Question

Construct a triangle with sides $5cm,6cm$ and $7cm$ and then another triangle whose sides are $\frac{7}{5}$ of the corresponding sides of the first triangle.

Answer 1

Given that

Construct a triangle of sides $AB=5cm,BC=6cm$, and $AC=7cm$ and then a triangle similar to it whose sides are ${\frac{7}{5}}^{th}$of the corresponding sides of$△ABC$.

We follow the following steps to construct the triangles as

Step of construction

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

Step: II- With A as center and radius $AC=7cm$, draw an arc.

Step: III- With B as center and radius $BC=6cm$, draw an arc, intersecting the arc drawn in step II at C.

Step: IV- Joins AC and BC to obtain $△ABC$.

Step: V- Below AB, makes an acute angle $\angle BAX={60}^{\circ }$.

Step: VI- Along AX, mark off seven points $A_1,A_2,A_3,A_4,A_5,A_6,$ and $A_7$ such that $A{A}_1=A_1{A}_2=A_2{A}_3=A_3{A}_4=A_4{A}_5=A_5{A}_6=A_6{A}_7$

Step: VII- Join $A_{5}B$.

Step: VIII- Since we have to construct a triangle each of whose sides is ${\frac{7}{5}}^{th}$ of the corresponding sides of $△ABC$.

So, we draw a line $A_7{B}^{\prime }$ on AX from point $A_7$, which is $A_7{B}^{\prime }\parallel A_{5}B,$ and meeting AB at B’.

Step: IX- From B’ point draw $B^{\prime }{C}^{\prime }\parallel BC$ and meeting AC at C’

Thus, $△A{B}^{\prime }{C}^{\prime }$ is the required triangle, each of whose sides is ${\frac{7}{5}}^{th}$of the corresponding sides of $△ABC$.