The other 3 answers are all correct. But there's a danger lurking in this type of question. Let's say the problem were slightly different, with perimeter = 38, C=A+8 and A=2B. Work it mechanically the same way substituting from 38=A+B+C to get 38=2B+B+2B+8=5B+8 and solve for B, B=6, A=2B=12 C=12+8=20. Check the math, it's all correct. Then the surprise, There is no such triangle with sides 6, 12 and 20. No such triangle exists. The given problem perimeter=35 with solution 6, 12, 17 involves a very flat or obtuse triangle. A little perspective sometimes helps. For the given problem with perimeter = 35, the longest side of a triangle would have to be somewhere between 1/3 and 1/2 of that perimeter, or between about 12 and 17. If the longest side were 17.5, it's no longer a triangle, just a flat line or a "triangle" with angles 0, 0 and 180 degrees. IF you encountered a problem like this that had no solution, maybe that's n unfair trick question. But some instructors will do it to you, sometimes accidentally, as they just make up some numbers that seem to lead to a solution, but where no real triangle can exist with those 3 sides.

Bryan K.

asked • 09/18/12# traingle has perimeter of 35 inches. The length

story problem - P = A +b+c P=35

A= length of B times2

C= A + 5

I have the answers but dont know how to work the equation. 35= 12+6+17

## 4 Answers By Expert Tutors

Jyothi C. answered • 11/07/12

Experienced Math Tutor

We know that the perimeter of triangle is sum of three sides .... step(1)

So P=Lengths of A+B+C =35

but we know A is Length of B times2 , techniqually A=2B and C=A+5

Plug the' A 'and C values in the above equation A+B+C=35 we get 2B+B+A+5=35

3B+2B+5=35

5B=30

B=6

plug the value for B in A=2B we get A=12

and C=A+5 which is 12+5=17

Maria W. answered • 09/18/12

Physicist with a Passion for Tutoring - Specializing in Physics, Math

We have 3 Equations:

Eq.1) 35=A+B+C

Eq. 2) A=2B

Eq. 3) C=A+5

Notice that all three equations have the variable A. What we want to do is to replace B and C in Eq. 1 in terms of A. If we solve Eq. 2 for B, we get that B=(1/2)A. Note that in Eq. 3, C is already solved in terms of A. Now we substitute B=(1/2)A and C=A+5 in to Eq. 1:

35=A+(1/2)A+A+5 (Add like terms)

35=2.5A+5 (subtract 5 from both sides)

30=2.5A (divide by 2.5 on both sides)

A=12

Now plug in A=12 in to Eq. 2, and we find that B=6.

Finally, plug A=12 in to Eq. 3, and we find that C=17.

So P=A+B+C, and 35=12+6+17

Additionally, you can also put Eq. 1 in terms of B as the other answer states, and solve for B.

I hope that helps!

Nigel H. answered • 09/18/12

You Name It Tutoring

Here's a hint. Put C in terms of B. Then you will have A, B, and C in terms of B. Then work with the original equation.

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