The perimeter of a triangle with integer side lengths is equal to 15.

Question

The perimeter of a triangle with integer side lengths is equal to 15. How many such triangles are possible?

A. 5
B. 6
C. 7
D. 8
E. 9

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GMATinsight · Accepted Answer

a+b+c = 15

if c = longest then a+b > c
 i.e.  (a+b)_{min} = 8  and  c_{max} = 7

If c = 7 
(a,b) = (1,7)(2,6)(3,5)(4,4)   i.e. 4 cases

If c = 6 
(a,b) = (3,6)(4,5)   i.e. 2 cases

If c = 5 
(a,b) = (5,5)   i.e. 1 case

Total Cases = 7

Answer: Option C

GMATWhizTeam · Answer

Solution: 
 
• All the sides of the triangle are integers• Perimeter of the triangle = 15 units• We will use the sum property of the triangle.  
o The sum of two sides of triangle > Length of the third sideo Let’s maximize the value of third side, which is only possible when the sum of two sides is minimum  
 The sum of two sides = 8 Third side = 7 
  
• Difference between the two sides < Third side 
  
o Difference between two sides < 7  
 The sum of two sides must be 8 and the difference should be minimum. 
o The possibility is difference between the two sides is 0. 
  
 0 < Third side

• The possible length of third side is between 0 and 8, 
  
o There are 7 possibilities of triangle.

Hence, the correct answer is  Option C.

freedom128 · Answer

The perimeter of a triangle with integer side lengths (x, y, z) is equal to 15. So, x+y+z=15

Let z be (one of) the longest side ,
x+y>z --> 15-z > z --> z < 7.5
z value can be 5,6,7

z=7  --> x+y=8
(x,y) = (1,7),(2,6),(3,5),(4,4) ... 4 nos.

z=6  --> x+y=9
(x,y) = (3,6),(4,5) ... 2 nos.

z=5  --> x+y=10
(x,y) = (5,5) ... 1 no.

So, there are 7 triangles possible

FINAL ANSWER IS (C)

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unraveled · Answer

chondro48 This is how i did it. Here a +b + c = 15. The easiest of the triangle that comes right at first is an equilateral triangle since 15 is a multiple of 3. From there we need to know the maximum and minimum length a side can take. We also know that sides will always be > 0 So, either of a,b or c can be > 0 Breaking the perimeter sum in two parts as per the sum property of the triangle and taking any two sides together and the other separately, we have (a+b) + c = 15; a + b > c With all sides being integers c can take 7 as its maximum value. The least value of c possible is 1. Hence 0 < c < 8 i.e. 7 different values. So the possible triangles are: (a,b,c) = (5,5,5) ;(4,5,6); (4,4,7); (3,5,7); (3,6,6); (2,6,7); (1,7,7) Answer C.

freedom128 · Answer

I think monikakumar wants to read your worthy explanation.

ScottTargetTestPrep · Answer

Since all the answer choices are less than 10, we can just list all the possible triangles. Recall that in order to be the lengths of the sides of a triangle, the sum of the lengths of the two shortest sides must be greater than the longest side. The possible integer side lengths of the triangles are:

1) {1, 7, 7}

2) {2, 6, 7}

3) {3, 6, 6}

4) {3, 5, 7}

5) {4, 4, 7}

6) {4, 5, 6}

7) {5, 5, 5}

Answer: C

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The perimeter of a triangle with integer side lengths is equal to 15.

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The perimeter of a triangle with integer side lengths is equal to 15.

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