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3 semicircles within a semicircle - area

by Andrew
(New York)

3 semicircles within a semicircle

3 semicircles within a semicircle










































Three semicircles of radius 1 are constructed on diameter AB of a semicircle of radius 2. The centers of the small semicircles divide AB into four line segments of equal length, as shown. What is the exact area of the shaded region that lies within the large semicircle but outside the smaller semicircles?

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Mar 10, 2012
3 semicircles within a semicircle - area
by: Staff


Question:

by Andrew
(New York)


Three semicircles of radius 1 are constructed on diameter AB of a semicircle of radius 2. The centers of the small semicircles divide AB into four line segments of equal length, as shown. What is the exact area of the shaded region that lies within the large semicircle but outside the smaller semicircles?


Answer:

As I’m sure you already know, the area of the shaded region:

Area_shaded = Area_large_semicircle - Area_small_semicircles


The area of the large semicircle is easy to compute:

Area_large_semicircle = ½ * π * (radius of large semicircle)²

Area_large_semicircle = ½ * π * (2)²

Area_large_semicircle = ½ * π * 4

Area_large_semicircle = 2π



Finding the area of the small semicircles is a little more involved.

Take a look at the following diagram.


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http://www.solving-math-problems.com/images/3_semicircles_within_a_semicircle-area-2012-03-10_01.jpg



The area of the small semicircles is equal to:

Area_small_semicircles = the area of “FIVE” 60° sectors of a semicircle with a radius of 1 (these sectors are shown in gold on the diagram) + the area of “TWO” equilateral triangles with sides equal to 1 (these are shown in green on the diagram)


the area of “FIVE” 60° sectors of a semicircle with a radius of 1 (these sectors are shown in gold on the diagram) = 5 * (1/6 ) * π * (radius of small semicircles)²
= 5 * (1/6 ) * π * (1)²

= (5/6) * π

= (5π /6)



The area of “TWO” equilateral triangles with sides equal to 1 (these are shown in green on the diagram) = 2 * (1/4) * √(3) * (side)²


= 2 * (1/4) * √(3) * (1)²

= √(3)/2


Area_shaded = Area_large_semicircle - Area_small_semicircles

Area_shaded = 2π - [(5π /6) + √(3)/2]

Area_shaded = 2π - [(5π /6) + 3√(3)/6]

Area_shaded = 12π/6 - [(5π + 3√(3)]/6)

Area_shaded = [12π - 5π - 3√(3)]/6

Area_shaded = [7π - 3√(3)]/6

Area_shaded = 7π/6 - √(3)/2

Area_shaded = (7/6) * π - (1/2) * √(3)

Area_shaded = (7/6)π - (1/2)√(3)


>>> the final answer is: (7/6)π - (1/2)√(3)



Thanks for writing.

Staff
www.solving-math-problems.com



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