QUESTION: Dear Azeem Hussain,
I am rarely in a situation where I have no idea what to do, but today is one of those rare moments. This problem has completely stumped me and I do not understand it at all. I have a test coming up with more problems like this and I would really appreciate it if you could explained how to solve these kind of problems.
An annulus has a 36 cm chord of the outer circle that is also tangent to the inner concentric circle. Find the area of the annulus.
A Struggling Student
ANSWER: Dear Valerie,
This is a cheeky question because naively you might think you need both radii, but they cannot be resolved. However, what is really necessary is the difference of their squares, and that can be determined.
Let R denote the larger radius (OB), r the smaller radius (OT). The area of the annulus is the difference of the circles' areas.
Note that T is the midpoint of AB. (AB is tangent at T, and if OT viewed as a ray then it contains a large radius, thereby bisecting the chord.) Therefore BT measures half of 36, so 18. Also, by tangency, OT is perpendicular to BT.
Consider right triangle OTB. Its hypotenuse is R, and its legs are r and 18. It follows from the Pythagorean Theorem that R²-r²=18². Multiply this by pi to obtain the annulus' area.
Good luck on your upcoming test!
---------- FOLLOW-UP ----------
QUESTION: Dear Azeem,
You were quite helpful, however I am a little confused by the final formula (R²-r²=18²). How do you find R and r? I'm sorry for the inconvenience, I'm just really confused.
A Somewhat Struggling Student
You don't find R and r. Consider (R²-r²) as a single unknown.
The area of the annulus is given by Area=π(R²-r²). From the Pythagorean Theorem, R²-r²=18². Substitute that in to get Area=π(18²).
This question can be very challenging because one's natural instinct is to determine the values of R and r, but you have insufficient information to do that here. Working backwards is very helpful. Set up your area equation first. This highlights the exact relation between the unknowns. Then, over the course of attempts to make equations containing R and r, you stumble upon exactly what you need.
Hope this was helpful. If not, let me know and I'll try to clarify further (and perhaps you can sign "A No-Longer Struggling Student"!).