Question

Construct a triangle PQR with PQ of 8 cm and QR of 7 cm and an included angle ∠PQR of 60°. Construct a triangle similar to a triangle PQR with its sides equal to $\frac{3}{4}$ of the corresponding sides of the triangle PQR.

Answer 1

Step 1:
Draw two line segments PQ and QR whose length is equal to 8 cm and 7 cm respectively and an included angle ∠PQR as 60°, thus draw a △PQR.

Step 2:
Draw a ray such that it makes an acute angle with line segment PQ on the opposite side of R. Name the inclined ray as PS.

Step 3:
Now locate 4 points and name them as
P1, P2, P3, and P4 on PS such that
PP1 = P1P2 = P2P3 = P3P4.

Step 4:
Join the last marked point
P4 with the point Q on the initial line segment PQ. So that P4Q, another straight line can be obtained.

Step 5:
Through the point P3 draw a line parallel to P4Q by copying the angle ∠PP4Q and intersecting PQ and name that point as Q’.

Reason for 3th point - Since we need a required smaller triangle which is which is $\frac{3}{4}$ times the corresponding sides of △PQR.

Step 6:
Join Q’ R′. R′P Q’ is the required triangle.