Two circles have the same radius. Complete the description for whether the combined area of the two circles is the same as the area of a circle with twice the radius.
The combined area of two circles with the same radius is __ ${πr}^{2}$ .
The area of a circle with twice the radius is ${πr}^{2}$ .
The combined area of two circles is as the area of a circle with twice the radius.

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Solution

Finding the relation between the combined area of two circles and the area of circle with twice the radius:
The formula for the area of circle having radius $r$ is ${πr}^{2}$ .
So the combined area of two circles is $2$ ${πr}^{2}$
The area of the circle with twice the radius:
$π {(2 r)}^{2} = 4 {πr}^{2}$
So, the area of a circle with twice the radius is $4$ ${πr}^{2}$ .
The combined area of two circles is half as the area of a circle with twice the radius.
Hence, The combined area of two circles with the same radius is $2 {πr}^{2}$ . The area of a circle with twice the radius is $4 {πr}^{2}$ . The combined area of two circles is $half$ as the area of a circle with twice the radius.

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Similar questions

If the sum of the areas of two circles with radii $R_{1}$ and $R_{2}$ is equal to the area of a circle of radius R then

(a) $R_{1} + R_{2} = R$

(b) $R_{1} + R_{2} > R$

(d) $R_{1}^{2} + R_{2}^{2} = R^{2}$

R_{1} + R_{2} = R

R_{1} + R_{2} < R

R_{1}^{2} + R_{2}^{2} < R^{2}

R_{1}^{2} + R_{2}^{2} = R^{2}

r_{1}

r_{2}

r

r = r_{1} + r_{2}

{r_{1}}^{2} + {r_{2}}^{2} = r^{2}

r_{1} + r_{2} < r

{r_{1}}^{2} + {r_{2}}^{2} < r^{2}

r_{1}^{2} + r_{2}^{2}

R_{1}

R_{2}

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