In triangle \(ABC, \angle B = 90^\circ\) Semicircles are constructed on sides \(\overline{AB}, \overline{AC},\) and \(\overline{BC},\) as shown below. Show that the total area of the shaded region is equal to the area of triangle \(ABC\).
Thank you!
+2094 Hi, Guest! This seems like a really hard problem. This is just a hint to help you.
I thought of this a little like the Pythagorean Theorem. If you've got some of those 3-D shapes at home, they help. I filled those in with water, and poured it into the triangle.
+128408 This leads to an interesting result ..... [ the shaded areas are known as "lunes" ]
The area of the triangle is (1/2)(BC)(BA)
The area of the whole semi-circle is (1/2)pi * (AC/2)^2 = (pi/8)AC^2 = (pi/8) (BC^2 + AC^2)
So....the sum of the areas between the semi-circle and the triangle = (pi/8)AC^2 - (1/2)(BC)(BA)
The area of the semi-circle constructed on BC = (1/2)(pi)(BC/2)^2 = (pi/8)(BC)^2
The area of the semi-circle constructed on AC is similar = (pi/8)(AC)^2
So......the sum of the shaded areas =
Area of the two-semi-circles - area between the semi-circle and the triangle =
(pi/8) (BC^2 + BA^2) - [ (pi/8)AC^2 - (1/2)(BC)(BA) ] =
(pi/8) ( AC)^2 - ( pi/8)(AC)^2 + (1/2)(BC)(BA) =
(1/2)(BC)(BA) = the area of the right triangle
+1486 AB => a = 3
BC => b = 4
CA => c = 5
Area of triangle is: A = ab/2 = 6 u²
Area of AB semicircle is: A = [(a/2)² * pi] /2 = 3.534291735 u²
Area of BC semicircle is: A = [(b/2)² * pi] /2 = 6.283185307 u²
Area of AC semicircle is: A = [(c/2)² * pi] /2 = 9.817477043 u²
Tota area is: At = A + A + A = 15.81747704 u²
Area of a shaded region is: As = At - A = 6 u²
These areas are equal: A = As 
