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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Solution

We assume ABCD be the quadrilateral having sides AB, BC, CD, DA and.

We take a diagonal DB, where DB divides ABCD into two triangles ΔBCD and ΔABD

In ΔBCD, we have

DC = 5 m; BC = 12 m

Use Pythagoras theorem

BD² = DC²+ BC²

Area of right angled triangle ΔBCD, say A₁ is given by

Where, Base = DC = 5 m; Height = BC = 12 m

Area of triangle ΔABD, say A₂having sides a, b, c and s as semi-perimeter is given by

Where, a = AD = 8 m; b = AB = 9 m; c = BD = 13 m

Area of quadrilateral $□$ ABCD, say A

A = Area of triangle DCB + Area of triangle ABD

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∠ C = 90^{\circ}, A B = 9 m, B C = 12 m, C D = 5 m a n d A D = 8 m

∠ 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

.

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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