Question
In Figure, $ABCD$ is a quadrilateral in which diagonal $AC = 84\ cm$; $\text{DL}\bot\text{AC},\text{BM}\bot\text{AC},$ $DL = 16.5\ cm$ and $BM = 12\ cm$. Find the area of quadrilateral $ABCD$.
✓
Answer
We have,
$\mathrm{AC}=84 \mathrm{~cm}, \mathrm{DL}=16.5 \mathrm{~cm}$ and $\mathrm{BM}=12 \mathrm{~cm}$
Area of triangle $A D C=\frac{1}{2}(A C \times D L)$
$=\frac{1}{2}(84 \mathrm{~cm} \times 16.5 \mathrm{~cm})=693 \mathrm{~cm}^2$
Area of triangle $\mathrm{ABC}=\frac{1}{2}(\mathrm{AC} \times \mathrm{BM})$
$=\frac{1}{2}(84 \mathrm{~cm} \times 12 \mathrm{~cm})=504 \mathrm{~cm}^2$
Hence, Area of quadrilateral $A B C D=$ Area of $A D C+$ Area of $A B C$
$=(693+504) \mathrm{cm}^2=1197 \mathrm{~cm}^2$
$\mathrm{AC}=84 \mathrm{~cm}, \mathrm{DL}=16.5 \mathrm{~cm}$ and $\mathrm{BM}=12 \mathrm{~cm}$
Area of triangle $A D C=\frac{1}{2}(A C \times D L)$
$=\frac{1}{2}(84 \mathrm{~cm} \times 16.5 \mathrm{~cm})=693 \mathrm{~cm}^2$
Area of triangle $\mathrm{ABC}=\frac{1}{2}(\mathrm{AC} \times \mathrm{BM})$
$=\frac{1}{2}(84 \mathrm{~cm} \times 12 \mathrm{~cm})=504 \mathrm{~cm}^2$
Hence, Area of quadrilateral $A B C D=$ Area of $A D C+$ Area of $A B C$
$=(693+504) \mathrm{cm}^2=1197 \mathrm{~cm}^2$
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