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!

Guest Mar 19, 2020

#1**+1 **

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.

CalTheGreat Mar 19, 2020

#2**+2 **

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

CPhill Mar 19, 2020

#3**+1 **

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: **A _{t} = **A + A + A

Area of a shaded region is: **A _{s} = **

**These areas are equal: **A = **A _{s} **

Dragan Mar 20, 2020