NCERT Solutions for Class 10 Maths Exercise 11.1 Chapter 11 Constructions – FREE PDF Download

NCERT Class 10 Maths Ch 11 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths
Chapter 11 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Constructions solutions will help you understand the chapter thoroughly.

NCERT Solutions for Class 10 Maths Chapter 11 – Constructions

In each of the following, give the justification of the construction also:

1. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.

Ans. Given: A line segment of length 7.6 cm.

To construct: To divide it in the ratio 5 : 8 and to measure the two parts.

Steps of construction:
(a) From a point A, draw any ray AX, making an acute angle with AB.
(b) Locate 13 (=5 + 8) points on AX such that
(c) Join.
(d) Through the point, draw a line parallel to intersecting AB at the point C.

Then, AC : CB = 5 : 8

On measurement we get, AC = 3.1 cm and CB = 4.5 cm

Justification:

[By construction]

[By Basic Proportionality Theorem]

But [By construction]

Therefore,

AC : CB = 5 : 8

2. 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. To construct: To 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.

Steps of construction:

(a) Draw a triangle ABC with sides AB = 4 cm, AC = 5 cm and BC = 6 cm.
(b) From point B, draw any ray BX, making an acute angle with BC on the side opposite to the vertex A.
(c) Locate 3 points on BX such that .
(d) Join and draw a line through the point, draw a line parallel to 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:

[By construction]

[By Basic Proportionality Theorem]

But [By construction]

Therefore,

………(i)

CA C’A’ [By construction]

BC’A’ BCA [AA similarity]

[From eq. (i)]

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

Ans. To construct: To construct a triangle of sides 5 cm, 6 cm and 7 cm and then a triangle similar to it whose sides are

of the corresponding sides of the first triangle.

Steps of construction:
(a) Draw a triangle ABC with sides AB = 5 cm, AC = 6 cm and BC = 7 cm.
(b) From the point B, draw any ray BX, making an acute angle with BC on the side opposite to the vertex A.
(c) Locate 7 points on BX such that.

(d) Join and draw a line through the point, draw a line parallel to 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:

C’A’CA [By construction]

ABCA’BC’ [AA similarity]

[By Basic Proportionality Theorem]

[By construction]

[AA similarity]

But [By construction]

Therefore,

4. 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 the point B, draw a ray BX, making an acute angle with BC on the side opposite to the vertex A.
(f) Locate 3 points on BX such that.
(g) Join and draw a line through the point, draw a line parallel to 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]

ABCA’BC’ [AA similarity]

[By Basic Proportionality Theorem]

[By construction]

[AA similarity]

But [By construction]

Therefore,

Hence, we get the new triangle similar to the given triangle whose sides are equal to $\frac{3}{2}$ i.e., $1 \frac{1}{2}$ times of corresponding sides of triangle ABC.

5. 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 the point B, draw a ray BX, making an acute angle with BC on the side opposite to the vertex A.
(c) Locate 4 points on BX such that .
(d) Join and draw a line through the point, draw a line parallel to 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:

[By construction]

[By Basic Proportionality Theorem]

But [By construction]

Therefore, ………(i)

CA C’A’ [By construction]

BC’A’ BCA [AA similarity]

[From eq. (i)]

Hence, we get the new triangle similar to the given triangle whose sides are equal to $\frac{3}{4}$ th of corresponding sides of triangle ABC.

6. 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 the point B, draw a ray BX, making an acute angle with BC on the side opposite to the vertex A.
(c) Locate 4 points on BX such that.
(d) Join and draw a line through the point, draw a line parallel to 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:

[By construction]

[AA similarity]

[By Basic Proportionality Theorem]

But [By construction]

Therefore, ………(i)

CA C’A’ [By construction]

BC’A’ BCA [AA similarity]

[From eq. (i)]

Hence, we get the new triangle similar to the given triangle whose sides are equal to $\frac{4}{3}$ times of corresponding sides of triangle ABC.

7. 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, right angled at B.
(b) From the point B, draw a ray BX, making an acute angle with BC on the side opposite to the vertex A.
(c) Locate 5 points on BX such that.
(d) Join and draw a line through the point, draw a line parallel to intersecting BC at the point C’.
(e) Draw a line through C’ parallel to the line CA to intersect BA at A’.