If radius increased by 25%, find percent increase in area

[australopithecus23]

If the radius of a circular garden plot is increased by 25% by what percentage does the area of the garden increase?

I have:
1+(25/100)=1.25
A=(pi)r[sup:37y3xd3r]2[/sup:37y3xd3r]

so, original area= A[sub:37y3xd3r]1[/sub:37y3xd3r]=(pi)r[sup:37y3xd3r]2[/sup:37y3xd3r]
final area=A[sub:37y3xd3r]2[/sub:37y3xd3r]=(pi)r[sup:37y3xd3r]2[/sup:37y3xd3r][sub:37y3xd3r]2[/sub:37y3xd3r]

so,

(delta A/A)X100%= ((pi)r[sup:37y3xd3r]2[/sup:37y3xd3r][sub:37y3xd3r]2[/sub:37y3xd3r] - (pi)r[sup:37y3xd3r]2[/sup:37y3xd3r]/(pi)r[sup:37y3xd3r]2[/sup:37y3xd3r])x100%= r[sup:37y3xd3r]2[/sup:37y3xd3r][sub:37y3xd3r]2[/sub:37y3xd3r] - r[sup:37y3xd3r]2[/sup:37y3xd3r]/r[sup:37y3xd3r]2[/sup:37y3xd3r])x100%=
(1.25[sup:37y3xd3r]2[/sup:37y3xd3r]r[sup:37y3xd3r]2[/sup:37y3xd3r] - r[sup:37y3xd3r]2[/sup:37y3xd3r]/r[sup:37y3xd3r]2[/sup:37y3xd3r])x100%= (1.25[sup:37y3xd3r]2[/sup:37y3xd3r] - 1/1)x100%= 56%

my question is, did they cross out the (pi), b/c it looks like the (pi) has been crossed out. Also, how did 1.25 become 1.25[sup:37y3xd3r]2[/sup:37y3xd3r]. Why does it carry a 2 superscript?

and also, how come when (r) is crossed out (1) is left?

a note the (r) by itself means (r[sub:37y3xd3r]1[/sub:37y3xd3r])

 

[mmm4444bot]

Re: how did they get this number?

If the radius of a circular garden plot is increased by 25% by what percentage does the area of the garden increase?

I have:
1+(25/100)=1.25
A=(pi)r[sup:88wawxhz]2[/sup:88wawxhz]

so, original area= A[sub:88wawxhz]1[/sub:88wawxhz]=(pi)r[sup:88wawxhz]2[/sup:88wawxhz]
final area=A[sub:88wawxhz]2[/sub:88wawxhz]=(pi)r[sup:88wawxhz]2[/sup:88wawxhz][sub:88wawxhz]2[/sub:88wawxhz]

so,

(delta A/A)X100%= ((pi)r[sup:88wawxhz]2[/sup:88wawxhz][sub:88wawxhz]2[/sub:88wawxhz] - (pi)r[sup:88wawxhz]2[/sup:88wawxhz]/(pi)r[sup:88wawxhz]2[/sup:88wawxhz])x100%= r[sup:88wawxhz]2[/sup:88wawxhz][sub:88wawxhz]2[/sub:88wawxhz] - r[sup:88wawxhz]2[/sup:88wawxhz]/r[sup:88wawxhz]2[/sup:88wawxhz])x100%=
(1.25[sup:88wawxhz]2[/sup:88wawxhz]r[sup:88wawxhz]2[/sup:88wawxhz] - r[sup:88wawxhz]2[/sup:88wawxhz]/r[sup:88wawxhz]2[/sup:88wawxhz])x100%= (1.25[sup:88wawxhz]2[/sup:88wawxhz] - 1/1)x100%= 56% ? This is rounded to the nearest percent

Actual increase is 56.25%

\(\displaystyle \mbox{Let r = small radius}\)

\(\displaystyle \mbox{Let 1.25r = large radius}\)

\(\displaystyle \mbox{Percent-Increase factor = } \frac{Large Area}{Small Area}\)

\(\displaystyle \frac{Pi \cdot (1.25r)^2}{Pi \cdot r^2}\)

Pi cancels; r^2 cancels; you're left with 1.25^2

~ Mark

 

[soroban]

Re: how did they get this number?

Hello, australopithecus23!

If the radius of a circular garden plot is increased by 25%,
by what percentage does the area of the garden increase?

\(\displaystyle \text{Let }R\text{ be the original radius.}\)

\(\displaystyle \text{The area of the garden is: }\:A_1\:=\:\pi R^2\)


\(\displaystyle \text{The radius is increased by 25 percent, so the new radius is: }\:R + \frac{1}{4}R \:=\:\frac{5}{4}R\)
\(\displaystyle \text{The area of the larger garden is: }\:A_2 \:=\;\pi\left(\frac{5}{4}R\right)^2 \:=\:\frac{25}{16}\pi R^2\)


\(\displaystyle \text{The area increased by: }\:A_2 - A_1 \;=\;\frac{25}{16}\pi R^2 - \pi R^2 \;=\;\left(\frac{25}{16} - 1\right)\pi R^2 \;=\;\frac{9}{16}\pi R^2\)


\(\displaystyle \text{The percent of increase is: }\;\frac{\frac{9}{16}\pi R^2}{\pi R^2} \;=\;\frac{9}{16} \;=\;0.5625 \;=\;56.25\text{ percent}\)