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A park, in the shape of a quadrilateral ABCD, has 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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A park, in the shape of a quadrilateral ABCD, has 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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First, construct a quadrilateral ABCD and join BD.

We know that

C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m

The diagram is:

Now, apply Pythagoras theorem in ΔBCD

BD2 = BC+ CD2

⇒ BD2 = 122 + 52

⇒ BD2 = 169

⇒ BD = 13 m

Now, the area of ΔBCD = (1/2 × 12 × 5) = 30 m2

The semi perimeter of ΔABD

(s) = (perimeter/2)

= (8 + 9 + 13)/2 m

= 30/2 m = 15 m

Using Heron’s formula,

Area of ΔABD

= \(\sqrt{s(s - a)(s - b)(s - c)}\)

= \(\sqrt{15(15 - 13)(15 - 9)(15 - 8)}m^2\)

= \(\sqrt{15 \times 2 \times 6 \times 7}m^2\)

= 6\(\sqrt{35}m^2\)

= 35.5 m2

∴ The area of quadrilateral ABCD = Area of ΔBCD+Area of ΔABD

= 30 m2 + 35.5m2 = 65.5 m2

This post was modified 3 years ago 2 times by Samar shah
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