SOLUTION: Let $\mathcal{R}$ be the circle centered at $(0,0)$ with radius $15.$ The lines $x = 8$ and $y = 1$ divide $\mathcal{R}$ into four regions $\mathcal{R}_1$, $\mathcal{R}_2$, $\math

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Let $\mathcal{R}$ be the circle centered at $(0,0)$ with radius $15.$ The lines $x = 8$ and $y = 1$ divide $\mathcal{R}$ into four regions $\mathcal{R}_1$, $\mathcal{R}_2$, $\math      Log On


   



Question 1204814: Let $\mathcal{R}$ be the circle centered at $(0,0)$ with radius $15.$ The lines $x = 8$ and $y = 1$ divide $\mathcal{R}$ into four regions $\mathcal{R}_1$, $\mathcal{R}_2$, $\mathcal{R}_3$, and $\mathcal{R}_4$. Let $[\mathcal{R}_i]$ denote the area of region $\mathcal{R}_i$. If
[R1] > [R2] > [R3] > [R4],
then find $[\mathcal{R}_1] + [\mathcal{R}_2] + [\mathcal{R}_3] + [\mathcal{R}_4]$.

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.

The sum of areas of 4 regions is the area of the entire circle, which is pi%2A15%5E2 = 3.14159265*225 = use your calculator,

otherwise you will learn nothing from my post.