Question 1
A park, in the shape of a quadrilateral ABCD, has $∠ C = 90^{\circ}, A B = 9 m, B C = 12 m, C D = 5 m a n d A D = 8 m$ . How much area does it occupy?

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Solution

Let us join BD.

In $Δ B C D$ , applying Pythagoras theorem,
$B D^{2} = B C^{2} + C D^{2} = (12)^{2} + (5)^{2} = 144 + 25 B D^{2} = 169 B D = 13 m A r e a o f Δ B C D = \frac{1}{2} \times B C \times C D = (\frac{1}{2} \times 12 \times 5) m^{2} = 30 m^{2} F o r Δ A B D, s = \frac{P e r i m e t e r}{2} = \frac{(9 + 8 + 13)}{2} = 15 m B y H e r o n^{'} s f o r m u l a A r e a o f t r i a n g l e = \sqrt{s (s - a) (s - b) (s - c)} A r e a o f Δ A B D = \sqrt{s (s - a) (s - b) (s - c)} = ⌊ \sqrt{15 (15 - 9) (15 - 8) (15 - 13)} ⌋ m^{2} = (\sqrt{15 \times 6 \times 7 \times 2}) m^{2} = 6 \sqrt{35} m^{2} = (6 \times 5.916) m^{2} = 35.496 m^{2} A r e a o f t h e p a r k = A r e a o f Δ A B D + A r e a o f Δ B C D = 35.496 + 30 m^{2} = 65.496 m^{2} = 65.5 m^{2} (a p p r o x i m a t e l y)$

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Similar questions

∠ C = 90, A B = 9 m, B C = 12 m, C D = 5 m

A D = 8 m

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

and AD = 8m

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How much area does it occupy ?

A B C D

∠ C = 90^{o}, A B = 9 m, B C = 12 m, C D = 5 m, A D = 8 m

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