If radius of circle is reduced by 10%,then area is diminishd by.
a) 10% b)19% c)20% d)36%
please derive dont just tell the answer :)
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Let the radius of circle be 'r'
Area = pi*r*r....(i)
Now, radius is decrease by 10%
r = r - r*10/100 = 90*r/100 = 9*r/10
Area = pi*9r*9r/10*10 = pi*81r^2/100
Change in area = pi*r^2 - pi*81r^2/100 = Pi*19r^2/100
% diminish(decrease) in area = (pi*19r^2/100)/pi*r^2 * 100
= 19/100 * 100 = 19 %
OR
Reduced % = 10%
10% = 10/100 = 1/10
r==1/10----- 10(original r)===== 9(diminished r)
Area = pi*r^2----- pi*100(original area)=====pi*81(diminished area)
Hence % diminish in area = 100*pi - 81*pi/100*pi * 100 = 19/100 * 100 = 19%...
Hope it helps.
Area = pi*r*r....(i)
Now, radius is decrease by 10%
r = r - r*10/100 = 90*r/100 = 9*r/10
Area = pi*9r*9r/10*10 = pi*81r^2/100
Change in area = pi*r^2 - pi*81r^2/100 = Pi*19r^2/100
% diminish(decrease) in area = (pi*19r^2/100)/pi*r^2 * 100
= 19/100 * 100 = 19 %
OR
Reduced % = 10%
10% = 10/100 = 1/10
r==1/10----- 10(original r)===== 9(diminished r)
Area = pi*r^2----- pi*100(original area)=====pi*81(diminished area)
Hence % diminish in area = 100*pi - 81*pi/100*pi * 100 = 19/100 * 100 = 19%...
Hope it helps.
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