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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
avatar+4394 
+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
avatar+98005 
+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

 

 

cool cool cool

 Dec 21, 2018

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