Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
A square $S_{1}$ encloses another square $S_{2}$ in such a manner that each corner of $S_{2}$ is at the mid-point of the side of $S_{1}$. If $A_{1}$ is the area of $S_{1}$ and $A_{2}$ is the area of $S_{2}$, then
(a)$A_{1}=4 A_{2}$
(b)$A_{1}=2 A_{2}$
(c)$A_{2}=2 A_{1}$
(d)$A_{1}=A_{2}$
Answer
Answer (as printed): B
Explanation
Let $A B C D$ be the square $S_{1}$ and $E F G H$ be the square $\mathrm{S}_{2}$. Let the length of each side of $S_{1}$ be $a$. Then, $A F=A G=\frac{a}{2}$. So $F G=\sqrt{(A F)^{2}+(A G)^{2}}=\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{a}{2}\right)^{2}}=\sqrt{\frac{2 a^{2}}{4}}=\frac{a}{\sqrt{2}}$. Length of the side of $S_{2}=\frac{a}{\sqrt{2}}$. $$\therefore \quad \frac{A_{1}}{A_{2}}=\frac{a^{2}}{\left(\frac{a}{\sqrt{2}}\right)^{2}}=2 \quad \text { or } \quad \mathrm{A}_{1}=2 A_{2} \text {. }$$
Explanation as extracted from the printed page; notation may be imperfect.