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Example of circle application
Question:
point O is the center of the circle and r is its radius. If the radius of the circle increase by 40%, by what percent will the area of the circle increase? Solution:
r is the radius of the original circle
R is the radius of the circle after r increase 40 percent
Areasmall is the area of the original circle
Arealarge is the area of the circle after the radius r increase by 40%
then R = r + (40/100)r = r[1 + 40/100] = r[1 + 4/10] = r[1 + 2/5] = r[5/5 + 2/5] = (7/5) r
the area of the large circle is
Arealarge = pi (R2)
= pi [(7/5) r]2
= pi (7/5)2 r2
= pi (72/52) r2
= (49/25) pi r2
remark: Areasmall = pi r2
so Arealarge = (49/25) Areasmall
so the percent increase of the Area is:
{[Arealarge - Areasmall]/Areasmall} × 100%
= {[(49/25)pi r2 - pi r2]/pi r2} × 100%
= (pi r2)/(pi r2){[(49/25) - 1]/1} × 100%
= {[(49/25) - 1]/1} × 100%
= {[(49/25) - (25/25)]/1} × 100%
= [(24/25)/1] × 100%
= (24/25) × 100%
= 96%
so when r increase by 40% the area of the circle increased by 96%.

The area of a circle is the pi multiply the square of the radius of the circle. The percent increase of the area is equal to the area of the large circle subtract the area of the small circle divided by the area of the small circle, then multiply 100%.