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Two identical quadrants were overlapped in a square of side 2 cm. The radius of the quadrants were 6 cm. Find the area of the figure. Take π as 3.14.

Question

Two identical quadrants were overlapped in a square of side 2 cm. The radius of the quadrants was 6 cm. Find the area of the figure. Take π \pi as 3.14.

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Solution

Sure, let's solve this step by step:

Step 1: First, let's find the area of the square. The formula for the area of a square is side^2. So, the area of the square is 2 cm * 2 cm = 4 cm^2.

Step 2: Now, let's find the area of one quadrant. The formula for the area of a circle is πr^2. Since a quadrant is a quarter of a circle, the area of a quadrant is (πr^2)/4. So, the area of one quadrant is (3.14 * 6 cm * 6 cm)/4 = 28.26 cm^2.

Step 3: Since there are two identical quadrants, the total area of the quadrants is 2 * 28.26 cm^2 = 56.52 cm^2.

Step 4: The quadrants are overlapped in the square, so the area of the figure is the area of the square plus the area of the quadrants. So, the area of the figure is 4 cm^2 + 56.52 cm^2 = 60.52 cm^2.

So, the area of the figure is 60.52 cm^2.

This problem has been solved

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