Question
In the given figure, $ABCD$ is a quadrilateral in which diagonal $BD = 24\ cm,\text{AL}\perp\text{BD}$ and $\text{CM}\perp\text{BD}$ such that $AL = 9\ cm$ and $CM = 12\ cm.$ Calculate the area of the quadrilateral.
✓
Answer
Given,
$BD = 24\ cm$
$AL = 9\ cm$
$CM = 12\ cm$
Area of $\triangle\text{BCD}=\frac{1}{2}\times\text{BD}\times\text{CD}$
$=\frac12\times24\times12=144\text{cm}^2$
Area of $\triangle\text{DAB}=\frac{1}{2}\times\text{BD}\times\text{AL}$
$=\frac12\times24\times9=108\text{cm}^2$
Now, area of quadrilateral $ABCD =$ Area of $\triangle\text{BCD}$ $+$ Area of $\triangle\text{DAB}$
$=(144+108) \mathrm{cm}^2 $
$=252 \mathrm{~cm}^2$
$BD = 24\ cm$
$AL = 9\ cm$
$CM = 12\ cm$
Area of $\triangle\text{BCD}=\frac{1}{2}\times\text{BD}\times\text{CD}$
$=\frac12\times24\times12=144\text{cm}^2$
Area of $\triangle\text{DAB}=\frac{1}{2}\times\text{BD}\times\text{AL}$
$=\frac12\times24\times9=108\text{cm}^2$
Now, area of quadrilateral $ABCD =$ Area of $\triangle\text{BCD}$ $+$ Area of $\triangle\text{DAB}$
$=(144+108) \mathrm{cm}^2 $
$=252 \mathrm{~cm}^2$
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