ABC is a triangle in which AB = AC = 4 cm and ∠A = 90°. Calculate :

(i) the area of Δ ABC,

(ii) the length of perpendicular from A to BC.

(i) Given:

AB = AC = 4 cm

∠A = 90°

Area = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x 4 x 4 cm²

= $\dfrac{1}{2}$ x 16 cm²

= 8 cm²

(ii) By using the Pythagoras theorem,

AB² + AC² = BC²

⇒ (4)² + (4)² = BC²

⇒ 16 + 16 = BC²

⇒ BC² = 32

⇒ BC = $\sqrt{32}$

⇒ BC = 4 $\sqrt{2}$ cm

Now considering BC as the base of the triangle, altitude AD will be its height.

Area of the triangle will remain the same as before i.e., 8 cm².

Let h be the altitude of triangle

Area = $\dfrac{1}{2}$ x base x height

∴ 8 = $\dfrac{1}{2}$ x BC x AD

⇒ 8 = $\dfrac{1}{2}$ x $4 \sqrt{2}$ x h

⇒ 8 = $2 \sqrt{2}$ x h

⇒ h = $\dfrac{8}{2 \sqrt{2}}$

⇒ h = $\dfrac{4}{\sqrt{2}}$

⇒ h = 2.83 cm

Hence, the length of perpendicular from A to BC is 2.83 cm.