Draw a triangle $P Q R$ , right angled at $Q$ such that $P Q = 3$ cm, $Q R = 4$ cm. Now construct $△ A Q B$ similar to $△ P Q R$ each of whose sides is $\frac{7}{5}$ times the corresponding side of $△ P Q R$ .

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Solution

1. Draw base $Q R$ of length $4 c m$ . Draw a perpendicular from $Q$ to side $Q R$ .
2. Mark $P$ at a distance of $3 c m$ from $Q$ on the perpendicular draw. Join $P - R$ . $△ P Q R$ is the required triangle.
3. Draw an acute angled ray $Q S$ from point $Q$ .
4. Divide $Q S$ into parts such that $Q Q_{1} = Q_{1} Q_{2} = Q_{2} Q_{3} = Q_{3} Q_{4} = Q_{4} Q_{5} = Q_{5} Q_{6} = Q_{6} Q_{7}$ .
5. Join $Q_{5}$ to $R$ . Draw $Q_{7} B$ parallel to $Q_{5} R$ .
6. Draw $B A$ parallel to $R P$ .

$△ A B Q$ is the required similar triangle.

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Draw a triangle PQR, right angled at Q such that PQ=3 cm, QR=4 cm. Now construct △AQB similar to △PQR each of whose sides is 75 times the corresponding side of △PQR.

Draw a triangle $P Q R$ , right angled at $Q$ such that $P Q = 3$ cm, $Q R = 4$ cm. Now construct $△ A Q B$ similar to $△ P Q R$ each of whose sides is $\frac{7}{5}$ times the corresponding side of $△ P Q R$ .