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NO LINKS!!! Find the arc measure and arc length of AB. Then find the area of the sector ABQ.​

NO LINKS!!! Find the arc measure and arc length of AB. Then find the area of the sector-example-1
User Lennaert
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1 Answer

13 votes
13 votes

Answer:

Arc Measure: equal to the measure of its corresponding central angle.

Formulas


\textsf{Arc length}=2 \pi r\left((\theta)/(360^(\circ))\right)


\textsf{Area of a sector of a circle}=\left((\theta)/(360^(\circ))\right) \pi r^2


\textsf{(where r is the radius and the angle }\theta \textsf{ is measured in degrees)}

Question 39

Given:

  • r = 7 in

  • \theta = 90°

Substitute the given values into the formulas:

Arc AB = 90°


\textsf{Arc length of AB}=2 \pi (7) \left((90^(\circ))/(360^(\circ))\right)=3.5 \pi=11.00\:\sf in\:(2\:d.p.)


\textsf{Area of the sector AQB}=\left((90^(\circ))/(360^(\circ))\right) \pi (7)^2=(49)/(4) \pi=38.48\:\sf in^2\:(2\:d.p.)

Question 40

Given:

  • r = 6 ft

  • \theta = 120°

Substitute the given values into the formulas:

Arc AB = 120°


\textsf{Arc length of AB}=2 \pi (6) \left((120^(\circ))/(360^(\circ))\right)=4\pi=12.57\:\sf ft\:(2\:d.p.)


\textsf{Area of the sector AQB}=\left((120^(\circ))/(360^(\circ))\right) \pi (6)^2=12 \pi=37.70\:\sf ft^2\:(2\:d.p.)

Question 41

Given:

  • r = 12 cm

  • \theta = 45°

Substitute the given values into the formulas:

Arc AB = 45°


\textsf{Arc length of AB}=2 \pi (12) \left((45^(\circ))/(360^(\circ))\right)=3 \pi=9.42\:\sf cm\:(2\:d.p.)


\textsf{Area of the sector AQB}=\left((45^(\circ))/(360^(\circ))\right) \pi (12)^2=18 \pi=56.55\:\sf cm^2\:(2\:d.p.)

User Unacorn
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