Solve the problem.
-An arch in the form of a semiellipse is 52 ft wide at the base and has a height of 20 ft. How wide is the arch at a height of 12 ft above the base?
A) 41.6 ft
B) 35.5 ft
C) 20.8 ft
D) 17.7 ft

Question

Quizplus · Accepted Answer

The equation of an ellipse centered at the origin with a horizontal major axis is $x^2/a^2 + y^2/b^2 = 1$, where 2a is the length of the major axis and 2b is the length of the minor axis. For this semiellipse, the major axis length (width of the base) is 52 ft, so $a = 26$ ft, and the height (minor axis length) is 20 ft, so $b = 20$ ft. At a height of 12 ft above the base, $y = 8$ ft (since the total height is 20 ft, and we're looking for the width 12 ft above the base, we subtract from the total height). Plugging into the ellipse equation: $x^2/26^2 + 8^2/20^2 = 1$. Solving for $x$ gives the half-width at that height, and multiplying by 2 gives the total width, which is approximately 41.6 ft.

Solve the Problem. -An Arch in the Form of a Semiellipse Is 52