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

Answer: Option A

Solution

Answer: Option A

Solution:

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