Practical Geometry - Class 8 Maths - Extra Questions

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.

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

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.

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

1. Draw $$AB =2.5$$ $$cm$$
2. Draw $$90°$$ at $$A$$
3. Mark $$d$$ such that $$AD=4.5$$ $$cm$$
4. Draw $$120°$$ at $$B$$
5. Mark $$C$$ such that $$BC=3.5$$ $$cm$$
6. Draw $$CD$$
7. $$ABCD$$ is the required quadrilateral

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,

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.

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 $$\angle L={{75}^{o}}$$

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 $$DC$$.

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 $$ABCD$$.

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.

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.

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.

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.

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.

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.

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.

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.

$$∠P + ∠Q + ∠R + ∠S = 360°$$ 

$$∠P = 360° – (120° + 90° + 60°)$$ 

$$=> ∠P = 90°$$ 

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.

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 $$RD$$.