Question

question 9 (1 point)
a fan blade rotates through a 150° angle. the blade reaches 18 inches from the center.
what is the area of the region swept by the blade?
______ in²
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question 10 (1 point)
a fireworks burst spreads in a 210° sector of a circle with radius 20 m.
find the area covered by the fireworks.
______ m²
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Explanation:

Question 9

Step1: Recall the formula for the area of a sector

The formula for the area of a sector of a circle with radius \( r \) and central angle \( \theta \) (in degrees) is \( A=\frac{\theta}{360}\times\pi r^{2} \). Here, \( \theta = 150^{\circ} \) and \( r = 18 \) inches.

Step2: Substitute the values into the formula

First, calculate \( r^{2}=18^{2} = 324 \). Then, \( \frac{\theta}{360}=\frac{150}{360}=\frac{5}{12} \). Now, multiply by \( \pi r^{2} \): \( A=\frac{5}{12}\times\pi\times324 \).

Step3: Simplify the expression

\( \frac{5}{12}\times324 = 5\times27=135 \). So, \( A = 135\pi \approx 135\times3.1416 = 424.116 \) (if we use \( \pi\approx3.1416 \)). But we can leave it in terms of \( \pi \) or as a decimal. If we use \( \pi\approx3.14 \), \( 135\times3.14 = 423.9 \).

Step1: Recall the sector area formula

The formula for the area of a sector is \( A=\frac{\theta}{360}\times\pi r^{2} \), where \( \theta = 210^{\circ} \) and \( r = 20 \) m.

Step2: Substitute the values

First, \( r^{2}=20^{2}=400 \). Then, \( \frac{\theta}{360}=\frac{210}{360}=\frac{7}{12} \). Now, \( A=\frac{7}{12}\times\pi\times400 \).

Step3: Simplify the expression

\( \frac{7}{12}\times400=\frac{2800}{12}=\frac{700}{3}\approx233.333 \). So, \( A=\frac{700}{3}\pi\approx\frac{700}{3}\times3.1416\approx733.04 \) (using \( \pi\approx3.1416 \)). If we use \( \pi\approx3.14 \), \( \frac{700}{3}\times3.14=\frac{2198}{3}\approx732.67 \).

Answer:

\( 135\pi \) (or approximately \( 424.12 \))

Question 10