Geometrical Analysis, or the Construction and Solution of Various Geometrical Problems from Analysis, by Geometrical, Algebra, and the Dieferential Calculus Also, the Ceometrical Construction of Algebraic Equations, and a Mode of Constructing Curves of Th
Geometrical Analysis, or the Construction and Solution of Various Geometrical Problems from Analysis, by Geometrical, Algebra, and the Dieferential Calculus Also, the Ceometrical Construction of Algebraic Equations, and a Mode of Constructing Curves of Th
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1872 Excerpt: ... Demonstration by geometry.--Join GL, AL, and GG, and let fall on BL the perpendicular GH; then AG2 + GL2 = AL2=AE2 +.# 2 = #2 + P= A#2 + # 2 + 22 = AB2 + I2== OL2-J-BI2. Taking GL2 from the first and last of these equals, we have A G2 = BI27 or A G = J57. Q. E. D. Calculation.--Weh&veELEI=/EB2+BF=/EB2+AW, BL = BE--EL, ...
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1872 Excerpt: ... Demonstration by geometry.--Join GL, AL, and GG, and let fall on BL the perpendicular GH; then AG2 + GL2 = AL2=AE2 +.# 2 = #2 + P= A#2 + # 2 + 22 = AB2 + I2== OL2-J-BI2. Taking GL2 from the first and last of these equals, we have A G2 = BI27 or A G = J57. Q. E. D. Calculation.--Weh&veELEI=/EB2+BF=/EB2+AW, BL = BE--EL, GG=GL = BL, GH=V OG2--GH2, OL--GH=HL = AE, EG=GB--BE=GL--BE, FG = EH = EG+GH, and GG=GF+FG, GL = AB = Vbe2 + AE2. Limit.--The line A G may be taken any length. Problem IY.--In a right-angled triangle are given the hypothenuse, and the side of the inscribed square, to determine the triangle. Analysis by algebra.--Let ABG represent the required triangle of which the hypothenuse AG and the inscribed square BGFE are given. Draw GL perpendicular to AG, meeting EF, produced, in L, and let fall the perpendicular GH, which will be equal to GF or EF; then (IY. 21) the triangles GHL and EAF are similar and equal, having GL = AF and HL = AE. Now, jput AG = b, EF = a, EL = x, and GL--y = AF. In similar triangles AFE and GFL, we have AF (y): FE (a):: FL (x--a): FG (b--y). Where fore by--y2=zax--a2, or--26?/ + 2y2 =--2ax + 2a2. Also, FL2--EG2 + GL2; that is, (x--a)2 = (6--y)2-f y2, or x2--2aa? + a2--62--2%-f-2?/2 = (from above) b2--2ax--2a2. From which we have x2 = a2-f-62, and this Gonstruction.--On 56, the side of the given square, produced, lay BI= the given hypothenuse. Join EI, and make EL = EI= SEB2--i?/2 = vV2-f-b2 = x. On i describe a semicircle of which the centre is 0, cutting BIin 0; draw GFA; then will 4(7 == BI, the given length of the hypothenuse. Demonstration by geometry.--Join GL, AL, and OG, and on EL let fall the perpendicular GH; then AG2+ G L2 = AL2 = AE2 + EL2 = J J?2 + EP = AE2 + J92 + P = 2 + EF2...
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Add this copy of Geometrical Analysis, or the Construction and Solution to cart. $24.03, new condition, Sold by Ingram Customer Returns Center rated 5.0 out of 5 stars, ships from NV, USA, published 2006 by University of Michigan Library.
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Add this copy of Geometrical Analysis, Or the Construction and Solution to cart. $43.32, good condition, Sold by Bonita rated 4.0 out of 5 stars, ships from Newport Coast, CA, UNITED STATES, published 2005 by Scholarly Publishing Office, U.
