Question

Construct a ΔABC with BC = 6 cm, ∠B = 60° and AB = 5 cm. Construct another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of ΔABC.

Answer 1

Given: In ΔABC,
BC = 6 cm,
∠B = 60°
AB = 5 cm

Steps of construction:
(1) Draw a line segment AB = 5 cm.
(2) From the point B, draw an ∠ABY = 60°
(3) Taking B as center, 6 cm radius, draw an arc on the ray BY,
Let the point where the arc intersects the ray named as C.
(4) Join AC.



Hence, ΔABC is the required triangle.

Now, we construct another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of ΔABC.

Steps of construction:
(1) Draw any ray BX making an acute angle with BA on the side opposite to the vertex C.
(2) Mark four points B₁, B₂, B₃ and B₄ on BX, so that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
(3) Join B₄A and draw a line through B₃ parallel to B₄A to intersect AB at A'.
(4) Draw a line through A' parallel to AC to intersect BC at C'.



Hence, Δ​A'BC' is the required triangle.