A Circle is Inscribed in a Square with the Diagonal of 4 Centimeters.

Question: A circle is inscribed in a square with the diagonal of 4 centimeters. What is the approximation area of the square that is not occupied by the circle?

1. 1.7
2. 2.7
3. 3.4
4. 5.4
5. 8

“A circle is inscribed in a square with the diagonal of 4 centimeters. What is the approximation area of the square that is not occupied by the circle?”– is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.0

Answer

Approach Solution 1 

Area of the square is calculated as:

Area = \(\frac{diagonal^2}{2}\)

As given in the question, diagonal of square = 4 cm 

Therefore, Area of square = \(\frac{4^2}{2}\) = \(\frac{16}{2}\) = 8 \(cm^2\)

So to find the value of the side of the square, we will use the Pythagoras Theorem as:

\(4^2=a^2+a^2\)

16 = \(2^{a^2}\)

\(a^2\)= 8

a = \(2\sqrt{2}\)

As the value of diagonal of the square is 4 cm then the side of the square will be equal to \(2\sqrt{2}\) .

As the circle is inscribed in it, so the radius of the circle will be half the side of the square.

So, radius of the circle = \(\sqrt2\)

Area of the circle = \(\pi{r^2}\) = \(\pi({\sqrt{2})^2}\) = \(2\pi\)

So the approximate area of the square that is not occupied by the circle = 8 – \(2\pi\) = 8 – 6.28 = 1.72

Correct option: A

Approach Solution 2

Side of the square = \(\frac{4}{\sqrt2}\)

So, the area of the square = \( (\frac{4}{\sqrt2})^2\) = 8

Area of the circle = \(\pi*(\frac{4}{2\sqrt2})^2\) = \(4\pi\)

So the approximate area of the square that is not occupied by the circle = \(8-4\pi\) = 8 – 4 (> 1.5) = 8 – (> 6)

Only answer choice A meets the substruction requirement to be less than 2, which is 1.7

Correct option: A

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