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.

The three sides of a scalene triangle have different measures. One measure \(\displaystyle BC\) cannot have is 12, but this is not a choice.

It cannot be true that \(\displaystyle BC = AC\). Since the perimeter is 

\(\displaystyle AB + BC + AC\), we can find out what other value can be eliminated as follows:

\(\displaystyle 12 + BC + BC = 30\)

\(\displaystyle 12 + 2 \cdot BC = 30\)

\(\displaystyle 2 \cdot BC = 18\)

\(\displaystyle BC = 9\)

Therefore, if \(\displaystyle BC = 9\), then \(\displaystyle AC = 9\), and the triangle is not scalene. 9 is the correct choice.

We are looking for ways to add three primes to yield a sum of 43. Two or all three (since an equilateral triangle is considered isosceles) must be equal (although, since 43 is not a multiple of three, only two can be equal).

We will set the shared sidelength of the congruent sides to each prime number in turn up to 19:

\(\displaystyle 3+3 + 37\)

\(\displaystyle 5+5+ 33\)

\(\displaystyle 7+7+ 29\)

\(\displaystyle 11+11+21\)

\(\displaystyle 13+13+15\)

\(\displaystyle 17+17+9\)

\(\displaystyle 19+19 + 5\)

By the Triangle Inequality, the sum of the lengths of the shortest two sides must exceed that of the greatest. We can therefore eliminate the first three. \(\displaystyle 11+11+21\), \(\displaystyle 13+13 + 15\), and \(\displaystyle 17+17+9\) include numbers that are not prime (21, 15, 9). This leaves us with only one possibility:

\(\displaystyle 19+19+5\) - greatest length 19

19 is the correct choice.

The three sides of \(\displaystyle \Delta DEF\) are the midsegments of \(\displaystyle \Delta A BC\), so \(\displaystyle \Delta DEF\) is similar to \(\displaystyle \Delta A BC\).

By the Triangle Midsegment Theorem, each is parallel to one side of \(\displaystyle \Delta A BC\). 

By the same theorem, each has length exactly half of that side, giving \(\displaystyle \Delta A BC\) twice the perimeter of \(\displaystyle \Delta DEF\).

But since the sides of \(\displaystyle \Delta A BC\) have twice the length of those of \(\displaystyle \Delta DEF\), the area of \(\displaystyle \Delta A BC\), which varies directly as the square of a sidelength, must be four times that of \(\displaystyle \Delta DEF\).

The correct choice is the one that asserts that the area of \(\displaystyle \Delta A BC\) is twice that of \(\displaystyle \Delta DEF\).