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Prob. 14. Given a chord of a circle and its distance from the cer tre, to find the radius of the circle.
To the square of the half chord, add the square of the distance frors the centre, and the square root of the sum will be the radius required
Er. Given the chord CD 48 feet, and its distance EF from ths centre 7 required the radius of the circle. See the figure to the Example Problem 13. L=24 and 24 x 24=576
7X 7= 49
625(25 the radius
Prob. 15. Given any two parallel chords in a circle, and the distance between them, to find the perpendicular height from the middle of either chord to the circumference.
Find the nearest distance of the greater chord from the centre, by Problem 13, and find the radius of the circle by Problem 14, add the distance between the two parallel chords, and the distance between the greater chord, and the centre of the circle together : this sum being taken from the radius, will give the perpendicular height from the middle of the lesser chord, to the circumference or height of the lesser segment; to the lesser segment, add the distance between the parallel chords, and the sum will be the height of the greater segment.
Ex. Given the greater chord CD 48 feet, and the lesser chord AB 30 feet, their distance EG 13 feet, required the distance GH perpendicular from the middle of AB to the circumference.
See the figure to the Example Problem 13.
The distance from the centre to the greater chord, will be found to be 7 feet, by Problem 13, and the radius 25 feet, by Problem 14.
13+7=20 and 25—20=5 feet, height of the lesser segment.
Then 13+5=18 the height of the greater segment. Prob. 16. To find the area of a circle, the diameter being given.
Method 1. Multiply half the circumference by half the diameter, and the product will be the area.
Er. 1. What is the area of a circle whose diameter is 28 feet, and its circumference 88 feet?
616 feet, the area required.
Method 2. Multiply the square of the diameter by 7854, and the product will be the area.
Fax. What is the area of a circle whose diameter is 3f. 6. :
12 25 square of the diameter.
4900 6125 9800 8575
9•621150 the answer.
In common practice, multiply the square of the diameter when given in feet, inches, &c. by gi. 5ii.
Ex. What is the area of a circle whose diameter is 3f. 6i.?
Method 3. When the circumference is given.
Multiply the square of the circumteretre by •07958, and the produci will be the area,
Ex, What is the area of a circle when the circumfereuce is 88 fect
7744 square of the circumference. *07958
61952 38720 69696 54208
616 26752 the area of the circle.
Prol. 17. To find the area of a sector of a circle.
Multiply the radius, or half the diameter, by half the length of the arc of the sector, and the product will be the area,
Fr. 1. What is the area of a sector ABC, the arc BC being 3f. 6i. and the radius AB or AC 6f. 2i.?
Prob. To find the area of the segment of a circle, t'e chord and height of the arc being given.
Find the length of the arc ABC by Prol. 11, and the radius o. the circle by Prob. 12, the area of the sector ABCE by Prob. 17.
Subtract the area of the triangle AEC, fomnd by Prob. 2, from the area of the sector, and the remainder will be the area of the segment.
Ex. What is the area of the segment of a circle ABC, the chord AC being 48 feet, and the versed sine DB 18 feet ?
The length of the arc will be found to be 64 feet, and the radius 25 feet, then 2)64
7 perpendicular DK.
632 area of the segment. Method 2. 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.
Er. What is the area of a circular segment ABC, whose height is 18 feet, and the chord 48 feet •626
Method 3 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.
Er. What is the area of a circular segment, the chord AB being 48 fiet, 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.
This last rule was invented by the Editor, of this course of Mathematics, and published, upwards of twenty-six years ago, in his Principles of Architecture. The demonstration was afterwards given in his Architectural Dic. tionary, under the article Mensuration. The method was afterwards copied into the new editions of Hawney's Mensuration by its last Editor, Mr. Keith, without a candid acknowledgment; and hence found its way into several other books on this subject.
Prob. 19. 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 Leight of each segment by Prob. 15, and the diameter by Prob. 14 ; uen the difference of the segments found by Prob. 18 will be the answer