Question

Each of the equal sides of an isosceles triangle is 4 cm greater than its height. If the base of the triangle is 24 cm; calculate the perimeter and the area of the triangle.

Answer 1

Each of equal sides of an isosceles triangle is 4 cm greater than its height.

Let h be the height of the triangle.

Equal sides: AB = AC = h + 4

Base: BC = 24 cm

In ΔABD and ΔACD,

AD = AD   ...[∵ Common Side]

∠ADB = ∠ADC   ...[∵ Both are 90°]

AB = AC   ...[∵ ΔABC is isosceles]

∴ ΔABD ≅ ΔACD   ...[RHS axiom]

∴ BD = CD   ...[C.P.C.T]

∴ BD = CD

= `"BC"/2`

= `24/2`

= 12 cm

By using the Pythagoras theorem in ΔABD,

BD² + AD² = AB²

⇒ 12² + h² = (h + 4)²

⇒ 144 + h² = h² + 4² + 2 × h × 4

⇒ 144 + `\cancel("h"^2)` = `\cancel("h"^2)` + 16 + 8h

⇒ 144 – 16 = 8h

⇒ 128 = 8h

⇒ h = `128/8`

⇒ h = 16 cm

⇒ a = h + 4

= 16 + 4

= 20 cm

Perimeter of triangle = Sum of all sides

= AB + AC + BC

= (20 + 20 + 24) cm

= 64 cm

Area of triangle = `1/2` × base × height

= `1/2 xx 24 xx 16`

= 12 × 16

= 192 cm²

Hence, the perimeter is 64 cm and the area is 192 cm².