Question 145433
The width of a rectangle gate is 2 meters larger than its height. The diagonal brace measures the square root of 6 m. Find the width and height.
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Draw the picture.
Label the diagonal as sqrt(6)
Label the width as "x+2"
Label the height as "x"
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Use Pythagoras to solve for "x":
x^2 + (x+2)^2 = (sqrt(6))^2
2x^2 + 4x + 4 = 6
x^2 + 2x - 1 = 0
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x = [-2 +- sqrt(4 - 4*1*-1)]/2

x = [-2 +- sqrt(8)]/2
x = [-1 +- sqrt(2)]
Positive solution:
x = -1 + 1.414.. = 0.414 m (height of the gate)
x+2 = 2.414 m (width of the gate)
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Cheers,
Stan H.