Question

As shown in the figure, there is a square of side 24 cm. A circle is inscribed inside the square. Inside the circle are four circles of equal radius which are inscribed.

The total area of the shaded region in the figure given below is ________

• A. \(576 - 196\pi\)
• B. \(584 - 196\pi\)
• C. \(864 - 196\pi\)
• D. None of the above

Hint:

Break the shaded area into square minus big circle plus small circles minus their overlaps.

Answer 1

The Correct Option is D

To solve this problem, let's go through the steps one by one: 

1. Firstly, we are given a square of side 24 cm. The area of the square is calculated as follows: \(Area_{\text{Square}} = \text{side}^2 = 24^2 = 576 \, \text{cm}^2\).
2. Inside this square, a circle is inscribed. The diameter of this circle is equal to the side of the square, so the radius of the circle is: \(r = \frac{24}{2} = 12 \, \text{cm}\). The area of the inscribed circle is: \(Area_{\text{Large Circle}} = \pi \times r^2 = \pi \times 12^2 = 144\pi \, \text{cm}^2\).
3. Inside the larger circle, there are four smaller equal circles inscribed. Each of these smaller circles touches each other and the larger circle, forming a square within the large circle. The side of the square that fits exactly inside the large circle is equal to the diagonal of the square formed by the centers of these four circles. This diagonal equals the diameter of the large circle, which is 24 cm.
4. Given that the diagonal of the square formed by the centers of four smaller circles is \(24 \, \text{cm}\), the side of this square is \(a\), where: \(a\sqrt{2} = 24\). Therefore, the side \(a = \frac{24}{\sqrt{2}} = 12\sqrt{2} \, \text{cm}\).
5. Each of the smaller circles has a radius equal to half the side of the square: \(r_{\text{small}} = \frac{12\sqrt{2}}{2} = 6\sqrt{2} \, \text{cm}\). Thus, the area of one smaller circle is: \(Area_{\text{Small Circle}} = \pi \times (6\sqrt{2})^2 = 72\pi \, \text{cm}^2\). The total area of the four smaller circles is: \(4 \times 72\pi = 288\pi \, \text{cm}^2\).
6. Now, to find the shaded area, subtract the area of the four smaller circles from the area of the large circle: \(Area_{\text{Shaded}} = 144\pi - 288\pi = 576 - 288\pi \, \text{cm}^2\).
7. None of the given options match this result, so the correct answer is "None of the above."

Answer 2

Step 1: Look at the figure and note the given sizes.
The outer square has side 24 cm. A circle is inscribed in the square, so its diameter is 24 cm and its radius is 12 cm. Inside that big circle, four equal small circles are inscribed in a 2 by 2 pattern, each touching the big circle and its neighbors, so each small circle has radius 6 cm.

Step 2: Work out the plain areas of the square and the two sizes of circle.
Area of the square = \(24 \times 24 = 576\) sq cm. Area of the big circle = \(\pi (12)^2 = 144\pi\) sq cm. Area of one small circle = \(\pi (6)^2 = 36\pi\) sq cm, so all four small circles together = \(4 \times 36\pi = 144\pi\) sq cm.

Step 3: Work out the lens shaped overlap between two neighboring small circles, then scale up to all 4 pairs.
Where two adjacent small circles of radius 6 cm cross, joining the two centers and the two crossing points makes a right triangle with legs 6 and 6. A quarter of one small circle has area \(\frac{1}{4}(36\pi)=9\pi\), and that right triangle has area \(\frac{1}{2}(6\times6)=18\), so half of the lens shape is \(9\pi-18\), and the full lens (the overlap area of one adjacent pair) is \(2(9\pi-18)=18\pi-36\). Around the 2 by 2 arrangement there are 4 such neighboring pairs, so the total overlap correction is \(4(18\pi-36)=72\pi-144\).

Step 4: Combine the pieces, then compare with the options.
Shaded area = (square) minus (big circle) plus (all 4 small circles) minus (the overlaps, which get counted twice): \(576 - 144\pi + 144\pi - (72\pi-144) = 576 + 144 - 72\pi = 720 - 72\pi\). This does not equal \(576-196\pi\), \(584-196\pi\) or \(864-196\pi\): neither the constant term nor the \(\pi\) coefficient matches any of them.

Final Answer:
The shaded area works out to \(720-72\pi\) sq cm, which is not listed, so option D, none of the above, is correct. \[ \boxed{720 - 72\pi \Rightarrow \text{None of the above}} \]