Question

Find the perimeter and area of the quadrilateral ABCD in which AB = 21 cm, ∠BAC = 90°, AC = 20 cm, CD = 42 cm and AD = 34 cm.

Answer 1

In right angled ∆ABC,
BC² = AB² + AC² (Pythagoras Theorem)
⇒ BC² = 21² + 20²
⇒ BC² = 441 + 400
⇒ BC² = 841
⇒ BC = 29 cm

Area of ∆ABC = $\frac{1}{2}\times AB\times AC$
= $\frac{1}{2}\times 21\times 20$
= 210 cm² ....(1)

In ∆ACD,
The sides of the triangle are of length 20 cm, 34 cm and 42 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{20+34+42}{2}=\frac{96}{2}=48\mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ACD=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{48\left(48-20\right)\left(48-34\right)\left(48-42\right)} \\ =\sqrt{48\left(28\right)\left(14\right)\left(6\right)} \\ =336{\mathrm{cm}}^2...\left(2\right) \end{gathered}$$

Thus,
Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD
= (210 + 336) cm²
= 546 cm²

Also,
Perimeter of quadrilateral ABCD = (34 + 42 + 29 + 21) cm
= 126 cm

Hence, the perimeter and area of quadrilateral ABCD is 126 cm and 546 cm², respectively.