Question

Using the measures given in the figure below, find the area of □EFGH.

Answer 1

The given map of the plot is of a quadrilateral that can be divided in two triangles,
∆HGF and ∆HEF.

Lengths of the sides of ∆HGF are 34 m, 50 m and 52 m

Semi-perimeter of a triangle = $s=\frac{a+b+c}{2}$

Semi-perimeter $=\frac{34+50+52}{2}=68\mathrm{m}$

Area of a triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

$$\begin{gathered} =\sqrt{68\left(68-34\right)\left(68-50\right)\left(68-52\right)} \\ =\sqrt{68\left(34\right)\left(18\right)\left(16\right)} \\ =\sqrt{665856} \\ =816\mathrm{sq}\mathrm{m} \end{gathered}$$

∴ Area of the triangle ∆HGF is 816 sq m.

Lengths of the sides of the triangle ∆HEF are 20 m, 48 m and 52 m.

Semi-perimeter of a triangle = $s=\frac{a+b+c}{2}$

Semi-perimeter = $\frac{20+48+52}{2}=60\mathrm{m}$

Area of a triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

$$\begin{gathered} =\sqrt{60\left(60-20\right)\left(60-48\right)\left(60-52\right)} \\ =\sqrt{60\left(40\right)\left(12\right)\left(8\right)} \\ =\sqrt{230400} \\ =480 \end{gathered}$$
∴ Area of the triangle ∆HEF is 480 sq m.

So, the area of $\square EFGH$ = A(∆HGF) + A(∆HEF)
= 816 sq m + 480 sq m =1296 sq m