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Bunuel
If the radius of a circle is decreased 20%, what happens to the area?

A. 10% decrease
B. 20% decrease
C. 36% decrease
D. 40% decrease
E. 50% decrease

Area = pi*r^2

Therefore New area = pi*(0.8r)^2 = pi*0.64r^2

%Decrease = 0.36*100 = 36% Option C
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Bunuel
If the radius of a circle is decreased 20%, what happens to the area?

A. 10% decrease
B. 20% decrease
C. 36% decrease
D. 40% decrease
E. 50% decrease

-20 -20 +{(-20)(-20)}/100

Or, -40 + (-4)

Or, -36%

Answer is (C)
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A orig = Pi(R squared)
A new = Pi(0.80r)^2 = 0.64Pi

Orig - new = 0.36
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Bunuel
If the radius of a circle is decreased 20%, what happens to the area?

A. 10% decrease
B. 20% decrease
C. 36% decrease
D. 40% decrease
E. 50% decrease

Another method -

Let radius be 5
So, Area = \(π5^2\) = \(25π\)

After reduction of radius by 20% , new radius = 4

So, New Areas = \(π4^2\) = \(16π\)

Change in area = \(9π\)

So, Percentage change = \(\frac{9π}{25π}\)*\(100\) = \(36\)%

Hence, answer will be (C)
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Bunuel
If the radius of a circle is decreased 20%, what happens to the area?

A. 10% decrease
B. 20% decrease
C. 36% decrease
D. 40% decrease
E. 50% decrease

Asked: If the radius of a circle is decreased 20%, what happens to the area?

New area = \(\pi (.8r)^2 = .64 \pi r^2 = (1 - .36) \pi r^2 \)

IMO C
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