Subject: General Aptitude · Chapter: Area · Exam: 2007 · Marks: · Difficulty:
The area of a square is twice that of a rectangle. The perimeter of the rectangle is 10 cm. If its length and breadth each is increased by 1 cm, the area of the rectangle becomes equal to the area of the square. The length of side of the square is
(a)$2 \sqrt{3} \mathrm{cm}$
(b)$3 \sqrt{2} \mathrm{cm}$
(c)$4 \sqrt{3} \mathrm{cm}$
(d)12 cm
Answer
Answer (as printed): A
Explanation
Let the length and breadth of the rectangle be $l \mathrm{cm}$ and $b$ cm respectively. Then, $2(l+b)=10 \Rightarrow l+b=5 \Rightarrow b=(5-l) \mathrm{cm}$. Area of the rectangle $=l(5-l) \mathrm{cm}^{2}=\left(5 l-l^{2}\right) \mathrm{cm}^{2}$. Area of the square $=2\left(5 l-l^{2}\right) \mathrm{cm}^{2}=\left(10 l-2 l^{2}\right) \mathrm{cm}^{2}$. $$\begin{aligned} & \therefore(l+1)(6-l)=\left(10 l-2 l^{2}\right) \Rightarrow l^{2}-5 l+6=0 \\ & \Rightarrow(l-3)(l-2)=0 \Rightarrow l=3 . \end{aligned}$$ Area of the square $=(10 \times 3-2 \times 9) \mathrm{cm}^{2}=12 \mathrm{cm}^{2}$. $$\text { ∴ } \text { Side of the square }=\sqrt{2} \mathrm{cm}=2 \sqrt{3} \mathrm{cm} .$$