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 ΔDAB = 90°. If the volume of the prism be 2070 cm3, then the area of the lateral surface is

In right $$\triangle$$DAB, using Pythagoras theorem,

=> $$BD = \sqrt{(AD)^2 + (AB)^2}$$

=> $$BD = \sqrt{12^2 + 9^2} = \sqrt{225}$$

=> $$BD = 15 cm$$

Now, area of $$\triangle$$DAB = $$\frac{1}{2} * 9 * 12 = 54 cm^2$$

Area of $$\triangle$$BCD = $$\sqrt{s(s-a)(s-b)(s-c)}$$

where, $$s = \frac{a+b+c}{2}$$

=> Area of $$\triangle$$BCD = $$\sqrt{21 * 6 * 7 * 8} = 84 cm^2$$

=> Area of quad ABCD = area of $$\triangle$$DAB + area of $$\triangle$$BCD

= 54+84 = $$138 cm^2$$

Volume of prism = base area * height

=> 2070 = 138 * height

=> height = 15 cm

Lateral surface area of prism = perimeter of base * height

= (12 + 9 + 14 + 13) * 15 = 48 * 15

= 720 $$cm^2$$