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sides of ∆ABC are integers

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by sanju09 » Mon Nov 18, 2013 11:43 pm
The perimeter of ∆ABC is 24. If all sides of ∆ABC are integers, then which of the following is the greatest possible value for the longest side of ∆ABC?
A. 9
B. 10
C. 11
D. 12
E. 13


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by pareekbharat86 » Tue Nov 19, 2013 12:39 am
sanju09 wrote:The perimeter of ∆ABC is 24. If all sides of ∆ABC are integers, then which of the following is the greatest possible value for the longest side of ∆ABC?
A. 9
B. 10
C. 11
D. 12
E. 13


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C?
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by pareekbharat86 » Tue Nov 19, 2013 12:44 am
We know the formula that the length of the longest side (a) of a triangle with sides a,b,c will be less than the sum of the lengths of the remaining sides.

a<b+c

Since the perimeter is 24, we can eliminate D and E.

If 12 is the greatest side then we will get a(=12)= b+c (=12).

If 13 is the greatest side then we will get a(=13)> b+c (=11).

However, C is appropriate since it will satisfy the equation a<b+c i.e. 11<13.

Answer is C.
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by sanju09 » Tue Nov 19, 2013 2:14 am
pareekbharat86 wrote:We know the formula (RULE) that the length of the longest side (a) of a triangle with sides a,b,c will be less than the sum of the lengths of the remaining sides.

a<b+c

Since the perimeter is 24, we can eliminate D and E.

If 12 is the greatest side then we will get a(=12)= b+c (=12).

If 13 is the greatest side then we will get a(=13)> b+c (=11).

However, C is appropriate since it will satisfy the equation a<b+c i.e. 11<13.

Answer is C.
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by Mathsbuddy » Tue Nov 19, 2013 9:36 am
The perimeter of ∆ABC is 24.

Choosing the biggest options:

Longest side + Sum of remaining sides = 24

L + S = 24

13 + 11 = 24 creates a gap, no triangle

12 + 12 = 24 is 1-dimensional, no triangle

11 + 13 = 24 makes a triangle as L < S

No need to try smaller values

Therefore L max = 11

Answer C
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