SSC CGL 19 Oct 2014 Shift 2 Paper

Area of base = Area of Δ ABD + Area of Δ BCD

In, Δ ABD
BD = $\sqrt{\mathit{A}{\mathit{B}}^2+\mathit{A}{\mathit{D}}^2}$ = $\sqrt{9^2+{12}^2}$
= $\sqrt{81+144}$ = $\sqrt{225}$ = 15 cm
Area of Δ ABD = $\frac{1}{2}$ × AB × AD
= $\frac{1}{2}$ × 9 × 12
= 54 sq. cm
For Δ BCD,
Semi-perimeter (s) = $\frac{13+14+15}{2}$
= $\frac{42}{2}$ = 21
∴ Area of Δ BCD
= $\sqrt{\mathit{s}\left(\mathit{s}-\mathit{a}\right)\left(\mathit{s}-\mathit{b}\right)\left(\mathit{s}-\mathit{c}\right)}$
= $\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)}$
= $\sqrt{21\times 8\times 7\times 6}$
= 21 × 4 = 84 sq. cm
Area of quadrilateral ABCD
= 54 + 84 = 138 sq, cm
∴ Height of prism
= $\frac{\mathrm{Volume}}{\mathrm{Area}\mathrm{of}\mathrm{base}}$ = $\frac{2070}{138}$
= 15 cm
Perimeter of base
= (9 + 14 + 13 +12) cm = 48 cm
∴ Area of lateral surfaces = perimeter × height
= 48 × 15 = 720 sq. cm.