Draw a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm.

Given that

Construct a right triangle of sides let $A B=5 \mathrm{~cm}, A C=4 \mathrm{~cm}$, and $\angle A=90^{\circ}$ and then a triangle similar to it whose sides are $(5 / 3)^{\text {th }}$ of the corresponding sides of $\triangle A B C$.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment.

Step: II- With A as centre and draw an angle.

Step: III- With A as centre and radius.

Step: IV -Join BC to obtain.

Step: V -Below AB, makes an acute angle.

Step: VI -Along $A X$, mark off five points $A_{1}, A_{2}, A_{3}, \mathrm{~A}_{4}$ and $\mathrm{A}_{5}$ such that $A A_{1}=A_{1} A_{2}=A_{2} A_{3}=A_{3} A_{4}=A_{4} A_{5}$

Step: VII -Join $A_{3} B$.

Step: VIII -Since we have to construct a triangle each of whose sides is $(5 / 3)^{\text {th }}$ of the corresponding sides of $\triangle A B C$.

So, we draw a line $A_{5} B$ on $A X$ from point $A_{5}$ which is $A_{5} B^{\prime} \| A_{3} B$, and meeting $A B$ at $B$ '.

Step: IX -From $B^{\prime}$ point draw $B^{\prime} C^{\prime} \| B C$, and meeting $A C$ at $C^{\prime}$

Thus, $\triangle A B^{\prime} C^{\prime}$ is the required triangle, each of whose sides is $(5 / 3)^{\text {th }}$ of the corresponding sides of $\triangle A B C$.