We can view the horizontal segment connecting \(\displaystyle (-7,0), (5,0)\), and \(\displaystyle (5,0)\) as the base; its length wiill be \(\displaystyle 5 - (-7) = 12\). The height will be the perpendicular (vertical) distance to this segment from the opposite point \(\displaystyle (M,N)\), which is \(\displaystyle N\), the \(\displaystyle y\)-coordinate; therefore, the area of the triangle will be half the product of these two numbers, or

\(\displaystyle A = \frac{1}{2} \cdot 12 \cdot N = 6N\).

This can be illustrated by showing this triangle inscribed inside a rectangle whose vertices are \(\displaystyle (0,0), (5,0),(0,6),(5.6)\):

The area of the white triangle \(\displaystyle A _{W}\) is the one whose area we calculate. To do this, we need the area of the square:

\(\displaystyle A_{\square } = 5 \cdot6 = 30\)

The area of the red triangle:

\(\displaystyle A_{1} = \frac{1}{2} \cdot 5 \cdot 2 = 5\)

The area of the green triangle:

\(\displaystyle A_{2} = \frac{1}{2} \cdot 6 \cdot 2 = 6\)

And the area of the beige triangle:

\(\displaystyle A_{3} = \frac{1}{2} \cdot 4 \cdot 3 = 6\)

The area of the white triangle will be as follows:

\(\displaystyle A _{W} = A_{\square } - \left (A_{1} + A_{2} + A_{3} \right )\)

\(\displaystyle = 30 - (5+6+6)\)

\(\displaystyle = 30 - 17 = 13\)

By the Triangle Inequality, the sum of the lengths of two shortest sides must exceed that of the third.

Case 1: \(\displaystyle C\) is the greatest of the three sidelengths. 

Then \(\displaystyle C < 5 + 11 = 16\)

Case 2: \(\displaystyle C\) is not the greatest of the three sidelengths - that is, 11 is.

Then \(\displaystyle 11 < C + 5\), or, equivalently, \(\displaystyle 6 < C\).

Therefore, \(\displaystyle 6 < C < 16\).

If two sides are 4 and 8, then the third side must be greater than  \(\displaystyle 8-4\) and less than \(\displaystyle 8+4\). This means \(\displaystyle z\) can be 5, 6, 7, 8, 9, 10, or 11. 

The triangle can exist by the Triangle Inequality, since the sum of the two smaller sides exceeds the greatest:

\(\displaystyle 15 + 17 = 32 > 21\)

To determine whether the triangle is acute, right, or obtuse, add the squares of the two smaller sides, and compare the sum to the square of the largest side.

\(\displaystyle 15^{2} + 17^{2} = 225 +289 =514 > 441 = 21^{2}\)

Since this sum is greater, the triangle is acute.

If these are the measures of the interior angles of a triangle, then they total \(\displaystyle 180^{\circ }\). Add the expressions, and solve for \(\displaystyle x\).

\(\displaystyle x + (x+40) + (x+80) = 180\)

\(\displaystyle 3x + 120 = 180\)

\(\displaystyle 3x = 60\)

\(\displaystyle x=20\)

One angle measures \(\displaystyle x = 20^{\circ }\) The others measure:

\(\displaystyle x+40 = 20 + 40 = 60^{\circ }\)

\(\displaystyle x+80 = 20 + 80 = 100^{\circ }\)

Since the largest angle measures greater than \(\displaystyle 90^{\circ }\), the angle is obtuse, and the triangle is as well. Since the three angles each have different measure, their opposite sides do also, making the triangle scalene.

The three sides of a scalene triangle have different measures, so 15 can be eliminated.

By the Triangle Inequality, the sum of the lengths of the two smaller sides must exceed the length of the third side. Since \(\displaystyle 7+8 =15\), 8 violates this theorem; since \(\displaystyle 7+15 = 22\), 22 does as well. 

10 is a valid measure of the third side, since \(\displaystyle 7 + 10 < 15\); it makes all three segments of different length, so it is the correct choice.

By trial and error, we get four ways to add distinct primes to yield sum 33:

\(\displaystyle 3 + 7 + 23\) 

\(\displaystyle 3+ 11 + 19\)

\(\displaystyle 3+ 13+ 17\)

\(\displaystyle 5 + 11 + 17\)

In each case, however the Triangle Inequality is violated - the sum of the two shortest lengths does not exceed the third. 

No triangle can exist as described.

A scalene triangle has three sides of different lengths, so we are looking for three distinct prime integers whose sum is a prime integer. 

One of the sides cannot be 2, since the sum of 2 and two odd primes would be an even number greater than 2, a composite number. Therefore, beginning with the least three odd primes, add increasing triples of distinct prime numbers, as follows, until a solution presents itself:

\(\displaystyle 3+5+7 =15\) - incorrect

\(\displaystyle 3+5+11 =19\) - correct

The correct answer, 19, presents itself quickly.

A scalene triangle has three sides of different lengths, so we are looking for three distinct prime integers whose sum is 47. 

There are ten ways to add three distinct primes to yield sum 47:

\(\displaystyle 3 + 5+39\)

\(\displaystyle 3+7+37\)

\(\displaystyle 3+13+31\)

\(\displaystyle 5+11+31\)

\(\displaystyle 5+13+29\)

\(\displaystyle 5+19+23\)

\(\displaystyle 7 + 11+29\)

\(\displaystyle 7 + 17+23\)

\(\displaystyle 11+13+23\)

\(\displaystyle 11+17+19\)

By the Triangle Inequality, the sum of the lengths of the shortest two sides must exceed that of the greatest. We can therefore eliminate all but four:

\(\displaystyle 5+19+23\)

\(\displaystyle 7 + 17+23\)

\(\displaystyle 11+13+23\)

\(\displaystyle 11+17+19\)

The greatest possible length of the longest side is 23.