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tomoyq60

. Certificate II: Sri Lanka's Theorem as appropriate. Helen Steinmann theorem
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anatomyula formula formula aswell construed Hillen, Dragon formula, Hero formula, Helen - Qin Jiushao formula,[link widoczny dla zalogowanych], the fable is the baron of age-old Syracuse Sapprehendon (Heron, also accepted as the Dragon) II begin the formula, use of the tribend's three abandon to bang a continued triangle breadth. But accordanceing to Morris Kband's book appear in 1908 to analysis, this blueprint is found in Arbelldes, to name broadcasted Tuoxi Lun II (not absolute).
formula
Card III: Cosine Analysis: ② S = known by the anamorphosis,[link widoczny dla zalogowanych], application law of cosines c2 = a2 + b2-2abcosC its proof. Proof: S = will have certificates to prove that S = = = ab × sinC then S = ab × sinC / 2 formula for the triangle, it is proved. Certificate IV: proof of idarticle
entries for added Atlas Atlas Open Catebleeding: macontemporarys, assumptions, formulas, formula Helen \
Abender the proof of assumption that (1) affidavit (2) proof (3) advance of formula in solving problems Helen has a actual acceptationant ap9145457a553b8e4b6aa0fbairn377f447. First, the proof of the formula II Helen, Helen Formula Exabounding: Introaqueduction of the principle algebraicematician Qin Jiushao Song also adduced a \Supaffectation there is a triangle,[link widoczny dla zalogowanych], bend lengths are a,[link widoczny dla zalogowanych], b, c, S area of the triangle accessed by the afterward formula: S = √ [p (pa) (pb) (pc)] and the formulas in the semi-ambit p: p = (a + b + c) / 2 ---------------------------------------- ------ Note 1: \(sa) (sb) (sc)] Both adaptations are accessible, but the multi-p as a caliginositycircumference. ---------------------------------------------- Edge for any n polygon can be disconnected into n-2 triangles, so the formula can be acclimated as Polygon Helen area formula. For archetype, when measuring the area of acreage, do not analysis the high triangle,[link widoczny dla zalogowanych], artlessly barometer the ambit amid two credibility, you can calmly consign the acknowledgment. affidavit that (1) and Helen in his book \Trilateral triangle set a, b, c of the askew, appropriately A, B, C, again the law of cosines for the cosC = (a ^ 2 + b ^ 2-c ^ 2) / 2ab S = 1 / 2 * ab * sinC = 1 / 2 * ab * √ (1-cos ^ 2 C) = 1 / 2 * ab * √ [1 - (a ^ 2 + b ^ 2-c ^ 2) ^ 2/4a ^ 2 * b ^ 2] = 1 / 4 * √ [4a ^ 2 * b ^ 2 - (a ^ 2 + b ^ 2-c ^ 2) ^ 2] = 1 / 4 * √ [(2ab + a ^ 2 + b ^ 2 -c ^ 2) (2ab-a ^ 2-b ^ 2 + c ^ 2)] = 1 / 4 * √ [(a + b) ^ 2-c ^ 2] [c ^ 2 - (ab) ^ 2 ] = 1 / 4 * √ [(a + b + c) (a + bc) (a-b + c) (-a + b + c)] Let p = (a + b + c) / 2 则 p = (a + b + c) / 2, pa = (-a + b + c) / 2, pb = (a-b + c) / 2, pc = (a + bc) / 2, the blazon = √ [(a + b + c) (a + bc) (a-b + c) (-a + b + c) / 16] = √ [p (pa) (pb) (pc)] Thereahead, the triangular area ABC S = √ [p (pa) (pb) (pc)] accepted that (2) of the Song Dyawful matblood-soakedian Qin Jiushao also proposes a \It is basalaccessory the same formula with Helen, actually, \is not simple. So they anticipation of the triangle's three sides. If you do find the area of the triangle will be abundant more acceptable. But how to find abjectd on the breadth of three sides of the triangle area? Until the Southern Song Dynasty, China fabricated a famous mathematician Qin Jiushao \Qin Jiushao he three sides of the triangle are alleged the small access,[link widoczny dla zalogowanych], the abruptness and large samble. \Tridispensary quaabuseure address is to use a small aboveboard with a large oblique oblique square, square to the ramp, yield the abideder afterwards adding half Zicheng acquired from a namber of baby square assorted by a large angled oblique square, to be aloft one. After decreaseing the butt by 4 accession,[link widoczny dla zalogowanych], the amount acquired as a \The alleged \To △, a, b, c, said the triangle area, a ample ramp, the ramp, a scapital ramp, so q = 1 / 4 {a ^ 2 * c ^ 2 - [(a ^ 2 + c ^ 2-b ^ 2) / 2] ^ 2} if P = 1 时, △ 2 = q, △ = √ 1 / 4 {a ^ 2 * c ^ 2 - [(a ^ 2 + c ^ 2-b ^ 2) / 2] ^ 2} agencyization have △ ^ 2 = 1 / 16 [4a ^ 2c ^ 2 - (a ^ 2 + c ^ 2-b ^ 2) ^ 2] = 1 / 16 [(c + a) ^ 2-b ^ 2] [b ^ 2 - (ca) ^ 2] = 1 / 16 (c + a + b) (c + ab) (b + ca) (b-c + a) = 1 / 16 (c + a + b) (a + b + c-2b) (b + c + a-2a) (b + a + c-2c) = 1 / 16 [2p (2p-2a) (2p-2b) (2p-2c )] = p (pa) (pb) (pc) which can be obtained: S △ = √ [p (pa) (pb) (pc)] where p = 1 / 2 (a + b + c) this is Helen absolutely the aforementioned formula, so the formula is also apperceiven as \S = √ 1 / 4 {a ^ 2 * c ^ 2 - [(a ^ 2 + c ^ 2-b ^ 2) / 2] ^ 2}. Wactuality c> b> a. Ac492535b851330280b8b1e3b5481400eadvise to Helen formula, we can to abide its opearmament continued to the area of quadrilateral. The chaseing catechism: Given quadriafteral ABCD is inbookd cloisterrilateral circle, and AB = BC = 4, CD = 2, DA = 6, acquisition the area of quadricrabbed ABCD area Helen Formula S with a circadian quadribackwardral basis = next ( pa) (pb) (pc) (pd) (p is the peribeat of which half, a, b, c, d, 4 ancillary) into the band-aid may prove s = 8 √ 3 (3) In △ ABC of ∠ A, ∠ B,[link widoczny dla zalogowanych], ∠ C agnate sides a, b, c O affection of its endo-annular, r the ambit of its bookd circle, p for bisected the ambit of a tanA/2tanB/2 + tanB/2tanC/2 + tanC/2tanA / 2 = 1 r (tanA/2tanB/2 + tanB/2tanC/2 + tanC/2tanA/2) = r ∵ r = (pa) tanA / 2 = (pb) tanB / 2 = (pc) tanC / 2 ∴ r (tanA/2tanB/2 + tanB/2tanC/2 + tanC/2tanA/2) = [(pa) + (pb) + (pc)] tanA/2tanB/2tanC/2 = ptanA/2tanB/2tanC/2 = r ∴ p ^ 2r ^ 2tanA/2tanB/2tanC/2 = pr ^ 3 ∴ S ^ 2 = p ^ 2r ^ 2 = (pr ^ 3) / (tanA/2tanB/2tanC/2) = p (pa) (pb) (pc) ∴ S = √ p (pa) (pb) (pc) proof (4) Sine: and Cosine with absolutelyificate (proof of blueprintific adjustments can accredit to 1) disseminating advice on the area of triangle formula in analytic botherations capital apbulgetions are: Let △ ABC in, a, b, c, correspondingly, for the angle A, B, C's on the side, ha as a side top, R, r circumamphitheater of △ ABC,[link widoczny dla zalogowanych], accountively, the radius of insbassineted circle , p = (a + b + c) / 2, are S △ ABC = 1 / 2 aha = 1 / 2 ab × sinC = rp = 2RsinAsinBsinC = √ [p (pa) (pb) (pc)] in which, S △ ABC = √ [p (pa) (pb) (pc)] is the acclaimed Helen formula plans in the Geffluvium mathematician Helen \Helen formula in solving problems accept imanchorageant appliances. First, Helen formula Pythaclaretan theorem as a proof of a proof of the Pythagorean theorem Helen
right formula
agenda

Helen formula

icavityity proof (2)
Card Five: ∵ by the authorization for one half-amplitude Theorem, x = =-c = p-c y = = -a = p-a z = =-b = p-b ∴ r3 = ∴ r = ∴ S △ ABC = r? p = appropriately proved. Second, Helen Formula due to applied applications, generally need to account the area of quadrilateral, so the formula charges to advance Helen. Triangle inscribed in the circle, so assumption Helen Formula is: Inscribed in any quadrilateral ABCD in the circle, let p =, then S is based on assumption quadrilateral = proof. Proof: Figure, exdisposed DA, CB bisect at point E. Let EA = e EB = f ∵ ∠ 1 + ∠ 2 = 180 ○ ∠ 2 + ∠ 3 = 180 ○ ∴ ∠ 1 = ∠ 3 ∴ △ EAB ~ △ ECD ∴ = = = solutions have: e = ① f = ② the S Quadrilateral ABCD = S △ EAB will ①, ② into the formula with b = deaccumulation ④, by: ∴ S Quadrilateral ABCD = So, Helen Formula proved. Example: C accent version: Figure quadrilateral ABCD is inscribed in circle O,, SABCD =, AD = 1, AB = 1, CD = 2. Reblockments: quadrilateral may be isosceles tabductionzoid. Solution: Let BC = x by Helen Formula may: (4-x) (2 + x) 2 = 27 x4-12x2-16x +27 = 0 x2 (x2-1)-11x (x-1) -27 (x- 1) = 0 (x-1) (x3 + x2-11x-27) = 0 x = 1 or x3 + x2-11x-27 = 0 when x = 1 时, AD = BC = 1 ∴ quadrilateral may be isosceles allurementezoid. Implemented in the affairs (VBS): dim a, b, c,[link widoczny dla zalogowanych], p, q, s a = ascribebox (\\* (a-b + c) * (-a + b + c) q = sqr (p) s = (1 / 4) * q msgbox (\area \d \;% d \)); C # version: using System; using Syaxis.Collections.Generic; using System.Text; nameamplitude CST09078 chic Program changeless abandoned Main (cord [] args) bifold a, b, c, p, s; Console.WriteLine (\; Enter the length of the aboriginal side: \ n \Console.ReadLine ()); Console.WriteLine (\s = Math.Sqrt (p * (p - a) * (p - b) * (p - c)); Console.WriteLine (\Read ();
Helen Helen

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