In the given figure ABCD is a rectangle in which length is twice of breadth. H and G divide the line CD into three equal parts. Similarly points E and F trisect the line AB. A circle PQRS is circumscribed by a square PQRS which passes through the points E, F, G and H. What is the ratio of areas of circle to that of rectangle? |
A) \[3\pi :7\]
B) \[3:4\]
C) \[25\pi :72\]
D) \[32\pi :115\]
Correct Answer: C
Solution :
Let \[AD=3a\]and \[DC=6a\] |
\[\therefore \] \[DH=HG=GC=\frac{6a}{3}=2a\] |
\[HM=MG=\frac{2a}{2}=a=SM\] |
\[NQ=a\](also) |
and \[SQ=SM+MN+NQ\] |
\[=a+3a+a=5a\] |
Since, diagonal of square, \[SQ=5a\] |
Diameter of circle, SQ = Diagonal of square, SQ |
Radius of the circle \[=\frac{5a}{2}\] |
Area of the circle \[=\pi \times {{\left( \frac{5a}{2} \right)}^{2}}\] |
\[\therefore \]\[\frac{\text{Area}\,\,\text{of}\,\,\text{circle}}{\text{Area}\,\,\text{of}\,\,\text{rectangle}}=\frac{25/4\,\,({{a}^{2}}\pi )}{3a\times 6a}=\frac{25\pi }{72}\] |
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