2. Which of the following groups of side lengths could form a triangle?  3, 4, and 5  7, 15, and 8  32, 12, and 10

3. Only the first group of sides (3, 4, and 5) could form a triangle, since 3+4 > 5, 4+5 > 3, and 3+5 > 4. The second group (7, 15, and 8), fails because 7+8 = 15: the sum of the two shorter sides must be strictly greater than, not equal to the length of the third side. The last group (32, 12, and 10) also fails because 12 + 10 < 32.

4.  In the figure to the right, what is the smallest possible integer value of x? What is the largest possible integer value?

5. Since the smallest value of x must still fulfill the triangle inequality, we can say that 10 + x > 12, since 12 will be the longest side for small values of x. Thus, x must be greater than 2, so the smallest possible integer value of x is 3. The largest value of x must also fulfill the triangle inequality, meaning that it must be small enough that the other two sides still add to be greater than x. Thus, we have 12+10 > x, or x<22. The largest possible integer value of x, then, is 21.

6. What is the largest possible perimeter of a triangle with side lengths of 5x, 3x+6, and x+2?

7. Using the triangle inequality, we can make three inequalities: 5x + 3x + 6 > x+2 3x+6 + x + 2 > 5x 5x + x + 2 > 3x+6 Solving them, we obtain: X > -1 X < 8 X > 4/3 This tells us that the triangle with the longest possible sides occurs when x is just less than 8. We can use this to estimate its perimeter. Its sides will have lengths of about 10, 30, and 40. Technically, these side lengths don’t form a triangle, but we can imagine a triangle with slightly smaller sides that barely fulfills the triangle inequality, but still has a perimeter that’s nearly the same. Thus, the perimeter of the largest possible triangle with these side lengths is about 80.