A park, in the shape of a quadrilateral $A B C D$ has $∠ C = 90^{o}$ , $A B = 9 m, B C = 12 m, C D = 5 m, A D = 8 m$ . How much area does it occupy?

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Solution

Join the diagonal AD of the quadrilateral ABCD.
$B D^{2} = B C^{2} + C D^{2} = (12)^{2} + (5)^{2} = 169$
$\Rightarrow B D = \sqrt{169} = 13 m$
Area of right angle $Δ B C D$
$= \frac{1}{2} \times B D \times C D = \frac{1}{2} \times 12 \times 5 m^{2} = 30 m^{2}$
Now, the sides of the $Δ A B D$ are 9 m,13 m and 8 m.
Semiperimeter of $Δ A B D$
$S = \frac{9 m + 13 m + 8 m}{2} = 15 m .$
The area of the $Δ A B D$
$= \sqrt{15 \times (15 - 9) \times (15 - 13) \times (15 - 8)} m^{2}$
$\Rightarrow \sqrt{15 \times 6 \times 2 \times 7} m^{2} = \sqrt{5 \times 3 \times 3 \times 2 \times 2 \times 7}$
$\Rightarrow 6 \times 5.916 m^{2} (a p p r o x .) = 35.495 m^{2} (a p p r o x .) = 35.5 (a p p r o x .)$
Thus, area of the quadrilateral ABCD $= a r (Δ B C D) + a r (Δ A B D)$
$= 30 m^{2} + 35.5 m^{2} (a p p r o x .) = 65.5 m^{2} (a p p r o x .)$

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A park, in shape of a quadrilateral ABCD has $∠ C = 90^{0}$ , AB= 9m, BC= 12m, CD= 5m and AD = 8m. How much area does it occupy?

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A park, in the shape of a quadrilateral ABCD, has ∠ {C =90}^{\circ}, AB = 9m, BC = 12m, CD = 5m

and AD = 8m

.

How much area does it occupy ?

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