The sides of a quadrilateral ABCD are 6 , cm, | vedclass

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(B) To find the area of quadrilateral $ABCD$,we divide it into two triangles: $\Delta ABC$ and $\Delta ACD$.Since $\angle B = 90^\circ$,$\Delta ABC$ i

Vedclass · Accepted Answer

$82.787$

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(B) To find the area of quadrilateral $ABCD$,we divide it into two triangles: $\Delta ABC$ and $\Delta ACD$.Since $\angle B = 90^\circ$,$\Delta ABC$ is a right-angled triangle.By Pythagoras theorem in $\Delta ABC$:$AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, cm$.Area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 6 	imes 8 = 24 \, cm^2$.Now,for $\Delta ACD$,the sides are $a = 10 \, cm$,$b = 12 \, cm$,and $c = 14 \, cm$.The semi-perimeter $s = \frac{a + b + c}{2} = \frac{10 + 12 + 14}{2} = \frac{36}{2} = 18 \, cm$.Using Heron's formula,Area of $\Delta ACD = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{18(18-10)(18-12)(18-14)} = \sqrt{18 	imes 8 	imes 6 	imes 4} = \sqrt{3456} \approx 58.787 \, cm^2$.Total area of quadrilateral $ABCD = 	ext{Area}(\Delta ABC) + 	ext{Area}(\Delta ACD) = 24 + 58.787 = 82.787 \, cm^2$.

The sides of a quadrilateral $ABCD$ are $6 \, cm$,$8 \, cm$,$12 \, cm$,and $14 \, cm$ (taken in order) respectively,and the angle between the first two sides is a right angle. Find its area in $cm^2$.

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