AD is altitude of an isosceles triangle ABC in which AB = AC = 30 cm and BC = 36 cm. A point O is marked on AD in such a way that ∠BOC = 90°. Find the area of quadrilateral ABOC.

ΔABC is shown in the figure below:

Area of isosceles triangle ABC =

$= \dfrac{b}{4} \sqrt{4a^2 - b^2}\\[1em] = \dfrac{36}{4} \sqrt{4 \times 30^2 - 36^2}\\[1em] = 9 \sqrt{4 \times 30^2 - 36^2}\\[1em] = 9 \sqrt{4 \times 900 - 1,296}\\[1em] = 9 \sqrt{3600 - 1,296}\\[1em] = 9 \sqrt{2,304}\\[1em] = 9 \times 48 \\[1em] = 432 \text{ cm}^2$

∠ BOC = ∠ COD = 45° (∵ AD divide ∠ BOC in 2 equal halves)

Let OB = OC = x.

In Δ BOC, by using the Pythagoras theorem,

OB² + OC² = BC²

⇒ x² + x² = (36)²

⇒ 2x² = 1,296

⇒ x² = $\dfrac{1,296}{2}$

⇒ x² = 648

⇒ x = $\sqrt{648}$

⇒ x = 18 $\sqrt{2}$

Now the area of triangle BOC = $\dfrac{1}{2}$ x base x height

$= \dfrac{1}{2} \times 18 \sqrt{2} \times 18 \sqrt{2}\\[1em] = \dfrac{1}{2} \times 18 \times 18 \times 2\\[1em] = 18 \times 18\\[1em] = 324 \text{ cm}^2$

Area of quadrilateral ABOC = Area of Δ ABC - Area of Δ BOC

= 432 - 324 cm²

= 108 cm²

Hence, the area of quadrilateral ABOC is 108 cm².