In the given figure, PQRS is square whose side is 8 cm. PQS and QPR are two quadrants. A circle is placed touching both the quadrants and the square as shown in the figure. What is the area (in cm2) of the circle?
A. $$\frac{{13}}{{17}}$$
B. $$\frac{{11}}{{14}}$$
C. $$\frac{{19}}{{31}}$$
D. $$\frac{{15}}{{19}}$$
Answer: Option B
Solution (By Examveda Team)
Let radius of circle = r
∴ OA = 8 - r and OP = 8 + r
Now ∠OAP is right angle triangle and use pythagoras theorem in ΔOAP
⇒ (8 + r)2 = (8 - r)2 + 42
64 + r2 + 16r = 64 + r2 - 16r + 16
32r = 16
r = $$\frac{1}{2}$$cm
And area of circle = πr2
$$\eqalign{ & = \frac{{22}}{7} \times \frac{1}{2} \times \frac{1}{2} \cr & = \frac{{11}}{{14}}{\text{ c}}{{\text{m}}^2} \cr} $$
This Question Belongs to Arithmetic Ability >> Mensuration 2D
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