Area and Perimeter Practice Test 2

First of all we will have to find the area of ΔBDF
Here, a = 17, b = 25, c = 12
∴ s = $\frac{\mathit{a}+\mathit{b}+\mathit{c}}{2}$ = $\frac{17+25+12}{2}$
= $\frac{54}{2}$ = 27 m

∴ Area of ΔBDF = $\sqrt{27\left(27-17\right)\left(27-25\right)\left(27-12\right)}$
= $\sqrt{27\times 10\times 2\times 15}$
= $\sqrt{3\times 3\times 3\times 2\times 5\times 2\times 5\times 3}$
= 3 × 3 × 5 × 2
= 90 sq. m.
Since Area of ΔBDF = $\frac{1}{2}$ × base × height ⇒ 90 = $\frac{1}{2}$ × 25 × height
∴ Height of the triangle, i.e., distance between the parallel sides of the trapezium
= $\frac{90\times 2}{25}$ = $\frac{36}{5}$ m
∴ Area of the trapezium = $\frac{1}{2}$ × (25 + 10) × $\frac{36}{5}$
= $\frac{1}{2}$ × 75 × $\frac{36}{5}$ = 270 sq. cm.