Question

The perimeter of a triangle is 70cm. The longest side is 4cm less than the sum of the other two sides. Twice the shortest side is 9cm less than the longest side. Find the length of each side of the triangle.

Answer 1

Let's get some variables in here and make an equation..
You have a given perimeter = 70. So P = 70.
Longest side is L = ( x + y) - 4
where x and y are the other sides.
2x = L - 9

Answer 2

2x+9 = L
we need to find y then substitute
L = x + y - 4
-y = - L + x -4
y = L - x + 4

Answer 3

L + x + y = 70

Answer 4

2x + 9 + x + 2x + 9 - x + 4 = 70

Answer 5

4x + 22 = 70
4x = 48
x = 12

Answer 6

Okay, so now you have the value of x figured out, do you know how to get the value of L and y?

Answer 7

L + 12 + y = 70
L + y = 58
Remember
2x + 9 = L
so solve for L now.

Answer 8

if x = 12, and
2x + 9 = L
L = ?

Answer 9

I'm still trying to process how you found x. It's so different compared to how I thought it was suppose to be worked out.

Answer 10

There are different ways... all depends on how you want to do it.

L = S + M - 4
L + S + M = 70
2S = L-9

L - S - M = - 4
L + S + M = 70
-L + 2S+0M =-9

Can be soved as a matrix or three equation system, but what shamil98 is showing works great too.

Answer 11

Alright.
The perimeter of a triangle is 70cm. The longest side is 4cm less than the sum of the other two sides. Twice the shortest side is 9cm less than the longest side

P = 70
Longest side is 4 less than the sum of the other sides is
Longest side = (x+y) - 4
or L = (x+y) - 4
twice the shortest side is 9 less than the longest side so
2x = L - 9
2x + 9 = L
you have two equations now.
you want to find out what y is equal to and then substitute y and L and solve for x. which is what I did.
L = x + y - 4
-L - y -L - y
-y = -L + x - 4
divide by -1
y = L - x + 4


so

y = L - x + 4
L = 2x + 9

and
L + x +y = 70


let's plug in those values into L + x + y = 70

2x + 9 + x + (L - x + 4) = 70
3x + 9 + 2x + 9 - x + 4 = 70
combine the like terms
4x + 22 = 70
4x = 48
x = 12
Follow so far?..

Answer 12

I somewhat see how you did that. Are the two equations: L=(x+y)-4 and 2x+9=L?

Answer 13

I think I understand how to get y now: substitute the 12 in for x in the equation: 2x=(x+y-4)-9, which gives me 25.

Answer 14

Yep, so know you want to get the Longest side.

Answer 15

L = 2x + 9

Answer 16

How do I get the longest side?

Answer 17

x = 12,
L = 2x + 9

Answer 18

Ok, I see! Thank you so much!

Answer 19

So, you got them all now?

Answer 20

I mentioned the matrix method of solving this. In case you wondered what I meant:

L = S + M - 4
L + S + M = 70
2S = L-9

L - S - M = - 4
L + S + M = 70
-L + 2S+0M =-9

So I have three rows, R1, R2, and R3. If I use elementary row operations to get them to pivots, a value of 1 on the left of the | in each row and column, it goes like this:

\(\left[\begin{array}{ccc|c}
1 & -1 & -1 & -4\\
1 & 1 & 1 & 70\\
-1 & 2 & 0 & -9
\end{array}\right]\)

Add R1 and R2 new R1

\(\left[\begin{array}{ccc|c}
2 & 0 & 0 & 66\\
1 & 1 & 1 & 70\\
-1 & 2 & 0 & -9
\end{array}\right]\)

R1 \(\times \dfrac{1}{2}\)

\(\left[\begin{array}{ccc|c}
1 & 0 & 0 & 33\\
1 & 1 & 1 & 70\\
-1 & 2 & 0 & -9
\end{array}\right]\)

R2 + (-1)R1 new R2
R3 + R1 new R1

\(\left[\begin{array}{ccc|c}
1 & 0 & 0 & 33\\
0 & 1 & 1 & 37\\
0 & 2 & 0 & 24
\end{array}\right]\)

R3 \(\times \dfrac{1}{2}\)

\(\left[\begin{array}{ccc|c}
1 & 0 & 0 & 33\\
0 & 1 & 1 & 37\\
0 & 1 & 0 & 12
\end{array}\right]\)
R2 + (-1)R3

\(\left[\begin{array}{ccc|c}
1 & 0 & 0 & 33\\
0 & 0 & 1 & 25\\
0 & 1 & 0 & 12
\end{array}\right]\)

Longest = 33cm
Medium = 25cm
Shortest = 12cm

And a reference on this:

http://www.epcc.edu/tutorialservices/valleverde/Documents/Gauss-Jordan_Method.pdf

Answer 21

Yes, I have all of the values now. Thank you, I appreciate you showing me the alternate way on how to do it.

Answer 22

np. have fun!