Please help me with this

Answer:
Step-by-step explanation:
The question is ambiguous. Is the longer side the hypotenuse of the length of the other leg making up the right angle.
The missing side is the hypotenuse
x + 2x + sqrt(x^2 + (2x)^2 ) = 28
3x + sqrt(x^2 + (2x)^2 ) = 28
sqrt(x^2 + (2x)^2 ) = 28 - 3x Square both sides
x^2 + 4x^2 = 784 - 168x + 9x^2
5x^2 = 784 - 168x + 9x^2
0 = 784 - 168x + 4x^2 Use the quadratic formula to solve this. It does not factor.
x = 5.34
2x = 10.68
hypotenuse = sqrt (5*5.35^2)
hypotenuse = 5.35*sqrt(5)
hypotenuse = 11.96
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Shortest leg = 5.35
Other leg = 10.7
Hypotenuse = 11.96
Total = 28.01
So the answer is mathematically correct for the hypotenuse being the unknown.
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If the leg is unknown, then I'll set up the equation and you can solve it.
x + sqrt(4x^2 - x^2) + 2x = 28 subtract x from both sides
sqrt(4x^2 - x^2) + 3x = 28 Square both sides
sqrt(3x^2) = 28 - 3x
3x^2 = 784 - 168 + 9x^2
6x^2 - 168x + 784 = 0
x = 5.91
You can carry on from here.
Answer:
Step-by-step explanation:
Let's call the sides a,b,c
P=28 so
a+b+c=28
twice as long means a = 2b substitute in equation above
2b+b+c=28
3b+c=28
but we also know is a right triangle so thhe Pythagorean Theorem can be apply yet we shoud figure what side is the hypothenuse
Since a=2b, b-cannot be the hypothenuse because is not the biger side so is eithher a or c
a² = b²+c² or c² = b²+a² substitute a² for (2b)²=4b²
4b² = b²+c² or c² = b²+ 4b²
3b² = c² or c² = 5b²
yet 3b+c=28, c=28-3b, so c²= 28²- 168b+9b² =9b²-168b+784
9b²-168b+784=3b² or 9b²-168b+784=5b²
6b²-168b+784=0 or 4b²-168b+784=0
and probably solve using quadratic equations