Diagonal brace. The width of a rectangular gate is 2 meters
(m) larger than its height. The diagonal brace measures
sqrt6m. Find the width and height.
solve for x
x^2 + (x+2)^2 = 6
2x^ + 4x + 4 = 6
x^2 + 2x - 1 = 0
I get x = appr. .4142
check
.4142^2 + 2.4142^2 = 5.999992 , not bad
To find the width and height of the rectangular gate, we can set up a system of equations based on the given information.
Let's assume that the height of the gate is "x" meters. Then, according to the given information, the width of the gate is "x + 2" meters.
We can use the Pythagorean theorem to relate the height, width, and diagonal length:
diagonal^2 = height^2 + width^2
Substituting the values we know:
(sqrt(6))^2 = x^2 + (x + 2)^2
6 = x^2 + (x^2 + 4x + 4)
6 = 2x^2 + 4x + 4
Rearranging the equation:
2x^2 + 4x - 2 = 0
Divide the entire equation by 2:
x^2 + 2x - 1 = 0
Now, we can solve this equation using the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / 2a
For our equation, a = 1, b = 2, and c = -1:
x = [-2 ± sqrt(2^2 - 4(1)(-1))] / 2(1)
Calculating the values within the square root:
x = [-2 ± sqrt(4 + 4)] / 2
x = [-2 ± sqrt(8)] / 2
x = [-2 ± 2√2] / 2
Simplifying the expression:
x = -1 ± √2
Since the height cannot be negative, we choose the positive value:
x = -1 + √2
So, the height of the rectangular gate is approximately 0.41 meters (rounded to two decimal places).
To find the width, we can substitute this value back into the equation for the width:
width = height + 2
width = (0.41) + 2
width = 2.41 meters (rounded to two decimal places).
Therefore, the width of the rectangular gate is approximately 2.41 meters, and the height is approximately 0.41 meters.