In a kite-shaped quadrilateral ABCD,AB = AD = | vedclass

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Question

(A) kite $ABCD$ is composed of two triangles,$	riangle ABC$ and $	riangle ADC$,joined at the diagonal $AC$. Since $AB = AD = 12 \, cm$ and $CB = CD

Vedclass · Accepted Answer

$180 \, cm^2, 14.4 \, cm$

Vedclass · Answer

(A) kite $ABCD$ is composed of two triangles,$	riangle ABC$ and $	riangle ADC$,joined at the diagonal $AC$. Since $AB = AD = 12 \, cm$ and $CB = CD = 17 \, cm$,$	riangle ABC \cong 	riangle ADC$ by $SSS$ congruence. For $	riangle ABC$,sides are $a = 12 \, cm, b = 17 \, cm, c = 25 \, cm$. Semi-perimeter $s = (12 + 17 + 25) / 2 = 54 / 2 = 27 \, cm$. Area of $	riangle ABC = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{27(27-12)(27-17)(27-25)} = \sqrt{27 	imes 15 	imes 10 	imes 2} = \sqrt{8100} = 90 \, cm^2$. Total area of kite $ABCD = 2 	imes 90 = 180 \, cm^2$. Let $BD$ intersect $AC$ at $O$. $AC \perp BD$. Area of $	riangle ABC = (1/2) 	imes AC 	imes BO = 90$. $(1/2) 	imes 25 	imes BO = 90 \implies BO = 180 / 25 = 7.2 \, cm$. Since $BD = 2 	imes BO$,$BD = 2 	imes 7.2 = 14.4 \, cm$.

In a kite-shaped quadrilateral $ABCD$,$AB = AD = 12 \, cm$,$CB = CD = 17 \, cm$,and $AC = 25 \, cm$. Find its area and the length of diagonal $BD$.

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