Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
Four equilateral triangles are described on the four sides of a rectangle with perimeter 12 cm. If the sum of the areas of the four triangles is $10 \sqrt{3} \mathrm{cm}^{2}$, what is the area of the rectangle?
(a)$5 \mathrm{cm}^{2}$
(b)$8 \mathrm{cm}^{2}$
(c)$9 \mathrm{cm}^{2}$
(d)$6.75 \mathrm{cm}^{2}$
Answer
Answer (as printed): B
Explanation
Let the length and breadth of the rectangle be $l$ cm and $b$ cm respectively. Then, $2(l+b)=12$ or $l+b=6$ or $b=(6-l)$. Sum of areas of the four triangles $$\begin{aligned} & \quad=\frac{\sqrt{3}}{4}\left[2 l^{2}+2(6-l)^{2}\right]=\frac{\sqrt{3}}{4}\left(4 l^{2}-24 l+72\right) \\ & \quad=\sqrt{3}\left(l^{2}-6 l+18\right) . \\ & \therefore \quad \sqrt{3}\left(l^{2}-6 l+18\right)=10 \sqrt{3} \Rightarrow l^{2}-6 l+18=10 \\ & \Rightarrow l^{2}-6 l+8=0 \Rightarrow(l-4)(1-2)=0 \\ & \Rightarrow l=4 \text { or } l=2 . \end{aligned}$$ Hence, length $=4 \mathrm{cm}$, breadth = 2 cm. Area of rectangle $=(4 \times 2) \mathrm{cm}^{2}=8 \mathrm{cm}^{2}$.