Question : 
In the given figure, the radius of a circle is $14\sqrt{2}$ cm. PQRS is a square. EFGH, ABCD, WXYZ, and LMNO are four identical squares. What is the total area (in cm2) of all the small squares?
Option 1: 31.36
Option 2: 125.44
Option 3: 62.72
Option 4: 156.8
Correct Answer: 125.44
Solution :
Given that the radius of the circle = $14\sqrt{2}$ cm
The diameter of the circle = $2 \times 14\sqrt{2} = 28\sqrt{2}$ cm.
The side of the larger square (PQRS) = $\frac{\operatorname{diameter}}{\sqrt2}=28$ cm.
TO = half the side of the larger square = $14$ cm.
The radius (OX = OU) of the circle = $14\sqrt{2}$ cm
Let the V as the midpoint of the side WX of one of the smaller squares.
Let the side of this smaller square = $a$ cm
⇒ XY = WX = VT = $a$ cm
⇒ VX = $\frac{a}{2}$ cm
⇒ VO = VT + TO = $14 + a$.
Using the Pythagorean theorem in triangle OVX,
⇒ OX
2
= VX
2
+ VO
2
⇒ $(14\sqrt{2})^2 = (\frac{a}{2})^2 + (14 + a)^2$
⇒ $ 392 = \frac{a^2}{4} + a^2 + 28a + 196 $
Multiply the entire equation by 4
$⇒ 1568 = a^2 + 4a^2 + 112a + 784 $
$⇒ 5a^2+112a-784=0$
$⇒ 5a^2+140a-28a-784=0$
$⇒ 5a(a+28)-28(a+28)=0$
$⇒ (a+28)(5a-28)=0$
$⇒ (5a-28)=0$
$⇒ a = 5.6$ cm
The total area of all the small squares = $4 \times (5.6)^2 = 125.44$ cm
2
Hence, the correct answer is 125.44.
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