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# geometry

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Three semicircles (with equal radii) are drawn inside the large semicircle so that their diameters all sit on the diameter of the large semicircle. What is the ratio of the red area to the blue area?

May 14, 2020

#1
+1015
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The radius of the blue circle is 2+1 = 3 times the radius of the red, since the total diameter is 6r, divide 2 = 3.

πr^2 is the area of a red, divided by 2, and multiplied by 3 for the 3, so $$\frac{3πr^2}{2}$$.

The radius of the blue is 3r, so (3r)^2 * π /2 is $$\frac{9πr^2}{2}$$.

Since they are both divided by two, that is of little importance. They are all (number) times r^2 π, so r and π doesn't matter. The red is 3rπ, second is 9rπ, so  The ratio is 3:9 = 1:3, red:blue

If you don't understand anything feel free to ask!

May 14, 2020

#1
+1015
0

The radius of the blue circle is 2+1 = 3 times the radius of the red, since the total diameter is 6r, divide 2 = 3.

πr^2 is the area of a red, divided by 2, and multiplied by 3 for the 3, so $$\frac{3πr^2}{2}$$.

The radius of the blue is 3r, so (3r)^2 * π /2 is $$\frac{9πr^2}{2}$$.

Since they are both divided by two, that is of little importance. They are all (number) times r^2 π, so r and π doesn't matter. The red is 3rπ, second is 9rπ, so  The ratio is 3:9 = 1:3, red:blue

If you don't understand anything feel free to ask!

hugomimihu May 14, 2020
#2
+118665
0

Excellent, Hugo!!!