Calculate the area of quadrilateral ABCD in which : AB = 24 cm, AD = 32 cm, ∠BAD = 90°, and BC = CD = 52 cm.

Given,

AB = 24 cm

AD = 32 cm

∠BAD = 90°

BC = 52 cm

CD = 52 cm

Diagonal BD divides the quadrilateral into two triangles: △ABD and △BCD

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

= $\dfrac{1}{2}$ × AB × AD

= $\dfrac{1}{2}$ × 24 × 32

= 12 × 32

= 384 cm².

Since ∠BAD = 90°, by applying pythagoras theorem

⇒ BD² = BA² + AD²

⇒ BD² = 24² + 32²

⇒ BD² = 576 + 1024

⇒ BD² = 1600

⇒ BD = $\sqrt{1600}$

⇒ BD = 40 cm.

Area of △BCD,

Let BC = a = 52 cm, CD = b = 52 cm, BD = c = 40 cm

$\Rightarrow s = \dfrac{a + b + c}{2} \\[1em] \Rightarrow s = \dfrac{52 + 52 + 40}{2} \\[1em] \Rightarrow s = \dfrac{144}{2} \\[1em] \Rightarrow s = 72 \text{ cm}.$

By formula,

$\Rightarrow \text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] = \sqrt{72 \times (72 - 52) \times (72 - 52) \times (72 - 40)} \\[1em] = \sqrt{72 \times (20) \times (20) \times (32)} \\[1em] = \sqrt{921600} \\[1em] = 960 \text{ cm}^2.$

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

= 384 + 960

= 1344 cm².

Hence, area of quadrilateral ABCD = 1344 cm².