Unformatted text preview: 96 Chapter 4 • Radian Measure §4.3 Example 4.10
Find the area of a sector whose arc is 6 cm in a circle of radius 9 cm.
Solution: Using s = 6 and r = 9 in formula (4.6) for the area A , we get A= 1
2 rs = = Note that the angle subtended by the arc is θ = 1
2 (9) (6) = 27 cm2 . s
r = 2
3 rad. Example 4.11
Find the area K inside the belt pulley system from Example 4.7 in Section 4.2.
Solution: Recall that the belt pulleys have radii of 5 cm and 8 cm, and their centers are 15 cm apart.
We showed in Example 4.7 that EF = AC = 6 6, ∠ D AE = 1.37 rad, and ∠ GBF = 1.77 rad. We see
from Figure 4.3.2 that, by symmetry, the total area K enclosed by the belt is twice the area above the
line DG , that is, K = 2 ((Area of sector D AE ) + (Area of rectangle AEFC ) + (Area of triangle △ ABC ) + (Area of sector GBF )) . F 66
E 5 66 5 1.77 3 1.37 D C B A G 15 Figure 4.3.2 Belt pulleys with radii 5 cm and 8 cm
Since AEFC is a rectangle with sides 5 and 6 6, its area is 30 6. And since △ ABC is a right
triangle whose legs have lengths 3 and 6 6, its area is 1 (3) (6 6) = 9 6. Thus, using formula (4.5)
2
for the areas of sectors D AE and GBF , we have K = 2 (Area of sector D AE ) + 30 6 + 9 6 + (Area of sector GBF )
=2 1
2 (5)2 (1.37) + 30 6 + 9 6 + = 338.59 cm2 . 1
2 (8)2 (1.77) ...
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 Fall '11
 Dr.Cheun
 Calculus, Pulley, gBf, belt pulleys, sector GBF, sector D AE

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