In quadrilateral ABCD,AB = 9 , cm,BC = 13 , c | vedclass

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Question

(C) The quadrilateral $ABCD$ is divided into two triangles: $	riangle ABD$ and $	riangle BCD$.For $	riangle ABD$,the sides are $a = 9 \, cm$,$b = 1

Vedclass · Accepted Answer

$138$

Vedclass · Answer

(C) The quadrilateral $ABCD$ is divided into two triangles: $	riangle ABD$ and $	riangle BCD$.For $	riangle ABD$,the sides are $a = 9 \, cm$,$b = 12 \, cm$,and $c = 15 \, cm$.The semi-perimeter $s_1 = (9 + 12 + 15) / 2 = 36 / 2 = 18 \, cm$.Area of $	riangle ABD = \sqrt{s_1(s_1-a)(s_1-b)(s_1-c)} = \sqrt{18(18-9)(18-12)(18-15)} = \sqrt{18 	imes 9 	imes 6 	imes 3} = \sqrt{2916} = 54 \, cm^2$.For $	riangle BCD$,the sides are $a = 13 \, cm$,$b = 14 \, cm$,and $c = 15 \, cm$.The semi-perimeter $s_2 = (13 + 14 + 15) / 2 = 42 / 2 = 21 \, cm$.Area of $	riangle BCD = \sqrt{s_2(s_2-a)(s_2-b)(s_2-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 	imes 8 	imes 7 	imes 6} = \sqrt{7056} = 84 \, cm^2$.Total area of quadrilateral $ABCD = 54 + 84 = 138 \, cm^2$.

In quadrilateral $ABCD$,$AB = 9 \, cm$,$BC = 13 \, cm$,$CD = 14 \, cm$,$DA = 12 \, cm$,and $BD = 15 \, cm$. Find the area of quadrilateral $ABCD$ in $cm^2$.

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