Question

Draw a ∆ABC in which base BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct another triangle whose sides are $\frac{3}{4}$ of the corresponding sides of ∆ABC.

Answer 1

ΔA'BC' whose sides are of the corresponding sides of ΔABC can be drawn as follows.

Step 1

Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.

Step 2

Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 3

Locate 4 points (as 4 is greater in 3 and 4), B₁, B₂, B₃, B₄, on line segment BX.

Step 4

Join B₄C and draw a line through B₃, parallel to B₄C intersecting BC at C'.

Step 5

Draw a line through C' parallel to AC intersecting AB at A'. ΔA'BC' is the required triangle.

Justification

The construction can be justified by proving

In ΔA'BC' and ΔABC,

∠A'C'B = ∠ACB (Corresponding angles)

∠A'BC' = ∠ABC (Common)

∴ ΔA'BC' ∼ ΔABC (AA similarity criterion)

… (1)

In ΔBB₃C' and ΔBB₄C,

∠B₃BC' = ∠B₄BC (Common)

∠BB₃C' = ∠BB₄C (Corresponding angles)

∴ ΔBB₃C' ∼ ΔBB₄C (AA similarity criterion)

From equations (1) and (2), we obtain

⇒