Question

Segment AB has midpoint C, and segment BC has midpoint D. Semi-circles are

constructed with diameters AB and BC to form the entire region shown. Segment CP
splits the region into two sections with ratio of areas 1 : 2 (left region to the right
region). What is the degree measure of angle ACP? Express your answer as a decimal
to the nearest tenth.

Answer 1

The measure of the angle ACP, obtained using the area of a sector of a circle is m∠ACP = 90°

The steps by which the measure of the angle ACP is found is as follows;

Let R represent the length of the large semicircle and let r represent the length of the smaller semicircle, we get;

R = 2·r

R/2 = r

Area of the large semicircle = π·R²/2

Area of the small semicircle = π·R²/8

Area of the sector BCP = (θ/360) × π·R²/2

Therefore; (θ/360) × π·R²/2 + π·R²/8 : π·R²/2 = 1 : 2

(θ/360) × π·R²/2 + π·R²/8 = ((θ/360) × 4 + 1) × π·R²/8

(((θ/360) × 4 + 1) × π·R²/8)/(π·R²/2) = 1/2

(((θ/360) × 4 + 1) × 1/4) = 1/2

((θ/360) × 4 + 1) = 4 × (1/2)

((θ/360) × 4 + 1) = 2

((θ/360) × 4 = 2 - 1

(θ/360) = 1/(4)

θ = 360/4

θ = 90

Angle BCP, θ is 90°

The degree measure of the angle ACP, can be obtained from the linear pair property as follows;

m∠BCP + m∠ACP = 180°

m∠ACP = 180° - m∠BCP

m∠ACP = 180° - 90°

m∠ACP = 90°