CBSE Class 10 Maths Chapter-11 Constructions – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 11 – Constructions prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 10 Maths Chapter-11 Constructions Important Questions
CBSE Class 10 Maths Important Questions Chapter 11 – Constructions
4 Mark Questions
1. Construct a 
in which 
and
. Also construct a triangle ABC similar to whose each side is 
times the corresponding side of the .
Ans. Steps of construction:
1. Draw a line segment AB = 6.5cm.
2. At B construct 
3. With B as centre and radius BC = 5.5cm draw an arc intersecting BX at C.
4. Join AC.
Triangle so obtained is the required triangle.
5. Construct an acute angle ÐBAY at A on opposite side of vertex C of 
6. Locate 3 points 
on AY such that 
7. Join 
to B and draw the line through 
parallel to 
intersecting the extended line segment AB at B’.
8. Draw a line through B’ parallel to BC intersecting the extended line seg.AC at C’.
9. 
AB’C’ so obtained is the required triangle
2. Draw a circle of radius 4 cm from a point P, 7 cm from the centre of the circle, draw a pair of tangents to the circle measure the length of each tangent segment.
Ans. Steps of construction:
1. Take a point O in the plane of a paper and draw a circle of the radius 4 cm.
2. Make a point P at a distance of 7 cm from the centre O and Join OP.
3. Bisect the line segment OP. Let M be the mid-point of OP.
4. Taking M as a centre and OM as radius draw a circle to intersect the given circle at the points T and T’.
5. Join PT and PT’, then PT and PT’ are required tangents.
PT = PT’ = 5.75 cm
3. Draw a right triangle in which the sides containing the right angle are 5cm and 4cm.Construct a similar triangle whose sides are 
times the sides of the above triangle.
Ans. Steps of construction:
1. Draw a line segment BC = 5 cm.
2. At B construct 
3. With B as centre and radius = 4 cm draw an arc intersecting the ray BX at A.
4. Join AC to obtain the required
5. Draw any ray BY making an acute angle with BC on the opposite side to the vertex A.
6. Locate 5 points 
on BY so that 
7. Join 
to C and draw a line through 
parallel to 
intersecting the extended line segment BC at C’.
8. Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’.
Thus, DA’BC’ is the required right triangle.
4. Construct a circle whose radius is equal to 4 cm. Let P be a point whose distance from its centre is 6 cm. Construct two tangents to it from P.
Ans. Steps of construction:
1. Take a point O in the plane of the paper and draw a circle of radius 4cm.
2. Make a point P at a distance of 6cm from the centre O and join OP.
3. Bisect the line segment OP. Let the point of bisection be M.
4. Taking M as centre and OM as radius, draw a circle to intersect the given circle at the point T and T’.
5. Join PT and PT’ to get the required tangents.
5. Draw a triangle ABC with sides BC = 6.3 cm, AB = 5.2 cm and 
. Then construct a triangle whose sides are 
times the corresponding sides of .
Ans. Steps of construction:
1. Draw a line segment BC = 6.3 cm.
2. At B make 
3. With B as centre and radius equal to 5.2 cm, draw an arc intersecting BX at A.
4. Join AC, then DABC is the required triangle.
5. Draw any ray by making an acute angle with BC on the opposite side to the vertex A.
6. Locate the points 
on BY so that 
7. Join to C and draw a line through 
parallel to intersecting the extended line segment BC at C’.
8. Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’.
Thus, DA’BC’ is the required triangle.
6. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at 
.
Ans. Steps of construction:
1. Draw a circle with center O and radius 5cm.
2. Draw any diameter AOC.
3. Construct 
meeting the circle at B.
4. At A and B draw perpendiculars to OA and OB intersecting at P.
5. PA and PB are required tangents and 
7. Draw the tangents at the extremities of a diameter AB of a circle of radius 2cm. Are these tangents parallel? Given reasons.
Ans. Steps of construction:
1. Draw a circle of radius 2 cm.
2. Draw any diameter AOB.
3. Draw 
4. AT and BM are tangents extremities of the diameter AB.
5. 
, they are alternate angles.
8. Construct a in which AB = 4 cm, BC = 5 cm and AC = 6 cm. Now construct a triangle similar to such that each of its sides is two-third of the corresponding sides of . Also prove your assertion.
Ans. Steps of construction:
1. Draw with sides BC = 5 cm, AB = 4 cm and AC = 6 cm.
2. Below BC make acute 
3. Along BX mark off three points 
such that 
4. Join 
5. Draw 
also 
Thus, D
is the required triangle
9. Construct a in which AB = 6.5 cm 
and BC = 5.5 cm. Also, construct a 
similar to whose each side is times the corresponding sides of the .
Ans. Steps of construction:
1. Construct a is which AB = 6.5cm, and BC = 5.5cm.
2. Draw a ray AX making any acute angle with AB on the opposite side of the vertex C.
3. Cut three equal parts from AX say 
4. Join 
to B.
5. From 
draw 
at 
.
6. At B’ draw
intersecting AY at C’.
7. is required triangle similar to
10. Draw a pair of tangents to a circle of radius 5cm which are inclined to each other at .
Ans. Steps of construction:
1. Draw a circle with centre O and radius 5cm.
2. Draw any radius OT.
3. Construct 
4. Draw 
and 
Then 
and PT are the two required tangents such that 
Here,
11. Draw a circle of radius 4 cm with centre O. Draw a diameter POQ. Through P or Q draw tangent to the circle.
Ans. Steps of construction:
1. Draw a circle of radius 4 cm.
2. Draw diameter POQ.
3. Construct 
4. Produce PQ to 
, then 
is the required tangent at the point Q.
12. Construct a triangle with sides 5cm, 6cm and 7cm and then another triangle whose sides are 
of the corresponding sides of first triangle.
Ans. Steps of construction:
1. Draw a line segment AB = 5cm.
2. With A as centre and radius 6cm draw an arc.
3. Again B as centre and radius 7cm draw another arc cutting the previous arc at C. Join AC and BC,thenDABC is required triangle.
4. Draw any ray AX making acute angle.
5. Locate 7 points 
on AX so that 
6. Join 
to B and draw a line through 
parallel to 
intersecting the extended line segment AB at .
7. Draw , then is the required triangle
13. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and 
. BD is the perpendicular from B on AC. The circle through B, C and D is drawn construct the tangents from A to this circle.
Ans. Steps of construction:
1. Draw with BC = 8 cm, AB = 6 cm and 
2. Draw perpendicular BD from B to AC.
3. Let O be the mid-point of BC. Draw a circle with centre O and radius OB = OC. This circle will pass through the point D.
4. Join AO and bisect AO.
5. Draw a circle with centre
and 
as radius cuts the previous circle at B and P.
6. Join AP, AP and AB are required tangents drawn from A to the circle passing through B,C and D.
14. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 
of the corresponding sides of the first triangle
Ans. Steps of construction:
1. Draw with AB = 4 cm, BC = 6 cm and AC = 5 cm.
2. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
3. Locate 3 points 
and on BX so that
4. Join and draw 
5. Draw a line through 
such that Then,
is the required triangle.
15. Construct an isosceles triangle whose base is 8cm and altitude 4cm and then another triangle whose sides are 
times the corresponding sides of the isosceles triangle.
Ans. Steps of construction:
1. Draw line segment AB = 8 cm.
2. Draw perpendicular bisector of AB which intersects AB at D.
3. Cut DC = 4 cm. Join AC and BC. Thus, is the required triangle.
4. Draw acute angle 
5. Locate 3 points 
on AX so that
6. Join and draw 
7. Draw 
8. 
isthe required triangle
16. Draw a with side BC = 7cm, 
and
, then construct a triangle whose sides are times the corresponding sides of .
Ans. Steps of construction:
1. Draw in which BC = 7cm, and 
2. Locate 4 points 
and on BZ such that
3. Join and draw 
4. Now Draw
5. is the required triangle.
17. Construct a circle whose radius is equal to 4 cm. Let P be a point whose distance from its centre is 6 cm. Construct two tangents to it from P.
Ans. Steps of construction:
1. Draw a circle with centre O and radius 4 cm.
2. Mark point P at a distance of 6 cm from the centre O and join OP.
3. Bisect the line segment OP. Let the point of bisection be M.
4. Take M as centre and OM as radius draw a circle to intersect the given circle at the points T and T’.
5. Join PT and PT’ to get the required tangents.
18. Draw a line segment AB of length 8 cm taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre to the other circle.
Ans. Steps of construction:
1. Draw a line segment AB = 8 cm.
2. Taking A as centre and radius = 4 cm draw a circle.
3. Taking B as centre and radius = 3 cm draw another circle.
4. Bisect the line segment AB, let the point of bisection be M.
5. Taking M as centre and MA as radius, draw a circle intersecting the given circles at the point 
19. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are of the corresponding sides of the first triangle.
Ans. Steps of construction:
1. Draw in which AB = 4 cm, BC = 6 cm and AC = 5 cm.
2. Draw acute
3. Locate 3 points and 
on BX so that
4. Join and draw 
5. Now Draw
Thus, is the required triangle.
20. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of .
Ans. Steps of construction:
1. Draw a circle with centre O and radius OA = 5 cm.
2. Extend OA to B such that OA = AB = 5 cm
3. With A as centre draw a circle of radius OA = AB = 5 cm. Suppose it intersect the circle drawn in step 1 at the points P and Q.
4. Join BP and BQ. Then BP and BQ are the required tangents.
Justification: In 
Also, AP = 5 cm
is an equilateral triangle
In 
,
AB = AP and 
Similarly,
21. From a point P two tangents are drawn to a circle with centre O. If OP = diameter of the circle, show that 
is equilateral
Ans. Join OP.
Suppose OP meets the circle at Q. Join AQ.
We have
i.e., OP = diameter
OQ + PQ = diameter
PQ = Diameter – radius [
]
PQ = radius
Thus, OQ = PQ = radius
Thus, OP is the hypotenuse of right triangle
OAP and Q is the mid-point of OP
OA = AQ = OQ
[
mid-point of hypotenuse of a right triangle is equidistant from the vertices]
is equilateral
So,
Also 
But 
Hence,
is equilateral.
22. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 
times the corresponding sides of the isosceles triangle.
Ans. To construct: To construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then a triangle similar to it whose sides are 
of the corresponding sides of the first triangle.
Steps of construction:
(a) Draw BC = 8 cm
(b) Draw perpendicular bisector of BC. Let it meets BC at D.
(c) Mark a point A on the perpendicular bisector such that AD = 4 cm.
(d) Join AB and AC. Thus
ABC is the required isosceles triangle.
(e) From any ray BX, making an acute angle with BC on the side opposite to the vertex A.
(f) Locate 3 points B1, B2and B3 on BX such that BB1 = B1B2 = B2B3.
(g) Join B2C and draw a line through the point B3, draw a line parallel to B2C intersecting BC at the point C’.
(h) Draw a line through C’ parallel to the line CA to intersect BA at A’.
Then, A’BC’ is the required triangle.
Justification:
C’A’
CA [By construction]
ABC
A’BC’ [AA similarity]
[By Basic Proportionality Theorem]
B3C’B2C [By construction]
BB3C’ BB2C [AA similarity]
But 
[By construction]
Therefore,
23. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and 
ABC = 
Then construct a triangle whose sides are 
of the corresponding sides of triangle ABC.
Ans. To construct: To construct a triangle ABC with side BC = 6 cm, AB = 5 cm and ABC = 
and then a triangle similar to it whose sides are of the corresponding sides of the first triangle ABC.
Steps of construction:
(a) Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ABC = .
(b) From any ray BX, making an acute angle with BC on the side opposite to the vertex A.
(c) Locate 4 points B1, B2, B3 and B4 on BX such that BB1 = B1B2 = B2B3= B3B4.
(d) Join B4C and draw a line through the point B3, draw a line parallel to B4C intersecting BC at the point C’.
(e) Draw a line through C’ parallel to the line CA to intersect BA at A’.
Then, A’BC’ is the required triangle.
Justification:
B4CB3C’ [By construction]
[By Basic Proportionality Theorem]
But 
[By construction]
Therefore, 
………(i)
CAC’A’ [By construction]
BC’A’ BCA [AA similarity]
[From eq. (i)]
24. Draw a triangle ABC with side BC = 7 cm, B = 
A = 
Then construct a triangle whose sides are 
times the corresponding sides of ABC.
Ans. To construct: To construct a triangle ABC with side BC = 7 cm, B = 
and C = 
and then a triangle similar to it whose sides are of the corresponding sides of the first triangle ABC.
Steps of construction:
(a) Draw a triangle ABC with side BC = 7 cm, B = and C = .
(b) From any ray BX, making an acute angle with BC on the side opposite to the vertex A.
(c) Locate 4 points B1, B2, B3 and B4 on BX such that BB1 = B1B2 = B2B3= B3B4.
(d) Join B3C and draw a line through the point B4, draw a line parallel to B3C intersecting BC at the point C’.
(e) Draw a line through C’ parallel to the line CA to intersect BA at A’.
Then, A’BC’ is the required triangle.
Justification:
B4C’B3C [By construction]
BB4C’ BB3C [AA similarity]
[By Basic Proportionality Theorem]
But 
[By construction]
Therefore, 
………(i)
CAC’A’ [By construction]
BC’A’ BCA [AA similarity]
[From eq. (i)]
25. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 
times the corresponding sides of the given triangle
Ans. To construct: To construct a right triangle in which sides (other than hypotenuse) are of lengths 4 cm and 3 cm and then a triangle similar to it whose sides are of the corresponding sides of the first triangle ABC.
Steps of construction:
(a) Draw a right triangle in which sides (other than hypotenuse) are of lengths 4 cm and 3 cm.
(b) From any ray BX, making an acute angle with BC on the side opposite to the vertex A.
(c) Locate 5 points B1, B2, B3, B4and B5 on BX such that BB1 = B1B2 = B2B3= B3B4 = B4B5.
(d) Join B3C and draw a line through the point B5, draw a line parallel to B3C intersecting BC at the point C’.
(e) Draw a line through C’ parallel to the line CA to intersect BA at A’.
Then, A’BC’ is the required triangle.
Justification:
B5C’B3C [By construction]
BB5C’ BB3C [AA similarity]
[By Basic Proportionality Theorem]
But 
[By construction]
Therefore, 
………(i)
CAC’A’ [By construction]
BC’A’ BCA [AA similarity]
[From eq. (i)]
26. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Ans. To construct: A circle of radius 3 cm and take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre and then draw tangents to the circle from these two points P and Q.
Steps of Construction:
(a) Bisect PO. Let M be the mid-point of PO.
(b) Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the points A and B.
(c) Join PA and PB.
Then PA and PB are the required two tangents.
(d) Bisect QO. Let N be the mid-point of QO.
(e) Taking N as centre and NO as radius, draw a circle. Let it intersects the given circle at the points C and D.
(f) Join QC and QD.
Then QC and QD are the required two tangents.
Justification: Join OA and OB.
Then PAO is an angle in the semicircle and therefore PAO= 
.
PA 
OA
Since OA is a radius of the given circle, PA has to be a tangent to the circle. Similarly, PB is also a tangent to the circle.
Again join OC and OD.
Then QCO is an angle in the semicircle and therefore QCO= .
Since OC is a radius of the given circle, QC has to be a tangent to the circle. Similarly, QD is also a tangent to the circle.
27. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Ans. To construct: A line segment of length 8 cm and taking A as centre, to draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Also, to construct tangents to each circle from the centre to the other circle.
Steps of Construction:
(a) Bisect BA. Let M be the mid-point of BA.
(b) Taking M as centre and MA as radius, draw a circle. Let it intersects the given circle at the points P and Q.
(c) Join BP and BQ.
Then, BP and BQ are the required two tangents from B to the circle with centre A.
(d) Again, Let M be the mid-point of AB.
(e) Taking M as centre and MB as radius, draw a circle. Let it intersects the given circle at the points R and S.
(f) Join AR and AS.
Then, AR and AS are the required two tangents from A to the circle with centre B.
Justification: Join BP and BQ.
Then APB being an angle in the semicircle is 
BP AP
Since AP is a radius of the circle with centre A, BP has to be a tangent to a circle with centre A. Similarly, BQ is also a tangent to the circle with centre A.
Again join AR and AS.
Then ARB being an angle in the semicircle is
AR BR
Since BR is a radius of the circle with centre B, AR has to be a tangent to a circle with centre B. Similarly, AS is also a tangent to the circle with centre B.
28. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Ans. To construct: A circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Steps of Construction:
(a) Draw a circle with the help of a bangle.
(b) Take two non-parallel chords AB and CD of this circle.
(c) Draw the perpendicular bisectors of AB and CD. Let these intersect at O. Then O is the centre of the circle draw.
(d) Take a point P outside the circle.
(e) Join PO and bisect it. Let M be the mid-point of PO.
(f) Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the points Q and R.
(g) Join PQ and PR.
Then PQ and PR are the required two tangents.
Justification: Join OQ and OR.
Then, PQO is an angle in the semicircle.
PQO =
PQ OQ
Since OQ is a radius of the given circle, PQ has to be a tangent to the circle. Similarly PR is also a tangent to the circle.
