Question
Trapezium given below; find its area.
✓
Answer
In the given figure, we can observe that the non$-$parallel sides are equal and hence it is an isosceles trapezium.
Therefore, let us draw $DE$ and $CF$ perpendiculars to $AB.$
Thus, the area of the parallelogram is given by
$AB = AE + EF + FB$ and $CD = EF = 18 \ cm,$ we have
$30 = AE + 18 + FB$
$\Rightarrow 30 = AE + 18 + AE$
$\Rightarrow 2AE + 18 = 30$
$\Rightarrow 2AE = 30 - 18$
$\Rightarrow 2AE = 12$
$\Rightarrow AE = 6 \ cm$
Now, consider the right angled triangle $\text{ADE}.$
$AD^2 = AE^2 + DE^2$
$\Rightarrow 12^2 = 6^2 + DE^2$
$\Rightarrow 144 = 36 + DE^2$
$\Rightarrow DE^2 = 144 - 36$
$\Rightarrow DE^2 = 108$
$\Rightarrow D E=\sqrt{36 \times 3}$
$\Rightarrow D E=6 \sqrt{3}$
Area $(\square \text{ABCD} )=$ Area $(\Delta \text{ADE} )+$ Area $(\square \text{DEFC} )+$ Area $(\Delta \text{CFB} )$
$\Rightarrow$ Area $(\square \text{ABCD} )=\frac{1}{2} \times 6 \times 6 \sqrt{3}+18 \times 6 \sqrt{3}+\frac{1}{2} \times 6 \times 6 \sqrt{3}$
$\Rightarrow$ Area $(\square \text{ABCD} )=6 \times 6 \sqrt{3}+18 \times 6 \sqrt{3}$
$\Rightarrow$ Area $(\square \text{ABCD} )=144 \sqrt{3}=249.41 \ cm ^2$
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