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I was asked for help on this question:

"A circle touches two adjacent sides of a rectangle, AB and AD, at points P and Q respectively. The fourth vertex of the rectangle, C, lies on the circle. The length of the perpendicular from C to chord PQ is 5. What is the area of the rectangle?"

I found the answer (25 square units) and made some sketches, but my solution is incomplete:

The positions of P and Q mean that C must lie on the arc C₁C₂.

https://www.flickr.com/photos/dwread/14432795701/

Suppose C lies at C₁, then PQ = 5 and the diameter of the circle is 5√2.

The rectangle is 5√2 units high and (5√2)/2 units wide, so its area is 25 square units.

https://www.flickr.com/photos/dwread/14249601758/

Similarly, if C lies at C₂, the area of the rectangle is 25 square units.

That's as far as the math goes. I haven't yet worked out a formula for the area of the rectangle when C lies between C₁ and C₂. From here, the solution devolves into guesswork about standardized tests: Since they ask for "the" area of the rectangle, I figured the area is the same no matter where C is on the arc. But I haven't proved it algebraically.

Hi Janet,

Your conclusion is indeed the correct one: the area of the rectangle is 25 square units, irrespective of where the point C lies on the arc. Proving this, however, is not a particularly easy task. I'll start with some motivating ideas, then I'll set up the proof.

(I will refer to the label in the diagrams you provided.) Because P and Q are points of tangency for perpendicular lines, it follows that triangle APQ in a right triangle. This means that the arc PQ is 90 degrees. PQ is the arc subtended by C. As C is formed on the circle, angle PCQ must measure half of the arc PQ, so angle PCQ measures 45 degrees.

It is clear that triangle APQ is a right isosceles, so its acute angles measure 45 degrees. Triangles PBC and QCD are also right triangles. PCQ is not a right triangle, but angle PCQ is known to be 45 degrees. So rectangle ABCD has been partitioned into three right triangles and one acute triangle, PCQ.

Here comes the sneaky bit. Consider angle QCD. (Note that if C lies strictly between C1 and C2, D will be to the right of Q, instead of directly on it.) The measure of angle QCD is not known, so call it x for now. Angle BCD is a right angle and angle PCQ measures 45, and angle QCD measures x, so angle BCP must measure 45-x (so that the angles sum to 90). Using supplementary angles and the fact that triangles' angles sum to 180, every angle measure in the diagram can now be found in terms of x. (Do this part yourself.)

Angles are great, but the question was to find the area of the rectangle, which requires the lengths of its sides. Here we invoke the one length given in the question. Draw the perpendicular from C to chord PQ. Label the point where it hits the chord E. Line CE measures 5 units. Triangle PCQ has now been subdivided into two right triangles, PCE and ECQ. Angles PCE and ECQ are found to be x and 45-x, respectively.

Angle BCD is now completely partitioned into right triangles. Line CE measures 5 units. By using two applications of basic trigonometry, BC can be found to measure (5/cos(x))cos(45-x) units. Similarly, line CD can be found to measure (5/cos(45-x))cos(x) units. Multiplying those two dimensions always yields 25 square units, because the cosines cancel.

So there you have it! I hope this was not too hard to follow. Ask a follow-up if you have need any clarification.

Thanks for asking,

Azeem

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