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A strip of paper is available for each whole number length from 1 inch to 15 inches. What is the probability that if a strip is randomly selected, it can be used as the third side of a triangle with two sides of 5 inches and 7 inches? Express your answer as a common fraction.

 Nov 3, 2019
 #1
avatar+8863 
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A strip of paper is available for each whole number length from 1 inch to 15 inches. What is the probability that if a strip is randomly selected, it can be used as the third side of a triangle with two sides of 5 inches and 7 inches? Express your answer as a common fraction.

 

Für jede ganzzahlige Länge von 1 Zoll bis 15 Zoll ist ein Papierstreifen erhältlich. Wie groß ist die Wahrscheinlichkeit, dass ein zufällig ausgewählter Streifen als dritte Seite eines Dreiecks mit zwei Seiten von 5 Zoll und 7 Zoll verwendet werden kann? Drücken Sie Ihre Antwort als gemeinsamen Bruch aus.

 

Hello Guest!

 

\(The\ probability\ is\ \frac{5+7-1}{15}=\color{blue}\frac{11}{15}\)

laugh  !

 Nov 3, 2019
edited by asinus  Nov 3, 2019
 #2
avatar+106539 
+1

We can use  the triangle  inequality to solve this

 

Let   S  be the unknown  side

 

So....

 

S + 5  >  7           S + 7  >   5             7  +   5   >   S

S  >  2                  S  >   -2                   12  >    S

                                                              S   <  12

 

Taking the most restrictive positive interval for S......  

 

S  = ( 2 , 12)

 

We have   9  integers  that  will satisfy this   (3, 4, 5, 6, 7, 8 , 9 , 10 , 11)

 

So....the probability  =     9  / 15    =    3   /  5

 

cool cool cool

 Nov 3, 2019

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