In quadrilateral ABCD,AB = 40 , cm,BC = 35 , | vedclass

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(D) The diagonal $BD$ of quadrilateral $ABCD$ divides it into two non-overlapping triangles,$\Delta ABD$ and $\Delta BCD$.For $\Delta ABD$,the sides a

Vedclass · Accepted Answer

$924$

Vedclass · Answer

(D) The diagonal $BD$ of quadrilateral $ABCD$ divides it into two non-overlapping triangles,$\Delta ABD$ and $\Delta BCD$.For $\Delta ABD$,the sides are $a = 40 \, cm$,$b = 29 \, cm$,and $c = 29 \, cm$.The semi-perimeter $s_1 = \frac{40 + 29 + 29}{2} = \frac{98}{2} = 49 \, cm$.Area of $\Delta ABD = \sqrt{s_1(s_1 - a)(s_1 - b)(s_1 - c)} = \sqrt{49(49 - 40)(49 - 29)(49 - 29)} = \sqrt{49 	imes 9 	imes 20 	imes 20} = 7 	imes 3 	imes 20 = 420 \, cm^2$.For $\Delta BCD$,the sides are $a = 29 \, cm$,$b = 48 \, cm$,and $c = 35 \, cm$.The semi-perimeter $s_2 = \frac{29 + 48 + 35}{2} = \frac{112}{2} = 56 \, cm$.Area of $\Delta BCD = \sqrt{s_2(s_2 - a)(s_2 - b)(s_2 - c)} = \sqrt{56(56 - 29)(56 - 48)(56 - 35)} = \sqrt{56 	imes 27 	imes 8 	imes 21} = \sqrt{(8 	imes 7) 	imes (9 	imes 3) 	imes 8 	imes (3 	imes 7)} = \sqrt{8^2 	imes 7^2 	imes 3^2 	imes 9} = 8 	imes 7 	imes 3 	imes 3 = 504 \, cm^2$.Area of quadrilateral $ABCD = 	ext{Area of } \Delta ABD + 	ext{Area of } \Delta BCD = 420 + 504 = 924 \, cm^2$.

In quadrilateral $ABCD$,$AB = 40 \, cm$,$BC = 35 \, cm$,$CD = 48 \, cm$,$DA = 29 \, cm$ and $BD = 29 \, cm$. Find the area of quadrilateral $ABCD$. (in $, cm^2$)

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