Question

Construct a triangle PQR with sides QR = 7 cm, PQ = 6 cm and $\angle$PQR = 60º. Then construct another triangle whose sides are $\frac{3}{5}$ of the corresponiding sides of ∆PQR. [CBSE 2014, 2015]

Answer 1

Steps of Construction

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

Step 2. At B, draw $\angle$XQR = 60º.

Step 3. With Q as centre and radius 6 cm, draw an arc cutting the ray QX at P.

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

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

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

Step 7. Join RR₅.

Step 8. From R₃, draw R₃R' || RR₅ meeting QR at R'.

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



Here, ∆P'QR' is the required triangle whose sides are $\frac{3}{5}$ of the corresponding sides of ∆PQR.