Question

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order) respectively and the angle between the first two sides is a right angle. Find its area.

• A. $18+24\sqrt{2}c{m}^2$
• B. $18+24\sqrt{6}c{m}^2$
• C. $36+24\sqrt{2}c{m}^2$
• D. $36+24\sqrt{6}c{m}^2$

Answer 1

The correct option is B $18+24\sqrt{6}c{m}^2$


Area of $△ABC=\frac{1}{2}\times \text{base}\times height$
$=\frac{1}{2}\times 6\times 8$
$=24c{m}^2$

Diagonal AC = $\sqrt{B{C}^2+A{B}^2}=10cm$

For $△ACD,$
$s=\frac{10+12+14}{2}=18$
Area $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{18(18-10)(18-12)(18-14)}$
$=\sqrt{18\times 8\times 6\times 4}$
$=24\sqrt{6}c{m}^2$

$\therefore$ area of quadrilateral
$=18+24\sqrt{6}c{m}^2$