Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
One diagonal of a parallelogram is 70 cm and the perpendicular distance of this diagonal from either of the outlying vertices is 27 cm. The area of the parallelogram (in sq. cm) is
(a)1800
(b)1836
(c)1890
(d)1980
Answer
Answer (as printed): C
Explanation
Let $A B C D$ be the given $\| \mathrm{gm}$. Let $A C=70 \mathrm{cm}$. Draw $B L \perp A C$ and $D M \perp A C$. Then, $D M=B L=27 \mathrm{cm}$. Area of $\| \mathrm{gm} \mathrm{ACBD}=\operatorname{ar}(\triangle \mathrm{ABC})+\operatorname{ar}(\triangle A C D)$ $$=\left[\left(\frac{1}{2} \times 70 \times 27\right)+\left(\frac{1}{2} \times 70 \times 27\right)\right] \text { sq. } \mathrm{cm}=1890 \text { sq. } \mathrm{cm} .$$