Question

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?

Answer 1

Let consider a quadrilateral ABCD

In ∆BCD;

BD = \(\sqrt{(12)^2+5^2}\) = 13 m

BC = a = 12 m, CD = b = 5 m, BD = c = 13 m

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}2\)[Heron’s Formula]

s = \(\frac{a+b+c}2\) = \(\frac{12+5+13}2\) = 15

A₁ = \(\sqrt{15(15-12)(15-5)(15-13)}\)

A₁ = \(\sqrt{15\times3\times10\times2}\) = 30 m²

In ∆ABD;

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

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}2\)

s = \(\frac{a+b+c}2\) = \(\frac{9+8+13}2\) = 15

A₂ = \(\sqrt{15(15-9)(15-8)(15-13)}\)

A₂ = \(\sqrt{15\times6\times7\times2}\) = 35.50 m²

Therefore area of quadrilateral ABCD = A₁ + A₂ = 30+35.50 = 65.50 m²