The medium side of a triangle is 9 more than 1/2 the shortest side

[mlsdayspa]

Hello, having some difficulty getting the correct answer. If someone could help me me step by step, I would greatly appreciate it

Here is the question...

The medium side of a triangle is 9 more than 1/2 the shortest side, and the longest side is 4 times the shortest side. If the perimeter is 31 in., find the lengths of the sides of the triangle.

Thank you in advance!

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[stapel]

The medium side of a triangle is 9 more than 1/2 the shortest side, and the longest side is 4 times the shortest side. If the perimeter is 31 in., find the lengths of the sides of the triangle.

The "medium" ("middle-length"?) side and the longest side are each defined in terms of the shortest side. What variable did you pick for the length of the shortest side? What expressions did you create (here) for the lengths of the "medium" and longest sides, in terms of this variable? How did you plug these into the "perimeter" formula, along with the perimeter value, to create an equation? Where are you stuck in solving (here) that equation?

Please be complete, starting with your variable. Thank you!

 

[Prove It]

Hello, having some difficulty getting the correct answer. If someone could help me me step by step, I would greatly appreciate it

Here is the question...

The medium side of a triangle is 9 more than 1/2 the shortest side, and the longest side is 4 times the shortest side. If the perimeter is 31 in., find the lengths of the sides of the triangle.

Thank you in advance!

Sent from my SGH-I337M using Tapatalk

From the information given, if we call "s" the shortest side, "m" the medium side, and "l" the longest side, we have

$\displaystyle \begin{align*} m &= \frac{s}{2} + 9 \\ \\ l &= 4\,s \end{align*}$

Now since the perimeter is 31 inches

$\displaystyle \begin{align*} s + m + l &= 31 \\ s + \frac{s}{2} + 9 + 4\,s &= 31 \\ \frac{2\,s}{2} + \frac{s}{2} + 9 + \frac{8\,s}{2} &= 31 \\ \frac{11\,s}{2} + 9 &= 31 \\ \frac{11\,s}{2} &= 22 \\ 11\,s &= 44 \\ s &= 4 \\ \\ m &= \frac{s}{2} + 9 \\ &= \frac{4}{2} + 9 \\ &= 2 + 9 \\ &= 11 \\ \\ l &= 4\,s \\ &= 4 \times 4 \\ &= 16 \end{align*}$

So the shortest side is 4 inches, the medium is 11 inches and the largest is 16 inches.

 

[stapel]

It is generally wise to "Preview" one's posts, especially in order to correct formatting (such as the invalid LaTeX markup). When errors are discovered, one may make corrections by clicking the "Edit Post" button and entering the desired changes in the box.

From the information given, if we call "s" the shortest side, "m" the medium side, and "l" the longest side, we have...

$\displaystyle...

\(\displaystyle s \,+\, m \,+\, l\,=\, 31\)

Since the original poster had not replied in the intervening month, the thread may be regarded as "solved" (since the poster would otherwise have replied with further questions). There is no need to resurrect old threads.

Also, since the poster added this question to the "Pre-Algebra" category, the context should be assumed to be (barely!) one-variable. Systems of equations in three variables are likely to be inappropriate (similar to posting fully-worked solutions, especially when the student has yet to show any effort).

Thank you for your consideration.