Question

Solve for x if the two legs of a right triangle are 2x and 2x+4, and the hypotenuse is 4x−4. How would you start this and do it step by step?

Answer 1

x = 6

What we know:

The sides of a triangle follow the Pythagorean Theorem of "a^2 + b^2 = c^2"

In this question:

• a & b = the two legs of the triangle (2x & 2x+4)

• c = hypotenuse (4x-4)

Solving:

1. Plug these given values into the equation:

• a^2 + b^2 = c^2
• (2x)^2 + (2x + 4)^2 = (4x - 4)^2

2. And now we continue on by squaring each of the terms:

• (2x)(2x) + (2x + 4)(2x + 4) = (4x - 4)(2x + 4)

• (4x^2) + (4x^2 + 16x + 16) = (16x^2 - 32x + 16)

3. Simplify:

• 8x^2 + 16x + 16 = 16x^2 - 32x + 16

4. Set equation equal to '0':

• 0 = 8x^2 - 48x

5. Factor:

• 0 = 8x^2 - 48x
• 0 = (8x)(x - 6)

6. Solve for each 'x':

• 0 = 8x --> 0/8 = x --> x = 0
• 0 = x - 6 --> 6 = x --> x = 6

7. Check each solution:

x = 0

• 4x - 4
• 4(0) - 4
• -4 = hypotenuse
• This is not possible because the hypotenuse cannot be a negative number

x = 6

• 4x - 4
• 4(6) - 4
• 24 - 4
• 22
• This is possible because the hypotenuse is a positive number