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

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

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### Solution:

Let us given that,

∠ C = 90º, AB = 9 m, BC = 12 m, CD = 5 m,  AD = 8 m

From above information we draw the figure,

we connected B and D such the the triangle is formed

So, the triangle BDC is a right angled triangle

By using the Pythagoras theorem

we get,

BD2 = BC2 $+$ CD2

BD2 = 122 $+$ 52

BD2 = 144 $+$ 25

BD = $\sqrt{169}$

BD = 13 m

Area of quadrilateral ABCD = area of ∆BCD$+$ area of ∆ABD

Now, Area of ∆BCD = 1/2 $×$ base $×$ height

= 1/2 $×$ 12 m $×$ 5 m

= 30 m2

Now, in ∆ABD, AB = a = 9 m, AD = b = 8 m, BD = c = 13 m

Semi Perimeter of ΔABD

s = (a $+$ b $+$ c)/2

= (9 $+$ 8 $+$ 13)/2

= 30/2

= 15 m

We know that, Heron's formula

Area of triangle =

where, s is the semi-perimeter = half of the perimeter

and  a, b and c are the sides of the triangle

Area of triangle ABD =

= 6$\sqrt{35}$

= 35.5 m

Area of  triangle ABD = 35.5 m2

Therefore,

Area of park ABCD = 30 m2 $+$ 35.5 m2 = 65.5 m2

Hence,  the park ABCD occupies an area of 65.5 m2.

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