[Guest]

So this guy is hired to put a carpet in a circular, ring-shaped corridor. At this point, he doesn't know the circumference of the corridor. He does know that a chord of the outer circle, which is a tangent of the inner circle, is 100 meters long.

What is the area of the corridor? Not including the inside "island" which doesn't need to be carpeted.

bdcarpet1.bmp

[Guest]

The apothem of the chord is equal in length to r, the radius of the inner circle.

R, the radius of the outer circle, using the Pythagorean Theorem, is equal to the square of the sum of apothem squared plus half the length of the chord squared:

R² = r² + (100/2)²

Solving for r,

r = SQRT(R² - 2500)

The area to be carpeted is the area of the outer circle minus the area of the inner circle:

Area_carpet = 2∙pi∙R - 2∙pi∙r

Substituting in r, the area to be carpeted is:

Area_carpet = 2∙pi∙(R - SQRT(R² - 2500))

Edited July 25, 2010 by Dej Mar

[Guest]

R = sqrt(r² + 50²])

A = πR² - πr²

A = π(r² + 50²) - πr²

A = πr² + π2500 - πr²

A = π2500 = 7850m²

Is that it?

[Guest]

The apothem of the chord is equal in length to r, the radius of the inner circle.

R, the radius of the outer circle, using the Pythagorean Theorem, is equal to the square of the sum of apothem squared plus half the length of the chord squared:

R² = r² + (100/2)²

Solving for r,

r = SQRT(R² - 2500)

The area to be carpeted is the area of the outer circle minus the area of the inner circle:

Area_carpet = 2∙pi∙R - 2∙pi∙r

Substituting in r, the area to be carpeted is:

Area_carpet = 2∙pi∙(R - SQRT(R² - 2500))

Tried to correct the post, but time ran out. Mistakenly used equation for circumference instead of area.

Here is the corrected version:

The apothem of the chord is equal in length to r, the radius of the inner circle.

R, the radius of the outer circle, using the Pythagorean Theorem, is equal to the square of the sum of apothem squared plus half the length of the chord squared:

R² = r² + (100/2)²

Solving for r,

r = SQRT(R² - 2500)

The area to be carpeted is the area of the outer circle minus the area of the inner circle:

Area_carpet = pi∙R² - pi∙r²

Substituting in r, the area to be carpeted is:

Area_carpet = pi∙R² - pi∙SQRT(R² - 2500)² = pi∙2500

Edited July 25, 2010 by Dej Mar

[Guest]

The apothem of the chord is equal in length to r, the radius of the inner circle.

R, the radius of the outer circle, using the Pythagorean Theorem, is equal to the square of the sum of apothem squared plus half the length of the chord squared:

R² = r² + (100/2)²

Solving for r,

r = SQRT(R² - 2500)

The area to be carpeted is the area of the outer circle minus the area of the inner circle:

Area_carpet = 2∙pi∙R - 2∙pi∙r

Substituting in r, the area to be carpeted is:

Area_carpet = 2∙pi∙(R - SQRT(R² - 2500))

Isn't Pythagorean Theorem Triangles?

[Guest]

Isn't Pythagorean Theorem Triangles?

He made a right triangle with the 2 legs being the small radius and 50 and hypotenuse being the big radius, so he could replace one with the other when calculating the area.

[Guest]

The carpet area must be related only to the chord figure, therefor it's independent of the center hole size.

Make the center hole diameter = zero. The carpet area will be pi R squared, where R = 50.

[Guest]

Area = pi R squared. Inner circle diameter can be any number, including zero. That makes R = 50.

[Guest]

Area = pi R squared. Inner circle diameter can be any number, including zero. That makes R = 50.

That's clever, but it's based on the assumption that the question provides all the necessary information to calculate the answer, which is true in this case, but not in every case. You might want to prove that with algebra.

[Guest]

Hope I did the "solved" thing correctly.

[Guest]

Isn't Pythagorean Theorem Triangles?

Yes, just not thinking clearly. Distracted by my 2-year-old daughter. I meant the Law of Cosines. lol

Edited July 25, 2010 by Dej Mar

[MissKitten]

Area = pi R squared. Inner circle diameter can be any number, including zero. That makes R = 50.

please, use spoilers when you answer so others can guess as well.