Question

In the adjoining 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 circles to that of the rectangle?

• A. $3\pi :7$
• B. $3\pi :4$
• C. $25\pi :72$
• D. $32\pi :115$

Answer 1

The correct option is C $25\pi :72$


Let AD = 3a and DC = 6a
$\therefore DH=HG=GC=\frac{6a}{3}=2aHM=MG=\frac{2a}{a}=a=SM\therefore NQ=a\therefore SQ=SM+MN+NQ$
= a + 3a + a = 5a
Since diagonal of square SQ = 5a
But, diameter of circle SQ = diagonal of square SQ
$\therefore$ Radius of the circle $=\frac{5a}{2}$
$\Rightarrow$ Area of the circle $=\pi \times {\left(\frac{5a}{2}\right)}^2$
$Hence,\frac{Areaofcircle}{Areaofrectangle}=\frac{\frac{25}{4}\left(a^2\pi \right)}{3a\times 6a}=\frac{25\pi }{72}$