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The side lengths of a triangle are 15 units, 47 units and $\frac{x}{2}$ units. How many integer values of $x$ are possible?

Guest Dec 21, 2018

Rom · Answer 1 · 2018-12-21T05:09:41

$32=(47-15)<\dfrac x 2 < (47+15) = 62 \\ 64 < x < 124 \\ x = 65, 66, \dots , 123\\ \text{which is 59 values}$

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CPhill · Answer 2 · 2018-12-21T05:23:15

By the Triangle inequality, we have these possible inequalities

x/2 + 15 > 47

x/2 > 32

x > 64

47 + 15 > x/2

62 > x/2

124 > x

x < 124

x/2 + 47 > 15

x/2 > -32

x > -64

We need to take the most restictive interval that will solve all three inequalities....this is

64 < x < 124

And the number of integers in this interval is 123 - 65 + 1 = 59