Constructions - Solutions

 CBSE Class–10 Mathematics

NCERT solution
Chapter - 11
 Constructions - Exercise 11.1

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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

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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)]

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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, 

 

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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 32$\frac{3}{2}$i.e., 112$1\frac{1}{2}$ times of corresponding sides of triangle ABC.

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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 34$\frac{3}{4}$th of corresponding sides of triangle ABC.

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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 43$\frac{4}{3}$ times of corresponding sides of triangle ABC.

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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’.

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 53$\frac{5}{3}$ times of corresponding sides of triangle ABC.