# A park, in the sh

Given: The figure is given below: The dimensions of the park as follows:

Angle C = 90°,

AB = 9m,

BC = 12m,

CD = 5m

Now, BD is joined.
Thus quadrilateral ABCD can now be divided into two triangles ABD and BCD
Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

Pythagoras Theorem: It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In
ΔBCD by Pythagoras theorem,

BD2 = BC2 + CD2

BD2 = 122 + 52

BD2 = 169

BD = 13m

As triangle BCD is right angled triangle,  Now, Semi perimeter of triangle (ABD) = (8 + 9 + 13)/2

= 30/2  = 15m

Using Heron’s formula,
ar(ABD) = where, s = semiperimeter of the triangle and a, b, and c are the sides of the triangle.

= = = = 6√35 m2 = 6 × 5.91
= 35.46 m2

Area of quadrilateral ABCD = 35.46 + 30

Area of quadrilateral ABCD = 65.46 m2

Thus, the park acquires an area of  65.5 m2(approx).

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