EXAMPLE.

What is the area of the segment of a circle ABC, the chord AC being 48 feet, and the height DB 18 feet?

The length of the arc will be found to be 64 feet, and the radius 25 feet, then

Multiply the height by 0-626 to the square of the product; add the square of half the chord; multiply twice the square root of the sum by two thirds of the height, and the product is the area, nearly.

EXAMPLE.

What is the area of a circular segment ABC, whose height is 18 feet, and the chord 48 feet?

METHOD III.

To two thirds of the product of the base multiplied by the height, add the cube of the height divided by twice the length of the segment, and the sum will be nearly

the area.

EXAMPLE.

What is the area of a circular segment, the chord AB being 48 feet, and the height CD 18 feet?

This method will be sufficiently near for all practical purposes, and is much shorter than the two first methods.

PROBLEM XIX.

To find the area of a circular zone, which is that part of a circle laying between two parallel chords, and the parts of the circle intercepted by the chords.

Find the height of each segment by Problem XV. and the diameter by Problem XIV.; then the difference of the segments found by Problem XVIII. will be the answer.

EXAMPLE.

The greater chord CD of a circular zone being 48 feet, and the lesser chord AB 30 feet, their distance FG 13 feet, required the area of the zone.

The distance EF will be found by Problem XIII. to be 7 feet.

The radius EF will be found by Problem XIV. to be 25 feet.

13+7=20, and 25—20=5 the height of the lesser segment. 13+5=18 the height of the greater segment.