Question

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

Answer 1

Given , a right angled $\Delta ABC$ , in which $BC=4cm,AB=3cm$ .

Here scale factor$=\frac{5}{3}>1$

Steps of construction:

1. Draw a line segment $BC=4cm.$

2. At $B$ draw $\angle B=90°$ and cut off $BA=3cm$ from it.

3. Join $AC$. Thus $\Delta ABC$ is a given right angled Triangle.

4. Now from $B$ draw a ray $BY$ making an acute $\angle CBY$ on the side opposite to the vertex $A$.

5. Mark $5$ points $B_1,B_2,B_3,B_4$ & $B_5$ on $BY$ such that $B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4=B_4{B}_5$.

6.Join $B_{3}C$ and from $B_5$ draw $B_5{C}^{\prime }||B_{3}C$ intersecting $BC$ at $C^{\prime }$.

7. From point $C^{\prime }$ draw $C^{\prime }{A}^{\prime }||CA$ intersecting $AB$ at $A^{\prime }$.

Thus, $\Delta A^{\prime }B{C}^{\prime }$ is the required triangle whose sides are $\frac{5}{3}$ of the corresponding sides of $\Delta ABC$.