A=.5(r)^2(theta angle) Area of a sector word problem......?

Question

An automobile windshield wiper 10 in. long rotates through an angle of 60deg. If the rubber part of the blade overs only the last 9 in. of the wiper, find the are of the windshield cleansed by the windshield wiper.

Answer 1

First calculate the area of an arc of 60° with a radius of 10 inches without regard to the gap. Put the angle in radians.
60° = π/6 radians.

A = (1/2)(r^2)θ = (1/2)(10^2)(π/6) = (1/2)(100)(π/6)
= 50π/6 = 25π/3

Now calculate the area of an arc of 60° with a radius of 1 inch.

a = (1/2)(r^2)θ = (1/2)(1^2)(π/6) = (1/2)(1)(π/6) = π/12

Subtract:

A - a = 25π/3 - π/12 = 100π/12 - π/12 = 99π/12 = 33π/4
= 25.9 inches^2

Answer 2

A part of a circle shaped like a pizza slice will have an area:

A = pi r^2 (angle in deg/360) or pi r^2 (angle in rad/2 pi)

The term in parenthesis is the fraction of the circle corresponding to the angle of the sector.

For the problem, the area swept by the1st one inch should be deducted from the area swept by the whole 10 inch.

A = pi (10^2) (60/360) - pi (1^2) (60/360)

Answer 3

In continuation of dax’s answer,

Remove the constant factors out side the bracket.

A = pi (60/360) {10^2 - 1^2}

A = pi (60/360) {10 + 1} {10 - 1}

A = pi (1/6) 99 ( inch)^2