# CBSE Class 10-Mathematics: Chapter – 11 Constructions Part 13 (For CBSE, ICSE, IAS, NET, NRA 2022)

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**Question 22**:

Construct an isosceles triangle whose base is and altitude and then another triangle whose sides are times the corresponding sides of the isosceles triangle.

**Answer**:

To construct: To construct an isosceles triangle whose base is and altitude 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 .

(d) Join AB and AC. Thus 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 points and 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:

[By construction]

[AA similarity]

[By Basic Proportionality Theorem]

[By construction]

[AA similarity]

But [By construction]

Therefore,

**Question 23**:

Draw a triangle ABC with side , and . Then construct a triangle whose sides are of the corresponding sides of triangle ABC.

**Answer**:

To construct: To construct a triangle ABC with side , and and 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 , and .

(b) From any ray BX, making an acute angle with BC on the side opposite to the vertex A.

(c) Locate 4 points and 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,

[AA similarity]

[From eq. (i) ]