Yes the correct answer is 1/2 because there two of them
B) 154, 171, 188, 205, 222
A constant value (17) is added to get the next number
![\qquad\qquad\huge\underline{{\sf Answer}}](https://tex.z-dn.net/?f=%5Cqquad%5Cqquad%5Chuge%5Cunderline%7B%7B%5Csf%20Answer%7D%7D)
Let's solve ~
Cost price for each Orange :
![\qquad \tt \dashrightarrow \: \dfrac{60}{200}](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctt%20%5Cdashrightarrow%20%5C%3A%20%5Cdfrac%7B60%7D%7B200%7D%20)
![\qquad \tt \dashrightarrow \:0.3 \: \: kobo](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctt%20%5Cdashrightarrow%20%5C%3A0.3%20%5C%3A%20%20%5C%3A%20kobo)
Now, Cost price for 8 oranges :
![\qquad \tt \dashrightarrow \:0.3 \times 8](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctt%20%5Cdashrightarrow%20%5C%3A0.3%20%5Ctimes%208)
![\qquad \tt \dashrightarrow \:2.4 \: \: kobo](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctt%20%5Cdashrightarrow%20%5C%3A2.4%20%5C%3A%20%20%5C%3A%20kobo)
And we have been given that selling price for 8 oranges is 12 kobo.
Therefore, the gain is :
![\qquad \tt \dashrightarrow \:12 - 2.4](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctt%20%5Cdashrightarrow%20%5C%3A12%20-%202.4)
![\qquad \tt \dashrightarrow \:9.6 \: \: kobo](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctt%20%5Cdashrightarrow%20%5C%3A9.6%20%5C%3A%20%20%5C%3A%20kobo)
Percentage gain is ~
![\qquad \tt \dashrightarrow \: \dfrac{9.6}{2.4} \times 100](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctt%20%5Cdashrightarrow%20%5C%3A%20%5Cdfrac%7B9.6%7D%7B2.4%7D%20%20%5Ctimes%20100)
![\qquad \tt \dashrightarrow \:4 \times 100](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctt%20%5Cdashrightarrow%20%5C%3A4%20%5Ctimes%20100)
![\qquad \tt \dashrightarrow \:400 \: \%](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctt%20%5Cdashrightarrow%20%5C%3A400%20%5C%3A%20%20%5C%25)
Answer:
![=(\frac{64}{3}\pi-16\sqrt{3}) in^2](https://tex.z-dn.net/?f=%3D%28%5Cfrac%7B64%7D%7B3%7D%5Cpi-16%5Csqrt%7B3%7D%29%20in%5E2)
Step-by-step explanation:
Area of segment equals area of sector minus area of isosceles triangle.
![=\frac{\theta}{360}\times \pi r^2 -\frac{1}{2} r^2 \sin(\theta)](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Ctheta%7D%7B360%7D%5Ctimes%20%5Cpi%20r%5E2%20-%5Cfrac%7B1%7D%7B2%7D%20r%5E2%20%5Csin%28%5Ctheta%29)
Given; the length of chord, ![d=8\sqrt{3} in.](https://tex.z-dn.net/?f=d%3D8%5Csqrt%7B3%7D%20in.)
and the angle of the sector,
.
We can use the formula for calculating the length of a chord to find the radius of the circle.
![d=2r\sin(\frac{\theta}{2})](https://tex.z-dn.net/?f=d%3D2r%5Csin%28%5Cfrac%7B%5Ctheta%7D%7B2%7D%29)
![8\sqrt{3}=2r\sin(60\degree)](https://tex.z-dn.net/?f=8%5Csqrt%7B3%7D%3D2r%5Csin%2860%5Cdegree%29)
![8\sqrt{3}=2r(\frac{\sqrt{3}}{2})](https://tex.z-dn.net/?f=8%5Csqrt%7B3%7D%3D2r%28%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%29)
![\Rightarrow r=8in.](https://tex.z-dn.net/?f=%5CRightarrow%20r%3D8in.)
Area of segment![=\frac{120}{360}\times \pi \times 8^2-\frac{1}{2}\times 8^2\sin(120\degree)](https://tex.z-dn.net/?f=%3D%5Cfrac%7B120%7D%7B360%7D%5Ctimes%20%5Cpi%20%5Ctimes%208%5E2-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%208%5E2%5Csin%28120%5Cdegree%29)
![=(\frac{64}{3}\pi-16\sqrt{3}) in^2](https://tex.z-dn.net/?f=%3D%28%5Cfrac%7B64%7D%7B3%7D%5Cpi-16%5Csqrt%7B3%7D%29%20in%5E2)
9) measure of angle 4 would be 89°
As a triangle is always 180 so
180-52-39 = 89