How can you solve this problem?

Question

1. The area of a circle of radius r is pi r^2. By how much must the radius be increased in order to increase its area by 44%?

2. A pavement 1-meter wide surrounds a circular pool. The area of the pool is 4/5 of the area of the pavement. Find the radius of the pool.

Answer 1

1) A/pi =r^2

1.44A/ pi= r^2

r= 1.2A/pi

the radius increases by 20%  

2) pi*r^2 = 4/5 (pi*(r+1)^2-pi*r^2)

pi*r^2 = 4/5 *pi* (r+1)^2 - 4/5 pi*r^2

9/5pi*r^2 = 4/5 pi(r+1)^2

2.25r^2 = (r+1)^2

2.25r^2 = r^2+2r+1

1.25r^2-2r-1=0

factor out quadratic using quadratic formula

r= 2 r= -0.4    -0.4 is impossible since radius isn't negative

so radius of the pool is 2 metres

hope this helps

Answer 2

1) 20%

If A = pir^2 and (1.44)A= pix^2 then (1.44)pir^2 = pix^2

clearly, 1.44 r^2 = x^2 or x = 1.2 r So new radius is 1.2 times the original or 20% bigger.

2) Solve pi[(r+1)^2 - r^2]/ pi r^2 = 4/5

(2r+1)/r^2 = 0.8

Can you solve it now?

Answer 3

1. Area, A= πr²

New area =A’ = πr’² = A * 144/100

=> πr’² = A * 144/100

=> πr’² = πr² * 144/100

=> r’² = r² * 144/100

=> r’ = ±√(r² * 144/100)

Since r is a length, it cannot be negative, so

=> r’ = +√(r² * 12²/10²)

=> r’ = r*12/10

Therefore increase in radius is r’ – r = r*12/10 – r

=> r(12/10 – 1)

=> r(.2)

So percentage increase  = 0.2r/r * 100

=> 20%

2. Let the radius of the pool be r.

Now the area of the pavement will be area of the circle (r+1) meter – area of the pool.

=> π(r+1) ² - πr²

=> π{ (r+1) ² - r²}

=> π{ r²+ 2r + 1 - r²}

=> π{ 2r + 1 }

Area of the pool is πr²

Now we know that the area of the pool is 4/5th the area of the pavement,

So πr² = π{ 2r + 1 } * 4/5

=> 5r² = { 2r + 1 } * 4

=> 5r² = 8r + 4

=> 5r²- 8r – 4 = 0

=> 5r² - 10r + 2r – 4 = 0

=> 5r(r – 2) + 2(r – 2) = 0

=> (5r + 2)(r-2) = 0

=> 5r + 2 = 0 or r-2 = 0

=> r = -2/5 or r = 2

Since r cannot be negative, r=  2 meters  

Answer 4

1.- Always wrtite every piece of information that you have in the form of an equation; these

questions are both geometry and algebra questions:

The area of your original circle is A1 = pi*(r1)^2

The area of your "new" circle is 44% bigger; that is A2=A1*1.44, and also

A2= pi*(r2)^2

Now put both formulas in front of each other:

pi*(r2)^2 = 1.44*(pi*(r1)^2)

(r2)^2 = (1.44*pi*(r1)^2) / pi

(r2)^2 = 1.44*(r1)^2

r2 = √(1.44*(r1)^2)

r2 = r1*√1.44

r2 = r1*1.2

The radius has to be increased by 20% in order to increase the area

by 44%

Example

r1 = 1m; A1 = pi*1^2 = 3.1416 m^2

r2 = r1*1.20; r2= 1.20 m

A2 = pi*(1.2)^2 = pi*1.44 = 4.5239

4.5239/3.1416 = 1.4399 ≈ 1.44

**************************************...

2.- r1 = radius of the inside circle

    r2 = radius of the largest circle made by the pavement

r2-r1 = 1m; the width of the pavement.

or r2 = r1 + 1m

A1 = pi*(r1)^2

The area of the pavement RING is the area of the largest circle made by the pavement minus the

area of the smaller circle inside of it.

A2= pi*(r2)^2 - pi*(r1)^2

A2= pi*[(r2)^2 - (r1)^2]

And you know that

A1 = 4/5*A2; so

pi*(r1)^2 = 4/5*pi*[(r2)^2 - (r1)^2] ;

now remember that r2=r1+1; so (r2)^2 = (r1+1)^2 = (r1)^2 + 2*r1 + 1; so

(r1)^2 = 4/5*[(r1+1)^2 - (r1)^2] = 4/5[(r1)^2 + 2*r1 + 1 - (r1)^2]

(r1)^2 = 4/5 [ 2*r1 + 1 ]

(r1)^2 = 8/5*r1 + 4/5

(r1)^2 - 8/5*r1 - 4/5 =0  multiply everything x5

5(r1)^2 -8r1 - 4 = 0   for simplicity r1 = x

this is a cuadratic equation; solve it by the general formula and you get:

x=  [-b±√(b^2 - 4ac)] / 2a

a=5, b=-8 c=-4

x = [-(-8)±√[(-8)^2 - 4*5*(-4)]]/ 2*5

x= [ 8 ±√(64+80) ] / 10

x= [8 ±√(144)] / 10 = [8 ±12] / 10

x1 = (8+12)/10 = 20/10 = 2

x2 = (8-12)/10 = -4/10 = -0.4;

since a physical dimension like a radius cannot be negative;

the radius of the pool is 2 meters