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Related If the area of a circle decreases by 36%, then the radius of the circle decreases bya)20%b)6%c)36%d)18%Correct answer is option 'A'. Can you explain this answer?
A = πr2
The area is decreased by 36%
=> remaining area = 0.64 A
Suppose the remaining radius = x times of r
∴ 0.64A = π(xr)2 = x2πr2
0.64 A = x2 * A
=> x2 = 0.64
i.e., x = 0.8 times of r is remaining
=> Radius decreases by 20%
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If the radius of a circular region were decreased by 20 percent, the area of the circular region would decrease by what percent? If R is the original radius then the area of the circle is pi*R^2 If we decrease the radius by 20% the new radius is .8*R so the new area is: pi*(.8*R)^2 = pi*(.64*R^2). Compute the percent of the new to the old: (pi*.64R^2)/(pi*R^2) = .64
The decrease then is 100 - 64 = 36%
Option 5 : Decreased by 40%
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10 Questions 10 Marks 10 Mins
Area of semi circle (A) = πr2/2 → 1
When Area reduces by 64%, new Area becomes
A' = A - 0.64A = πr'2/2 , where r' is new radius
⇒ A' = 0.36A = πr'2/2
From 1
⇒ 0.36 × πr2/2 = πr'2/2
⇒ 0.36r2 = r'2
Taking square root each side
⇒ 0.6r = r'
It can be written as
⇒ r - 0.4r = r'
We can see radius is decreased by 0.4 times or 40%
∴ Radius of a semi circle or circle (since both have equal radii) is decreased by 40%
Alternate Method
Area of semi-circle (A) = πr2/2
Let, the initial area of semi-circle be 1
Therefore, \(r_1^2=\frac{1 \times2}{\pi}\)
\(\Rightarrow r_1=\sqrt{\frac{1 \times2}{\pi}}=0.7978\)
When Area reduces by 64%, new Area becomes 0.36A
Therefore, \(r_2^2=\frac{0.36 \times2}{\pi}\)
\(\Rightarrow r_2=\sqrt{\frac{0.36 \times2}{\pi}}=0.4787\)
The new radius of a circle will become, \(R=\frac{r_1-r_2}{r_1}\)
\(\Rightarrow R=\frac{0.7978-0.4787}{0.7978}=0.399\)
So, from the above, the new radius is less than original one.
It is decreased by 40%
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