In a quadrilateral ABCD,AB = 9 , cm,BC = 40 , | vedclass

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(A) $1$. Join $AC$ to divide the quadrilateral $ABCD$ into two triangles: $	riangle ABC$ and $	riangle ADC$.$2$. In right-angled $	riangle ABC$,by

Vedclass · Accepted Answer

$306$

Vedclass · Answer

(A) $1$. Join $AC$ to divide the quadrilateral $ABCD$ into two triangles: $	riangle ABC$ and $	riangle ADC$.$2$. In right-angled $	riangle ABC$,by Pythagoras theorem: $AC^2 = AB^2 + BC^2 = 9^2 + 40^2 = 81 + 1600 = 1681$. Thus,$AC = \sqrt{1681} = 41 \, cm$.$3$. Area of $	riangle ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 40 	imes 9 = 180 \, cm^2$.$4$. For $	riangle ADC$,the sides are $a = 41 \, cm$,$b = 28 \, cm$,and $c = 15 \, cm$. Semi-perimeter $s = \frac{41 + 28 + 15}{2} = \frac{84}{2} = 42 \, cm$.$5$. Area of $	riangle ADC = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{42(42-41)(42-28)(42-15)} = \sqrt{42 	imes 1 	imes 14 	imes 27} = \sqrt{(14 	imes 3) 	imes 14 	imes (9 	imes 3)} = \sqrt{14^2 	imes 3^2 	imes 3^2} = 14 	imes 3 	imes 3 = 126 \, cm^2$.$6$. Total area of quadrilateral $ABCD = 	ext{Area}(	riangle ABC) + 	ext{Area}(	riangle ADC) = 180 + 126 = 306 \, cm^2$.

In a quadrilateral $ABCD$,$AB = 9 \, cm$,$BC = 40 \, cm$,$CD = 28 \, cm$,$DA = 15 \, cm$,and $\angle B = 90^{\circ}$. Find its area. (in $, cm^2$)

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