a quadrilateral ABCD given that BC = 6 cm, CD = 4 cm, B = $45^\cdot$ , C = $90^\cdot$ and D = $120^\cdot$

Steps of Construction:

(i) Construct BC = 6 cm.

(ii) At the point B, draw ∠CBP = $45^\cdot$

(iii) At the point C, draw ∠BCQ = $90^\cdot$

(iv) Cut off CD = 4 cm from CQ.

(v) At the point D, draw ∠CDR = $120^\cdot$

(iv) Let BP and DR meet at the point A.

Therefore, ABCD is the required quadrilateral.

1712138_1809537_ans_eaa98ffd99554f39aa9aa67bb055f055.PNG

Construct a quadrilateral ABCD such that BC = 5 cm, AD = 5.5 cm, CD = 4.5 cm, AC = 7 cm and BC = 5.5 cm.

Steps of Construction:

(i) Construct a line segment CD = 4.5 cm.

(ii) Taking C as centre and 5.5 cm radius and taking D as centre and 7 cm radius construct arcs which intersect each other at B.

(iii) Join BC and BD.

(iv) Taking C as centre and 5.5 cm radius and taking D as centre and 5.5 cm radius, construct arcs which intersect each other at A.

(v) Now join AC and AD.

(vi) Join AB.

Therefore, ABCD is the required quadrilateral.

1764118_1809518_ans_272989410a5f4068ad423c9e10f37bb1.PNG

Construct a quadrilateral PQRS where PQ = 4.5 cm, QR = 6 cm, RS = 5.5 cm, PS = 5 cm and PR = 6.5 cm.

Steps of Construction:

(i) Construct a line segment PR = 6.5 cm.

(ii) Taking P as centre and 4.5 cm radius and R as centre and 6 cm radius construct arcs which intersect each other at Q.

(iii) Now join PQ and QR.

(iv) Taking P as centre and 5 cm radius and R as centre and 5.5 cm radius, construct arcs which intersect each other at S.

(v) Join PS and SR.

Hence, PQRS is the required quadrilateral.

1712061_1809472_ans_b146ced5e9f542618d4de08b2bbffda3.PNG

Using ruler and compasses, construct a rectangle $ABCD$ with $AB = 5cm$ and $AD = 3 cm$

Steps of construction:
(i) Draw a straight-line

$AB = 5cm$
(ii) At

$A$ and

$B$ construct

$\angle XAB$ and

$\angle YBA = 90^{o}$ using compass
(iii) From

$A$ and

$B$ cut off

$AC$ and

$BD = 3cm$ each assuming

$A$ and

$B$ as center respectively
(iv) Join

$CD$
Thus,

$ABCD$ is the required rectangle.
1729484_1811501_ans_dc92282321aa4f7388c6332167f7e392.PNG

Construct a square ABCD, when :
One side = $4.5$ cm,.

Steps :
(i) Draw a line segment AB =

$4.5$ cm.
(ii) Draw

$AP \perp AB$ .
(iii) From AP cut off AD =

$44.5$ .
(iv) With B as a centre and radius 4.5 cm draw an are.
(v) With D as centre and radius 4.5 cm draw another are cutting the former arc at C.
(vi) Join BC and CD.
ABCD is the required square.
1800250_1841160_ans_8d2da8149ac2478191036f6a74febe1c.png

1800250_1841160_ans_8d2da8149ac2478191036f6a74febe1c.png

Construct a quadrilateral $PQRS$ , when $PQ= 4 \ cm, QR= 3 \ cm, QS= 4.8 \ cm, PS= 3.5 \ cm$ and $PR= 5 \ cm$ .

Step of construction.

$1.$ Draw

$PQ = 4 cm$ as base.

$2.$ With

$P$ as centre and radius

$5 cm$ draw an arc.

$3.$ With

$Q$ as centre, draw another arc

$QR = 3 cm.$ These arcs intersect at

$R.$

$4.$ With centre

$P,$ draw an arc

$SP = 3.5 cm$ and with centre

$Q$ draw an arc

$QS = 4.8 cm.$ These arcs intersect at point S. Join

$QR, RS, SP, PR$ and

$SQ.$

Hence,

$PQRS$ is required quadrilateral.

1882151_1879286_ans_22d27855ea9f405e8d305a491dd879b8.png

Construct a rhombus $PQRS$ whose diagonals $PR$ and $QS$ are $8cm$ and $6cm$ respectively.

Steps of construction:
(i) Draw a line segment

$PR = 8cm$
(ii) Draw its perpendicular bisector

$XY$ intersecting it at

$O$ .
(iii) From

$XY$ , cut off

$OQ = OS = 6/2 = 3\ cm$ each.(by drawing

$3 \ cm$ arc from

$O$ along

$XY$ line)
(iv) Join

$PQ, QR, RS$ and

$SP$
Thus,

$PQRS$ is the required rhombus.
1729518_1811524_ans_b377b3abc5ad44bfac36ff80605a1f77.PNG

Construct a square ABCD, when :
Perimeter = $24$ cm.

The perimeter of a square
P = 4a
Where a is the length of each side.
We have Perimeter = 24 cm
There fore 24 = 4a
a = 6
Therefore the sides of the squares are of length 6 cm.
Steps :
(i) Draw a line segment AB = 6cm
(ii) Draw

$AP \perp AB$ .
(iii) From AP cut off AD = 6 cm.
(iv) With B as a center and radius, 6 cm draw an are.
(iv) With B as a center and radius 6 cm draw another is cutting the former are at C.
(vi) join BC and CD,
ABCD is the required square.
1800252_1841164_ans_c0d57ab083ce4553b25708d5885a71e8.png

1800252_1841164_ans_c0d57ab083ce4553b25708d5885a71e8.png

Construct a rectangle ABCD, when :
Its sides are $6.0$ cm and $7.2$ cm.

Since each angle of a rectangle is

$90^{\circ}$ and opposite sides are equal. Therefore.
Steps :
(i) Draw BC =

$7.2$ cm
(ii)with B as a center dra a line BX taking as a

$90^{\circ}$ .
(iii) Now taking radius

$6$ cm draw an are at A.
(iv) From point A draw a line AY parallel to BC.
(v) With A as a center taking radius

$7.2$ cm draw an are at D.
(vi) Now join CD.
ABCD is the required rectangle.
1800232_1841100_ans_14a62b6d90dd4e7abe2fa86afe781004.png

Construct a quadrilateral $ABCD$ in which $AB = 3.3 cm, BC = 4.9 cm, CD = 5.8 cm, DA = 4 cm$ and $BD = 5.3 cm.$

Steps of construction:
(i) Draw a line segment

$AB = 3.3cm$
(ii) with centre

$A$ and radius

$4cm$ , and with centre

$B$ and radius

$5.3cm$ , draw arcs intersecting each other at

$D$ .
(iii) Join

$AD$ and

$BD$ .
(iv) With centre

$B$ and radius

$4.9 cm$ and with centre

$D$ and radius

$5.8cm$ , draw arcs intersecting each other at

$C$
(v) Join

$BC$ and

$DC$
Thus,

$ABCD$ is the required quadrilateral.
1729305_1811245_ans_20733d7f87604f3fa8814cb63f45c2f3.PNG

Construct a rhombus ABCD, when :
Diagonal AC = $6.3$ cm and diagonal BD = $5.8$ cm.

Steps :
(i) Draw AC =

$6.3$ cm.
(ii) Draw perpendicular bisector to AC which cuts AC at O.
(iii) From this perpendicular cut OD and OB such that.

$OD = OB = \dfrac{1}{2} BD = \dfrac{1}{2} \times 5.8 = 2.9 cm$ .
(iv) Join AB, BC, CDD and DA.
ABCD is the required rhombus.
1800245_1841141_ans_9d0e12c1763342e58049c5514118d42d.png

1800245_1841141_ans_9d0e12c1763342e58049c5514118d42d.png

Constant a quadrilateral ABCD, when :
Ab= $3.2$ cm, BC = $5.2$ cm, CD = $6.2$ cm, DA = $4.2$ cm and BD = $5.2$ cm

Steps :
(i) Draw AB = 3.2 cm
(ii) With A as a center draw an arc of radius 4.2 cm at D and with B as a center and radius

$5.2$ cm draw an are at D.
(iii) join AD and DB.
(iv) With D and B as a center taking radius

$6.2$ cm and

$5.2$ cm draw are at C. Now join BC and DC
ABCD is the required quadriateral.
1800192_1841009_ans_5ae265be196643b3818502cfbf78c3da.png

Construct a quadrilateral $ABCD$ , when $AB = 3.5 \ cm, BC= 4.8 \ cm, CD= 5.1 \ cm, AD= 4.4 \ cm$ and diagonal $AC= 5.9 \ cm$ .

1. Draw

$AB= 3.5 \ cm$ as base.
2. With centre

$A$ , draw an arc

$AC= 5.9 \ cm$ .
3. With

$B$ as centre, draw an arc

$BC= 4.8 \ cm$ . These arcs intersect at

$C$ .

4. With centre

$A$ , draw an arc

$AC= 4.4 \ cm$ .

5. With

$C$ as centre, draw an arc

$CD= 5.1 \ cm$ . These arcs intersect at

$D$ .

Now, Join the points to get required quadrilateral

1879531_1879227_ans_161aaf2776b94937b5fe1db18bfba285.png

Construct a quadrilateral $PQRS$ , when $PQ= 2.5 \ cm, QR= 3.7 \ cm, \angle Q= 120^\circ, \angle S = 60^\circ$ and $\angle R= 90^\circ$ .

$∠P + ∠Q + ∠R + ∠S = 360°$

$∠P = 360° – (120° + 90° + 60°)$

$=> ∠P = 90°$

1. Draw a line segment $PQ = 2.5 cm.$

2. At $P$ draw a ray $PX$ such that $∠XPQ = 90°$ and at $Q$ draw a ray $QY$ such that $∠PQY = 120°.$

3. With $Q$ as centre, take an arc of radius $3.7 cm$ and mark it as $R.$

4. Also at point $R$ construct an angle of $90°$ which will meet ray $PX$ at $S.$

Hence, $PQRS$ is the required quadrilateral.

1882209_1879307_ans_a6a4a0c072514e5eaeb9d2bfeb079f44.png

Construct a rhombus $ABCD$ , such that diagonals $AC= 7 \ cm$ nad diagonal $BD= 5 \ cm$ .

1. Draw one of the diagonal

$AC= 7 \ cm$ as base.
2. Draw perpendicular bisector of

$AC$ which intersects at

$O$ .
3. Take

$O$ as centre and cut an arc of

$OD= 2.5 \ cm$ above and

$OB= 2.5 \ cm$ below the bisector line.
4. Join

$AD, DC, AB$ and

$BC$ .
Hence,

$ABCD$ is the required rhombus.
1867029_1879335_ans_c1faa009c6d14e7a8f5793c4b1ab1b78.png

Construct a quadrilateral $ABCD$ , when $AB= BC= 3 \ cm, AD = 5 \ cm, \angle A = 90^\circ$ and $\angle B= 120^\circ$ .

$1.$ Draw

$AB= 3 \ cm$ as base.

$2.$ At

$A$ make an angle of

$90^\circ$ .

$3.$ With

$A$ as a centre and radius

$5cm$ cut off

$AD=5cm$

$4.$ At

$B$ make an angle of

$120^\circ$ .

$5.$ With

$B$ as a centre and radius

$3cm$ cut off

$BC=3cm$

$6.$ join

$CD$

Hence quadrilateral

$ABCD$ is required.

1882195_1879297_ans_70ea45cc52454771808decaa3fdeead5.png

Construct a quadrilateral $ABCD$ in which $AB= AD = 3.2 \ cm, BC= 2.5 \ cm, AC= 4 \ cm$ and $BD= 5 \ cm$ .

1. Draw

$AB= 3.2 \ cm$ .
2. With centre

$A$ , draw an arc of radius

$3 \ cm$ and by taking

$B$ as centre take an arc of

$2.5 \ cm$ and cut the previous arc at

$C$ .
3. Further, take

$B$ as centre and draw an arc of

$5 \ cm$ and cut an arc by

$3.2 \ cm$ from

$A$ and mark that point as

$D$ .
4. Join

$C$ to

$D$ .
Hence,

$ABCD$ is the required quadrilateral.
1880724_1879368_ans_d54f257c85a34dd3a6578fe54766d2fd.png

Construct a rhombus the length of whose diagonals are $7.2 \ cm$ and $5.6 \ cm$ .

1. Draw diagonal

$AC= 7.2 \ cm$ as base line.
2. Draw

$\perp$ bisector of

$AC$ which intersects

$AC$ at

$O$ .
3. Draw an arc of

$OD= OB = 2.8 \ cm$ above and below the

$\bot$ line and marked them as

$D$ and

$B$ .

4. Join

$DA,DC,AB$ and

$BC$

Hence

$ABCD$ is a required rhombus.

1883626_1879381_ans_0e59040f5c824f648ef9eac857a5abf8.png

Construct a quadrilateral $PQRS$ , when $PQ= 3 \ cm, QR= 3.5 \ cm, \angle Q= 90^\circ$ and $\angle P = 105^\circ, \angle R= 120^\circ$ .

1. Draw a line segment

$PQ= 3 \ cm$ as base.
2. At

$P$ draw a ray

$PX$ such that

$\angle QPS = 105^\circ$ .
3. At

$Q$ draw a ray

$QZ$ such that

$\angle PQZ = 90^\circ$ and mark an arc of radius

$3.5 \ cm$ on it. Mark this point as

$R$ .
4. At

$R$ draw

$\angle QRS= 120^\circ$ which cuts

$PX$ at

$S$ .
Hence,

$PQRS$ is the required quadrilateral.
1871721_1879305_ans_d924b1f1f12f480782256b5a5c3d2058.png

Construct a quadrilateral $PQRS$ where $PQ= 3.5 \ cm, QR= 3.5 \ cm, \angle P = 60^\circ, \angle Q = 105^\circ$ and $\angle S = 75^\circ$ .

$1.$ Draw

$PQ = 3.5 cm.$

$2.$ At

$P$ make an angle of

$60^o$ with the help of compass.

$3.$ At

$Q$ draw the angle of

$105^o$ with the help of protractor. Again by taking

$Q$ as centre, draw an arc of

$3.5 cm$ which intersects the line segment at

$R.$

$4.$ At R again draw an angle of

$120^o$ .

$i.e. 360^o – (60^o + 105^o + 75^o) = 360^o – 240^o = 120^o$

which will intersect the arm of the angle

$60^o$ at

$S.$

Hence,

$PQRS$ is the required quadrilateral.

1882916_1879369_ans_e3ef9f67ff0b4d4597e79c4291dcaeba.png

Construct a quadrilateral $ABCD$ , when $AB= 3.5 \ cm, BC = 3 \ cm, AD = 2.5 \ cm, AC= 4.5 \ cm$ and $BD= 4 \ cm$ .

$1.$ Draw

$AB = 3.5 cm$ as base.

$2.$ By taking

$A$ and

$B$ as centres, draw an arc of radius

$4.5 cm$ and

$4 cm$ respectively.

$3.$ Also from

$B,$ draw an arc of radius

$3 cm$ which intersects the previous arc of radius

$4.5 cm$ at

$C.$

$4.$ Again from

$A,$ draw an arc of

$2.5$ cm which intersects the arc of

$4 cm$ at

$D.$

$5.$ Join

$A$ to

$D, B$ to

$C, B$ to

$D$ and

$C$ to

$D.$

Hence,

$ABCD$ is the required quadrilateral.

1882173_1879293_ans_94b21834364a4461ac49b53fe91e88f5.png

To construct a quadrilateral $ABCD$ , when $AB= 3.8 \ cm, BC= 2.5 \ cm, CD= 4.5 \ cm, \angle B= 30^\circ$ and $\angle C= 150^\circ$ .

1. Draw

$BC = 2.5 \ cm$ as base.
2. At

$B$ and

$C$ , make

$\angle B= 30^\circ$ and

$\angle C= 150^\circ$ .
3. With

$B$ and

$C$ as centres, draw an arc of

$3.8 \ cm$ and

$4.5 \ cm$ respectively.
1870107_1879302_ans_9c6a36defac14be39076e8c506a73758.png

To construct a quadrilateral $PQRS$ when $PQ= 3.0 \ cm, QR = 4.0 \ cm, PS= 4.5 \ cm, QS= 5.5 \ cm$ and $PR = 6 \ cm$ .

$1.$ Draw

$PQ = 3.0$ cm as base.

$2.$ With

$P$ as centre and radius

$6 cm$ draw an arc.

$3.$ With

$Q$ as centre, draw another arcs of radius

$4 cm,$ these arcs intersect at

$R.$

$4.$ With centre

$P,$ draw an arc

$PS = 4.5 cm$ and with centre

$Q,$ draw an arc

$QS = 5.5 cm.$

$5.$ These arcs intersect at point

$S.$

$6.$ Join

$QR, RS, SP, PR$ and

$SQ.$

Hence,

$PQRS$ is the required quadrilateral.

1882187_1879296_ans_c17fe3a45a004466965c3959bd27a230.png

Parallelogram MORE
$OR =6 \ cm$
$RE=4.5 \ cm$
$EO =7.5\ cm$

2158174_1986543_ans_67ba3d902b7340b1bea93b543c5d927c.jpg

Construct quadrilaterals to the given measurements; the order of the letters in each case is A, B, C, D.
AB = 4.8 cm, BC = 4.4 cm, CD = 7 cm,DA = 3.4 cm, BD = 6.2 cm.

QUADRILATERALS - EXERCISE 4.2.3 - Class 9

Steps of construction

Draw

$AB = 4.8$ cm.

With

$A$ as centre and

$3.4$ cm radius draw an arc.

With

$B$ as centre and

$6.2$ cm as radius drawn an arc to intersect the first at

$D$ .

Join

$AD$ and

$BD$ .

With

$B$ as centre and

$4.4$ cm as radius drawn an arc.

With

$D$ as centre and

$7$ cm as radius draw an arc to intersect the above at

$C$ .

Join

$DC$ and

$BC$ .

Hence,

$ABCD$ is the required quadrilateral in the above drawn figure.

Construct a quadrilateral $ABCD$ in which $BC = 4$ cm, $AB = 4.6$ cm, $\angle {B}={ 70 }^{\circ }$ , $\angle{ A}= { 90 }^{ \circ }$ and $\angle {C}={120 }^{ \circ}.$

REF.Image

Draw AB=4.6 cm

Using a protractor, draw a line from A making

$90^{\circ}$ with the line segment AB.

Using a protractor, draw a line from B making

$70^{\circ}$ with the line segment BA.

Set the campus to the radius 3 cm and make an arc from the point B on the

$70^{\circ}$ line. Making the point as C where the arc intersects the line.

Using the protactor, draw a line from the point C making

$120^{\circ}$ with the line segmnt CB.

ABCD is the required quadrilateral.

1219948_1505522_ans_885d30ca718745f998e5998c88922c4b.png

Construct the following quadrilaterals:
Quadrilateral $DEAR,$ $DE=4\ cm,$ $EA=5\ cm,$ $AR=4.5\ cm,$ $\angle E=60^{o}$ and $\angle A=90^{o}$ .

Step 1. Draw a line segment $DE = 4\;{\rm{cm}}$ .

Step 2. Draw an arc from point $E$ with radius $5\ cm$ and having an angle of $60^\circ$ which intersects at $A$ .

Step 3. Draw an arc of radius $4.5\ cm$ with center $A$ having an angle of $90^\circ$ which intersects at $R$ .

Step 4. Join $RD$ .

1010115_1057143_ans_f91de4ec37ae4b189a068fa3ae465825.png

Construct quadrilateral to the given measurement, the order of the letterr in this case is $A, B, C, D$
$AC = 6 \ cm$ , $BD = 5.8 \ cm$ , $AB = 3.4 \ cm$ , $AD= 5.2 \ cm$ , $BC = 4.2 \ cm$

Steps of construction:

Draw

$AB = 3.4$ cm

With

$A$ as centre and

$6$ cm radius draw an arc.

With

$B$ as centre and

$4.2$ cm as radius drawn another arc to intersect the above at

$C$ .

Join

$AC$ and

$BC$ .

With

$A$ as centre

$5.2$ cm as radius draw an arc.

With

$B$ as centre and

$5.8$ cm as radius draw another arc to intersect the above at

$D$ .

Join

$AD, BD$ and

$CD$ .

Hence,

$ABCD$ is the required quadrilateral in the above drawn figure.

Construct a quadrilateral ABCD with $AB = 6.5\ \text{cm}$ , $AD = 5\ \text{cm}$ , $CD = 5\ \text{cm}$ , $\angle$ BAC = $40^\circ$ and $\angle$ ABC = $50^\circ$ , and also find its area.

Given:
AB = 6.5 cm, AD = 5 cm, CD = 5 cm,

$\angle BAC$ =

$40^\circ$ and

$\angle ABC$ =

$50^\circ$ .
Steps for construction:
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment

$AB = 6.5\ \text{cm}$ .
Step 3 : At A on

$\overline{AB}$ make

$\angle BAX$ whose measure is

$40^\circ$ and at B on

$\overline{AB}$ make

$\angle ABY$ whose measure is

$50^\circ$ . They meet at

$C$ .
Step 4 : With A and C as centres draw two arcs of radius

$5\ \text{cm}$ and let them cut at

$D$ .
Step 5 : Join

$\overline{AD}$ and

$\overline{CD}$ .

$ABCD$ is the required quadrilateral.
Step 6 : From

$D$ draw

$\overline{DE}\perp \overline{AC}$ and from

$B$ draw

$\overline{BC}\perp \overline{AC}$ . Then measure the lengths of

$BC$ and

$DE$ .

$BC = h_1 = 4.2\ \text{ cm}, DE = h_2 = 4.3\ \text{cm}$ and

$AC = d = 5\ \text{cm}.$

Calculation of area:
In the quadrilateral ABCD,

$d = 5\ \text{cm}, h_1 = 4.2\ \text{cm}$ and

$h_2 = 4.3\ \text{cm}$ .
Area of the quadrilateral

$ABCD$ =

$\dfrac {1}{2}d(h_1+h_2)$
=

$\dfrac {1}{2}(57.8)(4.2+4.3)$
=

$\dfrac {1}{2}\times5 \times8.5$
=

$21.25 \text{cm}^2$

Construct quadrilateral to the given measurement, the order of the letters in this case are $A, B, C, D$
$AC =5.8 \ cm$ , $BD = 5.6 \ cm$ , $AB = 3.5 \ cm$ , $AD= 4.9 \ cm$ , $CD = 4 \ cm$

Steps of construction:

Draw

$AB = 3.5$ cm

With

$A$ as centre and

$4.9$ cm radius draw an arc.

With

$B$ as centre and

$5.6$ cm as radius drawn another arc to intersect the above at

$D$ .

Join

$AD$ and

$BD$ .

With

$A$ as centre

$5.8$ cm as radius draw an arc.

With

$D$ as centre and

$4$ cm as radius draw another arc to intersect the above at

$C$ .

Join

$DC, AC$ and

$BC$ .

Hence,

$ABCD$ is the required quadrilateral in the above drawn figure.

Construct rectangle $ABCD$ with the following data:
$AB = 4 cm$ , $BC = 6 cm$

Steps of construction:

1. Draw

$AB = 4$ cm.

2. Construct a perpendicular to

$AB$ at

$B$ and cut off

$BC = 6$ cm.

3. With

$A$ as center and

$6$ cm as radius draw an arc.

4. With

$C$ as center and

$4$ cm as radius draw another arc to intersect the

$D$ .

5. Join

$AD$ and

$CD$ .

Hence,

$ABCD$ is the required rectangle is the above drawn figure.

Construct the following Quadrilateral of given measures.
In $\ell(MO)= 5.8\ cm, \ell(OR)=4.4\ cm, m\angle M=58^{o}, m\angle O=105^{o}, m\angle R=90^{o}$ ?

Steps of construction:

1) Draw a line segment MO of 5.8 cm.

2) Draw an angle of 58° from M and an angle of 105° from O with the help of protractor.

3) Measure a distance of 4.4 cm and make an arc on 105° angle line,mark that point R.

4) From R measure an angle of 90° extend the line till it reaches the line of 58° angle ,mark the point E where both lines intersect.

5) Quadrilateral MORE is drawn.

1112885_1109736_ans_9183b7a035624f21b846a849ccabc0d4.png

Is it possible to construct a quadrilateral ROAM in which RO $= 4 \,m$ , OA $= 5 \,cm , \angle$ O $= 120^{\circ} , \angle$ R $= 105^{\circ}$ and $\angle$ A $= 135^{\circ}$ ? If not why ?

1974331_1793291_ans_b539ec5a8a3049fc964e73be3b037807.jpg

Practical Geometry - Class 8 Maths - Extra Questions

Construct a rectangle ABCDABCD givenAB=6 cm,BC=5 cmAB=6\ cm, BC=5\ cm

Construct the quadrilaterals with the measurements given:Square LUDOLUDO with LU=4.5cmLU=4.5cm.

Construct a quadrilateral PQRSPQRS where PQ=4 cm,QR=6 cm,RS=5 cm.PS=5.5 cmPQ =4\ cm,QR=6\ cm,RS=5\ cm. PS=5.5\ cm and PR=7 cmPR =7\ cm.

Construct a rectangle ABCD, when:Its sides are 6.0 cm and 7.2 cm.

Construct a quadrilateral ABCD\text{ABCD}, when: AB=3.2 cm,BC=5.2 cm,CD=6.2 cm,DA=4.2 cm\text{AB} = 3.2\ \text{cm}, \text{BC} = 5.2\ \text{cm}, \text{CD} = 6.2\ \text{cm}, \text{DA} = 4.2\ \text{cm} and BD=5.2 cm\text{BD} = 5.2\ \text{cm}.

Construct a rhombus ABCD, given it's diagonal AC = 6.3 cm and diagonal BD = 5.8 cm.

Construct a square ABCD, given it's one diagonal = 4.5 cm

By constructing a rectangle WXYZWXYZ of base = 4.8cm4.8 \, cm and height =5 cm= 5\ cm, find the square root of its perimeter (where perimeter is in cm) upto the nearest integer.

Construct a square ABCD,ABCD, with perimeter is equal to 24 cm. 24\ cm.

Construct a quadrilateral MOREMORE where MO=6 cm,OR=4.5 cm,∠M=60o,∠O=105oMO=6\ cm,OR=4.5\ cm ,\angle M =60^o, \angle O=105^o and ∠R=105o\angle R=105^o.

Construct a square ABCDABCD, given it's one side = 4.5 cm.

Construct a quadrilateral ABCD,ABCD, given that BC=4.5 cm,AD=5.5 cmBC=4.5\ cm ,AD=5.5\ cm and CD=5 cmCD=5\ cm, the diagonal AC=5.5 cmAC=5.5\ cm and the other diagonal BD=7 cmBD=7\ cm.

Draw a square WXYZ with side 5.25.2 cm.

Construct a quadrilateral NILENILE such that NI=2.5 cm,IL=3.5 cm,NL=4 cm,IE=5 cm,EL=3 cm.NI=2.5\ cm , IL=3.5\ cm , NL= 4\ cm , IE=5\ cm , EL=3\ cm.

Construct a quadrilateral LIFTLIFT, when:LI=4 cm,IF=3 cm,TL=2.5 cm,LF=4.5 cm,IT=4 cmLI=4\ cm, IF=3\ cm ,TL =2.5\ cm ,LF=4.5\ cm ,IT=4\ cm.

Construct a quadrilateral PQRSPQRS such that PQ=3.5 cm,QR=4.5 cm,RS=4 cm,∠Q=110oPQ=3.5\ cm ,QR=4.5\ cm, RS=4\ cm, \angle Q=110^o and ∠R=80o\angle R =80^o.

Construct a quadrilateral JUMPJUMP, when: JU=3.5 cm,UM=4 cm,MP=5 cm,PJ=4.5 cm,PU=6.5 cmJU=3.5\ cm, UM =4\ cm, MP=5\ cm, PJ=4.5\ cm, PU=6.5\ cm

Construct a quadrilateral ABCDABCD such that AB=3 cm,BC=4 cm,AB=3\ cm , BC=4\ cm , ∠A=90O\angle A=90^O , ∠B=65o\angle B=65^o and ∠C=110o\angle C=110^o.

Construct a quadrilateral ABCDABCD with AB=6 cm,BC=3 cm,CD=8 cm,AB=6\ cm, BC=3\ cm, CD=8\ cm, ∠B=60∘,∠C=95∘.\displaystyle \angle B={ 60 }^{ \circ },\angle C={ 95 }^{ \circ }.

Construct a quadrilateral ABCDABCD, with BC=5BC=5 cm, CD=9CD=9 cm, DA=6DA=6 cm, ∠C=60∘,∠D=75∘\displaystyle \angle C={ 60 }^{ \circ },\angle D={ 75 }^{ \circ }

Construct a rectangle ABCDABCD with AB=8 cmAB = 8\ \mathrm{ cm} and BC=6 cmBC = 6\ \mathrm{cm}.

Construct a quadrilateral PQRSPQRS, with PQ=3PQ=3 cm, QR=9QR=9 cm, RS=5RS=5 cm, ∠Q=45∘,∠R=60∘\displaystyle \angle Q={ 45 }^{ \circ },\angle R={ 60 }^{ \circ }

Construct a quadrilateral ABCDABCD in which AB=6AB=6 cm, BC=8BC=8 cm, CD=5CD=5 cm, AD=4AD=4 cm and diagonal AC=4.5AC=4.5 cm

Construct a quadrilateral ABCDABCD with AB=6 cm,BC=8 cm,AB=6\ cm, BC=8\ cm, ∠A=75∘,∠B=90∘,∠C=45∘\displaystyle \angle A={ 75 }^{ \circ },\angle B={ 90 }^{ \circ },\angle C={ 45 }^{ \circ }.

Construct a quadrilateral PQRSPQRS with QR=8QR=8 cm, RS=4RS=4 cm, SP=9SP=9 cm, PQ=5.5PQ=5.5 cm and QS=6QS=6 cm.

Construct a quadrilateral ABCDABCD with AB=6 cm,BC=4 cm,CD=7 cm,AC=8 cmAB=6\ cm, BC=4\ cm, CD=7\ cm, AC=8\ cm and BD=7 cm.BD=7\ cm.

Construct a quadrilateral PQRS with PQ=9cm, QR=5cm, RS=8cm, PR=10cm, QS=13cm.

Construct a quadrilateral PQRSPQRS with PQ=7PQ=7 cm, QR=5QR=5 cm, RS=3RS=3 cm, ST=8ST=8 cm and PR=5PR=5 cm

Construct a quadrilateral ABCDABCD in which AB=5 cm,BC=6 cm,CD=4 cm,AD=3 cmAB=5\ cm, BC=6\ cm, CD=4\ cm, AD=3\ cm and diagonal AC=5 cm.AC=5\ cm.

Construct a quadrilateral ABCDABCD with AB=3AB=3 cm, BC=7BC=7 cm, CD=9CD=9 cm, CA=11CA=11 cm, BD=13BD=13 cm.

Construct a quadrilateral ABCDABCD with AB= 9cmAB=\ 9cm, BC=6 cmBC=6\ cm, CD=7 cmCD=7\ cm, AC=9 cmAC=9\ cm and BD=11 cmBD=11\ cm.

Construct a quadrilateral PQRS with PQ=7cm, QR=8cm, RS=6cm, PR=11cm, QS=12cm.

Construct a quadrilateral ABCDABCD with AB=5AB=5 cm, BC=9BC=9 cm, ∠A=30∘,∠B=90∘,∠C=15∘\displaystyle \angle A={ 30 }^{ \circ },\angle B={ 90 }^{ \circ },\angle C={ 15 }^{ \circ }.

Construct a quadrilateral PQRSPQRS with PQ=8PQ=8 cm, QR=6QR=6 cm, ∠P=60∘,∠Q=45∘,∠R=60∘\displaystyle \angle P={ 60 }^{ \circ },\angle Q={ 45 }^{ \circ },\angle R={ 60 }^{ \circ }.

Construct a quadrilateral MOREMORE where MO=6 cmMO=6\ cm,OR=4.5 cmOR=4.5 \ cm ,∠M=60o\angle M =60^o , ∠O=105o\angle O=105^o and ∠R=105o\angle R=105^oAlso, find the length of side EM (in cm)

Construct a quadrilateral PQRSPQRS with PQ=7PQ=7 cm, QR=6QR=6 cm, ∠P=45∘,∠Q=60∘,∠R=75∘\displaystyle \angle P={ 45 }^{ \circ },\angle Q={ 60 }^{ \circ },\angle R={ 75 }^{ \circ }.

Construct a quadrilateral ABCDABCD with AB=6AB=6 cm, BC=9BC=9 cm, ∠A=75∘,∠B=60∘,∠C=45∘\displaystyle \angle A={ 75 }^{ \circ },\angle B={ 60 }^{ \circ },\angle C={ 45 }^{ \circ }.

Construct a quadrilateral ABCDABCD with AB=4.5cm,BC=5.5cm,CD=4cm,AD=6cmAB=4.5cm, BC=5.5cm, CD=4cm, AD=6cm and diagonal AC=7cmAC=7cm.

Construct a square of side 5 cm5\ cm. Find its area also

Construct rhombus ABCDABCD such that:AC=7cmAC = 7 cm, BD=4cmBD = 4 cm

Construct rhombus ABCDABCD such that diagonal BD=6.8cmBD = 6.8 cm, AC=5.4cmAC = 5.4 cm.

Construct square ABCDABCD which has side-length 2 cm2\ cm.

Construct a quadrilateral ABCD with AB=2.5 cm, BC=3.5 cm, AD=4.5 cm, ∠A=900\angle A=90^0 and ∠B=1200 \angle B=120^0.

Construct the quadrilateral PQRSPQRS with PQ=3.5cm, QR=4cm, RS=5cm. PS=4.5cmPQ=3.5cm, QR=4cm, RS=5cm. PS=4.5cm and QS=6.5cmQS=6.5cm.

Construct a rectangle \text{ABCD}\text{ABCD} with \text{AB=5 cm, BC=3.5 cm}\text{AB=5 cm, BC=3.5 cm}

Construct a rhombus PQRSPQRS with PR = 8\ cmPR = 8\ cm and QS = 6\ cmQS = 6\ cm and find its area.

Construct a rectangle whose adjacent sides are of lengths 55 \text cm\text cm and 3.53.5 \text cm \text cm .

Construct a rhombus whose diagonals are 6 cm and 4 cm

Construct the following quadrilaterals.Quadrilateral MOREMOMO=6\ cm6\ cmOROR=4.5\ cm4.5\ cm\angle M=60^o\angle M=60^o\angle O=105^o\angle O=105^o\angle R=105^o\angle R=105^o

Construct the following quadrilaterals.Quadrilateral PLANPLPL=4cm4cmLALA=6.5cm6.5cm\angle P=90^o\angle P=90^o\angle A=110^o\angle A=110^o\angle N=85^o\angle N=85^o

Construct a quadrilateral ABCDABCD in which AB = 4.4AB = 4.4 cm, BC = 4 BC = 4 cm, CD = 6.4CD = 6.4 cm, DA = 3.8DA = 3.8 cm and BD = 6.6BD = 6.6 cm.

Construct \square DEFG\square DEFG such that ll(DE)=4.5=4.5 cm, ll(EF)=6.5=6.5cm, ll(DG)=5.5=5.5cm, ll(DF)=7.2=7.2 cm, ll(EG)=7.8=7.8 cm.

Construct a rhombus whose diagonals are 1616 cm and 1212 cm. What property of the diagonals is used?

Constant a quadrilateral ABCDABCD, where AB=4\ cm,BC=5\ cm,CD=6.5\ cmAB=4\ cm,BC=5\ cm,CD=6.5\ cm and \angle B=105^{o}\angle B=105^{o} and \angle C=80^{o}\angle C=80^{o}

Draw a quadrilateral LIFT, when LI = 4.0 cm, IF = 3.0 cm, TI = 4 cm, LF = 4.5 cm and LT = 2.5 cm.

Construct a quadrilateral ABCDABCD, where AB=4\ cm,BC=5\ cm,CD=6.5\ cmAB=4\ cm,BC=5\ cm,CD=6.5\ cm and angle B=75^{o}B=75^{o} and angle C=60^{o}C=60^{o}

Quadrilateral GOLDOL =7.5=7.5 cmGL =6=6 cmGD =6=6 cmLD =5=5 cmOD =10=10 cm.

Construct a quadrilateral ABCDABCD with the following measures: l(AB)=6.4\ cm,\ l(BC)=4.8\ cm,\ \angle A=70^o,\ \angle B=50^o,\ \angle C=140^o l(AB)=6.4\ cm,\ l(BC)=4.8\ cm,\ \angle A=70^o,\ \angle B=50^o,\ \angle C=140^o.

Construct a quadrilateral ABCDABCD in which AB=6\ cm,CD=5\ cm,DA=3.5\ cm,\angle C=120^{o}AB=6\ cm,CD=5\ cm,DA=3.5\ cm,\angle C=120^{o} and \angle D=75^{o}\angle D=75^{o}.

Quadrilateral HELP with HE=6cm,EL=4.5cm, \angle H=60^{0},\angle E=105^{0}\angle H=60^{0},\angle E=105^{0} and \angle P=120^{0}\angle P=120^{0}

Construct a quadrilateral ABCDABCD in which AB=4 \text{ cm}, BC=3.1 \text{ cm}, CD=3 \text{ cm}, AD=4.2 \text{ cm}AB=4 \text{ cm}, BC=3.1 \text{ cm}, CD=3 \text{ cm}, AD=4.2 \text{ cm} and AC=5 \text{ cm}AC=5 \text{ cm}. Measure the diagonal BDBD.

Construct a quadrilateral READREAD, when RE=3.51RE=3.51 cm, EA=4.3 EA=4.3 cm, AD=4.8AD=4.8 cm, \angle E=105^{\circ} \angle E=105^{\circ} & \angle R=55^{\circ}\angle R=55^{\circ}.

Construct a quadrilateral ABCDABCD in which AB = 4.2 cm, BC = 6 cm, CD = 5.2 cm, DA = 5 cmAB = 4.2 cm, BC = 6 cm, CD = 5.2 cm, DA = 5 cm and AC = 8 cmAC = 8 cm.

Construct a rectangle ABCDABCD given AB=7\ cm, BC=5.5\ cmAB=7\ cm, BC=5.5\ cm.

Construct a quadrilateral MIND, where MI = 3 cm, IN = 5 cm, ND = 4 cm, DM = 5 cm and MN = 7 cm.

Construct a quadrilateral ABCDABCD in which AB=3.6cm, BC=3.3cm, AD=2.7cmAB=3.6cm, BC=3.3cm, AD=2.7cm, diagonal AC=4.6cmAC=4.6cm and diagonalBD=4cmBD=4cm

Construct a quadrilateral PQRSPQRS with PQ=6.2cm, QR=5.6cm, RS=6.2cmPQ=6.2cm, QR=5.6cm, RS=6.2cm, \angle{Q}={90}^{o}\angle{Q}={90}^{o} and \angle {R}={120}^{o}\angle {R}={120}^{o}

Construct a quadrilateral ABCDABCD, when AB = 3\ cm, CD = 3\ cm, DA = 7.5\ cm, AC = 8\ cmAB = 3\ cm, CD = 3\ cm, DA = 7.5\ cm, AC = 8\ cm and BD = 4\ cmBD = 4\ cm.

Construct a quadrilateral ABCDABCD with AB=5\;cm,BC=3.6\;cm,\angle ABC=80^{\circ},\angle DAB=60^{\circ}AB=5\;cm,BC=3.6\;cm,\angle ABC=80^{\circ},\angle DAB=60^{\circ} and \angle ADC=120^{\circ}\angle ADC=120^{\circ}.

Construct a rectangle whose adjacent sides are 5.6\ cm5.6\ cm and 4\ cm4\ cm

Construct a quadrilateral PQRSPQRS with PQ=4.7\ cm, QR=3.9\ cm, \angle QPS=75^{o}, \angle PQR=120^{o}PQ=4.7\ cm, QR=3.9\ cm, \angle QPS=75^{o}, \angle PQR=120^{o} and \angle QRS=105^{o}\angle QRS=105^{o}. Measure \angle PSR\angle PSR.

Construct the following quadrilaterals :Quadrilateral RUNSRUNSRU=4RU=4 cmUN=5UN=5 cmNS=4.2 NS=4.2 cmSR=5.8 SR=5.8 cmRN=7.2 RN=7.2 cm

Quadrilateral JUMPJU=3.5 cmUM=4 cmMP=5 cmPJ=4.5 cmPU=6.5 cm

Construct a quadrilateral PQRSPQRS in which QR = 7.5 cm, PR = PS = 6 cm, RS = 5 cmQR = 7.5 cm, PR = PS = 6 cm, RS = 5 cm, and QS = 10 cmQS = 10 cm.

Construct a quadrilateral ABCDABCD in which \overline { AB } =\overline { BC } =\overline { CD } =\overline { AD } =5\overline { AB } =\overline { BC } =\overline { CD } =\overline { AD } =5 cm and diagonal \overline { AC } =6.5\overline { AC } =6.5 cm.

Construct a quadrilateral PQRSPQRS in which: PQ = 5.4 cm, QR = 4.6 cm, RS = 4.3 cm, SP = 3.5 cmPQ = 5.4 cm, QR = 4.6 cm, RS = 4.3 cm, SP = 3.5 cm, and Diagonal PR = 4 cmPR = 4 cm.

Construct a quadrilateral PQRSPQRS in which PQ=3.8PQ=3.8 cm, QR=6.8QR=6.8 cm and three angles \angle P=100^{o},\ \angle R=110^{o}\angle P=100^{o},\ \angle R=110^{o} and \angle S=75^{o}\angle S=75^{o} are known.

Construct a quadrilateral ABCDABCD with AB=2.5\ cm,BC=3.5\ cm,CD=4.5\ cm,\angle A=90^{o}AB=2.5\ cm,BC=3.5\ cm,CD=4.5\ cm,\angle A=90^{o} and \angle B=120^{o}\angle B=120^{o}.

Construct a Rhombus PQRSPQRS such that PR= 8 PR= 8 cm, QS= 5.4QS= 5.4 cm.

Construct a rectangle $ABCD$ given
$AB=6\ cm, BC=5\ cm$

Construct the quadrilaterals with the measurements given:
Square $LUDO$ with $LU=4.5cm$ .

Construct a quadrilateral $PQRS$ where $PQ =4\ cm,QR=6\ cm,RS=5\ cm. PS=5.5\ cm$ and $PR =7\ cm$ .

Construct a rectangle ABCD, when:

Its sides are 6.0 cm and 7.2 cm.

Construct a quadrilateral $\text{ABCD}$ , when: $\text{AB} = 3.2\ \text{cm}, \text{BC} = 5.2\ \text{cm}, \text{CD} = 6.2\ \text{cm}, \text{DA} = 4.2\ \text{cm}$ and $\text{BD} = 5.2\ \text{cm}$ .

By constructing a rectangle $WXYZ$ of base = $4.8 \, cm$ and height $= 5\ cm$ , find the square root of its perimeter (where perimeter is in cm) upto the nearest integer.

erimeter is equal to $24\ cm.$

Construct a quadrilateral $MORE$ where $MO=6\ cm,OR=4.5\ cm ,\angle M =60^o, \angle O=105^o$ and $\angle R=105^o$ .

Construct a square $ABCD$ , given it's one side = 4.5 cm.

Construct a quadrilateral $ABCD,$ given that $BC=4.5\ cm ,AD=5.5\ cm$ and $CD=5\ cm$ , the diagonal $AC=5.5\ cm$ and the other diagonal $BD=7\ cm$ .

Draw a square WXYZ with side $5.2$ cm.

Construct a quadrilateral $NILE$ such that $NI=2.5\ cm , IL=3.5\ cm , NL= 4\ cm , IE=5\ cm , EL=3\ cm.$

Construct a quadrilateral $LIFT$ , when:
$LI=4\ cm, IF=3\ cm ,TL =2.5\ cm ,LF=4.5\ cm ,IT=4\ cm$ .

Construct a quadrilateral $PQRS$ such that $PQ=3.5\ cm ,QR=4.5\ cm, RS=4\ cm, \angle Q=110^o$ and $\angle R =80^o$ .

Construct a quadrilateral $JUMP$ , when: $JU=3.5\ cm, UM =4\ cm, MP=5\ cm, PJ=4.5\ cm, PU=6.5\ cm$

Construct a quadrilateral $ABCD$ such that $AB=3\ cm , BC=4\ cm ,$ $\angle A=90^O$ , $\angle B=65^o$ and $\angle C=110^o$ .

Construct a quadrilateral $ABCD$ with $AB=6\ cm, BC=3\ cm, CD=8\ cm,$ $\displaystyle \angle B={ 60 }^{ \circ },\angle C={ 95 }^{ \circ }.$

Construct a quadrilateral $ABCD$ , with $BC=5$ cm, $CD=9$ cm, $DA=6$ cm, $\displaystyle \angle C={ 60 }^{ \circ },\angle D={ 75 }^{ \circ }$

Construct a rectangle $ABCD$ with $AB = 8\ \mathrm{ cm}$ and $BC = 6\ \mathrm{cm}$ .

Construct a quadrilateral $PQRS$ , with $PQ=3$ cm, $QR=9$ cm, $RS=5$ cm, $\displaystyle \angle Q={ 45 }^{ \circ },\angle R={ 60 }^{ \circ }$

Construct a quadrilateral $ABCD$ in which $AB=6$ cm, $BC=8$ cm, $CD=5$ cm, $AD=4$ cm and diagonal $AC=4.5$ cm

Construct a quadrilateral $ABCD$ with $AB=6\ cm, BC=8\ cm,$ $\displaystyle \angle A={ 75 }^{ \circ },\angle B={ 90 }^{ \circ },\angle C={ 45 }^{ \circ }$ .

Construct a quadrilateral $PQRS$ with $QR=8$ cm, $RS=4$ cm, $SP=9$ cm, $PQ=5.5$ cm and $QS=6$ cm.

Construct a quadrilateral $ABCD$ with $AB=6\ cm, BC=4\ cm, CD=7\ cm, AC=8\ cm$ and $BD=7\ cm.$

Construct a quadrilateral $PQRS$ with $PQ=7$ cm, $QR=5$ cm, $RS=3$ cm, $ST=8$ cm and $PR=5$ cm

Construct a quadrilateral $ABCD$ in which $AB=5\ cm, BC=6\ cm, CD=4\ cm, AD=3\ cm$ and diagonal $AC=5\ cm.$

Construct a quadrilateral $ABCD$ with $AB=3$ cm, $BC=7$ cm, $CD=9$ cm, $CA=11$ cm, $BD=13$ cm.

Construct a quadrilateral $ABCD$ with $AB=\ 9cm$ , $BC=6\ cm$ , $CD=7\ cm$ , $AC=9\ cm$ and $BD=11\ cm$ .

Construct a quadrilateral $ABCD$ with $AB=5$ cm, $BC=9$ cm, $\displaystyle \angle A={ 30 }^{ \circ },\angle B={ 90 }^{ \circ },\angle C={ 15 }^{ \circ }$ .

Construct a quadrilateral $PQRS$ with $PQ=8$ cm, $QR=6$ cm, $\displaystyle \angle P={ 60 }^{ \circ },\angle Q={ 45 }^{ \circ },\angle R={ 60 }^{ \circ }$ .

Construct a quadrilateral $MORE$ where $MO=6\ cm$ , $OR=4.5 \ cm$ , $\angle M =60^o$ , $\angle O=105^o$ and $\angle R=105^o$
Also, find the length of side EM (in cm)

Construct a quadrilateral $PQRS$ with $PQ=7$ cm, $QR=6$ cm, $\displaystyle \angle P={ 45 }^{ \circ },\angle Q={ 60 }^{ \circ },\angle R={ 75 }^{ \circ }$ .

Construct a quadrilateral $ABCD$ with $AB=6$ cm, $BC=9$ cm, $\displaystyle \angle A={ 75 }^{ \circ },\angle B={ 60 }^{ \circ },\angle C={ 45 }^{ \circ }$ .

Construct a quadrilateral $ABCD$ with $AB=4.5cm, BC=5.5cm, CD=4cm, AD=6cm$ and diagonal $AC=7cm$ .

Construct a square of side $5\ cm$ . Find its area also

Construct rhombus $ABCD$ such that:
$AC = 7 cm$ , $BD = 4 cm$

Construct rhombus $ABCD$ such that diagonal $BD = 6.8 cm$ , $AC = 5.4 cm$ .

Construct square $ABCD$ which has side-length $2\ cm$ .

Construct a quadrilateral ABCD with AB=2.5 cm, BC=3.5 cm, AD=4.5 cm, $\angle A=90^0$ and $\angle B=120^0$ .

Construct the quadrilateral $PQRS$ with $PQ=3.5cm, QR=4cm, RS=5cm. PS=4.5cm$ and $QS=6.5cm$ .

Construct a rectangle $\text{ABCD}$ with $\text{AB=5 cm, BC=3.5 cm}$

Construct a rhombus $PQRS$ with $PR = 8\ cm$ and $QS = 6\ cm$ and find its area.

Construct a rectangle whose adjacent sides are of lengths $5$ $\text cm$ and $3.5$ $\text cm$ .

Construct the following quadrilaterals.
Quadrilateral MORE
$MO$ = $6\ cm$
$OR$ = $4.5\ cm$
$\angle M=60^o$
$\angle O=105^o$
$\angle R=105^o$

Construct the following quadrilaterals.
Quadrilateral PLAN
$PL$ = $4cm$
$LA$ = $6.5cm$
$\angle P=90^o$
$\angle A=110^o$
$\angle N=85^o$

Construct a quadrilateral $ABCD$ in which $AB = 4.4$ cm, $BC = 4$ cm, $CD = 6.4$ cm, $DA = 3.8$ cm and $BD = 6.6$ cm.

Construct $\square DEFG$ such that $l$ (DE) $=4.5$ cm, $l$ (EF) $=6.5$ cm, $l$ (DG) $=5.5$ cm, $l$ (DF) $=7.2$ cm, $l$ (EG) $=7.8$ cm.

Construct a rhombus whose diagonals are $16$ cm and $12$ cm. What property of the diagonals is used?

Constant a quadrilateral $ABCD$ , where $AB=4\ cm,BC=5\ cm,CD=6.5\ cm$ and $\angle B=105^{o}$ and $\angle C=80^{o}$

Construct a quadrilateral $ABCD$ , where $AB=4\ cm,BC=5\ cm,CD=6.5\ cm$ and angle $B=75^{o}$ and angle $C=60^{o}$

Quadrilateral GOLD
OL $=7.5$ cm
GL $=6$ cm
GD $=6$ cm
LD $=5$ cm
OD $=10$ cm.

Construct a quadrilateral $ABCD$ with the following measures:
$l(AB)=6.4\ cm,\ l(BC)=4.8\ cm,\ \angle A=70^o,\ \angle B=50^o,\ \angle C=140^o$ .

Construct a quadrilateral $ABCD$ in which $AB=6\ cm,CD=5\ cm,DA=3.5\ cm,\angle C=120^{o}$ and $\angle D=75^{o}$ .

Quadrilateral HELP with HE=6cm,EL=4.5cm, $\angle H=60^{0},\angle E=105^{0}$ and $\angle P=120^{0}$

Construct a quadrilateral $ABCD$ in which $AB=4 \text{ cm}, BC=3.1 \text{ cm}, CD=3 \text{ cm}, AD=4.2 \text{ cm}$ and $AC=5 \text{ cm}$ . Measure the diagonal $BD$ .

Construct a quadrilateral $READ$ , when $RE=3.51$ cm, $EA=4.3$ cm, $AD=4.8$ cm, $\angle E=105^{\circ}$ & $\angle R=55^{\circ}$ .

Construct a quadrilateral $ABCD$ in which $AB = 4.2 cm, BC = 6 cm, CD = 5.2 cm, DA = 5 cm$ and $AC = 8 cm$ .

Construct a rectangle $ABCD$ given $AB=7\ cm, BC=5.5\ cm$ .

Construct a quadrilateral $ABCD$ in which $AB=3.6cm, BC=3.3cm, AD=2.7cm$ , diagonal $AC=4.6cm$ and diagonal $BD=4cm$

Construct a quadrilateral $PQRS$ with $PQ=6.2cm, QR=5.6cm, RS=6.2cm$ , $\angle{Q}={90}^{o}$ and $\angle {R}={120}^{o}$

Construct a quadrilateral $ABCD$ , when $AB = 3\ cm, CD = 3\ cm, DA = 7.5\ cm, AC = 8\ cm$ and $BD = 4\ cm$ .

Construct a quadrilateral $ABCD$ with $AB=5\;cm,BC=3.6\;cm,\angle ABC=80^{\circ},\angle DAB=60^{\circ}$ and $\angle ADC=120^{\circ}$ .

Construct a rectangle whose adjacent sides are $5.6\ cm$ and $4\ cm$

Construct a quadrilateral $PQRS$ with $PQ=4.7\ cm, QR=3.9\ cm, \angle QPS=75^{o}, \angle PQR=120^{o}$ and $\angle QRS=105^{o}$ . Measure $\angle PSR$ .

Construct the following quadrilaterals :
Quadrilateral $RUNS$
$RU=4$ cm
$UN=5$ cm
$NS=4.2$ cm
$SR=5.8$ cm
$RN=7.2$ cm

Quadrilateral JUMP
JU=3.5 cm
UM=4 cm
MP=5 cm
PJ=4.5 cm
PU=6.5 cm

Construct a quadrilateral $PQRS$ in which $QR = 7.5 cm, PR = PS = 6 cm, RS = 5 cm$ , and $QS = 10 cm$ .

Construct a quadrilateral $ABCD$ in which $\overline { AB } =\overline { BC } =\overline { CD } =\overline { AD } =5$ cm and diagonal $\overline { AC } =6.5$ cm.

Construct a quadrilateral $PQRS$ in which:
$PQ = 5.4 cm, QR = 4.6 cm, RS = 4.3 cm, SP = 3.5 cm$ , and Diagonal $PR = 4 cm$ .

Construct a quadrilateral $PQRS$ in which $PQ=3.8$ cm, $QR=6.8$ cm and three angles $\angle P=100^{o},\ \angle R=110^{o}$ and $\angle S=75^{o}$ are known.

Construct a quadrilateral $ABCD$ with $AB=2.5\ cm,BC=3.5\ cm,CD=4.5\ cm,\angle A=90^{o}$ and $\angle B=120^{o}$ .

Construct a Rhombus $PQRS$ such that $PR= 8$ cm, $QS= 5.4$ cm.

A construct rectangle $FLAG$ with sides $F L = 6 \mathrm { cm }$ and $L A = 4.2 \mathrm { cm }$

Construct a quadrilateral $ABCD$ in which $BC = 7.5\ cm, AC = AD = 6\ cm, CD = 5\ cm$ and $BD = 10\ cm$ .

Construct a quadrilateral $ABCD$ in which $AB = 3.5 cm, BC = 3.8 cm, CD = DA = 4.5 cm$ and Diagonal $BD = 5.6 cm$ .

In the figure, $ABCD$ is a quadrilateral.
How many pairs of adjacent sides are there?

Construct a quadrilateral $ABCD$ given $AD = 5\ cm, AB = 5.5\ cm, BC = 2.5\ cm, AC = 7.1\ cm$ and $BD = 8\ cm$ .

In the figure, $ABCD$ is a quadrilateral.
How many pairs of opposite sides are there?

Complete the following, so as to make a true statement:
A quadrilateral has ______ diagonals.

In the figure, $ABCD$ is a quadrilateral.
How many pairs of adjacent angles are there?