Mon Feb. 27

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

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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:

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

pi r^2 = A, so since r = 24 text{ cm} we have 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

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

360^circ cdot dfrac{5}{24} = 75^circ

So, the value of x, or the central angle of the sector is 75^circ

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