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Er. 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 DE.
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
96 48 576 square of the half chorder
626 11'268 X11.268
90144 67608 22536 U1268 11268
12 two thirds of the height.
702.967824)26.513 square root of the sum
53.026=26.513 x 20
636.312 area of the segment
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 segmenft, the chord AB being 48 fitt, 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
Er. The greater chord CD of a circular zone heing 48 feet, and the lesser chord AB 30 feet, their distance FG 13 feet, required the area of the zone.
See the figure to Erample Problem 13.
13+5=20, and 25—20=5 the height of the lesser segment.
2 x 48=96)5832(60 75
Prob. 20. To find the diameter of a circle whose area shall be in a given proportion to the area of a circle whose diameter is known.
If the area is required to be greater than the given circle, multiply the given diameter by the sq:17re root of the intended increase, and it will give the diameter of the required circle.
But if the area is intended to be less than the area of the given cir. cle, divide the given diameter by the square root of the intended lecrease, which will give the diameter of the given circle.
Er. What is the diameter of a circle, whose area is 9 times as much as one of 21 inches diameter ?
✓9=3, then 21 X3=63 inches. Prob. 21. To find the circumference of an ellipsis, the transverso and conjugate axis being given.
Multiply half l!ie zum of the two axes by 3; to the product add one-seventh part of the sum of the two axes, and this sum will give the circumference ner chough for most practical purposes.
Ex. 1. What is the circumference of an ellipsis, of which the major axis AB is 24 feet, and the minor axis CD 18 feet?
Er. 2. The width of an elliptical vault being 21 f. 7i. and the height 7f. 3ļi, what is the circumference ?
Prob, 22. To find the area of an ellipsis, the major and minor axes being given.
Multiply the two axes together, and the product by •7854, will give the area required.
Ex. What is the area of an ellipsis of which the major axis is 24 feet, and the minor 18 feet ?
Prob. 23. To find the area of a parabola, the base or double ordinate being given, and the axis or height.
Multiply the base by the height, and two-thirds of this product will be the area required.
Ex. What is the area of a parabola, the axis CD beir:g 12, and the double ordinad AB 18)
24. To find the area ABCD, of the frustum of a parabola, of which the parallel ends AB and CD are given , also their distance.
To the square of the greatest end, add the square of the less to the product of the ends : divide the sum by the sum of the ends, multiply the quotient by their distance, and two-thirds of the last product will be the 20swer.