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# help

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332
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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?

Dec 21, 2018

#1
+6185
+1

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

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Dec 21, 2018
#2
+111321
+1

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

Dec 21, 2018