Question

Find the area of the quadrilateral ABCD in which AB = 42 cm, BC = 12 cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.

Answer 1

$\text{In}△ABDs=\frac{a+b+c}{2}=\frac{42+20+34}{2}=48cm\text{Area of}△ABD=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{48(48-42)(48-20)(48-34)}=\sqrt{48\times 6\times 28\times 14}=\sqrt{112896}=336c{m}^2$

$\text{In}△BDCs=\frac{a+b+c}{2}=\frac{21+20+29}{2}=35cm\text{Area of}△BDC=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{35(35-29)(35-20)(35-21)}=\sqrt{35\times 6\times 15\times 14}=\sqrt{44100}=210c{m}^2$

Area of quadrilateral ABCD = Area of ΔABD + Area of ΔBDC $=336+210=546c{m}^2$