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

Answer 1

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.

Answer 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 = 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

Answer 3

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