Question

Correct Answer will get BRAINLIEST!

Answer 1

Answer:

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;

$\dfrac{A_1}{A_2} = \dfrac{2 \cdot \pi \cdot r^2}{\dfrac{1}{2} \cdot \pi \cdot r^2} = \dfrac{2}{\dfrac{1}{2} } = 2 \times \dfrac{2}{1} = 4$

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, $A_H$ = 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;

$A_H$/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;

$\dfrac{\theta}{360} \times \pi \times r^2$

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

Hence;

$\dfrac{5}{16} \times \pi \times r^2 = \dfrac{\theta}{360} \times \pi \times r^2$

$\dfrac{5}{16} = \dfrac{\theta}{360}$

$\theta \dfrac{}{} = \dfrac{360 \times 5}{16} = 112.5 ^ {\circ}$

Therefore the degree measure of ∠ACP is 112.5°.