In the figure, compute the area of the quadrilateral.
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Given:

In the quadrilateral $ABCD$,

$\angle A = 90^o, \angle CBD = 90^o, AD = 9\ cm, BC = 8\ cm$ and $CD = 17\ cm$

To do:

We have to compute the area of the quadrilateral.

Solution:

In $\triangle BCD$,

$CD^2 = BC^2 + BD^2$              (Pythagoras Theorem)

$17^2 = 8^2 + BD^2$

$289 = 64 + BD^2$

$BD^2 = 289 - 64$

$= 225$

$= (15)^2$

$\Rightarrow BD = 15\ cm$

In $\triangle ABD$,

$BD^2 = AB^2 + AD^2$

$(15)^2 = AB^2 + (9)^2$

$225 = AB^2 + 81$

$AB^2= 225 - 81$

$= 144$

$= (12)^2$

$\Rightarrow AB = 12\ cm$

Area of $\triangle \mathrm{ABD}=\frac{1}{2}\times$ base $\times$ height

$=\frac{1}{2} \mathrm{AB} \times \mathrm{AD}$

$=\frac{1}{2} \times 12 \times 9$

$=54 \mathrm{~cm}^{2}$

Area of $\Delta \mathrm{BCD}=\frac{1}{2} \times 8 \times 15$

$=60 \mathrm{~cm}^{2}$
Therefore,

Total area of quadrilateral $\mathrm{ABCD}=54+60$

$=144 \mathrm{~cm}^{2}$.

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Updated on: 10-Oct-2022

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