Question

What is the smallest integer x, for which x, x+5, and 2x−15 can be the lengths of the sides of a triangle?

Answer 1

Answer:

In a triangle with sides a, b and c, we must have the following constraints:

a > 0, b > 0, c > 0

Also, in order to be a valid triangle, every pair of sides must add to be more than the third side.

a + b > c, b + c > a, a + c > b

If we take the sides to be x, x + 5 and 2x - 15, we have the first set of three inequalities:

x > 0

x + 5 > 0

2x - 15 > 0

Rewriting these in terms of x, we have:

x > 0

x > -5

x > 7.5

In order for all of these to be true, x > 7.5

Now do the same thing for the other inequalities:

x + (x + 5) > 2x - 15

(x + 5) + (2x - 15) > x

x + (2x - 15) > x + 5

Rewriting each of these:

x + (x + 5) > 2x - 15

2x + 5 > 2x - 15

--> 5 > -15 (always true)

(x + 5) + (2x - 15) > x

3x - 10 > x

2x > 10

--> x > 5 (already covered by the earlier constraint of x > 7.5)

x + (2x - 15) > x + 5

3x - 15 > x + 5

2x > 20

x > 10

The most restrictive of all of the conditions listed is x > 10. It must be met, and when it is met, all of the others are too. The smallest integer value of x is therefore 11.  (It is the first integer greater than 10.)

Answer:

x = 11

Sides of the triangle = 11, 16, 7

Step-by-step explanation:

Answer 2

Given: x, x+5, and 2x−15.

To find: The smallest integer x for which x, x+5, and 2x−15 can be the lengths of the sides of a triangle.

Solution:

The value of the length of the side (2x-15) must be a positive integer greater than zero. This can be represented using the following inequation.

$2x - 15 > 0$

$2x > 15$

$x > 7.5$

Hence, for (2x-15) to be the length of one of the sides of the triangle, x must be greater than 7.5. The integer that is greater than 7.5 is 8. The other two sides would be a positive integer anyway because a constant is not being subtracted from them.

Therefore, the smallest integer x for which x, x+5, and 2x−15 can be the lengths of the sides of a triangle is 8.