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

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.

 

Solution:

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

$s=\frac{1}{2}(a+b+c)$

$s=\frac{42+20+34}{2}$

$s=48 \mathrm{~cm}$

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

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

$=\sqrt{112896}$

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

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

$s=\frac{1}{2}(a+b+c)$

$s=\frac{21+20+29}{2}$

$s=35 \mathrm{~cm}$

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

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

$=\sqrt{44100}$

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

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

$=336+210$

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

 

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