How to solve part b?

b)
Refer to the figure below.
P and Q are the centers of the bottom circle and the top right circle respectively.
PH = PK = QH = QK = 7 cm
∠HPK = ∠HQK = 120°

Area of segment HPK
= (Area of sector QHK) - (Area of ΔQHK)
= π × 7² × (120/360) - (1/2) × 7² × sin(120°)
= (49π/3) - (1/2) × 49 × (√3/2)

Area of minor segment PK (shaded with red line) = (1/2) × (Area of shaded part B)

Area of segment HPK
= (Area of ΔPHK) + (Area of minor segment PK) + (1/2) × (Area of shaded part B)
= (Area of ΔPHK) + (Area of shaded part B)

Area of shaded part B
= (Area of segment HPK) - (Area of ΔPHK)
= [(49π/3) - (1/2) × 49 × (√3/2)] - [(1/2) × 49 × (√3/2)]
= (49π/3) - (49√3/2)
≈ 8.877  (square units)

So you've figured out that Part a is 29 1/3 cm?


The area of the equilateral triangle, 63 cm² = (the base)² × (√3)/4.

The base² ≈ 145.497, and base ≈ 12.062

The height is 2 × 63 ÷ 12.062 ≈ 10.446

The distance from the top of B to the bottom of B is (2/3) × 10.446 ≈ 6.964

The area of B is twice the area of half of B (no duh!) so consider the "left" half of B,

and use the distance from the center of the circle on the right to the top of B, which is 7 cm.

The distance from the top of B to the middle of B is 3.482. 3.482/7 ≈ 0.4974.  The arcsin of this value × 2 ≈ 59.66°--the angle from the circle's center to the top and bottom of B

Half of area B is the difference between:

  the area of the sector formed by the curve to the left of B and the circle's radii

and

  the area of the triangle formed by the height of B and the circle's radii.

The area of the sector ≈  (1/2) × r² × the 59.66° angle in radians

= (1/2) × 49 × (59.66 * π / 180)

= 8.120π ≈ 25.52

The area of triangle ≈ (1/2) × r² × sin(59.66°) = 49/2 × .863 = 21.144.

So half the area of B is  25.52 - 21.144 = 4.376

The area of B is approximately 8.752 cm²