Question

Will mark BRAINLIEST! Correct Answer will get BRAINLIEST! Segment $AB$ has midpoint $C$, and segment $BC$ has midpoint $D$. Semi-circles are constructed with diameters $\overline{AB}$ and $\overline{BC}$ to form the entire region shown. Segment $CP$ splits the region into two sections of equal area. What is the degree measure of angle $ACP$? Express your answer as a decimal to the nearest tenth.

Answer 1

Answer:

112.5°

Step-by-step explanation:

Let the radius of the arc constructed with diameter AB be R while the radius of the arc constructed with diameter BC be r.

BC = AB/2, therefore, r = R/2

The area of the semicircle with arc AB is given as:

Area 1 = πR²/2 = π(2r)²/2= 2πr²

The area of the semicircle with arc BC is given as:

Area 2 = πr²/2

The sum of the two area is given as:

Total area = Area 1 + Area 2 = πr²/2 + 2πr²

Total area = 5πr²/2

Segment CP splits the region into two sections of equal area. Therefore half of the total area = (5πr²/2)/2 = 5πr²/4

The area of a sector is given by:

$Area \ of \ a\ sector =\frac{\theta}{360}*\pi r^2 \\but\ the\ radius\ of\ cp = R\\Area \ of \ sector\ cp =\frac{\theta}{360}*\pi R^2 \\\frac{5}{4}\pi r^2=\frac{\theta}{360}*\pi (2r )^2 \\\\ \frac{5}{4}\pi r^2=\frac{\theta}{360}*4\pi r^2 \\\frac{\theta}{360}=\frac{5}{16}\\ \theta=112.5^0$