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 $ABC.$

Asymptote code below
[asy]
unitsize(2 cm);

pair A = (1, 0);
pair B = (7/25, 24/25);
pair C = (-1, 0);

path AB = arc((16/25, 12/25), 3/5,-53.13, 180-53.13);
path BC =arc((-9/25, 12/25), 4/5,36.87, 180+36.87);
path AC = arc((0, 0), 1, 0, 180);

fill(AB--cycle, gray);
fill(BC--cycle, gray);
fill(AC--cycle, palegray);
draw(arc((0, 0), 1, 0, 180));
draw(arc((16/25, 12/25), 3/5,-53.13, 180-53.13));
draw(arc((-9/25, 12/25), 4/5,36.87, 180+36.87));
draw(A--B--C--cycle);
draw(rightanglemark(A, B, C, 3));

label("$A$", A, E);
label("$B$", B, N);
label("$C$", C, W);
[/asy]

Given a right triangle $ \triangle ABC $ with $ \angle B = 90^\circ $, semicircles are constructed on each of its sides. The goal is to prove that the total area of the shaded regions equals the area of the triangle $ \triangle ABC $.

### Step 1: Define the sides of the triangle

Let:
- $ AB = c $ (the side opposite $ \angle C $),
- $ BC = a $ (the side opposite $ \angle A $),
- $ AC = b $ (the hypotenuse).

### Step 2: Calculate the area of the triangle

The area of $ \triangle ABC $ can be calculated using the formula for the area of a right triangle:

\[
\text{Area of } \triangle ABC = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times c \times a.
\]

### Step 3: Find the area of each semicircle

1. **Semicircle on $ AB $:**
   - Radius $ r_1 = \frac{c}{2} $.
   - Area of the semicircle:

     \[
     \text{Area}_{AB} = \frac{1}{2} \pi \left(\frac{c}{2}\right)^2 = \frac{1}{2} \pi \frac{c^2}{4} = \frac{\pi c^2}{8}.
     \]

2. **Semicircle on $ BC $:**
   - Radius $ r_2 = \frac{a}{2} $.
   - Area of the semicircle:

     \[
     \text{Area}_{BC} = \frac{1}{2} \pi \left(\frac{a}{2}\right)^2 = \frac{1}{2} \pi \frac{a^2}{4} = \frac{\pi a^2}{8}.
     \]

3. **Semicircle on $ AC $:**
   - Radius $ r_3 = \frac{b}{2} $.
   - Area of the semicircle:

     \[
     \text{Area}_{AC} = \frac{1}{2} \pi \left(\frac{b}{2}\right)^2 = \frac{1}{2} \pi \frac{b^2}{4} = \frac{\pi b^2}{8}.
     \]

### Step 4: Determine the shaded area

The shaded area is formed by subtracting the areas of the semicircles on sides $ AB $ and $ BC $ from the semicircle on the hypotenuse $ AC $:

\[
\text{Shaded Area} = \text{Area}_{AC} - (\text{Area}_{AB} + \text{Area}_{BC}).
\]

Substituting the values from above:

\[
\text{Shaded Area} = \frac{\pi b^2}{8} - \left(\frac{\pi c^2}{8} + \frac{\pi a^2}{8}\right) = \frac{\pi b^2}{8} - \frac{\pi (a^2 + c^2)}{8}.
\]

### Step 5: Use the Pythagorean theorem

Since $ \triangle ABC $ is a right triangle, by the Pythagorean theorem:

\[
b^2 = a^2 + c^2.
\]

Substituting $ b^2 $ in the shaded area formula:

\[
\text{Shaded Area} = \frac{\pi (a^2 + c^2)}{8} - \frac{\pi (a^2 + c^2)}{8} = \frac{\pi (a^2 + c^2) - \pi (a^2 + c^2)}{8} = \frac{a \cdot c}{2}.
\]

Thus the shaded area $  ABC.$

### Step 5: Relative question
1. How to find the shaded part of semicircle?
2. How to find perpendicular triangle?
3. How to find x part of coordinate triangle
4. How semicircle defines a new theorem?


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 $ABC.$

Explore how to prove that the total area of the shaded regions formed by semicircles on the sides of a right triangle equals the area of the triangle itself. This problem utilizes the Pythagorean theorem and geometry concepts, suitable for Grades 9-12.

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