if the percentage increse in the area of circle is 44% what will be the percentage increase in its circumference?
Answers
Answer:
20%
Step-by-step explanation: Let the original radius be 'r' and new radius be 'R'.
Given, % increases in area = 44%
⇒ [New area - area]/area *100% = 44%
⇒ [πR² - πr²]/πr² * 100 = 44
⇒ [R² - r²]/r² = 44/100
⇒ [R² - r²]/r² = 11/25
⇒ 25[R² - r²] = 11r²
⇒ 25R² - 25R² = 11r²
⇒ 25R² = 36r²
Square-root on both sides,
⇒ 5R = 6r ⇒ R = 6r/5
∴ % increase in circumference is:
⇒ [2πR - 2πr]/2πr * 100%
= 2π(R - r)/2πr * 100%
= (R - r)/r * 100%
= [(6r/5) - r]/r *100%
= [ (6r - 5r)/5 ]/r * 100%
= [ 1/5 ] * 100%
= 20%
Answer:
Question :-
If the percentage increase in the area of tge circle is 44%.What will be the percentage increase in its circumstance.
Solution:-
Hey mate ,
First let verify your questions and next enter to calculations.
In the question given that if the percentage increase in the area of the circle is 44% .We should find (What will be the percentage increase in its circumstances. )
So,
to find this we know that,
Now ,
we know ,
- [New area- area/area ×100%]=44%
Now,
- lets put the values,
But here i will directly enter to main step.
By doing all the calculations we get that,
- 25 [R^2-r^2]=11r^2
- 25R^2-25r^2=11r^2
- 25R^2=11r^2+25r^2
- 25R^2=36r^2
- 5R=6r
- R=6r/5
Therefore,
Now,
The increase in % of circumstance .
- R-r/r×100%
Now,
Here put value of R.
we get that,
- 6r/5-r/r×100%
- =((6r-5r/(5r)-r)×100%
- 1/5×100%
Which can also be written as,
- 1/5×100%=20%.