Management Aptitude Test 2015 Paper

(b) Sides of a triangle
a = 5 cm, b = 12 cm and c = 13 cm
s = $\frac{\mathit{a}+\mathit{b}+\mathit{c}}{2}$
= $\frac{5+12+13}{2}$
= $\frac{30}{2}$ = 15
Area of triangle
= $\sqrt{\mathit{s}\left(\mathit{s}-\mathit{a}\right)\left(\mathit{s}-\mathit{b}\right)\left(\mathit{s}-\mathit{c}\right)}$
= $\sqrt{15\left(15-5\right)\left(15-12\right)\left(15-13\right)}$
= $\sqrt{15\times 10\times 3\times 2}$
= $\sqrt{5\times 3\times 5\times 2\times 3\times 2}$
= $\sqrt{5\times 5\times 3\times 3\times 2\times 2}$
= 5 x 3 x 2 = 30 $\mathit{c}{\mathit{m}}^2$
Area of rectangle = Area of triangle
Hence, area of rectangle = 30 $\mathit{c}{\mathit{m}}^2$
Width of rectangle = 10 cm
Area of rectangle = Length x Width
30 = l x 10
∴ l = 3 cm
Perimeter of rectangle =2(l + W)
= 2(35+ 10) =2 x 13 = 26 cm