#### Question

Draw a right triangle ABC in which AC = AB = 4.5 cm and ∠A = 90°. Draw a triangle similar to ΔABC with its sides equal to (5/4)^{th} of the corresponding sides of ΔABC.

#### Solution

Given that

Construct a right triangle of sides AC = AB = 4.5 cm and ∠A = 90° and then a triangle similar to it whose sides are (5/4)^{th} of the corresponding sides of ΔABC.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment AB = 4.5cm.

Step: II- With *A *as centre and draw an angle ∠A = 90°.

Step: III- With *A *as centre and radius AC = 4.5 cm*.*

Step: IV- Join *BC *to obtain ΔABC.

Step: V- Below *AB, *makes an acute angle ∠BAX = 60°.

Step: VI- Along *AX,* mark off five points A_{1}, A_{2}, A_{3}, A_{4} and A_{5}, such that AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5}

Step: VII-Join A_{4}B.

Step: VIII- Since we have to construct a triangle each of whose sides is (5/4)^{th} of the corresponding sides of ΔABC.

So, we draw a line A_{5}B' on *AX *from point A_{5} which is A_{5}B' || A_{4}B, and meeting *AB *at B*’.*

Step: IX- From B’ point draw B'C'||BC and meeting *AC *at *C’*

Thus, ΔAB'C' is the required triangle, each of whose sides is (5/4)^{th} of the corresponding sides of ΔABC*.*