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# A park, in the shape of a quadrilateral $ABCD$, has $\angle C=90°$, $AB=9 m$, $BC=12 m$, $CD=5 m$ and $AD=8 m$. How much area does it occupy?

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Solution

## Given -: In quadrilateral $ABCD$ where $\angle C=90°$, $AB=9m$, $BC=12m$, $CD=5m$ and $AD=8m$. Join BDStep 1: Find area of $∆BCD$By applying Pythagoras theorem in $∆BCD$, we get$B{D}^{2}=B{C}^{2}+C{D}^{2}$$⇒B{D}^{2}={12}^{2}+{5}^{2}$$⇒B{D}^{2}=144+25$$⇒B{D}^{2}=169$$⇒BD=\sqrt{169}=13m$Now, the area of $\Delta BCD$ $=\frac{1}{2}×12×5=6×5=30{m}^{2}$ ( area of triangle = $\frac{1}{2}×base×height$ )Step 2: Find area of $\Delta ABD$The semi perimeter of $\Delta ABD$ is $s=\frac{a+b+c}{2}$$⇒s=\frac{9+13+8}{2}=\frac{30}{2}=15m$Using Heron’s formula,Area of $\Delta ABD$ $=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ $=\sqrt{15\left(15-9\right)\left(15-13\right)\left(15-8\right)}$ $=\sqrt{15×6×2×7}$ $=\sqrt{3×5×2×3×2×7}$ $=2×3\sqrt{5×7}$ $=6\sqrt{35}{m}^{2}=35.5{m}^{2}\left(approx\right)$Step 3: Find area of quadrilateral $ABCD$∴ The area of quadrilateral $ABCD$ $=$ Area of $∆BCD$+Area of $\Delta ABD$ $=30{m}^{2}+35.5{m}^{2}=65.5{m}^{2}$Hence, the area of quadrilateral $ABCD$$=65.5{m}^{2}$

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