Calculate the area of quadrilateral ABCD in which ∠A = 90°, AB = 16 cm, AD = 12 cm and BC = CD = 12.5 cm.

Given: ∠A = 90°, AB = 16 cm, AD = 12 cm and BC = CD = 12.5 cm

Join BD to divide quadrilateral ABCD into two triangles: Δ ABD and Δ BCD.

Since ∠A = 90°, Δ ABD is a right-angled triangle at A.

Area of right angled triangle = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x 16 x 12

= $\dfrac{1}{2}$ x 192

= 96 cm²

Using Pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ BD² = AB² + AD²

= 16² + 12²

= 256 + 144

= 400

⇒ BD = $\sqrt{400}$ = 20 cm

Δ BCD is isosceles, with BC = CD = 12.5 cm and base BD = 20 cm.

s (semi-perimeter) = $\dfrac{a + b + c}{2}$

Substituting BC = a = 12.5 cm, CD = b = 12.5 cm, BD = c = 20 cm:

s = $\dfrac{12.5 + 12.5 + 20}{2}$

= $\dfrac{45}{2}$

= 22.5 cm

Area of triangle = $\sqrt{s(s - a)(s - b)(s- c)}$

$= \sqrt{22.5(22.5 - 12.5)(22.5 - 12.5)(22.5 - 20)}\\[1em] = \sqrt{22.5 \times 10 \times 10 \times 2.5}\\[1em] = \sqrt{5,625}\\[1em] = 75$

Area of quadrilateral ABCD = Area of Δ ABD + Area of Δ ABD

= 96 cm² + 75 cm²

= 171 cm²

Hence, the area of quadrilateral ABCD = 171 cm².