Work the following area application problem. you own a swimming pool elliptical in shape. you wish to cover it at night to retain its warmth. if the pool measures 18 ft. across at its minor axis and 25 ft. across at its major axis, how much plastic material will you need to cover the pool, assuming you need a 1.5 ft. overhang all around the pool? area (to the nearest tenth) = sq. ft.

SInce you need 1.5 feet of overhang, add 3 feet to each axis dimension 1.5 on each side):

Minor Axis = 18 + 3 = 21 feet

Major axis = 25 + 3 = 28 feet

The area of an ellipse is found by multiplying half the minor axis by half the major axis by PI.

1/2 minor axis = 21 / 2 = 10.5

1/2 major axis = 28 / 2 = 14

Using 3.14 for PI

Area = 10.5 x 14 x 3.14 = 147 x 3.14 = 461.6 sq ft

This problem is not as simple as it may appear at first. The area of an ellipse is ...

... A = πab

where a and b are the semi-axes.

Here, it looks like you're expected to choose these to be 1.5 ft longer than half the given axes, so the area of the pool cover is about ...

... A = π(9 ft + 1.5 ft)(12.5 ft +1.5 ft)

... A = 147π ft² ≈ 461.8 ft²

However, adding 1.5 ft of material to an ellipse results in a shape that is not an ellipse, but is slightly larger than the ellipse with the dimensions used above. The area of that may be about 462.5 ft² (found numerically).

y - y1 = m(x- x1)

y - (-5) = 2/3(x -0)

y + 5 = 2/3x

y = 2/3x - 5