Draw a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm. Then construct another triangle whose sides are 5/3^{th} times the corresponding sides of the given triangle.

#### Solution

Given that

Construct a right triangle of sides let AB = 5cm, AC = 4cm and ∠A = 90° and then a triangle similar to it whose sides are 5/3^{th} of the corresponding sides of .

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment AB = 5cm.

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

Step: III- With *A *as centre and radius AC = 4cm*.*

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_{3}B.

Step: VIII -Since we have to construct a triangle each of whose sides is 5/3^{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_{3}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 5/3^{th} of the corresponding sides of ΔABC*.*