question_answer

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 \[=\pi \times {{\left( \frac{5a}{2} \right)}^{2}}\]

\[\therefore \]\[\frac{\text{Area}\,\,\text{of}\,\,\text{circle}}{\text{Area}\,\,\text{of}\,\,\text{rectangle}}=\frac{25/4\,\,({{a}^{2}}\pi )}{3a\times 6a}=\frac{25\pi }{72}\]