NCERT Solutions For Class 10 Maths Chapter 11

NCERT Solutions Class 10 Maths Constructions

Maths is like a nightmare to the students, as it consist of various tricks and requires proper understanding. For solving questions, it is crucial to have good knowledge of the topic. Students can study various concepts of maths by visiting our site BYJU’S. To ease the fear of maths for the students, we at BYJU’S provide NCERT Solution for Class 10 Maths Chapter 11 Construction. Student can download the NCERT Solution for class 10 Maths Chapter 11 pdf or can view it online by following the link. These solution are provided in detailed manner, where one can find step-by-step solution to all the questions of NCERT Solutions for class 9 maths chapter 1.

NCERT solutions for class 10 maths chapter 11 pdf are prepared by our subject experts under the guidelines of NCERT to assists students in their board exam preparations.

NCERT Solutions Class 10 Maths Chapter 11 Exercises

Exercise 11.1

1) Draw a line segment of length 7.6 cm and divide it in 5: 8 ration. In addition, find the measure of two parts.

Solution:

1

Procedure for construction:

  1. Draw any ray AX, making an acute angle with AB.
  2. Locate 13(= 5 + 8) points A1, A2, A3 ……… A13 on AX so that AA1 = A1A2 Al2 A13.
  3. Join BA13.
  4. Through the point A5(m = 5), draw a line parallel to BA13 (by making an angle equal to L AA13 B at A5 intersecting AB at C. Then AC: CB = 5 : 8)

 

2) Construct a triangle with sides 4 cm, 5 cm and 6 cm and then a similar triangle to it whose sides are 2/3 of the corresponding sides of the first one.

Solution:

2

Procedure for construction:

  1. Draw a line segment BC with length 5 cm.
  2. With B as centre and radius of 4 cm draw an arc.
  3. With C as centre and radius of 6 cm draw an arc.
  4. Join AB and AC. Then, ∆ABC is the required triangle.
  5. Below BC, make an acute angle ∠CBX
  6. Along BX, mark up three points B1, B2, B3 such that BB1 = B1 B2 = B2B3
  7. Join B3C
  8. From B2, draw B2C’llB3c, meeting BC at C’
  9. From C’ draw C’ All CA, meeting BA at A’
  10. Then ∆A’BC’ is the required triangle, each of whose sides is two-third of the corresponding sides of ∆ABC.

 

3) Construct a triangle with side lengths 5 cm, 6 cm and 7 cm and then another triangle whose sides are  of the corresponding sides of the first triangle.

Solution:

3

Procedure for  construction :

  1. Draw a line segment BC with length 6 cm.
  2. With B as centre and keeping radius as 5 cm, draw an arc.
  3. With C as centre and keeping radius as 7 cm, draw another arc, intersecting the previously drawn arc at Point A.
  4. Join AB and AC. Then, ∆ABC is the required triangle.
  5. Below BC, make an acute angle∠CBX.
  6. Along BX, mark up seven points B1, B2, B3….. B7 such that BB1 = B1,B2, B6B7.
  7. Join B5 to C (5 being smaller of 5 and 7 in7/5) and draw a line through B7 parallel to B5C, 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’. Then A’BC’ is the required triangle.

 

4) Draw a triangle ABC with sides BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are ¾ of the corresponding sides of ∆ABC.

Solution:

4

Procedure for construction:

(i)            Draw a triangle ABC with BC = 6 cm, AB = 5 cm and ∠ABC = 60°.

(ii)           Draw any ray BX making an acute angle with BC on the side opposite to the vertex X.

(iii)          Locate 4(the greater of 3 and 4 in ¾) points B1, B2, B3, B4 on BX so that BB1 = B1B2 = B2B3 = B3B4.

(iv)         Join B4C and draw a line through B3(the 3rd point, 3 being smaller of 3 and 4 in ¾) parallel to B4C to intersect BC at C’.

(v)          Draw a line through C’ parallel to the line CA to intersect BA at A’. Then ∆A’BC’ is the required triangle.

Justification of construction

∆ABC ~ ∆A’BC’ , Therefore,

ABAB=ACAC=BCBC

But, BCBC=BB3BB4=34

So, ABAB=ACAC=BCBC=34
5) Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are4/3 times the corresponding sides of ∆ABC.

Solution:

5

Procedure for  construction :

(i)            Draw a triangle ABC with BC = 7cm, ∠B = 45° and ∠A = 105°.

(ii)           Draw any ray BX making an acute angle with BC on the side opposite to the vertex X.

(iii)          Locate 4(the greater of 3 and 4 in 4/3) points B1, B2, B3, B4 on BX so that 3 BB1 = B1 B2 = B2B3 = B3B4.

(iv)         Join NC’ and draw a line through B3(the 3rd point, 3 being smaller of 3 and 4 in 1) parallel to NC’ to intersect BC’ at C. 3

(v)          Draw a line through C’ parallel to the line CA to intersect BA at A’. Then A NBC’ is the required triangle.

 

6) Construct a triangle of isosceles type, whose base is 8 cm and height 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.

Solution:

6

Given:An isosceles triangle whose base is 8 cm and height 4 cm. Scale factor: 1  =

Required: To construct a similar triangle to above whose sides are 1.5 times the above triangle.

Procedure for construction:

(i)            Draw a line segment BC = 8 cm.

(ii)           Draw a perpendicular bisector AD of BC.

(iii)          Join AB and AC we get a isosceles ∆ABC.

(iv)         Construct an acute angle∠CBX downwards.

(v)          On BX make 3 equal parts.

(vi)         Join C to B2 and draw a line through B3 parallel to B2C intersecting the extended line segment BC at C’.

(vii)        Again draw a parallel line C’A’ to AC cutting BP at A’.

(viii)       ∆A’BC’ is the required triangle.

 

Exercise 11.2

7) Draw a circle with radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Solution:

7

Procedure for construction:

  1. Draw a line segment of length AB = 10 cm. Bisect AB by constructing a perpendicular bisector of AB. Let M be the mid-point of AB.
  2. With M as centre and AM as radius, draw a circle. Let it intersect the given circle at the points P and Q.
  3. Join PB and QB. Thus, PB and QB are the required two tangents.

Justification: Join AP. Here ∠APB is an angle in the semi-circle. Therefore, ∠APB = 90°. Since AP is a radius of a circle, PB has to be a tangent to a circle. Similarly, QB is also a tangent to a circle.

In a Right ∆APB, AB2 = AP2 + PB2 (By using Pythagoras Theorem)
PB2 = AB2 – AP2 = 102 — 62 = 100 – 36 = 64

PB = 8 cm.

 

8) Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation.

Solution:

8

Procedure for construction:

  1. Draw a line segment of length OA = 4 cm. With O as centre and OA as radius, draw a circle.
  2. With O as centre draw a concentric circle of radius 6 cm(0B).
  3. Let C be any point on the circle of radius 6 cm, join OC.
  4. Bisect OC such that M is the mid point of OC.
  5. With M as centre and OM as radius, draw a circle. Let it intersect the given circle of radius 4 cm at the points P and Q.
  6. Join CP and CQ. Thus, CP and CQ are the required two tangents.

Justification:

Join OP. Here ∠OPC is an angle in the semi-circle. Therefore, ∠OPC = 90°. Since OP is a radius of a circle, CP has to be a tangent to a circle. Similarly, CQ is also a tangent to a circle.

In ∆COP, ∠P = 90°

CO2=CP2+OP2 CP2=CO2OP2

=6242

CP=25cm

 

9) Draw a circle with radius 3 cm. On one of its extended diameter, take two points P and Q each at a distance of 7 cm from its centre. From two points P and Q, draw tangents to the circle.

Solution:

9

Given:

Two points P and Q on the diameter of a circle with radius 3 cm OP = OQ = 7 cm.

Aim:

To construct the tangents to the circle from the given points P and Q.

Procedure for construction:

  1. Draw a circle with radius 3 cm with centreO.
  2. Extend its diameter both the sides and cut OP = OQ = 7 cm.
  3. Bisect OP and OQ.Let mid-points of OP and OQ be M and N.
  4. With M as centre and OM as radius, draw a circle. Let it intersect (0, 3) at two points A and B. Again taking N as centre ON as radius draw a circle to intersect circle(0, 3) at points C and D.
  5. Join PA, PB, QC and QD. These are the required tangents from P and Q to circle (0, 3).

 

10) Draw a pair of tangents to a circle which is of radius 5 cm, such that they are inclined to each other at an angle of 60°.

Solution:

10

To determine: To draw tangents at the ends of two radius which are inclined to each other at 120°

Procedure for  construction :

  1. Keeping O as centre, draw a circle of radius 5 cm.
  2. Take a point Q on the circle and join it to O.
  3. From OQ, Draw∠QOR = 120°.
  4. Take an external point P.
  5. Join PR and PQ perpendicular to OR and OQ respectively intersecting at P.

The required tangents are RP and QP.