I need help?! When the radius of a circle was increased by 6 inches, the area of the circle increased by 125%?

Question

What was the length of the original radius?
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Answer 1

Let r be the original radius.

The two areas are given by the two formulae below:
A1 = pi r²
A2 = pi (r + 6)²

The ratio of the two areas is 125% (or 1.25) so divide:
A2 / A1 = 125% = 1.25

Substituting A2 and A1:
pi (r + 6)² / pi r² = 1.25

Canceling pi from numerator and denominator:
(r + 6)² / r² = 1.25

Expanding the numerator:
(r² + 12r + 36 ) / r² = 1.25

Multiply both sides by r²:
r² + 12r + 36 = 1.25r²

Subtract 1.25r² from both sides:
-(1/4)r² + 12r + 36 = 0

Multiply by -4:
r² - 48r - 144 = 0

Using the quadratic formula:
r = ( 48 ± √ 2880 ) / 2
r = 24 ± √ 720
r = 24 ± 12√5
r ≈ -2.8328... or 50.8328...

But for a radius, only a positive length makes sense, so you can throw out the negative value of r.

Thus the original radius was:
50.8328 inches.

Double-checking:
A1 = pi * 50.8328² ≈ 8117.8 sq. inches
A2 = pi * 56.8328² ≈ 10147.25 sq. inches
8117.8 x 125% = 10147.25 sq. inches

So the answer checks out... the original radius was approximately:
50.8328 inches

Answer 2

Let A be the area of the original circle and let r be its radius.
A = Pi*r^2
When the radius of a circle was increased by 6 inches, the area of the circle increased by 125% means:
If r is replaced by r+6, then the new area (Call it B) is
125%A = 1.25A
Now B/A = (Pi*(r+6)^2)/(Pi*r^2) = 1.25A/A = 1.25
The Pi's cancel and we're left with (r+6)^2/r^2 = 1.25
That is, 0.25r^2 -12r - 36 = 0
Using the quadratic formula to solve for r, we get:
r = [12 - sqrt(180)]/0.5 = 24 - 12sqrt(5) < 0 or
r = [12 + sqrt(180)]/0.5 = 24 +12sqrt(5) >0
Pick the second value of r because it's greater than zero. (Remember that the radius of a circle is always positve.)

Answer 3

A = pi * r^2

(pi * r^2) + 1.25(pi * r^2)
(pi * r^2)(1 + 1.25)
2.25pi * r^2

2.25pi * r^2 = pi * (r + 6)^2
divide both sides by pi

2.25r^2 = (r + 6)^2
2.25r^2 = (r + 6)(r + 6)
2.25r^2 = r^2 + 6r + 6r + 36
2.25r^2 = r^2 + 12r + 36
1.25r^2 - 12r - 36 = 0
(5/4)r^2 - 12r - 36 = 0
(1/4)(5r^2 - 48r - 144) = 0
5r^2 - 48r - 144 = 0

r = (-b ± sqrt(b^2 - 4ac))/(2a)

r = (-(-48) ± sqrt((-48)^2 - 4(5)(-144)))/(2(5))
r = (48 ± sqrt(2304 + 2880))/10
r = (48 ± sqrt(5184))/10
r = (48 ± 72)/10
r = (120/10) or (-24/10)
r = 12 or -2.4

since you can't have negative length

r = 12 inches

so the radius of the original circle was 12 inches.

Answer 4

1st off... no one likes shouters... kill the bold.

A_1 / A_2 = 1.25

Where A_1 is the area of the new circle and A_2 is the area of the old circle.

In terms of the radius r this is

[pi * (r+6)^2] / [pi * r^2] = 1.25

so multiplying across

4*(r+6)^2 = 5r^2

r^2 - 48r - 144 = 0

Solving we get

r = 48/2 +/- sqrt(48^2-4*(-144))/2

The negative root being invalid the only valid root is

r = 48/2 + sqrt(48^2-4*(-144))/2
= 50.83281572999748 approximately

Answer 5

A2=A1+125%A1=225%A1=2.25*A1 →
π(r1+6)^2 = 2.25π(r1)^2 → π[(r1)^2 + 12(r1) + 36] = 2.25π(r1)^2 → (r1)^2 + 12(r1) + 36 = 2.25(r1)^2 → 1.25(r1)^2 -12(r1) -36 = 0

when ax^2 + bx + c = 0 then x = [-b ±√¯¯(b^2 - 4ac)] ÷ (2a)
So r1 = [12 ±√¯¯(144 - 4*1.25*(-36))] ÷ (2*1.25) = [12 ±√¯¯(144+180)] ÷ 2.5 = (12 ± 18) ÷ 2.5 = 12 or -2.4
but r1 can not be negative. Thus, r1=12

to make sure you can test it. when r1=12, then r2=18
so A1 = π*(12^2) = π*144
and A2 = π*(18^2) = π*324
A2 ÷ A1 = 324π ÷ 144π = 2.25

Answer 6

A=pi*r^2
1.25A=pi*(r+6)^2

Divide Bottom by Top

1.25 = (r+6)^2 / r^2

Solve for r - have fun before someone gives you the answer.