Calculate the area of quadrilateral ABCD in which △BCD is equilateral with each side equal to 26 cm, ∠BAD = 90° and AD = 24 cm.

Given

△BCD is equilateral.

Each side = 26 cm

∠BAD = 90°

AD = 24 cm

$\Rightarrow \text{Area of equilateral △BCD} = \dfrac{\sqrt{3}}{4} \times \text{ (side)}^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 26^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 676 \\[1em] = \sqrt{3} \times 169 \\[1em] = 169 \times 1.732 \approx 292.71 \text{ cm}^2.$

Since ∠BAD = 90°

By applying pythagoras theorem in triangle BAD,

⇒ AB² + AD² = BD²

⇒ AB² + 24² = 26²

⇒ AB² + 576 = 676

⇒ AB² = 676 - 576

⇒ AB² = 100

⇒ AB = $\sqrt{100}$

⇒ AB = 10 cm.

⇒ Area of right angle triangle ABD = $\dfrac{1}{2}$ × Base × Height

= $\dfrac{1}{2}$ × AB × AD

= $\dfrac{1}{2}$ × 10 × 24

= 5 × 24 = 120.

⇒ Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

= 120 + 292.71

= 412.71 cm².

Hence, area = 412.71 cm².