A right triangle ABC is inscribed in a circle with centre O, as shown in the following diagram. A and C are endpoints of a diameter, and B is a point that lies on the circumference. AC measures #sqrt(277)# cm, and side BC measures 5 cm less than AB?->

As the shaded region, the triangle, and the semicircle not containing the triangle partition the circle, we know that the area of the shaded region is the difference between the area of the circle and the sum of the areas of the right triangle and the remaining semicircle. Thus, if we can calculate those areas, we are done.

First, note that as the triangle is a right triangle, we have #AB^2+BC^2=AC^2 = 277#.

Substituting in #BC=AB-5#, we get #AB^2+(AB-5)^2=277#

#=> 2AB^2 - 10AB - 252 = 0#

#=> AB^2-5AB-126 = 0#

#=> AB = (5+-sqrt((-5)^2-4(1)(-126)))/(2(1))#

#=(5+-23)/2#

As #AB>0#, this leaves us with #AB = (5+23)/2 = 14# and so #BC = 14-5 = 9#

Again, as the triangle is a right triangle, we can treat #AB# and #BC# as its base and height, meaning we can use the formula #"Area"_triangle = 1/2("base")("height")# to get the area of the triangle as

#1/2(AB)(BC) = 1/2(14)(9) = 63#

Next, as #bar(AC)# is a diameter of the circle, we know the radius of the circle is #(AC)/2 = sqrt(277)/2#. Plugging this into the formula #"Area"_"O"= pi("radius")^2#, we get the area of the circle as

#pi((AC)/2)^2 = pi(sqrt(277)/2)^2 = (277pi)/4#

Using our initial observation, we can now directly calculate the area #A# of the shaded region:

#A = (277pi)/4 - ((277pi)/4)/2 - 63 = (277pi-504)/8~~45.78#