Question

The sides of a quadrilateral ABCD taken in order are 6 cm, 8 cm, 12 cm and 14 cm respectively and the angle between the first two sides is a right angle. Find its area. (Given, $\sqrt{6}=2.45$).

Answer 1

In the given figure, ABCD is a quadrilateral with sides of length 6 cm, 8 cm, 12 cm and 14 cm respectively and the angle between the first two sides is a right angle.



Join AC.

In right angled ∆ABC,
AC² = AB² + BC² (Pythagoras Theorem)
⇒ AC² = 6² + 8²
⇒ AC² = 36 + 64
⇒ AC² = 100
⇒ AC = 10 cm

Area of ∆ABC = $\frac{1}{2}\times AB\times BC$
= $\frac{1}{2}\times 6\times 8$
= 24 cm² ....(1)

In ∆ACD,
The sides of the triangle are of length 10 cm, 12 cm and 14 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{10+12+14}{2}=\frac{36}{2}=18\mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ACD=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{18\left(18-10\right)\left(18-12\right)\left(18-14\right)} \\ =\sqrt{18\left(8\right)\left(6\right)\left(4\right)} \\ =24\sqrt{6}{\mathrm{cm}}^2 \\ =24\left(2.45\right){\mathrm{cm}}^2 \\ =58.8{\mathrm{cm}}^2...\left(2\right) \end{gathered}$$

Thus,
Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD
= (24 + 58.8) cm²
= 82.8 cm²

Hence, the area of quadrilateral ABCD is 82.8 cm².