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 $\mathrm{ABCD} =336+210$

$=546 \mathrm{~cm}^{2}$.