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 circle to that of rectangle?

• A. $3\pi :7$
• B. 3 : 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 = $\frac{6a}{3}$ = 2a
HM = MG $=\frac{2a}{2}$ = a = SM
$\therefore$ NQ = a (also)
$\therefore$ 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 $=\pi \times {\left(\frac{5a}{2}\right)}^2$
Here $\frac{Areaofcircle}{Areaofrectangle}=\frac{\frac{25}{4}\left(a^2\pi \right)}{3a\times 6a}=\frac{25\pi }{72}$