Find the area of the quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm,

Find the area of the quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.



Area of $\triangle A B D=\sqrt{s(s-a)(s-b)(s-c)}$



$s=48 \mathrm{~cm}$

Area of $\triangle A B D=\sqrt{48(48-42)(48-20)(48-34)}$

$=\sqrt{48 \times 6 \times 28 \times 14}$


$=336 \mathrm{~cm}^{2}$

Area of $\triangle B D C=\sqrt{s(s-a)(s-b)(s-c)}$



$s=35 \mathrm{~cm}$

Area of $\Delta B D C=\sqrt{35(35-29)(35-20)(35-21)}$

$=\sqrt{35 \times 6 \times 15 \times 14}$


$=210 \mathrm{~cm}^{2}$

$\therefore$ Area of quadrilateral $A B C D=$ Area of $\triangle A B D+$ Area of $\triangle B D C$


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



Leave a comment

Please enter comment.
Please enter your name.