The base of a right prism is a quadrilateral | vedclass

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(A) In right $\Delta DAB$,by the Pythagorean theorem,$BD = \sqrt{AD^2 + AB^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\, cm$.Area of $\

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$720$

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(A) In right $\Delta DAB$,by the Pythagorean theorem,$BD = \sqrt{AD^2 + AB^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\, cm$.Area of $\Delta DAB = \frac{1}{2} 	imes AB 	imes AD = \frac{1}{2} 	imes 9 	imes 12 = 54\, cm^2$.For $\Delta BCD$,the sides are $BC = 14\, cm$,$CD = 13\, cm$,and $BD = 15\, cm$. The semi-perimeter $S = \frac{14 + 13 + 15}{2} = \frac{42}{2} = 21\, cm$.Using Heron's formula,Area of $\Delta BCD = \sqrt{S(S-a)(S-b)(S-c)} = \sqrt{21(21-14)(21-13)(21-15)} = \sqrt{21 	imes 7 	imes 8 	imes 6} = \sqrt{7056} = 84\, cm^2$.Area of quadrilateral $ABCD = 	ext{Area of } \Delta DAB + 	ext{Area of } \Delta BCD = 54 + 84 = 138\, cm^2$.Volume of the prism = Base Area $	imes$ Height.$2070 = 138 	imes h \implies h = \frac{2070}{138} = 15\, cm$.Lateral Surface Area $(LSA)$ = Perimeter of base $	imes$ Height.Perimeter = $AB + BC + CD + DA = 9 + 14 + 13 + 12 = 48\, cm$.$LSA$ = $48 	imes 15 = 720\, cm^2$.

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 $\angle DAB = 90^{\circ}$. If the volume of the prism is $2070\, cm^3$,then the area of the lateral surface is........$cm^2$.

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