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.

The degree measure of ∠ACP is 112.5°

Step-by-step explanation:

The total area of the given diagram is the sum of two semicircles with arc AB and arc CB having radius R and r respectively

Where:

R = AC

r = DB

R = 2 × r

Therefore the area of the semicircles are given as follows;

For the semicircle with arc AB, we have;

Area, A₁ = π × R²/2 = π × (2×r)²/2 = 2×π×r²

For the semicircle with arc CB, we have;

Area, A₂ = π × r²/2 =  1/2×π×r²

The ratio of the two semicircles is presented in the following relation;

Therefore, the area of A₁ is four times that of A₂ or A₁ =  4 × A₂

The total area of the given diagram = A₁ + A₂ = 4 × A₂ + A₂ = 5·A₂

∴ Half of the area of the diagram, = 5·A₂/2 = 2.5·A₂ = 2.5 × 1/2×π×r² = 1.25×π×r²

The ratio of half of the diagram of the figure to the area of the semicircle with arc AB is found as follows;

/A₁ = (1.25×π×r²)/(2×π×r²) = 5/8

Therefore, the half of the diagram of the figure given by segment PAC is equivalent to 5/8 of the semicircle with arc AB

Given that the arc AB subtends an angle of 180° at the center (angle subtended by a semicircle), the arc AP will subtend 5/8×180 = 112.5°

To verify we have;

Area of a segment of a circle is presented in the following relation;

As segment PAC is 5/8 of a semicircle, it is therefore 5/(8×2) or 5/16 of the whole circle

Hence;

Therefore the degree measure of ∠ACP is 112.5°.

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