Question

Find the area of a quadrilateral ABCD whose sides are 9m, 40m, 28m and 15m, and the angle between the first two sides is a right angle.

Answer 1

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Answer 2

The area of the quadrilateral is $\bold{306\ m^{2}}$

Given: $\text{The area of the quadrilateral is}\ 306\ m^{2}$

9m, 40m, 28m, 15m are the sides of quadrilateral ABCD  

To find:

The area of a quadrilateral

Solution:

In $\triangle \mathrm{ABC}$

AB = 9m, BC = 40m, $\angle A B C=90^{\circ}$

By Pythagoras Theorem,  

$\begin{aligned} \mathrm{AC} &=\sqrt{40^{2}+9^{2}} m \\ &=\sqrt{1600+81} m \\ &=\sqrt{1681} m \\ &=41 m \end{aligned}$  

And, for $\triangle \mathrm{ADC} Semi-perimeter (s)=\frac{15+28+41}{2}\ m=42\ m$

$\begin{aligned} \text{Area\ of}\ \Delta \mathrm{ABC} &=\frac{1}{2} \times \text { Base } \times \text { Height } \\ &=\frac{1}{2} \times A B \times B C \\ &=\frac{1}{2} \times 9 m \times 40 m \\ &=180 \mathrm{m}^{2} \end{aligned}$

$\begin{aligned} \text{Area\ of} \Delta \mathrm{ADC} &=\sqrt{s(s-a)(s-b)(s-c)}\ m^{2} \\ &=\sqrt{42(42-15)(42-28)(42-41)}\ m^{2} \\ &=\sqrt{42 \times 27 \times 14 \times 1}\ m^{2} \\ &=126\ m^{2} \end{aligned}$

∴ Area of quadrilateral ABCD

$\begin{array}{l}{=\text { Area of } \Delta \mathrm{ABC}+\text { Area of } \Delta \mathrm{ADC}} \\ {=180\ \mathrm{m}^{2}+126\ \mathrm{m}^{2}} \\ {=306\ \mathrm{m}^{2}}\end{array}$