Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
A parallelogram has sides 30 m and 14 m and one of its diagonals is 40 m long. Then, its area is
(a)$168 \mathrm{m}^{2}$
(b)$336 \mathrm{m}^{2}$
(c)$372 \mathrm{m}^{2}$
(d)$480 \mathrm{m}^{2}$
Answer
Answer (as printed): B
Explanation
Let $A B C D$ be the given $\| \mathrm{gm}$. Area of $\| \mathrm{gm} A B C D=2 \times$ (area of $\triangle A B C$ ). Now, $a=30 \mathrm{m}, b=14 \mathrm{m}, c=40 \mathrm{m}$. $$\therefore \quad s=\frac{1}{2}(30+14+40) \mathrm{m}=42 \mathrm{m} . \begin{aligned} \therefore \quad \text { Area of } \triangle \mathrm{ABC} & =\sqrt{s(s-a)(s-b)(s-c)} \\ & =\sqrt{42 \times 12 \times 28 \times 2} \mathrm{m}^{2}=168 \mathrm{m}^{2} . \end{aligned}$$ Hence, area of $\| \mathrm{gm} A B C D=(2 \times 168) \mathrm{m}^{2}=336 \mathrm{m}^{2}$.