The base of a right prism is a quadrilateral ABCD, given that AB = 9 cm, BC = 14 cm, CD = 13 cm, DA = 12 cm and ∠DAB = 90°. If the volume of the prism be 2070 cm³, then the area of the lateral surface is

Solution (By Examveda Team)

$$\eqalign{ & {\text{In }}\Delta ABD, \cr & BD = \sqrt {A{B^2} + A{D^2}} \cr & = \sqrt {{9^2} + {{12}^2}} \cr & = \sqrt {81 + 144} \cr & = \sqrt {225} \cr & = 15\,{\text{cm}} \cr & {\text{Area of }}\Delta ABD = \frac{1}{2} \times AB \times AD \cr & = \frac{1}{2} \times 9 \times 12 \cr & = 54{\text{ c}}{{\text{m}}^2} \cr & {\text{In }}\Delta BCD \cr & {\text{Semiperimeter}} = \frac{{13 + 14 + 15}}{2} = \frac{{42}}{2} = 21 \cr & {\text{Area of }}\Delta BCD = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr & = \sqrt {21\left( {21 - 13} \right)\left( {21 - 14} \right)\left( {21 - 15} \right)} \cr & = \sqrt {21 \times 8 \times 7 \times 6} \cr & = 21 \times 4 \cr & = 84{\text{ c}}{{\text{m}}^2} \cr & {\text{Area }}ABCD = 84 + 54 = 138{\text{ c}}{{\text{m}}^2} \cr & {\text{Height of prism}} = \frac{{{\text{Volume}}}}{{{\text{Area of base}}}} \cr & = \frac{{2070}}{{138}} \cr & = 15{\text{ cm}} \cr & {\text{Perimeter of base}} = 9 + 14 + 13 + 12 = 48{\text{ cm}} \cr & {\text{Area of lateral surface}} = {\text{Perimeter}} \times {\text{Height}} \cr & = 48 \times 15 \cr & = 720{\text{ c}}{{\text{m}}^2} \cr} $$