Question 1200184
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Part A


If a triangle has sides a,b,c then all of the following conditions must be true
a < b+c
b < a+c
c < a+b
Any two sides must add to be larger than the third side.
See the <a href = "https://www.mathsisfun.com/geometry/triangle-inequality-theorem.html">triangle inequality theorem</a>.


In option 1, we have: a = 4, b = 4, c = 8
And a+b = 4+4 = 8 which does not exceed c = 8
In other words, c < a+b is not true.
A triangle isn't possible here.
Option 3 is a similar story.


Option 2 on the other hand works
b+c = 8+10 = 18 exceeds a = 6 so a < b+c is true.
a+c = 6+10 = 16 exceeds b = 8 so b < a+c is true.
a+b = 6+8 = 14 exceeds c = 10 so c < a+b is true.
A triangle is possible with sides 6, 8, and 10.


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Part B


Use the pythagorean theorem {{{a^2+b^2 = c^2}}}
Note that a = 6, b = 8, c = 10 satisfy this equation. 
Therefore, option 2 represents a right triangle.


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Part C


Let's look at option 1
Let's say we knew the first two sides a = 4 and b = 4
For a triangle to be possible, the third side c is given by this range 
b-a < c < b+a
where {{{b >= a}}}


See this my answer in this post for more info
<a href = "https://www.algebra.com/algebra/homework/Triangles/Triangles.faq.question.1200226.html">https://www.algebra.com/algebra/homework/Triangles/Triangles.faq.question.1200226.html</a>


The range for c in this case would be
b-a < c < b+a
4-4 < c < 4+4
0 < c < 8


This means we could pick something like c = 5
A triangle is possible with sides a = 4, b = 4, c = 5
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