Question

A park in the shape of a quadrilateral ABCD has AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m and ∠C = 90°. Find the area of the park.

Answer 1

We know that $△$BCD is a right triangle.
∴ $BD=\sqrt{B{C}^2+C{D}^2}=\sqrt{{12}^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\mathrm{m}$

$$\begin{gathered} \mathrm{Now,} \\ Area\mathrm{of}\mathrm{triangle}BCD=\frac{1}{2}\times \mathrm{base}\times \mathrm{height} \\ =\frac{1}{2}\times BC\times CD \\ =\frac{1}{2}\times 12\times 5 \\ =30{\mathrm{m}}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{Let}: \\ a=9\mathrm{m},b=8\mathrm{m}\mathrm{and}c=13\mathrm{m} \\ s=\frac{a+b+c}{2}=\frac{9+8+13}{2}=15\mathrm{m} \\ \mathrm{By}\mathrm{Heron}\text{'}\mathrm{s}\mathrm{formula},\mathrm{we}\mathrm{have}: \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}ABD=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{15(15-9)(15-8)(15-13)} \\ =\sqrt{15\times 6\times 7\times 2} \\ =\sqrt{5\times 3\times 3\times 2\times 7\times 2} \\ =3\times 2\sqrt{35} \\ =6\sqrt{35} \\ =6\times 5.9 \\ =35.4{\mathrm{m}}^2 \end{gathered}$$
Now,
Area of quadrilateral ABCD = Area of $△$BCD + Area of $△$ABD
= (30 + 35.4) m² = 65.4 m²