Construct the following with the help of straight-edge and compass only:
Draw ΔPQR with m∠P = 60, m∠Q = 45 and PQ = 6 cm. Then construct ΔPBC whose sides have lengths 
times the lengths of the corresponding sides of ΔPQR.
Steps of Construction:
1. Draw triangle PQR of given dimensions.
3. Draw a ray PX making an acute angle with PQ.
3. We have to construct triangle having ratio of 5:3. So we make 5 equal arcs on ray PX.
4. Taking P as center, cut an arc on ray PX. Mark the point as P1.
5. Now, with the same radius cut an arc on PX with P1 as center.
6. Mark till P5.
7. Join P3 with Q.Extend PQ.
8. As we have to divide in 5:3, we draw a line from P5 parallel to P3Q.
9. The point where it cuts extended PQ is S.
10. From S, draw a line parallel to QR.
11. PST is the required triangle.
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ABC = 90, BC = 4 cm and AC = 5 cm. Then construct ΔBXY, where scale factor is 
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of length 6.5 cm and divide it in the ratio 4 : 7. Measure the two parts.