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.