A windshield wiper assembly has a total length of 20 inches from its pivot point to its ti… — Armed Services Vocational Aptitude Battery (ASVAB)

A windshield wiper assembly has a total length of $20$ inches from its pivot point to its tip. The innermost $4$ inches of the assembly is a bare metal arm, and the outer $16$ inches is the rubber wiper blade. When turned on, the assembly sweeps back and forth through an angle of $135^\circ$ . What is the total area of the windshield that is wiped by the rubber blade during one sweep? (Leave your answer in terms of $\pi$ )

$144\pi$ sq in.Cevap
B
$96\pi$ sq in.
C
$36\pi$ sq in.
D
$12\pi$ sq in.

Cevap

The correct answer is

144\pi

sq in.

The region wiped by the blade forms a shape called an annulus sector. To find its area, we calculate the area of the larger sector created by the full assembly (radius = 20) and subtract the area of the smaller un-wiped sector created by the metal arm (radius = 4). Both sectors cover a

135^\circ

angle, which is

\frac{3}{8}

of a full circle. Large sector area is

\frac{3}{8}\pi(20^2) = 150\pi

, and small sector area is

\frac{3}{8}\pi(4^2) = 6\pi

. The difference is

144\pi

Adım Adım Çözüm

Identify the geometric shapes involved in the problem.

The total area swept is a large circular sector with a radius of

20

inches. The inner un-wiped area is a smaller circular sector with a radius of

4

inches. The wiped area is the difference between them.

The rubber blade starts

4

inches away from the pivot, leaving a small sector un-wiped.

Determine the fraction of a full circle the wiper sweeps.

\frac{135^\circ}{360^\circ} = \frac{3}{8}

of a circle.

Sector area requires knowing what fraction of the total

360^\circ

circle is being used.

Calculate the area of the large sector (total sweep).

\text{Area} = \frac{3}{8} \times \pi \times (20)^2 = \frac{3}{8} \times 400\pi = 150\pi \text{ sq in.}

This finds the entire area enclosed by the 20-inch assembly.

Calculate the area of the small un-wiped sector.

\text{Area} = \frac{3}{8} \times \pi \times (4)^2 = \frac{3}{8} \times 16\pi = 6\pi \text{ sq in.}

This finds the area swept only by the bare metal arm.

Subtract the un-wiped area from the total area.

150\pi - 6\pi = 144\pi \text{ sq in.}

Subtracting gives the area exclusively wiped by the rubber blade.

Anahtar Kavram

Sector Area of an Annulus

Soru

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