An arch is in the form of a semi ellipse .lt is 50 meter wide at the base and has a height of 20m .how wide is the arch at the height of 10m above the base
place the center of the ellipse at (0,0) and you have its equation as
x^2/25^2 + y^2/20^2 = 1
so now just find x at y=10, and double it.
a = semi major axis = 50 / 2 = 25 m
b = semi minor axis = 20 m
Equation of ellipse:
x² / a² + y² / b² = 1
x² / 25² + y² / 20² = 1
When y = 10 m then:
x² / 25² + 10² / 20² = 1
x² / 625 + ( 10 / 20 )² = 1
x² / 625 + 0.5² = 1
x² / 625 + 0.25 = 1
x² / 625 = 1 - 0.25
x² / 625 = 0.75
x² = 625 ∙ 0.75
x² = 468.75 = 468 + 3 / 4 75 = 1872 / 4 + 3 / 4 = 1875 / 4
x = ± √ ( 1875 / 4 ) = ± √ ( 625 ∙ 3 / 4 ) = ± √625 ∙ √3 / √4 = ± 25 √3 / 2
a = semi major axis = 50 / 2 = 25 m
b = semi minor axis = 20 m
Equation of ellipse:
x² / a² + y² / b² = 1
x² / 25² + y² / 20² = 1
When y = 10m then:
x² / 25² + 10² / 20² = 1
x² / 625 + ( 10 / 20 )² = 1
x² / 625 + 0.5² = 1
x² / 625 + 0.25 = 1
x² / 625 = 1 - 0.25
x² / 625 = 0.75
x² = 625 ∙ 0.75
x² = 468.75 = 468 + 3 / 4 75 = 1872 / 4 + 3 / 4 = 1875 / 4
x = ± √ ( 1875 / 4 ) = ± √ ( 625 ∙ 3 / 4 ) = ± √625 ∙ √3 / √4 = ± 25 √3 / 2
x = ± 25 √3 / 2 m
Your arch has equation
x^2/25^2 + y^2/20^2 = 1
x^2/625 + y^2/400 = 1
so when y = 10
x^2/625 + 100/400 = 1
x^2/625 = 1 - 1/4 = 3/4
Take square root of both sides
x/25 = √3/2
2x = 25√3
x = 25√3/2 = appr 21.65 metres
so it is 21.65 m from the centre of the arch
will let you decide what is meant by "how wide is the arch" ?
It will be 10 m high 21.65 m from the centre in either direction,
so from the 10 m high on the left to the 10 m height on the right would
be 43.3 m
To find the width of the arch at a specific height above the base, we need to use the formula for the equation of a semi-ellipse.
The equation of a semi-ellipse in standard form is:
[(x-a)^2 / h^2] + [(y-b)^2 / k^2] = 1
Where (a, b) represents the center of the ellipse, h represents the width of the ellipse, and k represents the height of the ellipse.
In this case, since the center of the ellipse coincides with the origin (0, 0) and the semi-ellipse is 50 meters wide at the base and has a height of 20 meters, we can write the equation as:
[(x-0)^2 / (25)^2] + [(y-0)^2 / (10)^2] = 1
Simplifying the equation gives us:
x^2 / 625 + y^2 / 100 = 1
To find the width of the arch at a height of 10 meters above the base, we need to find the x-coordinate when y = 10.
Substituting y = 10 into the equation:
x^2 / 625 + (10)^2 / 100 = 1
x^2 / 625 + 100 / 100 = 1
x^2 / 625 + 1 = 1
x^2 / 625 = 0
Taking the square root of both sides:
x / 25 = 0
Therefore, the width of the arch at a height of 10 meters above the base is 0 meters. This means that the arch narrows down to a point at that height.