Question

activity 2. solve what is being asked then round your answers to the nearest tenths. use π = 3.1415 to approximate your answers. show your complete solution in your answer sheet. 1) if the diameter of a circle is 6 units, find the area of a sector with a sector - angle of 80°. 2) find the area of a sector with a radius of 10 cm and a central angle of 135°. 3) a circle with center at e has a radius of 7 inches. if ∠rea forms a 121° angle, what would be the length of ra? 4) a circle has a central angle measuring 108° and a radius of 11 units. solve for the area of the sector as well as the length of the arc. for number 5 to 7, use circle o. 5) what is the length of me? 6) what is the area of sector moe? 7) what fraction of the circumference of the circle is me? 8) an arc length of 21.5 cm is intercepted by a sector with an area of 150.5 cm². what is the length of the circles diameter? 9) find the area of a segment of a circle with a radius of 6 meters and a central angle of 150°. 10) the central angle of a circle measures 60° and its radius is 4 cm, find the area of the segment formed by this central angle.

Explanation:

Step1: Recall sector - area formula

The formula for the area of a sector of a circle is $A=\frac{\theta}{360}\times\pi r^{2}$, where $\theta$ is the central - angle of the sector in degrees and $r$ is the radius of the circle.

Step2: Solve problem 1

Given that the diameter $d = 6$ units, then the radius $r=\frac{d}{2}=3$ units and $\theta = 80^{\circ}$.
$A=\frac{80}{360}\times3.1415\times3^{2}=\frac{80}{360}\times3.1415\times9=\frac{80\times3.1415\times9}{360}=6.283\approx6.3$ square units.

Step3: Solve problem 2

Given $r = 10$ cm and $\theta=135^{\circ}$.
$A=\frac{135}{360}\times3.1415\times10^{2}=\frac{135}{360}\times3.1415\times100=\frac{135\times3.1415\times100}{360}=117.80625\approx117.8$ square cm.

Step4: Recall arc - length formula

The formula for the length of an arc of a circle is $s=\frac{\theta}{360}\times2\pi r$, where $\theta$ is the central - angle of the arc in degrees and $r$ is the radius of the circle.

Step5: Solve problem 3

Given $r = 7$ inches and $\theta = 121^{\circ}$.
$s=\frac{121}{360}\times2\times3.1415\times7=\frac{121\times2\times3.1415\times7}{360}\approx14.7$ inches.

Step6: Solve problem 4 (area of sector)

Given $r = 11$ units and $\theta = 108^{\circ}$.
$A=\frac{108}{360}\times3.1415\times11^{2}=\frac{108}{360}\times3.1415\times121 = 114.0426\approx114.0$ square units.

Step7: Solve problem 4 (length of arc)

$s=\frac{108}{360}\times2\times3.1415\times11=\frac{108\times2\times3.1415\times11}{360}=20.73456\approx20.7$ units.

Step8: Solve problem 5

Assume the central - angle of arc $ME$ is $\theta$ (not given in the problem - statement, if we assume it is a quarter - circle $\theta = 90^{\circ}$ and $r = 11$ in).
$s=\frac{\theta}{360}\times2\pi r$, if $\theta = 90^{\circ}$, then $s=\frac{90}{360}\times2\times3.1415\times11=\frac{1}{4}\times2\times3.1415\times11 = 17.27825\approx17.3$ inches.

Step9: Solve problem 6

If $\theta = 90^{\circ}$ and $r = 11$ in, $A=\frac{90}{360}\times3.1415\times11^{2}=\frac{1}{4}\times3.1415\times121 = 94.98575\approx95.0$ square inches.

Step10: Solve problem 7

The fraction of the circumference that arc $ME$ represents is $\frac{\theta}{360}$. If $\theta = 90^{\circ}$, the fraction is $\frac{90}{360}=\frac{1}{4}$.

Step11: Recall the relationship between arc - length, area of sector and radius

We know that $A=\frac{1}{2}rs$ and $s$ is the arc - length and $A$ is the area of the sector. Given $s = 21.5$ cm and $A = 150.5$ cm².
First, from $A=\frac{1}{2}rs$, we can find $r=\frac{2A}{s}=\frac{2\times150.5}{21.5}=14$ cm. Then the diameter $d = 2r=28$ cm.

Step12: Recall the formula for the area of a segment

The area of a segment $A_{segment}=A_{sector}-A_{triangle}$.
For a central - angle $\theta$ and radius $r$, $A_{sector}=\frac{\theta}{360}\times\pi r^{2}$ and for a triangle with two sides equal to the radius $r$ and included - angle $\theta$, $A_{triangle}=\frac{1}{2}r^{2}\sin\theta$.

Step13: Solve problem 9

Given $r = 6$ m and $\theta = 150^{\circ}$.
$A_{sector}=\frac{150}{360}\times3.1415\times6^{2}=\frac{150}{360}\times3.1415\times36 = 47.1225$ square m.
$A_{triangle}=\frac{1}{2}\times6\times6\times\sin150^{\circ}=\frac{1}{2}\times6\times6\times\frac{1}{2}=9$ square m.
$A_{segment}=47.1225 - 9=38.1225\approx38.1$ square m.

Step14: Solve problem 10

Given $r = 4$ cm and $\theta = 60^{\circ}$.
$A_{sector}=\frac{60}{360}\times3.1415\times4^{2}=\frac{1}{6}\times3.1415\times16\approx8.3773$ square cm.
$A_{triangle}=\frac{1}{2}\times4\times4\times\sin60^{\circ}=\frac{1}{2}\times4\times4\times\frac{\sqrt{3}}{2}=4\sqrt{3}\approx6.9282$ square cm.
$A_{segment}=8.3773 - 6.9282 = 1.4491\approx1.4$ square cm.

Answer:

1. $6.3$ square units
2. $117.8$ square cm
3. $14.7$ inches
4. Area of sector: $114.0$ square units, Length of arc: $20.7$ units
5. $17.3$ inches (assuming $\theta = 90^{\circ}$)
6. $95.0$ square inches (assuming $\theta = 90^{\circ}$)
7. $\frac{1}{4}$ (assuming $\theta = 90^{\circ}$)
8. $28$ cm
9. $38.1$ square m
10. $1.4$ square cm