Question

Difficulty: MediumCircle Geometry: Arc Length and Sector Area

A circular grinding wheel has a radius of 99 inches. A section of the wheel bounded by a central angle of 140140^\circ is coated with a coarse abrasive layer. What is the area, in square inches, of the remaining non-coated section of the grinding wheel?

  1. A
    11π11\pi
  2. B
    31.5π31.5\pi
  3. 49.5π49.5\piAnswer
  4. D
    63π63\pi
  5. E
    99π99\pi

Answer

The area of the non-coated section of the grinding wheel is 49.5π49.5\pi square inches.
The central angle of the entire circle is 360360^\circ. Since the coated portion occupies 140140^\circ, the non-coated portion occupies 360140=220360^\circ - 140^\circ = 220^\circ. The total area of the circle is π(9)2=81π\pi (9)^2 = 81\pi. Multiplying the total area by the fraction 220360\frac{220^\circ}{360^\circ} yields 111881π=49.5π\frac{11}{18} \cdot 81\pi = 49.5\pi square inches.

Step-by-Step Solution

1
Find the central angle of the non-coated section
360140=220360^\circ - 140^\circ = 220^\circ
A full circle measures 360360^\circ, so subtracting the coated angle gives the non-coated angle.
2
Calculate the total area of the circular grinding wheel
Areatotal=πr2=π(92)=81π sq in\text{Area}_{\text{total}} = \pi r^2 = \pi (9^2) = 81\pi\text{ sq in}
The total area formula for a circle with radius rr is πr2\pi r^2.
3
Calculate the area of the non-coated sector
Areasector=220360×81π=1118×81π=49.5π sq in\text{Area}_{\text{sector}} = \frac{220^\circ}{360^\circ} \times 81\pi = \frac{11}{18} \times 81\pi = 49.5\pi\text{ sq in}
Multiply the total area by the fraction of the circle formed by the central angle.

Key Concept

Sector Area of a Circle
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