A park,in the shape of a quadrilateral ABCD,h | vedclass

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(D) Join $B$ and $D$ to divide the quadrilateral $ABCD$ into two triangles: $\Delta BCD$ and $\Delta ABD$.$1$. In the right-angled $\Delta BCD$ (since

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$65.5$

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(D) Join $B$ and $D$ to divide the quadrilateral $ABCD$ into two triangles: $\Delta BCD$ and $\Delta ABD$.$1$. In the right-angled $\Delta BCD$ (since $\angle C = 90^{\circ}$):Area of $\Delta BCD = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 12 	imes 5 = 30 \, m^{2}$.$2$. Calculate the length of the diagonal $BD$ using the Pythagoras theorem:$BD^{2} = BC^{2} + CD^{2} = 12^{2} + 5^{2} = 144 + 25 = 169$.$BD = \sqrt{169} = 13 \, m$.$3$. For $\Delta ABD$,the sides are $a = 9 \, m$,$b = 8 \, m$,and $c = 13 \, m$.Semi-perimeter $s = \frac{9 + 8 + 13}{2} = \frac{30}{2} = 15 \, m$.Using Heron's formula,Area of $\Delta ABD = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{15(15-9)(15-8)(15-13)} = \sqrt{15 	imes 6 	imes 7 	imes 2} = \sqrt{1260} = 6\sqrt{35} \, m^{2}$.Using $\sqrt{35} \approx 5.916$,Area $\approx 6 	imes 5.916 = 35.496 \approx 35.5 \, m^{2}$.$4$. Total area of quadrilateral $ABCD = 	ext{Area}(\Delta BCD) + 	ext{Area}(\Delta ABD) = 30 + 35.5 = 65.5 \, m^{2}$.

$A$ park,in the shape of a quadrilateral $ABCD$,has $\angle C = 90^{\circ}$,$AB = 9 \, m$,$BC = 12 \, m$,$CD = 5 \, m$,and $AD = 8 \, m$. How much area does it occupy (in $, m^{2}$)?

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