Constructions - Class 9 Maths - Extra Questions
Construct a scalene triangle $$ABC,$$ given base $$AB=3\ cm,$$ base angle $$=90^o$$ and sum of the lengths $$BC+AC=9\ cm$$.
Given base side as $$AB=3cm$$
Base angle $$=90^o=\cfrac {\angle B}{\angle C}$$
Sum of lengths of other two sides, $$BC+AC=9cm$$
Steps:
1. Draw the base $$AB$$, make an angle $$PAB$$ equal to the given angle
2. Cut a line segment $$AD$$(equal to $$BC$$ and $$AC$$) on ray $$PA$$
3. Join $$BD$$. Draw a $$\angle DBQ$$ equal to $$\angle ADB$$
4. Let ray $$BD$$ intersect $$AD$$ at point $$C$$
Then, $$\triangle ABC$$ is the required triangle
i.e, $$BC=3.7cm; AC=5.3cm$$
$$\angle AC+BC=3.7+5.3=9cm$$
$$\angle CAB=45^o$$
$$\angle ABC=80^o$$
Draw $$\angle ABC$$ of measure $$110^{\circ}$$ and bisect it.
1. Draw a ray $$BC$$. Take a protractor and draw an angle of $$110^\circ$$. Mark a point $$A$$ on the $$110^\circ$$ ray.
2. Place the compass end on $$B$$ and mark two arcs of same length on $$BC$$ and $$BA$$ respectively.
3. From the two arcs drawn above, draw two intersecting arcs in the interior of $$\angle ABC$$.
4. Draw a ray from $$B$$ passing through the intersection point of the above arcs.
Ray $$BM$$ is the angle bisector of $$\angle ABC = 110^\circ$$.
Construct a bisector of an angle of $$60^o$$.
1) Draw a line segment PQ
2) Draw an arc of any length from P, which cuts PQ at S
3) Draw an arc T of length PS from S. Angle TPS is $$60^o$$.
4) Draw two arcs from S and T, and mark their point of intersection as R. Join PR. Angle RPS is 30 degrees.
Draw $$\angle ABC$$ of measure $$115^{\circ}$$ and bisect it.
1. Draw a ray $$BC$$.
2. With the help of a protractor, draw an $$\angle ABC$$ of $$115^\circ$$.
3. With one end of the compass at $$B$$, mark two equidistant arcs on rays $$BA$$ and $$BC$$.
4. From the two points on the rays, draw two intersecting arcs. Join $$B$$ to the intersection point of those arcs.
Ray $$BM$$ is the angle bisector of $$\angle ABC$$.
Construct an angle of $$45^o$$ using compass.
First, draw a line segment AB.
Through compass, draw curves from point A and B to the upper and lower sides of AB
These curves cut each other point P and Q
After join P and Q, The PQ line cut the line AB at point C.
Join point P from A
So Angle PAB is the required angle $$45^{0}$$
Draw the following angles using ruler and compasses. Also label them.
$$45^{o}$$
$$!(QAR)=45^0$$
We can construct a $${45}^{\circ}$$ angle either by bisecting a rightangle
Step $$1:$$ Draw a line $$PQ$$.
Step $$2:$$ Bisect the line $$PQ$$ using compass
Step $$3:$$ Join $$SA$$
Step$$4:$$ Bisect the arc $$SQ$$
Step $$5:$$ Join $$RA$$
Result:$$\\angle{RAQ}={45}^{\circ}$$ is the required angle.
Construct $$\Delta PQR$$, such that $$QR = 6.5cm$$, $$\angle PQR = {60^0}$$ and $$PQ - PR = 2.5cm$$.
Step 1: Construct a straight line QR = 6.5 cm.
Step 2: At point Q, construct an angle of $$60^o$$
Step 3: On the arm of this angle mark point X at a distance of 2.5 cm from Q
Step 4: Join X to R.
Step 5: Now the copy the angle formed at X on R.
Step 6: Extend the arms of both the angles and mark their intersection as P.
The construction of $$\Delta P Q R$$ given that $$Q R = 5.2 \mathrm\ { cm }$$ angle $$Q= 50^{\circ}$$. Is it possible when the difference of $$PQ$$ and $$PR$$ is $$3.5\ cm$$? justify.
Draw the following angles using ruler and compasses. Also label them.
$$90^{o}$$
We can construct a $${90}^{\\circ}$$ angle either by bisecting a straight angle.
Step $$1:$$ Draw a line $$PQ$$.
Step $$2:$$ Bisect the line $$PQ$$ using compass
Step $$3:$$ Join $$TA$$
Result:$$\angle{PAT}=\angle{QAT}={90}^{\circ}$$ is the required angle.
How many of the given letters have perpendicular lines?
MATHS
Draw an angle of measure $$45^{o}$$ and bisect it.
Draw a right angle.
$$1)$$ draw a line segment AB and draw an arc cutting AB at C
$$2)$$ with C at center draw an arc cutting the previous arc at D and with D
as center draw and arc cutting the same arc at E
$$3)$$ with D as center draw an arc with E as center cutting the previous arc at F
$$4)$$ join FA
Draw $$\angle POQ$$ of measure $$75^{o}$$ and find its line of symmetry.
R.E.F image
a. Draw a line | and mark a point o on it.
b. place the pointer of the compasses at o and
draw an are of any radius which
intersects the l at A.
c.Take same radius, with center A.
cut the previous arc at B.
d. Join OB, then $$\angle BOA = 60 $$ Degree
e. Taking same radius, with centre B.
cut the previous arc at C.
f. Draw bisector of $$\angle BOC $$. The angle is of 90 degrees.
Mark it at D. Thus $$\angle DOA = 90$$ Degree.
g. Draw OP bar as bisector of $$\angle DOB$$.
Thus $$ \angle POA = 75 $$ degrees.
Draw the following angles using ruler and compasses. Also label them.
$$30^{\circ}$$
We know that:
So, to construct an angle of $${30}^{\\circ}$$, first construct a $${60}^{\circ}$$ angle and then bisect it. Often, we apply the following steps.
Step $$1:$$ Draw the arm $$PQ$$.
Step $$2:$$ Place the point of the compass at $$P$$ and draw an arc that passes through $$Q$$.
Step $$3:$$ Place the point of the compass at $$Q$$ and draw an arc that cuts the arc drawn in Step $$2$$ at $$R$$.
Step $$4:$$ With the point of the compass still at $$Q$$, draw an arc near $$T$$ as shown.
Step $$5:$$ With the point of the compass at $$R$$, draw an arc to cut the arc drawn in Step $$4$$ at $$T$$.
Step $$6:$$ Join $$T$$ to $$P$$. The $$\angle{QPT}$$ is $${30}^{\circ}$$.
Draw a line segment $$XY=7 cm$$ and draw $$\angle AXY={ 90 }^{ \circ }$$
$$1.$$Draw a line $$XY=7$$cm. Taking $$X$$ as center, and radius more than $$\dfrac{1}{2}XY$$,draw arcs on top and bottom of $$XY$$
$$2.$$Taking $$X$$ as center, and radius as before,draw arcs on top and bottom of $$XY$$
$$3.$$Let the point where arcs intersect on top of $$XY$$ be $$P$$.And the point where arcs intersect at the bottom be $$Q$$.Join $$PQ$$
$$4.$$Let $$PQ$$ intersect $$XY$$ at the point $$M$$.
Thus, line $$PMQ$$ is the required perpendicular bisector of $$XY$$.
Construct the angle of $$30^{o}$$ using a ruler and compass.
So, to construct an angle of $$30^{\circ}$$, first construct a $$60^{\circ}$$ angle and then bisect it. For this, we would apply the following steps:
Step 1: Draw the arm $$PQ.$$
Step 2: Place the point of the compass at $$P$$ and draw an arc that passes through $$Q$$.
Step 3: Place the point of the compass at $$Q$$ and draw an arc that cuts the arc drawn in Step 2 at $$R$$.
Step 4: With the point of the compass still at $$Q$$, draw an arc near $$T$$ as shown.
Step 5: With the point of the compass at $$R$$, draw an arc of the same measure to cut the arc drawn in Step 4 at $$T$$.
Step 6: Join $$T$$ to $$P.$$ The angle $$QPT$$ is $$30^{\circ}$$.
Construct an angle of the measure $$45^{\circ}$$ with ruler and compasses.
Steps to construct an angle of measure $$45^\circ:$$
Step 1: Draw a ray $$OA$$
Step 1: Draw a ray $$OA$$
Step 2: Taking $$O$$ as the center and any radius, draw an arc cutting $$OA$$ at $$B$$
Step 3: With $$B$$ as centre and with the same radius as before, construct an arc intersecting the previous arc at $$C.$$
Step 3: With $$B$$ as centre and with the same radius as before, construct an arc intersecting the previous arc at $$C.$$
Step 4: With $$C$$ as centre and with the same radius, draw an arc intersecting the arc at $$D.$$
Step 5: With $$C$$ and $$D$$ as centres and radius more than $$\dfrac{1}{2}CD,$$ draw two arcs intersecting at $$P.$$
Step 6: Join $$OP.$$
$$\therefore \angle{AOP}={90}^{\circ}$$
Step 7: Let $$OP$$ intersect the original arc at $$Q$$
Step 8: With $$B$$ and $$Q$$ as centres and radius greater than $$\dfrac{1}{2}BQ,$$ draw two arcs intersecting at $$R$$
Step 9: Join $$OR$$
Thus $$\angle AOR=45^\circ$$
Step 7: Let $$OP$$ intersect the original arc at $$Q$$
Step 8: With $$B$$ and $$Q$$ as centres and radius greater than $$\dfrac{1}{2}BQ,$$ draw two arcs intersecting at $$R$$
Step 9: Join $$OR$$
Thus $$\angle AOR=45^\circ$$
Construct with ruler and compasses, angle of the following measure:
$$30^{o}$$
Step 1: A ray from $$O$$ is drawn.
Step 2: An arc of $${60}^{\circ}$$ is drawn using compass.
Step 3:Construct a perpencicular bisector to the arc from $${0}^{\circ}$$ Join $$OA$$
$$\therefore \angle{AOX}={30}^{\circ}$$
Thus,$$\angle{AOX}$$ is the required angle making $${30}^{\circ}$$
Construct angles of following measures using ruler and compasses.
$$90^{o}$$
We can construct a $$90^{\circ}$$ angle either by bisecting a straight angle or using the following steps.
Step $$1:$$ Draw the arm $$PA$$.
Step $$2:$$ Place the point of the compass at $$P$$ and draw an arc that cuts the arm at $$Q.$$
Step $$3:$$ Place the point of the compass at $$Q$$ and draw an arc of radius $$PQ$$ that cuts the arc drawn in Step $$2$$ at $$R$$.
Step $$4:$$ With the point of the compass at $$R$$, draw an arc of radius $$PQ$$ to cut the arc drawn in Step $$2$$ at $$S$$.
Step $$5:$$ With the point of the compass still at $$R$$, draw another arc of radius $$PQ$$ near $$T$$ as shown.
Step $$6$$: With the point of the compass at $$S$$, draw an arc of radius $$PQ$$ to cut the arc drawn in step $$5$$ at $$T$$.
Step $$7:$$ Join $$T$$ to $$P. \angle APT$$ is $$90^{\circ}$$.
What is an angle bisector ?
Construct with ruler and compasses, angle of the following measure:
$$90^{o}$$
Step 1: A ray $$OA$$ is drawn.
Step 2: An arc from $$A$$ is drawn with the same length as $$OA$$
Step 3: An arc from $$B$$ is drawn with the same length as $$OB$$
Step 4:Construct a perpencicular bisector to the arc from $$B$$ to the another arc.Join $$OC$$
$$\therefore \angle{AOC}={90}^{\circ}$$
Thus,$$\angle{AOC}$$ is the required angle making $${90}^{\circ}$$
Construct angles of following measures using ruler and compasses.
$$60^{o}$$
We know that the angles in an equilateral triangle are all $$60^{\circ}$$ in size. This suggests that to construct a $$60^{\circ}$$ angle we need to construct an equilateral triangle as described below.
Step $$1:$$ Draw the arm $$PQ$$.
Step $$2:$$ Place the point of the compass at $$P$$ and draw an arc that passes through $$Q$$.
Step $$3:$$ Place the point of the compass at $$Q$$ and draw an arc that passes through $$P$$. Let this arc cut the arc drawn in Step $$2$$ at $$R$$.
Step $$4:$$ Join $$P$$ to $$R$$ .The $$\angle QPR=60^{\circ}$$ as the $$\triangle PQR $$ is equilateral triangle
Construct a bisector of an angle.
$$1.$$Draw an arcs of any radius intersecting $$BA$$ and $$BC$$ at points $$E$$ and $$D$$
$$2.$$Next,taking $$D$$ and $$E$$ as centers and with the radius more than $$\dfrac{1}{2}DE$$, draw arcs to intersect each other.
$$3.$$Mark the point as $$F$$
$$4.$$Join $$BF$$
So, $$BF$$ is the bisector of the $$\angle{ABC}$$
Construct a triangle ABC whose perimeter 12 cm and who base angles are $${ 50 }^{ \circ }$$ and $${ 80 }^{ \circ }$$.
STEPS OF CONSTRUCTION:
1) Draw a line segment XY of 12 cm.
2)AT X draw angle $$50^o$$ and at y draw angle $$80^0$$
3)Draw a bisector of angle $$50^o$$ as well as angle$$80^o$$.
4)Mark the point where both the bisectors meet as A.
5)Draw the perpendicular bisector of AX and AY Meeting XY at B and C respectively.
6)Join AB and AC .
7) ABC is the required triangle.
Construct with ruler and compasses, angle of following measures.
$$135^o$$.
The given steps will be followed to construct an angle of $$135^o$$.
$$(1)$$ Draw a line and mark a point P on it. Now taking P as centre and with a convenient radius, draw a semicircle which intersects line l at Q and R.
$$(2)$$ Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc to S.
$$(3)$$ Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T(see figure).
$$(4)$$ Taking S and T as centre, draw arcs of same radius to intersect each other at U.
$$(5)$$ Join PU. Let it intersect the arc at V. Now taking Q and V as centres and with radius more than $$2$$ QV, draw arcs to intersect each other at W.
$$(6)$$ Join PW which is the required ray making $$135^o$$ with line.
$$(1)$$ Draw a line and mark a point P on it. Now taking P as centre and with a convenient radius, draw a semicircle which intersects line l at Q and R.
$$(2)$$ Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc to S.
$$(3)$$ Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T(see figure).
$$(4)$$ Taking S and T as centre, draw arcs of same radius to intersect each other at U.
$$(5)$$ Join PU. Let it intersect the arc at V. Now taking Q and V as centres and with radius more than $$2$$ QV, draw arcs to intersect each other at W.
$$(6)$$ Join PW which is the required ray making $$135^o$$ with line.
Construct an angle of $$45^\circ$$ at the initial point of a given ray OA, using a ruler and compasses.
Construct with ruler and compasses, angle of following measures.
$$60^o$$.
The given steps will be followed to construct an angle of $$60^o$$.
$$(1)$$ Draw a line l and mark a point P on it. Now, taking P as centre and with a convenient radius, draw an arc of a circle which intersects line l at Q.
$$(2)$$ Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point R.
$$(3)$$ Join PR which is the required ray making $$60^o$$ with line l.
$$(1)$$ Draw a line l and mark a point P on it. Now, taking P as centre and with a convenient radius, draw an arc of a circle which intersects line l at Q.
$$(2)$$ Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point R.
$$(3)$$ Join PR which is the required ray making $$60^o$$ with line l.
Construct with ruler and compasses, angle of following measures.
$$90^o$$.
The given steps will be followed to construct an angle of $$90^o$$.
$$(1)$$ A line l and mark a point p on it. Now taking p as centre and with a convenient radius, draw an arc of a circle which intersects line l at Q.
$$(2)$$ Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R.
$$(3)$$ Taking R as centre and with the same radius as before, draw an arc intersecting the arc at S(see figure).
$$(4)$$ Taking R and S as centre, draw an arc of same radius to intersect each other at T.
$$(5)$$ Join PT, which is the required ray making $$90^o$$ with line l.
$$(1)$$ A line l and mark a point p on it. Now taking p as centre and with a convenient radius, draw an arc of a circle which intersects line l at Q.
$$(2)$$ Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R.
$$(3)$$ Taking R as centre and with the same radius as before, draw an arc intersecting the arc at S(see figure).
$$(4)$$ Taking R and S as centre, draw an arc of same radius to intersect each other at T.
$$(5)$$ Join PT, which is the required ray making $$90^o$$ with line l.
Draw an angle of measure $$135^o$$ and bisect it.
The given steps will be followed to construct an angle of $$135^o$$ and its bisector.
$$(1)$$ DPOQ of $$135^o$$ measure can be formed on a line I by using the protractor.
$$(2)$$ Draw an arc of a convenient radius, while taking point O as centre. Let it intersect both rays of angle $$1356^o$$ at point A and B.
$$(3)$$ Taking A and B as centres, draw arcs of radius more than $$1/2$$ AB in the interior of angle of $$135^o$$. Let those intersect each other at C. Join OC.
OC is the required bisector of $$135^o$$ angle.
$$(1)$$ DPOQ of $$135^o$$ measure can be formed on a line I by using the protractor.
$$(2)$$ Draw an arc of a convenient radius, while taking point O as centre. Let it intersect both rays of angle $$1356^o$$ at point A and B.
$$(3)$$ Taking A and B as centres, draw arcs of radius more than $$1/2$$ AB in the interior of angle of $$135^o$$. Let those intersect each other at C. Join OC.
OC is the required bisector of $$135^o$$ angle.
Construct with ruler and compasses, angle of following measures.
$$30^o$$.
The given steps will be followed to construct an angle of $$30^o$$.
$$(1)$$ Draw a line l and mark a point p on it. Now taking p as centre and with convenient radius, draw an arc of a circle which intersects line l at Q.
$$(2)$$ Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point R.
$$(3)$$ Now, taking Q and R as centre and with radius more than $$2$$ RQ, draw arcs to intersect each other at S. Join PS which is the required ray making $$30^o$$ with line l.
$$(1)$$ Draw a line l and mark a point p on it. Now taking p as centre and with convenient radius, draw an arc of a circle which intersects line l at Q.
$$(2)$$ Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point R.
$$(3)$$ Now, taking Q and R as centre and with radius more than $$2$$ RQ, draw arcs to intersect each other at S. Join PS which is the required ray making $$30^o$$ with line l.
Construct with ruler and compasses, angle of following measures.
$$120^o$$.
The given steps will be followed to construct an angle of $$120^o$$.
$$(1)$$ Draw a line l and mark(a point P on it. Now taking P as centre and with a convenient radius, draw an arc of a circle which intersects line at Q.
$$(2)$$ Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R.
$$(3)$$ Taking R as centre and with the same radius as before, draw an arc intersecting the arc at S.(see figure)
$$(4)$$ Join PS, which is the required ray making $$120^o$$ with line l.
$$(1)$$ Draw a line l and mark(a point P on it. Now taking P as centre and with a convenient radius, draw an arc of a circle which intersects line at Q.
$$(2)$$ Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R.
$$(3)$$ Taking R as centre and with the same radius as before, draw an arc intersecting the arc at S.(see figure)
$$(4)$$ Join PS, which is the required ray making $$120^o$$ with line l.
Draw a right angle and construct its bisector.
The given steps will be followed to construct a right angle and its bisector.
$$(1)$$ Draw a line l and mark a point P on it. Draw an arc of convenient radius, while taking point P as centre. Let it intersect line I at R.
$$(2)$$ Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
$$(3)$$ Taking s as centre and with the same radius as before, draw an arc intersecting the arc at T.(see figure)
$$(4)$$ Taking S and T as centres, draw arcs of same radius to intersect each other at U.
$$(5)$$ Join PU. PU is the required ray making $$900$$ with line l. Let it intersect the major arc at point V.
$$(6)$$ Now, taking R and V as centres, draw arcs with radius more than $$1/2$$ RV to intersect each other at W. Join PW.
PW is the required bisector of this right angle.
$$(1)$$ Draw a line l and mark a point P on it. Draw an arc of convenient radius, while taking point P as centre. Let it intersect line I at R.
$$(2)$$ Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
$$(3)$$ Taking s as centre and with the same radius as before, draw an arc intersecting the arc at T.(see figure)
$$(4)$$ Taking S and T as centres, draw arcs of same radius to intersect each other at U.
$$(5)$$ Join PU. PU is the required ray making $$900$$ with line l. Let it intersect the major arc at point V.
$$(6)$$ Now, taking R and V as centres, draw arcs with radius more than $$1/2$$ RV to intersect each other at W. Join PW.
PW is the required bisector of this right angle.
Construct with ruler and compasses, angle of following measures.
$$45^o$$.
The given steps will be followed to construct an angle of $$45^o$$.
$$(1)$$ Draw a line l and mark a point P on it. Now taking P as centre and with a convenient radius, draw an arc of a circle which intersects line l at Q.
$$(2)$$ Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R.
$$(3)$$ Taking R as centre and with the same radius as before, draw an arc intersecting the arc at S(see figure).
$$(4)$$ Taking R and s as centres, draw arcs of same radius to intersect each other at T.
$$(5)$$ Join PT. Let it intersect the major arc at point U.
$$(6)$$ Taking O and as centres draw arcs with radius more than $$1/2$$ QU to intersect each other at V. Join PV. PV is the required ray making $$45^o$$ with the given line l.
$$(1)$$ Draw a line l and mark a point P on it. Now taking P as centre and with a convenient radius, draw an arc of a circle which intersects line l at Q.
$$(2)$$ Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R.
$$(3)$$ Taking R as centre and with the same radius as before, draw an arc intersecting the arc at S(see figure).
$$(4)$$ Taking R and s as centres, draw arcs of same radius to intersect each other at T.
$$(5)$$ Join PT. Let it intersect the major arc at point U.
$$(6)$$ Taking O and as centres draw arcs with radius more than $$1/2$$ QU to intersect each other at V. Join PV. PV is the required ray making $$45^o$$ with the given line l.
Draw an angle of $$ 45 ^ { \circ } , $$ using a pair of compasses.
Draw a line PR
Let Q be the point on PR. With Q as centre and any suitable radius, draw an arc cutting AC at point M
and N
With N as centre and radius more than half MN, draw an arc
With M as centre and the same radius as before, draw another arc to cut the previous arc at X
Join QX, meeting the arc at Z and produce it to W
With Z as centre and radius more than half of ZN, draw an arc
With N as centre and the same radius as in step in $$(6)$$, draw another arc, cutting the previously drawn arc
at a point Y
Join QY and produce it to point S
$$\angle PQS$$ is the required angle whose measure is $$45^o$$
In this question, use a straight edge and compasses only and show all your construction arcs. Construct the bisector of angle ABC.
In this question, use a straight edge and compasses only and show all your construction arcs. Construct the perpendicular bisector of PQ.
Construct a $$\triangle ABC$$ in which: $$AB= 5.4\ cm$$, $$\angle CAB= 45^{0}$$ and $$AC\, +\, BC= 9\ cm$$.Then the length of $$AC$$ (in $$cm.$$) is:
1. Draw $$AB$$ of $$5.4\ cm.$$
2. Construct angle $$45^\circ$$ at $$A$$.
3. With radius as $$9\ cm$$ at $$A$$, cut an arc along the $$45^\circ$$ line and mark the point as $$D$$. Join $$BD$$.
4. Draw a perpendicular bisector of $$BD$$ and mark its intersection with $$AD$$ as $$C$$.
5. Join $$AC$$.
6. $$ABC$$ is the required triangle.
Construct a triangle $$PQR$$ in which $$\angle Q = 60^o$$ and $$\angle R = 45^o$$ and $$PQ + QR + PR = 11\ cm$$.
Step 1: Draw a line segment $$AB$$ of $$11 cm$$
$$ XY+YZ+ZX=11 cm $$
Step 2: Using a protractor mark $$ { 30 }^{ \circ } $$ at $$A$$, i.e $$\angle PAB$$ and $$\angle QBA$$ as $${ 90 }^{ \circ }$$
Step 3: Bisect $$ { 30 }^{ \circ } $$ at $$A$$, i.e $$\angle PAB$$ and $$\angle QBA$$ at point X.
Step 4: Join $$XZ$$ and make $$XZ$$ perpendicular to $$AB$$.
Step 5: Draw perpendicular bisector of $$AX$$ i.e $$ST$$.
Step 6: Mark the point $$Y$$ where $$ST$$ and $$AB$$ intersect. Join $$XY$$
$$\therefore$$ We get $$\triangle XYZ$$
Construct a $$\triangle ABC$$ in which $$AC= 5\ cm$$, and $$\angle BAC= 60^{\circ}$$ and $$AB\, -\, BC= 1.2\ cm$$.
1. Draw a line $$AC = 5\ cm.$$
2. Construct $$\angle A =60^{ \circ }$$ and mark it as $$ Y.$$
3. Cut off line segment $$AD = 1.2\ cm$$ on $$AY$$.
4. Join $$C-D.$$
5. Draw a perpendicular bisector of $$CD$$ such that it intersects $$AY$$ at $$B$$.
6. Join $$B-C$$.
7. $$ABC$$ is the required triangle.
Construct the bisector of an angle $$75^o$$.
$$(i)$$ Draw a line segment $$AP$$ of any length.
$$(ii)$$ Draw an angle of $$75^{o}$$ from $$A$$ and name the new point as B.
$$(iii)$$ Draw an arc that is centered at the vertex($$A$$) of the angle. This arc can have a radius of any length.
$$(iv)$$ However, it must intersect both sides of the angle. We will call these intersection points as $$C$$ and $$D$$. This provides a point on each line that is an equal distance from the vertex of the angle.
$$(v)$$ Draw arcs from $$C$$ and $$D$$ and name the point of intersection of these arcs as $$Q$$.
$$(vi)$$ Draw a line segment joining points $$A$$ and $$Q$$. Then line $$AQ$$ forms angle bisector of $$A$$.
Construct an angle of $$60^o$$ using compass.
Steps of construction-
1) Draw a line segment $$AB$$.
2) From point $$A$$ construct an arc $$PO$$ of any length .
3) Set compass of length $$AP$$ and cut the arc on $$PO$$ from point $$P$$.
4) Join $$AO$$.
Angle $$OAP$$ is the required $$60^o$$ angle.
Construct a triangle whose perimeter is $$9.8\ cm$$ and the base angles are $$45^\circ$$ and $$60^\circ$$.
1. Draw a line $$PQ$$ such that $$PQ = AB +BC+AC=9.8\ cm.$$
2. Construct angle $$P= 60^\circ$$ and angle $$Q= 45^\circ$$.
3. Bisect both angles. Let the angle bisectors meet at $$A.$$
4. Join $$A-P$$ and $$A-Q.$$
5. Draw perpendicular bisectors of $$AP$$ and $$AQ.$$
6. Perpendicular bisector of $$AP$$ meet $$PQ$$ at $$B$$ and that of $$AQ$$ at $$C$$.
7. Join $$A-B$$ and $$A-C$$.
8. $$ABC$$ is the required triangle.
Construct a triangle ABC in which $$ \displaystyle BC=5.6 $$ $$cm$$, $$ \displaystyle \angle B=45^{\circ} $$ and $$ \displaystyle AB+AC=8\ cm $$.
1. Draw $$BC = 5.6\ cm$$.
2. Construct $$\angle B={ 45 }^{ \circ }$$.
3. From $$BY$$ cut off $$BD = 8\ cm.$$
4. Join $$C-D.$$
5. Draw perpendicular bisector of $$CD$$ to intersect $$BD$$ at $$A$$.
6. Join $$AC.$$
7. $$ABC$$ is the required triangle.
Construct a $$\triangle$$ $$PQR$$, such that $$\angle$$ Q=$$70^o$$, $$\angle R=70^o$$ and $$PQ+QR+RP$$ $$=10$$ cm.
Construct a triangle ABC in which $$BC = 6\ cm, \angle A = \displaystyle 45^{\circ}$$ and median $$AD = 5\ cm$$.
Steps of Construction :
1. Draw a line segment $$BC = 6\ cm$$
2. Make an angle PCB = $$\displaystyle 45^{\circ}$$ with the help of a protractor
3. Draw the perpendicular bisector $$RQ$$ of $$AB$$
4. Draw $$\displaystyle BE\perp BP $$ at $$B$$
Let $$RQ$$ and $$BE$$ intersect at $$O$$
5. Draw the circle with centre O and radius OB
Then any angle in the major segment = $$\displaystyle \angle PBC=45^{\circ} $$
6. Let $$RQ$$ intersect $$BC$$ to $$D$$ is the mid-point of $$BC$$. Taking $$D$$ as centre and $$5\ cm$$ radius draw arcs intersecting the circle in $$A, A'$$.
7. Join $$AB, AC$$ and $$A'B, A'C$$
Then the required triangle is $$ABC$$ or $$A'BC$$.
1. Draw a line segment $$BC = 6\ cm$$
2. Make an angle PCB = $$\displaystyle 45^{\circ}$$ with the help of a protractor
3. Draw the perpendicular bisector $$RQ$$ of $$AB$$
4. Draw $$\displaystyle BE\perp BP $$ at $$B$$
Let $$RQ$$ and $$BE$$ intersect at $$O$$
5. Draw the circle with centre O and radius OB
Then any angle in the major segment = $$\displaystyle \angle PBC=45^{\circ} $$
6. Let $$RQ$$ intersect $$BC$$ to $$D$$ is the mid-point of $$BC$$. Taking $$D$$ as centre and $$5\ cm$$ radius draw arcs intersecting the circle in $$A, A'$$.
7. Join $$AB, AC$$ and $$A'B, A'C$$
Then the required triangle is $$ABC$$ or $$A'BC$$.
Construct $$\triangle MNO$$ such that $$NO = 6.2\ cm$$, $$\angle N = 50^\circ$$ and $$MO - MN = 2.4\ cm.$$
1. Draw $$NO = 6.2\ cm$$.
2. Construct $$\angle N = 50^{ \circ } $$ and produce it to $$X$$
3. Cut off line segment $$NP= 2.4\ cm$$ on $$NX.$$
4. Join $$P-O$$.
5. Draw a perpendicular bisector of $$PO$$ such that it intersects $$NX$$ at $$M.$$
6. Join $$M-O.$$
7. $$MNO$$ is the required triangle.
Perimeter of $$\triangle ABC$$ is 14cm, AB=4.5 cm and $$\angle A = 80^o$$. Construct $$\triangle ABC$$.
1. Draw AB= 4.5 cm
2. Draw angle 80 deg at A and cut an arc of AD= 9.5 cm. (14- 4.5 = 9.5)
3. Join BD.
4. Draw perpendicular bisector of BD which meets AD at C.
5. Join BC
ABC is the required triangle.
Construct an angle of $$30^o$$ using compass.
1) Draw a line segment PQ
2) Draw an arc at point P of any length, the arc cuts the line segment PQ at S
3) Draw an arc from point S
4) This cut the previous at point T, angle TPS is 60 degrees
5) Draw two arcs from points S and T
6) These arcs cut each other at point R
7) Join P to R
8) PR is the bisector of Angle TPS
$$\implies \angle RPS$$ is $$30^{\circ}$$
Hence the required construction is completed.
Construct $$\triangle MNO$$ where base $$NO = 6.7\ cm$$, $$\angle MNO = 45^\circ$$ and $$MO-MN = 2.8\ cm.$$
1. Draw $$NO = 6.7 \ cm.$$
2. Construct $$\angle N = 45^{ \circ } $$ and produce it to $$X.$$
3. Cut off line segment $$NP= 2.8 \ cm$$ on $$NX.$$
4. Join $$P-O$$.
5. Draw a perpendicular bisector of $$PO$$ such that it intersects $$NX$$ at $$M.$$
6. Join $$M-O$$.
7. $$MNO$$ is the required triangle.
Construct $$\triangle LMN$$ in which base $$MN= 7\ cm$$, $$\angle LMN=75^\circ$$ and $$LM+LN = 9\ cm.$$
1. Draw $$MN = 5\ cm$$.
2. Construct $$\angle M = 75^{ \circ }$$ and produce it to $$Y.$$
3. Cut off line segment $$MO = 9\ cm $$ on $$MY.$$
4. Join $$N-O.$$
5. Draw a perpendicular bisector of $$NO$$ such that it intersects $$MY$$ at $$L.$$
6. Join $$L-N.$$
7. $$LMN$$ is the required triangle.
Construct a triangle $$XYZ$$ in which $$\angle {Y}={30}^{o}$$, $$\angle {Z}={90}^{o}$$ and $$XY+ZX+YZ=11cm$$.
Steps of Construction:
1. Draw a line segment $$AB = 11\ cm$$. $$( = PQ+QR+RP)$$.
2. At $$A$$, construct an angle of $$30^\circ$$ and at $$B$$, an angle of $$90^\circ$$.
3. Bisect these angles. Let the bisectors of these angles intersect at a point $$X$$.
4. Draw perpendicular bisectors $$ST$$ of $$AX$$ and $$UV$$ of $$BX$$ .Let $$ST$$ intersect $$AB$$ at $$Y$$
5. Join $$XY$$ and $$XZ$$.
Then, $$XYZ$$ is the required triangle.
Construct a triangle $$ABC$$ in which $$BC=7cm$$, $$\angle {B}={75}^{o}$$ and $$AB+AD=13\ cm$$.
1. Draw $$BC= 7\ cm$$
2. Draw angle $$75^\circ$$ at $$B$$ and cut an arc of $$BD= 13\ cm$$.
3. Join $$CD$$.
4. Draw a perpendicular bisector of $$CD$$ which meets $$BD$$ at $$A$$.
5. Join $$AC$$.
$$ABC$$ is the required triangle.
Construct $$\triangle PQR$$ such that perimeter of $$\triangle PQR$$ is $$16\ cm$$ and $$\angle Q=90^\circ$$ and $$QR = 5\ cm.$$
Steps of Construction:
1. Draw a line segment $$QR = 5\ cm$$.
2. At $$Q$$, construct an angle of $$90^\circ$$.
3. Cut an arc $$QD$$ of $$11\ cm (16 - 5 = 11).$$
4. Draw perpendicular bisector of $$DR$$ which intersects $$QD$$ at $$P$$.
5. Join $$P-R$$. Then, $$PQR$$ is the required triangle.
Construct an angle of $${90}^{\circ}$$ at the initial point of a given ray and justify the construction.
2. Cut an arc from point $$O$$ of any length.
3. Cut two arcs $$A$$ and $$B$$ on the previous arc (which are at the angle of $$60^{\circ}$$ and $$120^{\circ}$$).
4. Draw two arcs from points $$A$$ and $$B$$ and their point of intersection is $$C$$.
5. Join $$O-C$$.
Hence, $$\angle COX$$ is $$90^{\circ}$$
Construct an equilateral triangle, given its side $$=3 cm$$ and justify the construction.
Draw $$AB =3 cm$$
Taking $$A$$ and $$B$$ as centers, radius $$= AB = 3 cm$$
draw two areas intersecting each other at $$C$$
join $$AB$$ and $$AC$$
thus $$\triangle ABC$$ is constructed
for justification just measure the length of each side
Construct a triangle $$PQR$$ in which $$QR=6cm$$, $$\angle {Q}={60}^{o}$$ and $$PQ-PO=2cm$$.
1. Draw $$QR= 6 \ cm$$
2. Draw angle $$60^\circ$$ at $$B$$ and cut an arc of $$QO= 2\ cm$$.
3. Join $$OR$$.
4. Draw a perpendicular bisector of $$OR$$ which meets $$QO$$ at $$P.$$
5. Join $$P-R$$.
$$PQR$$ is the required triangle.
Construct the angles of the following measurements:
$$(i) {30}^{\circ}$$
$$(ii) {22\cfrac{1}{2}}^{\circ}$$
$$(iii) {15}^{\circ}$$
1)
1. Draw a ray $$OX$$.
2. Cut an arc from $$O$$, cut another arc of same length as $$OA$$, which intersects at $$B$$.
3. Cut two arcs from $$A$$ and $$B$$ which intersects at $$C$$.
4. Join $$O-C$$. $$\angle COA$$ is $$30^\circ$$.
2)
1. Draw a ray $$OX.$$
2. Cut an arc from $$O$$, cut $$2$$ another arcs of same length as $$OA$$, which intersects as $$B$$ and $$F$$.
3. Cut two arcs from $$B$$ and $$F$$ which intersects at $$C$$. ($$\angle COX=90$$ deg)
4. In the similar way, bisect $$\angle COX$$ to get $$DOX$$ which is $$45$$ deg angle and then again bisect $$\angle DOX$$ to get to get $$EOX$$ .
6. Join $$O-E$$. $$\angle EOA$$ is $$22.5^\circ$$.
3)
1. Draw a ray $$OX$$.
2. Cut an arc from $$O$$, cut another arc of same length as $$OA$$, which intersects at $$B$$.
3. Cut two arcs from $$B$$ and $$A$$ which intersects at $$C$$. ($$\angle COX=30$$ deg)
4. In the similar way, bisect $$\angle COX$$ to get $$DOX$$.
6. Join $$O-D$$. $$\angle DOA$$ is $$15^\circ$$.
Construct a triangle $$ABC$$ whose perimeter $$12 cm$$ and whose base angles are $$65^{\circ}$$ and $$85^{\circ}$$
1. Draw a line segment $$PQ=AB+BC+CA=12cm$$
2. Mark $$\angle LPQ=65^o$$ and $$\angle MQP=80^o$$
3. Now bisect $$\angle LPQ$$ and $$\angle MQP$$ such that their bisector meet at $$A$$
4. Draw a perpendicular bisectors of $$AP$$ and $$AQ$$ i.e $$XY$$ and $$ST$$ intersecting $$PQ$$ at $$B$$ and $$C$$ respectively.
5. Join $$AB$$ and $$AC$$
Construct a triangle $$ABC$$ whose perimeter $$12 cm$$ and whose base angles are $$50^{\circ}$$ and $$80^{\circ}$$
1. Draw a line segment $$PQ = AB+BC+CA =12cm$$
2. Make $$\angle LPQ = 50^0$$ and $$\angle MQP = 80^0$$
3. Now bisect $$\angle LPQ $$ and $$\angle MQP $$ such that their bisection meet at A.
4. Draw perpendicular bisectors of $$AP$$ and $$AQ$$ as $$XY$$ and $$ST$$ intersecting $$PQ$$ at $$B$$ and $$C$$ respectively.
5. Join $$AB$$ and $$AC$$.
Construct triangle $$ABC$$ in which $$BC = 3.4 cm, AB - AC = 1.5 cm$$ and $$\angle B = 45^{\circ}$$
1. Draw a line segment $$BC=3.4cm$$
2. Construct $$45^o$$ at $$B$$ and extend line $$BP$$
3. Take $$1.5cm$$ in compass and cut an arc at $$BP$$. Join $$CX$$. Take the bisector of $$CX$$.
4. It intersects with a point on $$BP$$. That point is $$A$$.
5. Join $$AC$$
Construct a triangle $$ABC$$, whose perimeter is $$12 cm$$ and whose sides are in the ratio $$3 : 4 : 5$$
Steps of construction
1. Draw a line segment and locate points $$X,Y$$ such that $$XY=12cm$$.
2. Draw a ray $$XZ$$, making an acute angle with $$XY$$ and drawn in the down-ward direction.
3. From $$X$$, locate $$(3+4+5)=12$$ points at equal distances along $$XZ$$.
4. Mark point $$L,M,N$$ on $$XZ$$ such that $$XL=3$$ parts, $$LM=4$$ parts and $$MN=5$$ parts.
5. Join $$NY$$. Through $$L$$ and $$M$$, draw $$LB∥NY$$ and $$MC∥NY$$, intersecting $$XY$$ in $$B$$ and $$C$$ respectively.
6. With $$B$$ as center and $$BX$$ as radius, draw an arc; with $$C$$ as a centre and $$CY$$ as radius, draw an arc cutting the previous arc at $$A$$.
7. Join $$AB$$ and $$AC$$.
Then $$ABC$$ is the required triangle.
Construct triangle $$ABC$$ in which $$BC = 5 cm, AB - AC = 2.8 cm$$ and $$\angle B = 40^{\circ}$$
1. Draw a line $$BC=5cm$$
2. Construct $$40^o$$ at $$B$$. Extend $$BP$$.
3. Take $$2.8cm$$ on a compass and cut an arc on $$BP$$. Join $$CX$$.
4. Take a perpendicular bisector of $$CX$$. It intersects with $$BP$$ at $$A$$.
5. Join $$AC$$
Construct triangle $$ABC$$ in which $$BC = 6 cm, AB - AC = 3.1 cm$$ and $$\angle B = 30^{\circ}$$
1. Draw a line $$BC=6cm$$
2. Construct $$30^o$$ at $$B$$. Extend $$BP$$.
3. Take $$3.1cm$$ on compass, cut an arc on $$BP$$. Join $$CX$$.
4. Take perpendicular bisector of $$CX$$. It intersects with $$BP$$ at $$A$$. Join $$AC$$
Construct a triangle $$ABC$$ in which $$BC = 3.6 cm, AB + AC = 4.8 cm$$ and $$\angle B = 60^{\circ}$$
Construct a triangle $$ABC$$ in which $$AB=5.6 cm, BC = 5.4 cm$$ and $$\angle B = 40^{\circ}$$
1. Construct a line segment $$BC=5.4cm$$
2. Construct a $$40^o$$ angle at $$B$$
3. Take the length $$AB$$ as $$5.6cm$$
4. Join $$AC$$
Construct a triangle $$ABC$$, whose perimeter is $$12 cm$$ and whose sides are in the ratio $$2 : 3 : 4$$.
Steps of construction
$$1$$. Draw a line segment and locate points $$X, Y$$ such that $$XY = 12 cm$$.
$$2$$. Draw a ray $$XZ$$, making an acute angle with $$XY$$ and drawn in the down-ward direction.
$$3$$. From $$X$$, locate $$(2 + 3 + 4) = 9$$ points at equal distances along $$XZ$$.
$$4$$. Mark point $$L, M, N$$ on $$XZ$$ such that $$XL = 2$$ parts, $$LM = 3$$ parts and $$MN = 4$$ parts.
$$5$$. Join $$NY$$. Through $$L$$ and $$M$$, draw $$LB \parallel NY$$ and $$MC \parallel NY$$, intersecting $$XY$$ in $$B$$ and $$C$$ respectively.
$$6$$. With $$B$$ as center and $$BX$$ as radius, draw an arc; with $$C$$ as a centre and $$CY$$ as radius, draw an arc cutting the previous arc at $$A$$.
$$7$$. Join $$AB$$ and $$AC$$.
Then $$ABC$$ is the required triangle.
$$1$$. Draw a line segment and locate points $$X, Y$$ such that $$XY = 12 cm$$.
$$2$$. Draw a ray $$XZ$$, making an acute angle with $$XY$$ and drawn in the down-ward direction.
$$3$$. From $$X$$, locate $$(2 + 3 + 4) = 9$$ points at equal distances along $$XZ$$.
$$4$$. Mark point $$L, M, N$$ on $$XZ$$ such that $$XL = 2$$ parts, $$LM = 3$$ parts and $$MN = 4$$ parts.
$$5$$. Join $$NY$$. Through $$L$$ and $$M$$, draw $$LB \parallel NY$$ and $$MC \parallel NY$$, intersecting $$XY$$ in $$B$$ and $$C$$ respectively.
$$6$$. With $$B$$ as center and $$BX$$ as radius, draw an arc; with $$C$$ as a centre and $$CY$$ as radius, draw an arc cutting the previous arc at $$A$$.
$$7$$. Join $$AB$$ and $$AC$$.
Then $$ABC$$ is the required triangle.
Construct a triangle $$XYZ$$ whose perimeter $$15\ cm$$ and whose base angles are $$60^{\circ}$$ and $$70^{\circ}$$
1. Draw a line segment $$PQ = XY+YZ+XZ =15\ cm$$
2. Make $$\angle LPQ = 60^{\circ}$$ and $$\angle MQP = 70^{\circ}$$
3. Now bisect $$\angle LPQ $$ and $$\angle MQP $$ such that their bisection meet at X.
4. Draw perpendicular bisectors of $$XP$$ and $$XQ$$ as $$AB$$ and $$CD$$ intersecting $$PQ$$ at $$Y$$ and $$Z$$ respectively.
5. Join $$XY$$ and $$XZ$$.
Draw $$\angle ABC$$ of measure $$120^{\circ}$$ and bisect it.
1. Draw a ray $$BC$$. Using a protractor, draw an angle of $$120^\circ$$ from end $$B$$ to ray $$BC$$. Mark any point on the ray as $$A$$.
2. Draw two equidistant arcs on ray $$BC$$ and ray $$BA$$ respectively.
3. From the two points on ray $$BC$$ and $$BA$$, draw two intersecting arcs and mark it as $$M$$. Join $$B-M$$.
Ray $$BM$$ is the required angle bisector.
Construct a triangle $$ABC$$ in which $$AB = 5.8\ cm, BC + CA = 8.4\ cm$$ and $$\angle B = 60^{\circ}$$
Steps of construction
$$1$$. Draw a line segment $$AB$$ of length $$5.8 cm$$.
$$2$$. Draw a line segment $$BX$$, sufficiently large such that $$\angle ABX = 60^{\circ}$$.
$$3$$. From the segment $$BX$$, cut off line segment $$BD$$ of length $$8.4 cm$$.
$$4$$. Join $$AD$$.
$$5$$. Draw the perpendicular bisector of $$AD$$ and let it meet $$BD$$ at $$C$$.
$$6$$. Join $$AC$$.
Then $$ABC$$ is the required triangle.
$$1$$. Draw a line segment $$AB$$ of length $$5.8 cm$$.
$$2$$. Draw a line segment $$BX$$, sufficiently large such that $$\angle ABX = 60^{\circ}$$.
$$3$$. From the segment $$BX$$, cut off line segment $$BD$$ of length $$8.4 cm$$.
$$4$$. Join $$AD$$.
$$5$$. Draw the perpendicular bisector of $$AD$$ and let it meet $$BD$$ at $$C$$.
$$6$$. Join $$AC$$.
Then $$ABC$$ is the required triangle.
Using a ruler and compass only.
(i) Construct a $$\triangle ABC$$ with the following data.
$$AB=3.5\ cm, BC=6\ cm$$ and $$\angle ABC=120^\circ$$
(ii) In the same diagram, draw a circle with $$BC$$ as diameter. Find a point $$P$$ on the circumference of the circle which is equidistant from $$AB$$ and $$BC$$.
(iii) Measure $$\angle BCP$$.
Steps of Construction.
(i) Draw a line $$BC=6\ cm$$.
(ii) With the help of the point $$B$$, draw $$\angle ABC=120^\circ$$
(iii) Taking radius $$3.5\ cm$$, cut $$BA=3.5\ cm$$
(iv) Join $$A - C$$.
(v) Draw $$\perp^r$$ bisector $$MN$$ of $$BC$$.
(vi) Draw a circle $$O$$ as centre and $$OC$$ as radius.
(vii) Draw angle bisector of $$\angle ABC$$ which intersects circle at $$P$$.
(viii) Join $$B-P$$ and $$C-P$$
(ix) Now, $$\angle BCP=30^\circ$$.
(i) Draw a line $$BC=6\ cm$$.
(ii) With the help of the point $$B$$, draw $$\angle ABC=120^\circ$$
(iii) Taking radius $$3.5\ cm$$, cut $$BA=3.5\ cm$$
(iv) Join $$A - C$$.
(v) Draw $$\perp^r$$ bisector $$MN$$ of $$BC$$.
(vi) Draw a circle $$O$$ as centre and $$OC$$ as radius.
(vii) Draw angle bisector of $$\angle ABC$$ which intersects circle at $$P$$.
(viii) Join $$B-P$$ and $$C-P$$
(ix) Now, $$\angle BCP=30^\circ$$.
Construct a triangle $$ABC$$ in which base $$AB = 5 cm, \angle A = 30^{\circ}$$ and $$AC - BC = 2.5 cm$$
Steps of construction
$$1$$. Draw a line segment $$AB$$ of length $$5 cm$$.
$$2$$. Draw another line segment $$AX$$, sufficiently large, such that $$\angle BAX = 30^{\circ}$$.
$$3$$. From the segment $$AX$$, cut off line segment $$AD = 2.5 cm$$, which is equal to $$(AC - BC)$$.
$$1$$. Draw a line segment $$AB$$ of length $$5 cm$$.
$$2$$. Draw another line segment $$AX$$, sufficiently large, such that $$\angle BAX = 30^{\circ}$$.
$$3$$. From the segment $$AX$$, cut off line segment $$AD = 2.5 cm$$, which is equal to $$(AC - BC)$$.
$$4$$ Join $$BD$$
$$5$$. Draw the perpendicular bisector of $$BD$$ and let it cut $$AX$$ at $$C$$.
$$6$$. Join $$BC$$.
Then $$ABC$$ is the required triangle.
$$5$$. Draw the perpendicular bisector of $$BD$$ and let it cut $$AX$$ at $$C$$.
$$6$$. Join $$BC$$.
Then $$ABC$$ is the required triangle.
Construct a triangle $$ABC$$ whose perimeter is $$12.5 cm$$ and whose base angles are $$60^{\circ}$$ and $$75^{\circ}$$
Steps of construction
$$1$$. Draw a line segment and locate points $$P, Q$$ such that $$PQ = 12.5 cm$$.
$$2$$. Construct rays $$\vec {PR}$$ such that $$\angle QPR = 60^{\circ}$$ and $$\vec {QS}$$ such that $$\angle PQS = 75^{\circ}$$.
$$3$$. Draw the bisectors $$PL$$ and $$QM$$ of $$\angle QPR$$ and $$\angle PQS$$ respectively. Let these intersect at $$A$$.
$$4$$. Draw the perpendicular bisector of $$AP$$ and $$AQ$$ and let these intersect $$PQ$$ at $$B$$ and $$C$$ respectively.
$$5$$. Join $$AB$$ and $$AC$$.
Then $$ABC$$ is the required triangle.
$$1$$. Draw a line segment and locate points $$P, Q$$ such that $$PQ = 12.5 cm$$.
$$2$$. Construct rays $$\vec {PR}$$ such that $$\angle QPR = 60^{\circ}$$ and $$\vec {QS}$$ such that $$\angle PQS = 75^{\circ}$$.
$$3$$. Draw the bisectors $$PL$$ and $$QM$$ of $$\angle QPR$$ and $$\angle PQS$$ respectively. Let these intersect at $$A$$.
$$4$$. Draw the perpendicular bisector of $$AP$$ and $$AQ$$ and let these intersect $$PQ$$ at $$B$$ and $$C$$ respectively.
$$5$$. Join $$AB$$ and $$AC$$.
Then $$ABC$$ is the required triangle.
Construct an angle of $${45}^{o}$$ and bisect it.
Procedure :-
(1) Draw a line segment AB
(2) Taking A as centre draw an semi-circle of radius less that AB.
(3) Taken P as centre mark an arc Q and from Q mark an arc R.
(4) From R draw an arc & from Q draw an arc, intersecting at point & both arc.
Join S & A.
$$iii/{T_1}$$ bisect angle $$90^\circ$$, we get $$45^\circ$$
% bisect angle $$45^\circ$$ we get $$22\frac{1}{2}$$ 40 through figure above mentioned
Draw $$\angle$$ ABC of measure $$125^o$$ and bisect it.
Given that angle measure of $$\angle AOB$$ is $$125^{0}$$
We have constructed it using a protractor.
Now constructing its bisector,
Using compass, mark two arcs, then join $$C$$ to $$O$$, we get $$OC$$ as a bisector of $$\angle AOB$$.
Construct a triangle $$PQR$$ in which $$QR=5\ cm,\angle R=40^{o}$$ and $$PR-PQ=1\ cm$$.
It is given that $$QR = 5$$ cm, the difference between other two sides $$PR - PQ = 1$$ cm and $$\angle R = 40^{\circ}$$.
Steps of construction.
1. Draw a line segment $$QR = 5\ cm$$.
2. At $$R$$, construct at angle of $$40^{\circ}$$ and produce it.
3. By taking $$R$$ as centre, draw an arc of radius $$1$$ cm cutting $$RT$$ at $$S$$.
4. Join $$QS$$.
5. Also, draw $$\perp $$ bisector of $$QS$$ which meets $$RT$$ at $$P$$
6. Join $$PQ$$.
Hence, $$\Delta PQR$$ is the required triangle.
Steps of construction.
1. Draw a line segment $$QR = 5\ cm$$.
2. At $$R$$, construct at angle of $$40^{\circ}$$ and produce it.
3. By taking $$R$$ as centre, draw an arc of radius $$1$$ cm cutting $$RT$$ at $$S$$.
4. Join $$QS$$.
5. Also, draw $$\perp $$ bisector of $$QS$$ which meets $$RT$$ at $$P$$
6. Join $$PQ$$.
Hence, $$\Delta PQR$$ is the required triangle.
Construct a triangle $$ABC$$ in which $$BC=7cm,\angle{B}={75}^{o}$$ and $$AB+AC=13cm$$
Given:In $$\triangle{ABC},BC=7$$cm,$$\angle{B}={75}^{\circ}$$ and $$AB+AC=13$$cm
Steps:
$$1.$$Draw the base $$BC=7$$cm
$$2.$$At the point $$B$$ make an $$\angle{XBC}={75}^{\circ}$$
$$3.$$Cut a line segment $$BD$$ equal to $$AB+AC=13$$cm from the ray $$BX$$
$$4.$$Join $$DC$$
$$5.$$Make an $$\angle{DCY}=\angle{BDC}$$
$$6.$$Let $$CY$$ intersect $$BX$$ at $$A$$
Result:$$\triangle{ABC}$$ is the required triangle.
Construct an angle of $${90}^{o}$$ at the initial point of a given ray and justify the construction.
$$\textbf{Step-1: Draw ray}$$ $$\mathbf{OP}$$ 
$$\text{Take O as centre, draw an arc cutting}$$ $$OP$$ $$\text{at}$$ $$Q.$$
$$\textbf{Step-2: Taking Q as centre and same radius, cut arc R, then from R, cut arc S.}$$
$$\textbf{Step-3: Taking centre R and S, draw point A, taking distance more than half of RS}$$
$$\textbf{ and then join OA.}$$
$$\textbf{Hence ,we have constructed right angled angle}$$ $$\mathbf{\angle{POA}=90^0.}$$
If the measure of an angle is equal to $${225}^{\circ}$$, then find the fraction of one complete turn?
Construct $$\triangle ABC$$ in which $$BC=6.3\ cm$$ $$\angle B=75^{o}$$ and $$AB+AC=9\ cm$$.
(1) Draw a base BC of length 6.3 cm
(2) Draw angle $$B=75^\circ$$
(3) Open the compass for length $$AB+AC=9cm$$
(4) Cot an arc on ray BX at D
(5) Join CP
(6) Draw $$ \bot $$ bisector of CD.
(7) Mark point A where $$ \bot $$ bisector cut BD.
(8) Join AC
Construct $$\triangle XYZ$$ in which $$\angle Y=30^{o}, \angle Z=60^{o}$$ and $$XY+YZ+ZX=10\ cm$$.
REF.Image
Given $$xy+yz+zx= 10 cm, \angle y= 30^{\circ},\angle z=60^{\circ}$$
Step 1 :- Draw a line segment AB
$$AB= xy+yz+zx$$
Step 2 :- make angle $$30^{\circ}$$ that is $$\angle y=30^{\circ}$$ from point A.
Step 3 :- make angle $$\angle z=60^{\circ}$$ from point B
Step 4 :- Draw bisector of $$\angle LAB$$ & $$\angle MBA$$
these bisect at x.
Step 5 :- Draw $$\perp er$$ bisector from XA which intersect
AB at y.
Step 6 :- Draw $$\perp er$$ bisector from XB which is also
intersect AB at z.
Step 7 :- Join $$xy\ \&\ xz$$.
$$\therefore \triangle xyz$$ is required and is formed.
Write the steps to construct $$\triangle ABC$$, in which $$BC=5.2\ cm$$, $$\angle ACB={45}^{\circ}$$ and perimeter of $$\triangle ABC$$ is $$10\ cm$$.
$$\begin{array}{l} BC=5.2\, \, cm\, \, P=10\, \, cm \\ AB+BC+CA+10 \\ \Rightarrow AB+CA=4.8\, \, and\, \, \angle C=45^{ \circ } \end{array}$$
$$ \rightarrow$$ Draw a line segment BC = 5.2 cm
$$ \to$$ Draw $$ \angle C=45^\circ$$
$$ \rightarrow$$ with centre C and radius 4.8 cm, makes an AX which intersect CX at D
$$ \rightarrow$$ join A to C
$$ \rightarrow$$ Draw a perpendicular bisector at segment DB it intersect the
line segment CD at point A.
$$ \rightarrow$$ Join C to A.
Construct an angle of $${45}^{o}$$ at the initial point of a given ray and justify the construction.
$$\textbf{Step -1: Construction.}$$ 
$$\text{Draw a ray}$$ $$OA$$
$$\text{With}$$ $$O$$ $$\text{as center and any radius, draw arc}$$ $$DCB.$$
$$\text{With the same radius and center as}$$ $$B,C$$ $$\text{respectively, mark points}$$ $$C,D.$$
$$\text{From points}$$ $$C,D$$ $$\text{draw two arcs intersecting each other at}$$ $$P.$$
$$\text{Now, we have constructed a right angle}$$ $$\angle{POB}.$$
$$\text{Considering B and Q as center, draw a perpendicular bisector intersecting at point}$$ $$R.$$
$$\text{Join}$$ $$O,R.$$
$$\text{The angle formed}$$ $$\angle{ROA}=45^0.$$
$$\textbf{Step -2: Justification.}$$
$$\angle{POA}=90^0$$
$$\text{A perpendicular bisector divides an angle into two equal parts.}$$
$$\text{Hence,}$$ $$\angle{ROA}=\dfrac{1}{2}\angle{POA}.$$
$$\angle{ROA}=45^0$$
$$\textbf{Hence, we constructed}$$ $$\mathbf{\angle{ROA}=45^0.}$$
Construct $$\Delta ABC$$, such that $$BC = 6cm$$,$$\angle ABC = {100^0}$$ and $$AC - AB = 2.5cm$$.
Step 1: Construct a straight line BC = 6 cm.
Step 2: At point B, construct an angle of $$100^o$$
Step 3: Below the line BC,on the arm of this angle mark point X at a distance of 2.5 cm from B
Step 4: Join X to C.
Step 5: Now the copy the angle formed at X on C.
Step 6: Extend the arms of both the angles and mark their intersection as A.
Construct an angle of $$45^{\circ}$$ at the initial point of a given point of a given ray and justify the construction.
.
1 .First, draw a ray OA with intial point O.
2.Taking O as a centre and some radius, draw an arc of a circle, which intersects OA, at a point B.
3.Taking B as centre and with the same radius as before, draw an arc intersecting the previous drawn arc, at a point C.
4.Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 2, say at D.
5.Draw the ray OE passing through C. Then ∠EOA = 60∘ .
6.Draw the ray OF passing through D. Then ∠ FOE = 60∘ .
7. Next, taking C and D as centres and with radius more than ½ CD, draw arcs to intersect each other, at G.
8. Draw the ray OG, which is the bisector of the angle FOE, i.e., ∠FOG = ∠EOG = 1/2 ∠FOE = 1/2 (60∘ ) = 30∘ .
Thus, ∠GOA = ∠GOE + ∠ EOA = 30∘ + 60∘ = 90∘.
9.Now taking O as centre and any radius more than OB, draw an arc to intersect the rays OA and OG, at H and I.
10. Next, taking H and I as centres and with the radius more than 1/2 HI, draw arcs to ntersect each other, at J.
11. Draw the ray OJ. This ray OJ is the required bisector of the ∠ GOA. Thus, ∠GOJ = ∠AOJ = 1/2 ∠GOA = 1/2(90∘) = 45∘.
Justification:
(i)Join BC.
Then, OC = OB = BC triangle. (By construction)
∴ ∠COB is an equilateral triangle.
∴ ∠COB = 60∘.
∴ ∠EOA = 60∘.
(ii)Join CD.
Then, OD = OC = CD (By construction)
∆DOC is an equilateral triangle.
∴ ∠DOC = 60∘.
∴ ∠ FOE = 60∘.
(iii)Join CG and DG.
In ΔODG and ΔOCG,
OD = OC[ Radii of the same arc]
DG=CG [Arcs of equal radii]
OG=OG [Common]
∴ Δ ODG = ΔOCG [SSS Rule]
∴ ∠ DOG= ∠COG [CPCT]
∴ ∠FOG = ∠ EOG = 1/2 ∠FOE = 1/2 (60∘) = 30∘
Thus, ∠GOA = ∠GOE + ∠EOA = 30∘ + 60∘ = 90∘
∴ ∠AOJ= ∠GOJ= 1/2 ∠GOA = ½(90°)=45
Construct an angle of $$90^{o}$$ at the point of a given ray and justify the construction.
Steps of Construction:
$$1.$$Draw a ray $$OA$$
$$2.$$Taking $$O$$ as center and any radius, draw an arc cutting $$OA$$ at $$B.$$
$$3.$$Now, taking $$B$$ as center and with the same radius as before, draw an arc intersecting the previously drawn arc at point $$C$$.
$$4.$$With $$C$$ as center and the same radius,draw an arc cutting the arc at $$D$$.
$$5.$$With $$C$$ and $$D$$ as centers and radius more than $$\dfrac{1}{2}CD,$$ draw two arcs intersecting at $$P$$.
$$6.$$Join $$OP$$.
Thus, $$\angle{AOP}={90}^{\circ}$$
Justification:We need to prove $$\angle{AOP}={90}^{\circ}$$
Join $$OC$$ and $$BC$$
Thus, $$OB=BC=OC$$ radius of equal arcs by construction
$$\therefore \triangle{OCB}$$ is an equilateral triangle.
$$\angle{BOC}={60}^{\circ}$$
Join $$OD,OC$$ and $$CD$$
Thus, $$OD=OC=DC$$ radius of equal arcs by construction
$$\therefore \triangle{DOC}$$ is an equilateral triangle.
$$\angle{DOC}={60}^{\circ}$$
join $$PD$$ and $$PC$$
Now, in $$\triangle{ODP}$$ and $$\triangle{OCP}$$
$$OD=OC$$ radius of same arcs
$$DP=CP$$ arc of same radii
$$OP-OP$$(common)
$$\therefore\triangle{ODP}\cong\triangle{OCP}$$ by SSS congruency
$$\therefore \angle{DOP}=\angle{COP}$$ by CPCT
So, we can say that
$$\angle{DOP}=\angle{COP}=\dfrac{1}{2}\angle{DOC}$$
$$\angle{DOP}=\angle{COP}=\dfrac{1}{2}\times {60}^{\circ}={30}^{\circ}$$
Now, $$\angle{AOP}=\angle{BOC}+\angle{COP}$$
$$\angle{AOP}={60}^{\circ}+{30}^{\circ}={90}^{\circ}$$
Hence justified.
Draw the following angles using ruler and compasses. Also label them.
$$60^{o}$$
Step $$1:$$Draw a line $$l$$ and mark a point $$O$$ on it.
Step $$2:$$ Place the pointer of the compasses at $$O$$ and draw an arc of convenient radius which cuts the line $$PQ$$ at a point say, $$A$$.
Step $$3:$$ With the pointer at $$A$$ (as center), now draw an arc that passes through $$O$$.
Step $$4:$$ Let the two arcs intersect at $$B$$. Join $$OB$$. We get $$\angle{BOA}$$ whose measure is $${60}^{\circ}$$
'
Draw the following angles using ruler and compasses. Also label them.
$$180^{o}$$
Step$$1:$$Draw a line $$PQ$$
Step$$2:$$Mark a point $$A$$ at the center.
Step$$3:$$Draw a semi circle using compass from point $$A$$.
Result: $$\angle{PAQ}={180}^{\circ}$$ is the required angle.
Draw an angle of measure $${147^ \circ }$$ and construct its bisector.
1.Draw a line OA.
2.Using protractor and center 'O'draw an angle $$AOB=147^{\circ}$$
3.Now Taking 'O'as center and any radius draw an arc that intersects 'OA' and 'OB' at P and Q.
4.Now take P and Q as center and radius more than half of PQ,draw arcs.
5.both the arcs intersects at 'R'.
6.Join OR and produce it.
7.OR is the required bisector of angle AOB.
Draw an angle of measure $${147}^{\circ}$$ and construct its bisector
Steps of construction:-
- Draw a ray $$OA.$$
- Place the center of the protractor on point $$O,$$ and coincide line $$OA$$ and protractor line.
- Mark point $$B$$ on $${147}^{\circ}.$$
- Join $$OB$$ and mark points $$C$$ and $$D$$ where arc intersects $$OA$$ and $$OB.$$
- Now, taking $$C$$ and $$D$$ as centers and with a radius more than $$OC,$$ draw arcs to intersect each other.
- Let them intersect at point $$E$$
- Join $$O$$ and $$E.$$
$$\therefore\,\,OE$$ is bisector of $$\angle{AOB}={147}^{\circ}.$$
Draw a line segment $$PQ=6.2\ cm$$. Draw the perpendicular bisector of $$PQ$$.
To draw a perpendicular bisector of $$PQ$$ follow the below steps:
$$\Rightarrow$$ Draw a line of length $$6.2\,cm$$ and name the segment as $$PQ.$$
$$\Rightarrow$$ Now take a compass and measure any length greater than half the length of the segment.
$$\Rightarrow$$ Now, place the compass on any point $$P$$ or $$Q$$ and draw one arc above the segment and one below the segment.
$$\Rightarrow$$ Now measuring the same length in the compass, mark same arc intersecting the arc from other end of the segment.
$$\Rightarrow$$ Draw $$AB$$ meeting $$PQ$$ at $$R.$$
Construct a triangle $$ABC$$ in which $$BC=8$$cm,$$\angle{B}={45}^{\circ}$$ and $$AB-AC=3.5$$cm
$$1.$$Draw base $$BC$$ of length $$8$$cm
$$2.$$Now, let's draw $$\angle{B}={45}^{\circ}$$.Let the ray be $$BX$$
$$3.$$Open the compass to length $$AB-AC=3.5$$cm.
Since $$AB-AC=3.5$$cm is positive.
So, $$BD$$ will be above line $$BC$$
From point $$B$$ as center, cut an arc on ray $$BX$$.
Let the arc intersect $$BX$$ at $$D$$
$$4.$$Join $$CD$$
$$5.$$Now, we will draw perpendicular bisector of $$CD$$
$$6.$$Mark point $$A$$ where perpendicular bisector intersects $$BD$$
$$7.$$Join $$AC$$
Result:$$\therefore \triangle{ABC}$$ is the required triangle.
Construct a $$\triangle ABC$$ in which $$BC=5.6\ cm, \angle B=30^{o}$$ and the difference between the other sides is $$3\ cm$$.
\begin{array}{l} Step\, of\, construction \\ 1)\, Draw\, the\, base\, BC=5.6cm\, and\, at\, po{ { int } }\, B\, make\, an\, angle\, \angle \times BC=30^{ \circ } \\ 2)\, Cut\, the\, line\, segment\, BD\, equal\, to\, 3cm\, from\, ray\, BX \\ 3)\, Join\, CD\, and\, draw\, the\, perpendicular\, bi\sec tor,\, say\, PQ\, of\, CD \\ 4)\, Let\, it\, { { int } }er\sec t\, BX\, at\, a\, po{ { int } }\, A\, Join\, AC \\ \, \, \, \, \, \, \, The\, ABC\, is\, the\, required\, triangle.\, \end{array}
Construct triangle $$ABC$$ having given $$BC=7\ cm,\ AB-AC=1\ cm$$ and $$\angle\ ABC=45^{o}$$.
Construct a perpendicular line from point $$p$$ to any line $$AB$$.
1. Place your compass point on P and swing an arc of any size that crosses the line $$AB$$ twice.
2. Place the compass point on one of the two locations where the arc crossed the line $$AB$$ and make a small arc below the line (on the side where P is not located).
3. Without changing the span on the compass, place the compass point on the other location where the first arc crossed the line $$AB$$ and make another small arc below the line. The two small arcs should be intersecting (on the side of the line opposite of point $$P$$).
4. Using a straightedge, connect the intersection of the two small arcs to point $$P.$$
Use ruler and compasses only for this question.
Construct a $$\triangle ABC,\,AB=3.5cm,\,BC=6cm,\,\&\,\angle ABC=60^\circ$$
Mark the point P which is equidistant from $$AB$$, $$BC$$ and also equidistant from $$B$$ and $$C$$.
1. Draw a line segment $$BC=6cm$$
2. Draw $$\angle XBC=60^\circ$$ at $$B$$
3. Taking $$B$$ as centre and radius $$3.5cm$$ cut a line segment $$AB=3.5cm $$ on $$BX$$
4. Join $$A$$ and $$C$$
5. $$\triangle ABC$$ is the required triangle
6. Draw the angle bisector $$BY$$ of $$\angle ABC$$
7. Draw the perpendicular bisector $$LM$$ of side $$BC$$
8. Let $$BY$$ and $$LM$$ intersect at $$P$$
7. Draw the perpendicular bisector $$LM$$ of side $$BC$$
8. Let $$BY$$ and $$LM$$ intersect at $$P$$
9. $$P$$ is equidistant from $$AB, \,BC$$ and also equidistant from $$B$$ and $$C.$$
Construct a triangle ABC in which $$BC=8cm,\angle B={ 45 }^{ \circ }andAB-AC=3.5cm.$$
Draw an angle of $$60^\circ,$$ using a pair of compasses. Bisect it to make an angle of $$30^\circ.$$
Steps of construction:-
Draw a ray $$QP.$$ With $$Q$$ as the center and any suitable radius, draw an arc cutting $$QP$$ at a point $$N.$$
With $$N$$ as the center and the same radius as before, draw another arc to cut the previous arc at $$M.$$
Join $$QM$$ and produce it to $$R.$$ Hence, $$\angle PQR=60^\circ.$$
With $$M$$ taking as center and radius more than half $$MN,$$ draw an arc.
With $$N$$ as the center and the same radius as in the previous step, draw another arc, draw another arc cutting the previously drawn arc at point $$X.$$
Join $$QX$$ and produce it to point $$S.$$ Now, ray $$QS$$ is the bisector of $$\angle PQR.$$
Draw any angle with vertex $$O$$. Take a point $$A$$ on one of its arms and $$B$$ on another such that $$OA=OB$$. Draw the perpendicular bisectors of $$\overline{OA}$$ and $$\overline{OB}$$. Let them meet at $$P$$. Is $$PA=PB$$?
- Draw any angle whose vertex is $$O$$.
- With a convenient radius, draw arcs on both rays of this angle while taking $$O$$ as the center. Let these points be $$A$$ and $$B$$.
- Taking $$O$$ and $$A$$ as centers and with a radius of more than half of $$OA$$, draw arcs on both sides of $$OA$$. Let these be intersecting at $$C$$ and $$D$$. Join $$CD$$.
- Similarly, we can find the perpendicular bisector $$\overline{EF}$$ of $$\overline{OB}$$. These perpendicular bisectors, $$\overline{CD}$$ and $$\overline{EF}$$ will intersect each other at $$P$$.
Without using set-squares or protractor construct:
Triangle $$ABC$$, in which $$AB = 5.5$$ cm, $$BC = 3.2$$ cm and $$CA = 4.8$$ cm.
Draw the locus of a point which moves so that it is equidistant from the sides $$BC$$ and $$CA.$$
Steps of construction
1. Construct $$BC = 3.2$$ cm long.
2. draw an arc taking $$B$$ as centre and $$radius=5.5cm$$
3. draw another arc taking $$C$$ as centre and $$radius=4.8cm$$ to cut the previous arc at $$A$$
4. Join $$AB$$ and $$AC$$
5. Draw the angle bisector of $$\angle ACB$$
6. any point on the bisector of angle $$\angle ACB$$ is equidistant from $$BC$$ and $$CA$$
6. any point on the bisector of angle $$\angle ACB$$ is equidistant from $$BC$$ and $$CA$$
BM and CN are perpendiculars to a line passing through the vertex A of a triangle ABC. If L is the mid-point of BC, prove that $$LM=LN$$.
Construct a $$\Delta ABC$$ in which BC=5 cm, $$\angle B=45$$ and AB-AC=2.8cm
$$ABC$$ is a triangle. $$D$$ is a point on $$AB$$ such that $$AD=\dfrac{1}{4}AB$$ and $$E$$ is a point on $$AC$$ such that $$AE=\dfrac{1}{4}AC$$. Prove that $$DE=\dfrac{1}{4}BC$$.
Then, $$PQ\parallel BC$$ such that $$PQ=\dfrac{1}{2}BC$$
In $$\Delta APQ, D$$ and $$E$$ are mid-points of $$AP$$ and $$AQ$$$ respectively.
$$\therefore DE\parallel PQ$$ and $$DE=\dfrac{1}{2}PQ$$ ...(ii)
From (i) and (ii), we get
$$DE=\dfrac{1}{4}BC$$
Draw a right triangle having hypotenuse of length $$5.4$$cm, and one of the acute angles of measure $$30^o$$.
$$ Let\text{ ABC be the right triangle such that hypotenuse BC=5}\text{.4cm,} $$
$$ \text{Let }\angle C={{30}^{0}} $$
$$ Therefore $$
$$ \angle A+\angle B+\angle C={{180}^{0}} $$
$$ \Rightarrow \angle B=180-30-90={{60}^{0}} $$
$$ Steps\,of\,construction: $$
$$ 1.\,Draw\text{ a line segment BC=5}\text{.4cm} $$
$$ \text{2}\text{.Draw angle CBY=6}{{\text{0}}^{0}} $$
$$ 3.\,Draw\text{ angle BCX of measure 3}{{\text{0}}^{0}}\text{ with X on the same side of BC as Y} $$
$$ \text{4}\text{.Let BY and CX intersect at A} $$
$$ \text{Then ABC is the required triangle } $$
$$ \text{Let }\angle C={{30}^{0}} $$
$$ Therefore $$
$$ \angle A+\angle B+\angle C={{180}^{0}} $$
$$ \Rightarrow \angle B=180-30-90={{60}^{0}} $$
$$ Steps\,of\,construction: $$
$$ 1.\,Draw\text{ a line segment BC=5}\text{.4cm} $$
$$ \text{2}\text{.Draw angle CBY=6}{{\text{0}}^{0}} $$
$$ 3.\,Draw\text{ angle BCX of measure 3}{{\text{0}}^{0}}\text{ with X on the same side of BC as Y} $$
$$ \text{4}\text{.Let BY and CX intersect at A} $$
$$ \text{Then ABC is the required triangle } $$
Draw an angle of measure $$127^{\circ}$$ and construct its bisector.
Steps of construction :
(i) Draw ¯OQ.
(i) Draw ¯OQ.
Without using set-squares or protractor construct:
construct a $$\triangle ABC,\,AB=5.5cm,\,BC=3.2cm,\,\&\,AC=4.8cm$$
Find the locus which is equidistant from sides $$BC\,\&\,AC$$ and also at a distance of $$2.5cm$$ from $$B$$
Mark the point of intersection of the loci with the letter $$P$$ and measure $$PC$$.
Steps of construction;-
1. draw a line segment $$BC=3.2cm$$
2. taking $$B$$ as centre and $$radius=5.5cm$$ construct an arc
3. taking $$C$$ as centre and $$radius=4.8cm$$ draw an arc to cut the previous arc at $$A$$
4. join $$AB\,\&\,AC$$
5. Construct the bisector of $$\angle BCA.$$
6. Taking $$B$$ as centre and $$2.5$$ cm radius, construct an arc which intersects the angle bisector of $$\angle BCA.$$ at $$P$$ and $$P’$$.
Here $$P$$ and $$P’$$ are the two loci which satisfy the given condition.
By measuring $$CP$$ and $$CP’$$
$$CP = 3.6$$ cm and $$CP’ = 1.1$$ cm.
1. draw a line segment $$BC=3.2cm$$
2. taking $$B$$ as centre and $$radius=5.5cm$$ construct an arc
3. taking $$C$$ as centre and $$radius=4.8cm$$ draw an arc to cut the previous arc at $$A$$
4. join $$AB\,\&\,AC$$
5. Construct the bisector of $$\angle BCA.$$
6. Taking $$B$$ as centre and $$2.5$$ cm radius, construct an arc which intersects the angle bisector of $$\angle BCA.$$ at $$P$$ and $$P’$$.
Here $$P$$ and $$P’$$ are the two loci which satisfy the given condition.
By measuring $$CP$$ and $$CP’$$
$$CP = 3.6$$ cm and $$CP’ = 1.1$$ cm.
Bisect the $$ \angle XYZ $$ and write the steps of construction as well.
Steps for construction:
Step $$I$$ - Taking $$Y$$ as the center, draw an arc $$AB$$ of any radius intersecting $$ZY$$ and $$XY$$ at $$A$$ and $$B$$ respectively.
Step $$II$$ - Taking $$A$$ as the center, draw an arc of a radius greater than the length of arc $$AB$$.
Step $$III$$ - Taking $$B$$ as the center, draw an arc of the same radius as in the previous step.
Step $$IV$$ - Mark the intersecting point of the two arcs as point $$C$$.
Step $$V$$- Draw a ray $$YD$$ through the point $$C$$.
Thus, ray $$YD$$ bisects $$ \angle XYZ$$
Draw a line segment $$ AB=8\ cm. $$ Mark a point $$ P $$ in $$ AB $$ so that $$ AP =5\ cm . $$ At $$ P $$, construct $$\angle APQ=30^\circ.$$
Steps of Construction:
$$\angle APQ$$ is the required angle of measure $$30^\circ. $$
- Construct a line segment $$ AB =8\ cm.$$
- Now mark the point $$ P $$ on $$ AB $$ such that $$ AP=5\ cm.$$
- Taking $$ P $$ as center and some suitable radius construct an arc which meets $$ AB $$ in $$ L.$$
- Taking $$ L $$ as center and same radius cut the arc $$ LM. $$
- Now bisect the arc $$ LM $$ at the point $$ N .$$
- Join $$ PN $$ and produce it to point $$ Q .$$
Construct angle $$ABC=45^o$$ in which $$BC=5cm$$ and $$AB=4.6cm$$.
Steps constructions:
1. Construct a line segment $$BC=5cm$$
2. With center $$B$$, construct an arc of any suitable radius, which cuts $$BC$$ at the point $$D$$.
3. Taking $$D$$ as center and the radius, as taken in step $$2$$, construct an arc which cuts the previous arc at point $$E$$.
4. Taking $$E$$ as center and the same radius, construct one more arc which cuts the first arc at point $$F$$.
5. Taking $$E$$ and $$F$$ as center and radii equal to more than half the distance between $$E$$ at $$F$$, construct arc which cut each other at point $$P$$.
6. Now join $$BP$$ to meet $$EF$$ at $$L$$ and produce to point $$0$$. Then $$\angle OBC=90^o$$
7. Construct $$BA$$, the bisector of angle $$OBC$$. [with $$D,L$$ as centers and suitable radius construct two arc meeting each other at $$Q$$ produced it to $$R$$]
$$=>\angle ABC=45^o[\therefore BA$$ is bisector of $$\angle ABC==45^o]$$
8. From $$BR$$ cut arc $$AB=4.6cm$$
1. Construct a line segment $$BC=5cm$$
2. With center $$B$$, construct an arc of any suitable radius, which cuts $$BC$$ at the point $$D$$.
3. Taking $$D$$ as center and the radius, as taken in step $$2$$, construct an arc which cuts the previous arc at point $$E$$.
4. Taking $$E$$ as center and the same radius, construct one more arc which cuts the first arc at point $$F$$.
5. Taking $$E$$ and $$F$$ as center and radii equal to more than half the distance between $$E$$ at $$F$$, construct arc which cut each other at point $$P$$.
6. Now join $$BP$$ to meet $$EF$$ at $$L$$ and produce to point $$0$$. Then $$\angle OBC=90^o$$
7. Construct $$BA$$, the bisector of angle $$OBC$$. [with $$D,L$$ as centers and suitable radius construct two arc meeting each other at $$Q$$ produced it to $$R$$]
$$=>\angle ABC=45^o[\therefore BA$$ is bisector of $$\angle ABC==45^o]$$
8. From $$BR$$ cut arc $$AB=4.6cm$$
Construct an angle of $${45}^{\circ}$$ from a horizontal line and justify the construction.
1. Draw a ray $$OX$$.
2. Cut an arc from point $$O$$ of any length.
3. Cut two arcs $$A$$ and $$B$$ on the previous arc (which are at the angle of $$60^\circ$$ and $$120^\circ$$).
4. Cut two arc from points $$A$$ and $$B$$ and their point of intersection is $$C$$.
5. Join $$O-C$$. $$\angle COX$$ is $$90^\circ$$.
6. Bisect $$\angle COX$$ through cutting two arcs from $$D$$ and $$E$$, their point of intersection is $$F$$.
7. Join $$F-O$$,$$\angle FOX$$ is $$45^\circ$$
Using a protractor, draw $$ \angle BAC $$ of measure $$ 45^{\circ}$$. Take a point P in the interior of $$ \angle BAC.$$ From P draw line segments PM and PN such that $$ PM \perp AB$$ and $$ PN \perp AC$$, Measure $$ \angle MPN.$$
Step I: Draw a line segment $$A$$ on the line $$L$$ .
Step II: Take a protractor and place it on the segment $$AC$$ such that $$AC$$ coincides with the line of the diameter of the protractor and the middle point of the line coincides with point $$A$$.
Step III: Counting from the right side, mark a point as $$B$$ at the point of $$45^°$$ of protractor and draw a line segment $$AB$$.
Step IV: Take a convenient radius with $$P$$ as centre, construct an arc intersecting the line segments $$AB$$ at $$T$$ and $$Q$$ and $$AC$$ at $$R$$ and $$S$$.
Step V: Using the same radius and with $$T$$ and $$Q$$ as centres, construct two arcs intersecting at $$G$$ on the other side.
Step VI: Using the same radius and with $$R$$ and $$S$$ as centres, construct two arcs intersecting at $$H$$ on the other side.
Step VII: Join $$PG$$ and $$PH$$ which intersects $$AB$$ and $$AC$$ at $$M$$ and $$N$$, respectively.
On measuring $$∠MPN$$ using a protractor, we get it equal to $$135^°$$.
Construct a triangle $$ABC$$ with $$AB=5.5\ cm,AC=6\ cm$$ and $$\angle BAX=105^{o}$$
Hence :
Construct the locus of point equidistant from $$B$$ and $$C$$.
Construct a Triangle Abc with Ab = 5.5 Cm, Ac = 6 Cm and ∠Bac = 105° Construct the Locus of Points Equidistant from Ba and Bc Construct the Locus of Points Equidistant from B and C. Mark the Point Which Satisfies the Above Two Loci as P. Measure and Write the Length of Pc. Concept: Constructions Under Loci.
In triangle $$ABC$$; angle $$A=90^{o}$$, side $$AB=x\ cm,\ AC=(x+5)\ cm$$ and area $$=150\ cm^{2}$$. Find the sides of the triangle.
Area of $$\triangle{ABC}=\dfrac{1}{2}\times b\times h$$
$$\Rightarrow 150=\dfrac{1}{2}\times x\times \left(x+5\right)$$
$$\Rightarrow {x}^{2}+5x-300=0$$
$$\Rightarrow {x}^{2}+20x-15x-300=0$$
$$\Rightarrow x\left(x+20\right)-15\left(x+20\right)=0$$
$$\Rightarrow \left(x+20\right)\left(x-15\right)=0$$
$$\Rightarrow x\neq -20$$ and $$x=15$$
$$AB=x=15$$cm
$$AC=x+15=15+5=20$$cm
$$BC=\sqrt{{AB}^{2}+{BC}^{2}}=\sqrt{{20}^{2}+{15}^{2}}=\sqrt{400+225}=\sqrt{625}=25$$cms
Hence the sides of the triangle are $$15$$cm,$$20$$cm and $$25$$cm.
Construct a triangle $$ABC$$ with $$AB=5.5\ cm,AC=6\ cm$$ and $$\angle BAX=105^{o}$$
Hence :
Construct the locus of points equidistant from $$BA$$ and $$BC$$ .
1) Draw a line segment AB of length 5.5 cm.
2) Make an angle MZBAX = 105° using a
protractor
3) Draw an arc AC with radius AC = 6 cm on AX with centre at A.
4) Join BC.
Thus ABC is the required triangle.
a) Draw BR, the bisector of ABC, which is the locus of points equidistant from BA and BC.
b) Draw MN, the perpendicular bisector of BC, which is the locus of points
equidistant from B and C c) The angle bisector of ABC and the perpendicular bisector of BC meet at
point P. Thus, P satisfies the above two
loci.
Length of PC = 4.8 cm
Construct a $$ \triangle A B C $$ in which B C = 3.6 cm, A B = 5 cm and A C = 5.4 cm. Draw the perpendicular bisector of the side B C.
Construct a triangle XYZ in which $$\angle Y$$ = $${30}^0$$ , $$\angle Z$$ = $${90}^0$$ and XY + YZ + ZX = 11 cm .
Draw a line segment AB of 11 cm
Construct 30° from A and 90° from B
Construct angle bisectors for both the angles
They will meet at X
Draw perpendicular line bisector for AX and BX .
The both lines will meet AB at X and Y
Join XY and XZ
XYZ is our required triangle.
Draw a triangle ABC in which BC$$=4$$cm, AB$$=3$$cm and $$\angle B=45^o$$. Also, draw a perpendicular from A on BC.
Construct $$\Delta$$ABC in which AB$$=6.4$$cm, $$\angle A=45^o$$ and $$\angle B=60^o$$.
Construct the following angles with the help of ruler and compasses only:$$150^{\circ}$$
Draw a line $$AB$$ and take point $$O$$ at the middle of $$AB$$.
With a convenient radius and centre at $$O$$, draw an arc, which cuts the line $$AB$$ at $$P$$ and $$Q$$.
With the same radius and centre at $$Q$$, draw an arc, which cuts the first arc at $$R$$.
With the same radius and centre at $$R$$, draw an arc, which cuts the first arc at $$S$$.
With the centres $$P$$ and $$S$$ and radius more than half of $$PS$$, draw two arcs, which cut each other at $$T$$.
Draw $$OT$$ and extend it to $$C$$ to form the ray $$OC$$.
$$∠BOC$$ is required angle of $$150^0$$.
Construct the following angles with the help of ruler and compasses only:$$105^{\circ}$$
Draw a ray $$OA$$ and make an angle $$∠AOB = 90^0$$ and $$∠AOC = 120^0$$
Now bisect $$∠BOC$$ and get the ray $$OD$$.
$$∠AOD$$ is the required angle of $$105^0$$.
Draw $$\Delta$$ABC in which AC$$=6$$cm, $$\angle A=90^o$$ and $$\angle B=60^o$$.
Draw $$\Delta$$ABC in which $$\angle C=90^o$$ and $$AC=BC=4$$cm.
Using ruler and compasses only, draw a right angle.
Draw a ray $$OA$$.
With a convenient radius and centre at $$O$$, draw an arc $$PQ$$ with the help of a compass intersecting the ray $$OA$$ at $$P$$.
With the same radius and centre at $$P$$, draw another arc intersecting the arc $$PQ$$ at $$R$$.
With the same radius and centre at $$R$$, draw an arc cutting the arc $$PQ$$ at $$C$$, opposite $$P$$.
Taking $$C$$ and $$R$$ as the centre, draw two arcs of radius more than half of $$CR$$ that intersect each other at $$S$$.
Join $$O$$ and $$S$$ and extend the line to $$B$$.
The diagram is shown in the above image.
$$∠AOB$$ is the required angle of $$90^0$$.
Draw $$\Delta$$PQR in which $$\angle Q=80^o$$, $$\angle R=55^o$$ and $$QR=4.5$$cm. Draw the perpendicular bisector of side QR.
Draw $$\Delta$$ABC in which $$\angle A=120^o$$, AC$$=$$AB$$=3$$cm. Measure $$\angle B$$ and $$\angle C$$.
Class 9 Maths Extra Questions
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