| In the given figure, a square is inscribed in a circle of diameter d and another square is circumscribing the circle. |
| Then, the ratio between area of inner square to outer square. |
 |
A) 2:1
B) 1:4
C) 1:3
D) 1:2
Correct Answer: D
Solution :
Given diameter of a circle is d. \[\therefore \]Diagonal of a inner square = Diameter of a circle = d Let side of a inner square EFGH be x. In right angled \[\Delta EFG\], \[E{{G}^{2}}=E{{F}^{2}}+F{{G}^{2}}\] [by Pythagoras theorem] \[\Rightarrow \,\,\,{{d}^{2}}={{x}^{2}}+{{x}^{2}}\] \[\Rightarrow \,\,\,{{d}^{2}}=2{{x}^{2}}\Rightarrow {{x}^{2}}=\frac{{{d}^{2}}}{2}\] \[\therefore \]Area of a inner square \[EFGH={{\left( Side \right)}^{2}}={{x}^{2}}=\frac{{{d}^{2}}}{2}\] But side of a outer square ABCD = Diameter of a circle = d \[\therefore \]Area of outer square \[={{d}^{2}}\] \[Ratio=\frac{Area\text{ }of\text{ }inner\text{ }square}{Area\,of\,outer\,squarq}\] \[=\frac{{{d}^{2}}}{2\times {{d}^{2}}}=1:2\]
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