# Mon Feb. 27

In Class today we solved area problems involving sectors, segments and anuluses.

The trick in many area problems is to make a plan and think of ways to add and subtract calculable areas to find difficult to find areas. For example, a segment is just a sector minus a triangle.

One of the tasks students are assigned today is to find the measure of an angle given information about the area of a sector:

So, We would compare the area of the sector to the area of the circle

$\dpi{300}\inline \pi r^2 = A$, so since $\dpi{300}\inline r = 24 \text{ cm}$ we have $\dpi{300}\inline \pi \cdot (24)^2 = \pi \cdot 576 = 576 \pi \text{cm}^2$

Now, this means that the shaded area, as a fraction of the total area of the circle is

$\dpi{300}\inline \dfrac{120 \pi \text{ cm}^2}{576 \pi \text{ cm}^2} = \dfrac{5}{24}$

So, since the shaded area is 5/24 of the area of the circle, the central angle must make up 5/24 of 360 degrees, so we find

$\dpi{300}\inline 360^\circ \cdot \dfrac{5}{24} = 75^\circ$

So, the value of $\dpi{300}\inline x$, or the central angle of the sector is $\dpi{300}\inline 75^\circ$

Tonight's assignment is to complete 8.6 on p. 439: 1-12, 17-22