Question

The adjacent sides of a parallelogram are 10 cm and 12 cm. The diagonal joining the ends of these sides is 14 cm. Its area is

• A. $48\sqrt{6}$ $c{m}^2$
• B. $48\sqrt{5}$ $c{m}^2$
• C. $48\sqrt{3}$ $c{m}^2$
• D. none of these

Answer 1

The correct option is A $48\sqrt{6}$ $c{m}^2$

Area of the parallelogram $ABCD=2\times \text{area}△BCD$

Now, in $△BCD$

$S=\frac{12+10+14}{2}=18\text{cm}$

Area of $△BCD=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{18(18-12)(18-10)(18-14)}$

$=\sqrt{18\times 6\times 8\times 4}$

$=\sqrt{3\times 6\times 6\times 2\times 4\times 4}$

$=4\times 6\sqrt{6}$

$=24\sqrt{6}{\text{cm}}^2$

$\therefore$ Area of the parallelogram $ABCD=2\times 24\sqrt{6}=48\sqrt{6}{\text{cm}}^2$