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# MATHCOUNTS State

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The region shown is bounded by the arcs of circles having radius 4 units, having a central angle measure of 60 degrees and intersecting at points of tangency. The area of the region can be expressed in the form $$a\sqrt{b}+c\pi$$ square units, where $$\sqrt{b}$$ is a radical in simplest form. What is the value of $$a+b+c$$?

benjamingu22  Sep 5, 2017

#2
+5256
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First, let's connect the points of tangency to each other to make a triangle.

Then.....the area in question  =  the area of this triangle - the areas of the segments

To find the area of each segment, let's draw a circle that is a part of one of these arcs.

Then draw two radii to the points of tangency.

Since the central angle = 60° , and the two blue sides = 4    ,

this is an equilateral triangle where each angle is 60° .

So, the purple sides also are  4  units long.

area of segment  =     area of sector    -   area of triangle

area of segment  =  (60º)(π 42) / 360º  -  (1/2)(4)(4 sin 60° )

area of segment  =           8π / 3         -             4√3

Now, we just found the area of the triangle inside the circle = 4√3 .  The area of the triangle outside the circle is also an equilateral triangle with sides 4 units long, so its area also  =  4√3  .

the area in question  =  4√3  -  3(8π / 3 -  4√3)

=  4√3  -  8π  +  12√3

= 16√3 + -8π

And.....    16 + 3 + -8  =  11

hectictar  Sep 5, 2017
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#1
+78755
+2

Here's one possibility.......connect the "vertices" of this figure.....this will form an equilateral triangle with sides = 4 units

And the area bounded by one side of this triangle and one side of the figure will be =

[ (1/2) (4)^2 [pi]/3   -   16 sqrt (3) / 4 ] =   [ (8/3)pi  - 4sqrt (3) ]

And since there are three of these areas, the sum of these =  [8pi - 12 sqrt (3) ]     (1)

And the area formed formed by the equilateral triangle is   (4)^2 sqrt (3) / 4  =  4sqrt (3)    (2)

So......the area of the figure shown  =   (2)  - (1)  =    4sqrt (3)  - [ 8pi  - 12sqrt (3) ]  =  [16sqrt (3) - 8pi] square units

So   a = 16, b = 3  and c = -8  ....so a + b + c  = 11

Thanks to hectictar for catching my error....!!!!

CPhill  Sep 5, 2017
edited by CPhill  Sep 5, 2017
#2
+5256
+3

First, let's connect the points of tangency to each other to make a triangle.

Then.....the area in question  =  the area of this triangle - the areas of the segments

To find the area of each segment, let's draw a circle that is a part of one of these arcs.

Then draw two radii to the points of tangency.

Since the central angle = 60° , and the two blue sides = 4    ,

this is an equilateral triangle where each angle is 60° .

So, the purple sides also are  4  units long.

area of segment  =     area of sector    -   area of triangle

area of segment  =  (60º)(π 42) / 360º  -  (1/2)(4)(4 sin 60° )

area of segment  =           8π / 3         -             4√3

Now, we just found the area of the triangle inside the circle = 4√3 .  The area of the triangle outside the circle is also an equilateral triangle with sides 4 units long, so its area also  =  4√3  .

the area in question  =  4√3  -  3(8π / 3 -  4√3)

=  4√3  -  8π  +  12√3

= 16√3 + -8π

And.....    16 + 3 + -8  =  11

hectictar  Sep 5, 2017
#3
+78755
+1

Ah....thanks, hectictar.... I was in a rush.....!!!!

CPhill  Sep 5, 2017

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