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Find the area of a quadrilateral $ABCD$ in which $AB = 42\ cm, BC = 21\ cm, CD = 29\ cm, DA = 34\ cm$ and diagonal $BD = 20\ cm$.
Given:
A quadrilateral $ABCD$ in which $AB = 42\ cm, BC = 21\ cm, CD = 29\ cm, DA = 34\ cm$ and diagonal $BD = 20\ cm$.
To do:
We have to find the area of the quadrilateral.
Solution:
Area of $\triangle \mathrm{ABD}$,
$s=\frac{a+b+c}{2}$
$=\frac{42+20+34}{2}$
$=\frac{96}{2}$
$=48$
Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{48(48-42)(48-20)(48-34)}$
$=\sqrt{48 \times 6 \times 28 \times 14}$
$=\sqrt{4 \times 4 \times 3 \times 3 \times 2 \times 2 \times 14 \times 14}$
$=4 \times 3 \times 2 \times 14$
$=336 \mathrm{~cm}^{2}$
Area of triangle $BCD$,
$s=\frac{a+b+c}{2}$
$=\frac{20+21+29}{2}$
$=\frac{70}{2}$
$=35$
Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{35(35-20)(35-21)(35-29)}$
$=\sqrt{35 \times 15 \times 14 \times 6}$
$=\sqrt{7 \times 5 \times 5 \times 3 \times 3 \times 2 \times 2 \times 7}$
$=7 \times 5 \times 3 \times 2$
$=210 \mathrm{~cm}^{2}$
Area of quadrilateral $\mathrm{ABCD} =336+210$
$=546 \mathrm{~cm}^{2}$.