In a quadrilateral ABCD,AB = 29 , cm,BC = 35 | vedclass

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(D) The quadrilateral $ABCD$ is divided into two triangles,$	riangle ABC$ and $	riangle ADC$,by the diagonal $AC = 48 \, cm$.For $	riangle ABC$,the

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$840$

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(D) The quadrilateral $ABCD$ is divided into two triangles,$	riangle ABC$ and $	riangle ADC$,by the diagonal $AC = 48 \, cm$.For $	riangle ABC$,the sides are $a = 29 \, cm$,$b = 35 \, cm$,and $c = 48 \, cm$.The semi-perimeter $s_1 = (29 + 35 + 48) / 2 = 112 / 2 = 56 \, cm$.Area of $	riangle ABC = \sqrt{s_1(s_1-a)(s_1-b)(s_1-c)} = \sqrt{56(56-29)(56-35)(56-48)} = \sqrt{56 	imes 27 	imes 21 	imes 8} = \sqrt{254016} = 504 \, cm^2$.For $	riangle ADC$,the sides are $a = 50 \, cm$,$b = 14 \, cm$,and $c = 48 \, cm$.The semi-perimeter $s_2 = (50 + 14 + 48) / 2 = 112 / 2 = 56 \, cm$.Area of $	riangle ADC = \sqrt{s_2(s_2-a)(s_2-b)(s_2-c)} = \sqrt{56(56-50)(56-14)(56-48)} = \sqrt{56 	imes 6 	imes 42 	imes 8} = \sqrt{112896} = 336 \, cm^2$.Total area of quadrilateral $ABCD = 	ext{Area}(	riangle ABC) + 	ext{Area}(	riangle ADC) = 504 + 336 = 840 \, cm^2$.

In a quadrilateral $ABCD$,$AB = 29 \, cm$,$BC = 35 \, cm$,$CD = 14 \, cm$,$DA = 50 \, cm$ and $AC = 48 \, cm$. Find its area. (in $, cm^2$)

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