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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!

Mar 19, 2020

#1
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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.

Mar 19, 2020
#2
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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   Mar 19, 2020
edited by CPhill  Mar 19, 2020
#3
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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 Mar 20, 2020
edited by Dragan  Mar 20, 2020