Question

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

Answer 1

Answer:

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

Formulas

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

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

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

Question 39

Given:

• r = 7 in
• $\theta$ = 90°

Substitute the given values into the formulas:

Arc AB = 90°

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

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

Question 40

Given:

• r = 6 ft
• $\theta$ = 120°

Substitute the given values into the formulas:

Arc AB = 120°

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

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

Question 41

Given:

• r = 12 cm
• $\theta$ = 45°

Substitute the given values into the formulas:

Arc AB = 45°

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

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