Draw a triangle ABC with side BC = 7 cm, and AB = 6 cm. Then construct another triangle whose sides are times the corresponding sides of .

To construct a right-angled triangle such that its sides are 3/4 times of the corresponding sides of the given triangle:

(a). Draw a line BC = 7 cm as base.

(b). Taking BC as base draw 60° angle as ∠CBA = 60°, where AB = 6 cm.

(c). Connect A to C as hypotenuse.

Now we shall construct another triangle whose sides are 3/4 times the corresponding sides of the triangle.

(d). Set the compass to 1 cm, rest one of the legs of the compass at point B and start making arcs B₁ starting from B, B₂ from B₁, B₃ from B₂ and B₄ from B₃, since we have to construct a triangle having side equal to 3/4 times of corresponding sides of the given triangle.

(e). Join B₄ to C.

(f). Now, draw a line to BC from B₃, parallel to CB₄.

(g). Say this point on BC is C’.

(h). Similarly, make an arc from C’ and cut the arc at 60° from its foot, then join C^’A^’.

(i). We get BA’ automatically.

And hence, we have the required triangle BA’C’.