Find the area of quadrilateral ABCD in which ∠B = 90° , AB = 6 cm, BC = 8 cm and CD = AD = 13 cm.

In △ABC,

Using Pythagoras theorem,

AC² = AB² + BC²

Substituting the values we get,

⇒ AC² = 6² + 8²

⇒ AC² = 36 + 64 = 100

⇒ AC² = 10²

⇒ AC = 10 cm

Calculating area of △ADC,

In △ADC,

Let a = AD = 13 cm, b = DC = 13 cm and c = AC = 10 cm.

Semi-perimeter (s) = $\dfrac{a + b + c}{2} = \dfrac{13 + 13 + 10}{2} = \dfrac{36}{2}$ = 18 cm.

By Heron's formula,

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

Substituting values we get,

$\text{Area of △ADC} = \sqrt{18(18 - 13)(18 - 13)(18 - 10)} \\[1em] = \sqrt{18 \times 5 \times 5 \times 8} \\[1em] = \sqrt{3600} \\[1em] = 60 \text{ cm}^2.$

Calculating area of △ABC,

$\text{Area of △ABC } = \dfrac{1}{2} \times \text{ base × height} \\[1em] = \dfrac{1}{2} \times AB \times BC \\[1em] = \dfrac{1}{2} \times 6 \times 8 \\[1em] = 24 \text{ cm}^2$

From figure,

Area of quadrilateral ABCD = Area of △ADC + Area of △ABC

= 60 + 24

= 84 cm².

Hence, area of quadrilateral ABCD = 84 cm².