Find the area and the perimeter of quadrilateral ABCD, given below; if, AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and ∠DBC = 90°.

Given:

AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and ∠DBC = 90°

In Δ BCD,

By using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ BC² + 12² = 13²

⇒ BC² + 144 = 169

⇒ BC² = 169 - 144

⇒ BC² = 25

⇒ BC = $\sqrt{25}$

⇒ BC = 5 cm

Area of Δ BCD = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x 12 x 5 cm²

= $\dfrac{1}{2}$ x 60 cm²

= 30 cm²

For Δ ABD,

Let AD = a = 10 cm, BD = b = 12 cm and AB = c = 8 cm.

$∵ s = \dfrac{a + b + c}{2}\\[1em] = \dfrac{10 + 12 + 8}{2}\\[1em] = \dfrac{30}{2}\\[1em] = 15$

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

= $\sqrt{15(15 - 10)(15 - 12)(15 - 8)}$ cm²

= $\sqrt{15 \times 5 \times 3 \times 7}$ cm²

= $\sqrt{1,575}$ cm²

= 39.7 cm²

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

= 39.5 + 30 cm²

= 69.5 cm²

Perimeter of quadrilateral ABCD = Sum of all sides of quadrilateral

= AB + BC + CD + DA

= 8 + 10 + 13 + 5

= 36 cm

Hence, the area of quadrilateral is 69.7 cm² and the perimeter is 36 cm.