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**+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 4^{2}) / 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

#1**+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

#2**+3 **

Best Answer

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 4^{2}) / 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