We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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