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What is the area of the shaded region?

 

 Nov 10, 2019
 #1
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It is a triangle inside a circle.. 

Notice both sides of the triangle (Left and right) are radii of the circle so they are equal in length. 

The angle 120 is between both sides (4cm each) 

If we want the area, there is 2 ways to find it.. 

We can really just constract a right angle triangle from the 120 degrees and use trig to find the rest. 

Other method (I prefer) Is to find the opposite side of 120 degrees, and then apply Heron's formula. 

First of all let's label the sides 

Let the side opposite to angle 120 be c, 

Let the side on the left be b,

Let the side on the right be a. 

Using the law of cosines, we know that:

c^2=a^2+b^2-2ab*cos(x) , Let the angle 120=x 

So just applying the formula we get,

c^2=16+16-32*cos(120)

cos(120)=-0.5 

32-32(-0.5)=48 

c^2=48

c=sqrt(48) Positive or negative but since we are finding a side it must be positive. 

Now let's apply heron's formula 

which is:

Area=sqrt((s(s-a)(s-b)(s-c)) where s=(a+b+c)/2 

we know that,

a=4

b=4

c=sqrt(48)

adding these together we get:

4+4+sqrt(48) =14.928203 approx.=15

so s=15

just subsituiting, 

sqrt((15(15-4)(15-4)(15-sqrt(48))=121.038469 approx=121

So the area is 121 cm^2

I am pretty sure there is a faster solution.

Maybe using inscribed theorem etc..  

 Nov 10, 2019
 #2
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Thanx, Guest...... you should check to see if you think your answer is 'reasonable'....121 cm^2  seems wwwwaaaayyyy high doesn't it?  (even a square of 4cm sides has but 16 cm^2 area)

 

Area of isoceles triangle = 1/2 s2 sin (theta)  =  1/2 42 sin 120 = 6.93 cm^2

 Nov 10, 2019
 #3
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Notice that if you divided that isosceles triangle into two congruent parts, they form 30-60-90 triangles.

 

You can easily solve from there.

 Nov 10, 2019
 #4
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Much better solution!! 

Guest Nov 10, 2019
 #5
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This is why there is a controversy to whether someone should learn calculus at a young age.

 

In my opinion, it is not good to learn calculus. Because the extra knowledge my blur your vision on seeing simple solutions.

CalculatorUser  Nov 10, 2019
 #7
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Guest's solution doesn't use calculus, it uses triginometry.

 

But you get what I mean.

 

A simple knowledge of 30 60 90 well help!

CalculatorUser  Nov 10, 2019
 #6
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Using the method that CalculatorUser said, you split the triangle into two 30-60-90 triangles. We get that the height of the isosceles triangle is simply just 2, and the base is 4sqrt3. Using the formula for triangle area, we get that the area of the triangle 4sqrt3.

 

(That calculus method looks hard!)

 Nov 10, 2019
 #8
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What is the area of the shaded region?

laugh

 Nov 10, 2019
edited by Omi67  Nov 10, 2019
 #9
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I don't know.

 

Omi's solution might be even simpler using trig!

CalculatorUser  Nov 10, 2019
 #10
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Drop a line from the apex of the triangle to form 2 triangles then use law of sines:

Sin30 / height    = sin 90 / 4         

4 sin 30 = height  = 2

 

sin 60 / base = sin 90 / 4

4 sin 60 = base  = 2 sqrt 3

 

area of each triangle = 1/2 base * height

1/2 (2) * 2 sqrt 3 = 2 sqrt 3        for each triangle (remember, there are two)         Area = 4 sqrt 3

 Nov 10, 2019

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