Question

In the given figure, if AB = 32 cm, BC = 30 cm, AD = 10 cm and DC = 24 cm, then the area of ABCD is equal to

• A. $(24\sqrt{231}+5)sq.cm$
• B. $24(\sqrt{231}+5)sq.cm$
• C. $24(\sqrt{231}-5)sq.cm$
• D. $34(\sqrt{231}-5)sq.cm$

Answer 1

The correct option is C $24(\sqrt{231}-5)sq.cm$

Given: Reflex $\angle ADC={270}^0$
$\Rightarrow \angle ADC={90}^0$
$\therefore \Delta ADC$ is a right-angled triangle.
$\Rightarrow AC=\sqrt{{24}^2+{10}^2}$ (By pythagoras theorem)
$=\sqrt{576+100}$
$=26cm$

$\therefore$ Area of $\Delta ADC=\frac{1}{2}\times CD\times AD$
$=\frac{1}{2}\times 24\times 10$
$=120c{m}^2$
Now, consider ΔABC with sides AB = 32 cm, BC = 30 cm and AC = 26 cm.
$\therefore$ Semi-perimeter,(s) = $\frac{32+30+26}{2}=44cm$
$\therefore$ Area of $\Delta ABC=\sqrt{44(44-32)(44-30)(44-26)}$
$=\sqrt{44\times 12\times 14\times 18}$
$=\sqrt{11\times 4\times 4\times 3\times 7\times 2\times 9\times 2}$
$=4\times 2\times 3\times \sqrt{11\times 3\times 7}$
$=24\sqrt{231}c{m}^2$ .....(ii)
$\therefore$ Area of ABCD = Area of $\Delta ABC$ – Area of $\Delta ADC$
$=24\sqrt{231}-120$ [From (i) and (ii)]
$=24(\sqrt{231}-5)c{m}^2$
Hence, the correct answer is option (c).