Question

Report Error Suppose that the area between a pair of concentric circles is $49\pi$. Find the length of a chord in the larger circle that is tangent to the smaller circle.

Answer 1

Answer:

14

Step-by-step explanation:

Visualize this situation as in the attachment (it is not scaled).

First, denote by r the radius of the inner circle, R the radius of the outer circle and A the area between both circles. The area of the inner circle is πr² and the area of the outer circle is πR². The area of the inner circle and the area between the circle adds up to the area of the outer circle, that is, πr²+A=πR², then A=πR²-πr².

We are given that A=49π, then 49π=πR²-πr². Divide pi from this equation to obtain 49=R²-r². We will use this later on the problem.

Following the figure, suppose that A is the center of both circles and the chord ED is tangent to the first circle on point C. Construct the triangles ACE and ACD. Both are right triangles because a tangent line is perpendicular to the radius, in this case ED⊥AC.

Now, note that AE=AD=R and AC=r because E,D are points of the outer circle and C is a point of the inner circle. Applying the Pythagorean theorem (on both triangles, we get that CE²=AE²-AC²=R²-r²=49 and CD²=AD²-AC²=R²-r²=49, so that CE=7=CD.

Finally, we compute the length of the chord as ED=EC+CD=CE+CD=7+7=14.