



 
Hi James. Here is a diagram I drew of your problem. I've also drawn the radius of the outer circle as R and the radius of the inner circle as r. Remember that at the point of tangency, the tangent line is perpendicular to the radius of a circle, so this triangle is a rightangled triangle. As well, symmetry requires that the side of the right triangle be half of the chord length, so I've labelled it 7 cm.
The area of the outer circle is πR^{2 }. A = π(R^{2 }  r^{2}) Since that is a righttriangle, Pythagorus applies, so R^{2 } = 7^{2 } + r^{2 } Thus, A = π(7^{2}) Hope this helps, Stephen La Rocque.
 


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