phoenixio wrote:
If x>0, and two sides of a certain triangle have lengths 2x+1 and 3x+4 respectively, which of the following could be the length of the third side of the triangle?
Indicate all possible lengths.
A) 4x+5
B) x+2
C) 6x+1
D) 5x+6
E) 2x+17
IMPORTANT PROPERTY: If two sides of a triangle have lengths A and B, then . . .
DIFFERENCE between A and B < length of third side < SUM of A and BASIDE: If x > 0, then we can be certain that 3x+4 is greater than 2x+1 Let's plug the two given lengths into the above
property to get: (3x+4) - (2x+1) < length of 3rd side < (3x+4) + (2x+1)
Simplify to get:
x + 3 < length of 3rd side < 5x + 5So any length that
could satisfy the above inequality will be a possible length.
So let's start testing!!!
A) 4x+5
Plug this value into our inequality:
x+3 < 4x+5 < 5x+5If x = 1, the inequality holds.
So answer choice A COULD be the length of the third side.
B) x+2
Plug this value into our inequality:
x+3 < x+2 < 5x+5Let's focus on this part of the inequality:
x+3 < x+2Subtract x from both sides to get:
3 < 2At this point, we can conclude that there are no possible values of x that will satisfy the inequality.
So answer choice B
CANNOT be the length of the third side.
C) 6x+1
Plug this value into our inequality:
x+3 < 6x+1 < 5x+5If x = 1, the inequality holds.
So answer choice C COULD be the length of the third side.
D) 5x+6
Plug this value into our inequality:
x+3 < 5x+6 < 5x+5Let's focus on this part of the inequality:
5x+6 < 5x+5Subtract 5x from both sides to get:
6 < 5At this point, we can conclude that there are no possible values of x that will satisfy the inequality.
So answer choice D
CANNOT be the length of the third side.
E) 2x+17
Plug this value into our inequality:
x+3 < 2x+17 < 5x+5Don't forget that all we need is one value of x that satisfies the above any quality in order to show that the answer choice COULD be the length of the third side.
After a bit of testing I found that when x = 10, the inequality holds.
So answer choice E COULD be the length of the third side.
Answer: A, C, E