if the perimeter of a circle decrease by 50 percent then what is the percentage decrease in area
Answers
Answer:
If the circumference change is 50 percent then the new new circumference will be π*r….
So the new radius is r/2 ….Where r is the original radius……
Area of the new circle is ( π*r*r)/4….
The area of the original circle is π*r*r…..
Change in area is (3*π*r*r)/4
Change in area by the original area and multiplying the whole by 100 we get 75 percent….
So the reduction in area is 75 percent
product- Google
copied by Google
Answer:
75 %
Step-by-step explanation:
To make the explanation easier, I'm gonna use simple numbers.
Lets take radius to be 7 units.
Circumference (perimeter) = 2πr
= 2×22/7×7
= 2×22
= 44
Half of 44 (50%) = 22
Therefore radius when perimeter is half will be;
22 = 2πr
22 = 2×22/7×r
11 = 22/7×r
77/22 = r
7/2 = r
r = 3.5 units
(I went by the long method here just for a clear explanation, but you can just divide the radius by 2 or the half of it to directly make the perimeter half or decrease by 50 %. Here, radius when perimeter is half is 3.5, which is basically 7/2 = 3.5. 7 was the original radius which is divided into half)
Area when radius is 7 ( whole perimeter ):
πr² = 22/7×7×7
= 22×7
= 154 units
Area when perimeter is half ( radius = 3.5 units):
πr² = 22/7×3.5×3.5
= 22/2×3.5
= 11×3.5
= 38.5 units
Area 1 = 154 units
Area 2 = 38.5 units
Percentage decrease = 38.5/154×100
= 1/4×100
= 25%
Therefore, Area 2 is 25% of that of Area 1.
Therefore, Area 1 is greater by (100%-25%) 75% than Area 2.
Thus, we cann conclude by saying that when the perimeter of a circle is decreased by 50%, then in the area of the same circles, we see a decrease in area by 75%.
HOPE IT HELPS ;)