Question

Construct a ∆PQR, in which PQ = 6 cm, QR = 7 cm and PR = 8 cm. Then, construct another triangle whose sides are $\frac{4}{5}$ times the corresponding sides of ∆PQR. [CBSE 2013, 14]

Answer 1

Steps of Construction

Step 1. Draw a line segment QR = 7 cm.

Step 2. With Q as centre and radius 6 cm, draw an arc.

Step 3. With R as centre and radius 8 cm, draw an arc cutting the previous arc at P.

Step 4. Join PQ and PR. Thus, ∆PQR is the required triangle.

Step 5. Below QR, draw an acute angle $\angle$RQX.

Step 6. Along QX, mark five points R₁, R₂, R₃, R₄ and R₅ such that QR₁ = R₁R₂ = R₂R₃ = R₃R₄ = R₄R₅.

Step 7. Join RR₅.

Step 8. From R₄, draw R₄R' || RR₅ meeting QR at R'.

Step 9. From R', draw P'R' || PR meeting PQ in P'.



Here, ∆P'QR' is the required triangle, each of whose sides are $\frac{4}{5}$ times the corresponding sides of ∆PQR.