longer leg of right triangle is 1 unit less than twice the lngth of the shortr leg. The hypotnuse is 1 unit more than twice the lngth of the shorter leg. find the lngth of the 3 sides of the triangle.

Answer 1

You can plug your description above into the Pythagorean Theorem and solve for x

a² + b² = c²

a = longer leg = 2x -1

b = shorter leg = x

c = hypotenuse = 2x + 1

(2x - 1)² + x² = (2x + 1)²

PEMDAS

4x² - 4x + 1 + x² = 4x² + 4x + 1

5x² -4x +1 = 4x² + 4x + 1

Combine like terms by moving every thing to one side of the equal sign this sets the resulting quadratic equal to 0

5x² -4x² -4x -4x +1 -1 = 0

x² - 8x = 0

Add 8x to both sides of the equation

x² = 8x

Divide both sides by x to give

x²/x = 8

x = 8

Just plug this value back into the substitutions above to check

x = 8 = The Shorter Leg

2x - 1 = 2(8) - 1 = 16 -1 = 15 = The Longer Leg

2x + 1 = 2(8)+1 = 16 + 1 = 17 = The Hypotenuse

(2(8) -1)² + 8² = (2(8) + 1)²

(16 -1)² + 8² = (16 + 1)²

15² + 8² = 17²

225 + 64 = 289

The set up

4x² - 4x +1 + x² = 4x² + 4x + 1 Can also be arranged to isolate x² right away as

x² = 4x² + 4x + 1 - (4x² - 4x +1)

x² = 4x² + 4x + 1 - 4x² + 4x -1

x² = 4x² - 4x² + 4x + 4x + 1 - 1

The 4x² and 1 cancel out to give

x² = 8x

Dividing both sides by x gives

x²/x = 8

x = 8

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