Question

Find the area of a circle whose area equals the half of the sum of the areas of two circles with radii 2 and 3.

Answer 1

$A_O=\pi r^2\\\\r_1=2\to A_{O1}=\pi\cdot2^2=4\pi\\\\r_2=3\to A_{O2}=\pi\cdot3^2=9\pi\\\\A=\dfrac{1}{2}(A_{O1}+A_{O2})\\\\A=\dfrac{1}{2}(4\pi+9\pi)=\dfrac{1}{2}\cdot13\pi=\boxed{\frac{13\pi}{2}=6.5\pi}$

Answer 2

The required area of the circle is 6.5 π

Area of the Circle:

The area of the circle can be calculated as

Area = π $r^2$

where r is the radius of the circle.

How to calculate area of the circle?

Here we have given that two circles having 2 and 3 radii

Therefore their area will be

Area of first circle = π (2)^2 = 4π

Area of second circle = π (3)^2 = 9π

The area of required circle is = half of sum of these two areas

Area = $\frac{1}{2} (4\pi+9\pi)=\frac{13\pi}{2} =6.5\pi$

This is the conclusion to the answer.

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