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One side of a triangle is 9 cm shorter than the base, x. The other side is 17 longer than the base.

What lengths of the base will allow the perimeter of the triangle to be at least 32 cm? Write your
answer as an interval.

1 Answer

2 votes

Answer:

(26, ∞)

Explanation:

(I am assuming the question has been copied and pasted properly because the answer is not as simple as computing inequalities. It also involves using triangle properties)

Let the sides of the triangle be
x, y and z where x is the base

Let y be the side which is 9cm shorter than x:
y = x - 9

Let z be the side which is17cm longer than x:
z = x + 17

Perimeter = x + y + z

= x + (x-9) + (x + 17)

= x + x + x -9 + 17

= 3x + 8

We are told that the perimeter should be at least 32 cm

So 3x + 8 ≥ 32

Subtract 8 from both sides of the inequality:
==> 3x + 8 - 8 ≥ 32 - 8

==> 3x ≥ 24

Divide by 3 both sides:
==> 3x/3 ≥ 24/3

==> x ≥ 8

If these were inequalities in x, then this would be the solution and the interval will be [8, ∞ ]

However, we can see that some of the values for x are impossible within this inequality

For example, for x = 8, y = -1
for x = 9 y = 0 and both these are not possible in a triangle

Since these are the sides of a triangle. In any triangle, the sum of any two sides must be greater than the third side

In this particular triangle, the shortest side is y= x - 9 and the longest side is z = x + 17

So the shortest side + x > longest side

x + y > z

==> x + x -9 > x +17

2x - 9 > x + 17

Moving x to the left side and -9 to the right side

2x - x > 17 + 9

x > 26

So x must be greater than 26 and it can go all the way to infinity. Remember x need not be an integer

The interval is (26, ∞)

User Vijay Pujar
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