If I found 52 feet of board and build a triangual garden using all of the board. If the largest side is 4 more than 4 times the smallest and the other side 3 times the smallest, how long will each side be?
let the smallest side be s
s + 4s+4 + 3s = 52
8s + 4 = 52
8s = 48
s = 6
so, the sides are 6,18,28
but....
sides of 6, 18, and 28 cannot be a triangle.
The sum of any two sides must be greater than the third side,
Problem is flawed even though an algebraic solution is possible.
ouch. I thought I'd checked that. Good catch.
To find the lengths of the sides of the triangular garden, let's denote the length of the smallest side as "x".
According to the given information, the largest side is 4 more than 4 times the smallest side. So it can be expressed as (4x + 4).
The other side is 3 times the smallest side, which can be expressed as 3x.
Since the sum of the lengths of the three sides is equal to the total length of the board (52 feet), we can set up the following equation:
x + (4x + 4) + 3x = 52
Now, let's solve the equation to find the value of "x".
Combine like terms: 8x + 4 = 52
Subtract 4 from both sides: 8x = 48
Divide both sides by 8: x = 6
Therefore, the smallest side of the triangular garden is 6 feet.
Now, let's find the lengths of the other two sides:
Largest side = 4x + 4 = 4(6) + 4 = 24 + 4 = 28 feet
Other side = 3x = 3(6) = 18 feet
So, the lengths of the sides of the triangular garden will be: 6 feet, 18 feet, and 28 feet.