Question

Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are $\frac{2}{3}$ of the corresponding sides of the first triangle.

Answer 1

Let's first construct $△ABC$ with sides $4cm,5cm,6cm$

Steps to draw $△ABC$

1. Draw base AB of side $4cm.$

2. With A as centre , and $5cm$ as radius, draw an arc.

3. With B as centre , and $6cm$ as radius ,draw an arc.

4. Let C be the point where two points intersect.

5. Join AC and BC

Thus $△ABC$ is the required triangle

Now, lets make the triangle similar to this named as $△PQR$ whose sides are $\frac{2}{3}$ of the corresponding sides so the sides will become $\frac{8}{3}cm,\frac{10}{3}cm,4cm$

Steps to draw $△PQR$

1. Draw base PQ of side $\frac{8}{3}cm.$

2. With P as centre , and $\frac{10}{3}cm.$ as radius, draw an arc.

3. With Q as centre , and 4cm as radius ,draw an arc.

4. Let R be the point where two points intersect.

5. Join PQ and PR

Thus $△PQR$ is the required triangle