. The Decorator's assistant. 72 6283182199113 = circumf. a b c d a. = circumf. e f g h e. 2) = half sum of two circumf. 7240 2) 32 16 = half difF. of the two diam. sq. in. = area. Example 2.—The diameter a c is 30 inches,and the breadth of the ring a e 2^ inches : de-termine its area. 3030 90O .7854900 = area of circle a b c d a. 305 25 25 125 50 625 = F G. .7854 392701570847124 = area of circle e f g h a, sq. in. = area of ring. Example 2 is done by
. The Decorator's assistant. 72 6283182199113 = circumf. a b c d a. = circumf. e f g h e. 2) = half sum of two circumf. 7240 2) 32 16 = half difF. of the two diam. sq. in. = area. Example 2.—The diameter a c is 30 inches,and the breadth of the ring a e 2^ inches : de-termine its area. 3030 90O .7854900 = area of circle a b c d a. 305 25 25 125 50 625 = F G. .7854 392701570847124 = area of circle e f g h a, sq. in. = area of ring. Example 2 is done by the following rule :—Find the areas of tlie two circles separatelythen the difference of these areas gives thearea of the ring. PROBLEM XVII. To find the area of an ellipse. 1. Find the sum of the areas of the sectorsA F G B, F K E L, B I c M, and E G c D, fromwhich subtract the area of the lozenge a i h k,and the difference will be the required 2. Multiply continually together the twodiameters a b, c d, and the number 11. Dividethe last product by 14, and the quotient will bethe area nearly true. Example.—What will be the area of theellipse A D E c A, its transverse a b being 15feet, and its conjugate c d 10 feet?
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