Area of minor and major segment question (1 Viewer)

BlueGas · Sep 24, 2015

I need help for this question

rand_althor · Sep 24, 2015

$A_1$ is the major segment and $A_2$ is the minor segment. $A_1$ is given by the formula $\frac{1}{2}r^2(\theta-\sin{\theta})$. $A_2$ is the the area of the circle less $A_1$. So $A_2=\pi r^2-\frac{1}{2}r^2(\theta-\sin{\theta})$. Simplifying $\frac{A_2}{A_1}$ will give you the answer.$

BlueGas · Sep 24, 2015

rand_althor said:
$A_1$ is the major segment and $A_2$ is the minor segment. $A_1$ is given by the formula $\frac{1}{2}r^2(\theta-\sin{\theta})$. $A_2$ is the the area of the circle less $A_1$. So $A_2=\pi r^2-\frac{1}{2}r^2(\theta-\sin{\theta})$. Simplifying $\frac{A_2}{A_1}$ will give you the answer.$

I got confused in the simplifying part, can you do it?

rand_althor · Sep 24, 2015

BlueGas said:
I got confused in the simplifying part, can you do it?

$\begin{align*}\frac{A_2}{A_1}&=\frac{\pi r^2-\frac{1}{2}r^2(\theta-\sin{\theta})}{\frac{1}{2}r^2(\theta-\sin{\theta})} \\&=\frac{\frac{2\pi-(\theta-\sin{\theta})}{2}}{\frac{\theta-\sin{\theta}}{2}} \\&=\frac{2\pi-\theta+\sin{\theta}}{\theta-\sin{\theta}} \\\end{align*}$

Area of minor and major segment question (1 Viewer)

BlueGas

Well-Known Member

rand_althor

Active Member

BlueGas

Well-Known Member

rand_althor

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)