the area of a circle is increased by 22 cm square when its radius is increased by 1m the original radius of the circle is
Answers
Given :-
The area of a circle is increased by 22 cm² when its radius is increased by 1 m.
To find :-
The original radius of the circle .
Solution :-
Let the original radius of the circle be r cm
We know that
Area of the circle = πr² sq.cm
If the radius is increased by 1 m then or will be
r cm +1 m
= r cm + 100 cm
Since , 1 m = 100 cm
Radius of the new circle = (r+100) cm
Area of the new circle
=> π(r+100)² cm²
According to the given problem
The area of the new circle = πr²+22 cm²
Therefore, π(r+100)² = πr²+22
=> π(r²+200r+10000) = πr²+22
Since , (a+b)² = a²+2ab+b²
Where, a = r and b = 100
=> πr² +200πr+10000π = πr²+22
=> πr² +200πr+10000π-πr²-22 = 0
=> 200πr+10000π-22 = 0
=> 2(100πr+5000π-11) = 0
=> 100πr+5000π-11 = 0/2
=> 100πr+5000π-11 = 0
=> 100π(r+50)-11 = 0
=> 100π (r+50) = 11
=> r+50 = 11/100π
=> r = (11/100π)-50
=> r = (11-5000π)/100π cm
The radius of the original circle is
(11-5000π)/100π cm
Note :-
If the increase in the radius is 1 cm then the area will be 22 cm² more than its original area then
Let the original radius of the circle be r cm
We know that
Area of the circle = πr² sq.cm
If the radius is increased by 1 cm then or will be
r cm +1 cm
= (r+1) cm
Radius of the new circle = (r+1) cm
Area of the new circle
=> π(r+1)² cm²
According to the given problem
The area of the new circle = πr²+22 cm²
Therefore, π(r+1)² = πr²+22
=> π(r²+2r+1) = πr²+22
Since , (a+b)² = a²+2ab+b²
Where, a = r and b = 1
=> πr²+2πr+π = πr²+22
=> πr²+2πr+π -πr² = 22
=> 2πr+π = 22
=> π(2r+1) = 22
=> (22/7)(2r+1) = 22
=> 2r+1 = 22×(7/22)
=> 2r+1 = 7
=> 2r = 7-1
=> 2r = 6
=> r = 6/2
=> r = 3 cm
The original radius of the circle is 3 cm
Used formulae:-
♦ Area of a circle = πr² sq.units
- r = radius
- π = 22/7