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

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65.49 m 2

54.49 m 2

32.49 m 2

10.49 m 2

answer is A.

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Detailed Solution

Let the diagram of the situation be,
seo

Given that,

∠ C = 90 ∘, A B = 9 m, B C = 12 m, C D = 5 m, A D = 8 m

.
Apply the Pythagoras theorem, find the value of BD, and we have

B D 2 = B C 2 + C D 2 B D 2 = 122 + 52 B D 2 = 144 + 25 B D = 169 B D = 13 m

The area of the right-angled triangle BCD is given by,

A B C D = 12 × B C × C D = 12 × 12 × 5 = 30 m 2

To find the area of the other triangle ABD, its perimeter is half of the perimeter.

s = P 2 = A B + A D + B D 2 = 9 + 8 + 13 2 = 302 = 15

The area of the triangle is given by,

A A B D = s (s − a) (s − b) (s − c) = 15 (15 − 9) (15 − 8) (15 − 13) = 15 (6) (7) (2) = 1260 = 35.49 m 2

Hence, the area of the quadrilateral is given by,

A = A B C D + A A B D = 30 + 35.49 = 65.49 m 2

Hence, the correct option is 2.

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