In a quadrilateral ABCD,the sides AB, BC, CD, | vedclass

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Question

(A) The area of quadrilateral $ABCD$ is the sum of the areas of $\Delta ABC$ and $\Delta ADC$.For $\Delta ABC$,the sides are $a = 20\, m, b = 13\, m,

Vedclass · Accepted Answer

$210$

Vedclass · Answer

(A) The area of quadrilateral $ABCD$ is the sum of the areas of $\Delta ABC$ and $\Delta ADC$.For $\Delta ABC$,the sides are $a = 20\, m, b = 13\, m, c = 21\, m$.Semi-perimeter $s_1 = \frac{20 + 13 + 21}{2} = \frac{54}{2} = 27\, m$.Area of $\Delta ABC = \sqrt{s_1(s_1 - a)(s_1 - b)(s_1 - c)} = \sqrt{27(27 - 20)(27 - 13)(27 - 21)} = \sqrt{27 	imes 7 	imes 14 	imes 6} = \sqrt{15876} = 126\, m^2$.For $\Delta ADC$,the sides are $a = 10\, m, b = 17\, m, c = 21\, m$.Semi-perimeter $s_2 = \frac{10 + 17 + 21}{2} = \frac{48}{2} = 24\, m$.Area of $\Delta ADC = \sqrt{s_2(s_2 - a)(s_2 - b)(s_2 - c)} = \sqrt{24(24 - 10)(24 - 17)(24 - 21)} = \sqrt{24 	imes 14 	imes 7 	imes 3} = \sqrt{7056} = 84\, m^2$.Total area of quadrilateral $ABCD = 126\, m^2 + 84\, m^2 = 210\, m^2$.

In a quadrilateral $ABCD$,the sides $AB, BC, CD, DA$ measure $20\, m, 13\, m, 17\, m$ and $10\, m$ respectively,and the diagonal $AC$ is $21\, m$. The area of the quadrilateral is ........ $m^2$.

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