In quadrilateral ABCD,AB = 20 , cm,BC = 15 , | vedclass

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(A) $1$. Divide the quadrilateral $ABCD$ into two triangles by drawing diagonal $AC$.$2$. In $	riangle ABC$,since $\angle B = 90^{\circ}$,it is a rig

Vedclass · Accepted Answer

$240$

Vedclass · Answer

(A) $1$. Divide the quadrilateral $ABCD$ into two triangles by drawing diagonal $AC$.$2$. In $	riangle ABC$,since $\angle B = 90^{\circ}$,it is a right-angled triangle.$3$. Area of $	riangle ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 20 	imes 15 = 150 \, cm^2$.$4$. Using Pythagoras theorem in $	riangle ABC$,$AC^2 = AB^2 + BC^2 = 20^2 + 15^2 = 400 + 225 = 625$. Thus,$AC = 25 \, cm$.$5$. Now,in $	riangle ADC$,the sides are $a = 25 \, cm$,$b = 12 \, cm$,$c = 17 \, cm$. The semi-perimeter $s = \frac{25 + 12 + 17}{2} = \frac{54}{2} = 27 \, cm$.$6$. Area of $	riangle ADC = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{27(27-25)(27-12)(27-17)} = \sqrt{27 	imes 2 	imes 15 	imes 10} = \sqrt{8100} = 90 \, cm^2$.$7$. Total area of quadrilateral $ABCD = 	ext{Area}(	riangle ABC) + 	ext{Area}(	riangle ADC) = 150 + 90 = 240 \, cm^2$.

In quadrilateral $ABCD$,$AB = 20 \, cm$,$BC = 15 \, cm$,$CD = 12 \, cm$,$DA = 17 \, cm$ and $\angle B = 90^{\circ}$. Find the area of the quadrilateral in $cm^2$.

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