Circular Measure

Question

A semicircle of radius 10 cm has diameter AB. AP and AQ are chords of lengths 6 cm and 8 cm respectively.Calculate the are bounded by these two chords and the arc PQ.

Answer 1

There is likely more than one geometry-based way to do this, but the way that seems most obvious and straight-forward to me would be to calculate the area of the sector of the circle subtended by arc AQ, subtract the area between chord AP and arc AP, and subtract the area of the triangle between chord AQ, the radius to A, and the radius to Q.

 

Of those, the area between chord AP and arc AP is the trickiest, you could calculate the area of sector subtended by arc AP and subtract the triangle using chord AP, radius to A, and radius to P.

 

Is that enough help?

 

Then, as Richard posted before I finished writing my answer, there's always the calculus approach. Good one Richard!

Answer 2

There is a straightforward calculus approach to this problem.   Since 6,8,10 are the side lengths of a right triangle (double the 3,4,5 triangle) , there is probably a purely geometric approach as well.

 

The calculus approach is to note that the area bounded by a chord (length c) and a diameter,d, can be written as an angular integral:

 

   A =  (d²/2) ∫ (cos(θ))² dθ  {lower limit = 0,  upper limit = arccos(c/d)}

 

    A = (d² /2) (arccos(c/d) /2 + sin(2 arccos(c/d)) /4  )

 

   The desired area is the difference between the area calculated with d = 10 , c =6 and     the area calculated with  d =10 , c =8.

 

    After plugging in the numbers the answer comes out to be 7.095.  It is interesting to note that the contributions from the   sin( 2 arccos(c/d)) terms cancel out.    This is a strong hint that there is a geometric approach to this problem.