Question

Please help me with this

Answer 1

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

Answer 2

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.