Find the area of a quadrilateral ABCD in whic | vedclass

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Question

(A) The quadrilateral $ABCD$ is divided into two triangles,$\Delta ABC$ and $\Delta ACD$,by the diagonal $AC$.For $\Delta ABC$:The sides are $a = 3 \,

Vedclass · Accepted Answer

$15.2$

Vedclass · Answer

(A) The quadrilateral $ABCD$ is divided into two triangles,$\Delta ABC$ and $\Delta ACD$,by the diagonal $AC$.For $\Delta ABC$:The sides are $a = 3 \, cm, b = 4 \, cm, c = 5 \, cm$.Semi-perimeter $s_1 = \frac{3 + 4 + 5}{2} = 6 \, cm$.Area of $\Delta ABC = \sqrt{s_1(s_1 - a)(s_1 - b)(s_1 - c)} = \sqrt{6(6 - 3)(6 - 4)(6 - 5)} = \sqrt{6 	imes 3 	imes 2 	imes 1} = \sqrt{36} = 6 \, cm^2$.For $\Delta ACD$:The sides are $a = 5 \, cm, b = 4 \, cm, c = 5 \, cm$.Semi-perimeter $s_2 = \frac{5 + 4 + 5}{2} = 7 \, cm$.Area of $\Delta ACD = \sqrt{s_2(s_2 - a)(s_2 - b)(s_2 - c)} = \sqrt{7(7 - 5)(7 - 4)(7 - 5)} = \sqrt{7 	imes 2 	imes 3 	imes 2} = \sqrt{84} = 2\sqrt{21} \, cm^2$.Using $\sqrt{21} \approx 4.58$,Area $\approx 2 	imes 4.58 = 9.16 \, cm^2$.Total area of quadrilateral $ABCD = 	ext{Area}(\Delta ABC) + 	ext{Area}(\Delta ACD) = 6 + 9.16 = 15.16 \, cm^2 \approx 15.2 \, cm^2$.

Find the area of a quadrilateral $ABCD$ in which $AB = 3 \, cm, BC = 4 \, cm, CD = 4 \, cm, DA = 5 \, cm$ and $AC = 5 \, cm$. (in $, cm^2$)

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