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In mathematics, the Pythagoreantheorem or Pythagoras theorem is a relationin Euclidean geometry among the three sides ofa right triangle (right-angled triangle). It states that:In any right-angled triangle, the Square of theHypotenuse of a Right Angled Triangle Is Equal ToThe Sum of Squares of the Other Two Sides.
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The twentieth president of the United States gave the following proof to the Pythagoras Theorem. He discovered this proof five years before he became President. He hit upon this proof in 1876 during a mathematics discussion with some of the members of Congress. It was later published in the New England Journal of Education.
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In the figure shown below, wehave taken an arbitrary righttriangle with sides of lengtha and b and hypotenuse oflength c and have drawn asecond copy of this sametriangle (positioned aspictured) and have thendrawn an additional segmentto form atrapezium.
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The parallel sides of the trapezium(which are the top and bottom sides inthe figure) have lengths a and b. Theheight of the trapezium (which is thedistance from top to bottom in thefigure) is a + b. Thus the area of thetrapezium isA = ½ (a + b)(a + b) = ½ (a + b)²However, the area of the trapezium isalso the sum of the areas of the threetriangles that make up the trapezium.Note that the middle triangle is also aright triangle .The area of thetrapezium is thusA = ½ ab + ½ ab + ½ cc = ab + ½ c²
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We thus conclude that½ (a + b)² = ab + ½ c²Multiplying both sides of thisequation by 2 gives us(a + b)² = 2ab + c²Expanding the left hand sideof the above equation thengivesa² + b² + 2ab = 2ab + c²from which we arrive at theconclusion thata² + b² = c²Hence Proved.
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yIn the figure, ∆ACB is a right angle E Dtriangle, with angle ACB = 90 ⁰ withhypotenuse c cTo prove: a² + b² = c² xConstruction: Extend AC to D such that AAD = AB = c. c bDraw ED perpendicular to CD with ED = yDraw AE as the angle bisector of angle BAD. F u B a CLet EB and EA meet at E.Draw EF perpendicular to CF with EF = x.Proof: In ∆EAD and ∆EAB, AD = AB (by construction) Angle EAD = angle EAB (AE bisectsangle BAD) EA is commonSo, by SAS property ∆EAD is congruent to∆EAB = EB = y (by CPCTE) and EDSo, angle ADE = angle ABE = 90⁰ (byCPCTE)
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Now, angle EBF + angle EBA + angle ABC = y E D180⁰i.e. angle EBF + angle ABC = 90⁰ cAlso, in ∆EFB,angle EBF + angle BEF = 90⁰ x y ASo, angle ABC = angle BEF c bIn ∆ACB and ∆BFE, F u B a Cangle ABC = angle BEFangle ACB = angle BFE = 90⁰So, by AA similarity ∆ACB is similar to ∆BFEThus, AC/BF = CB/FE = AB/BEi.e. b/u = a/x = c/yThis implies u = bx/a = b(b+c)/a --------(1) and y = cx/a = c(b+c)/a --------(2) but y = u+a (as EFCD is a rectangle) -----(3) So, by using (2), c(b+c)/a = u+aUsing (1) we get, c(b+c)/a = b(b+c)/a + awhich on simplifying gives a² + b² = c².
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Proof of Pythagoras Theorem (III) We start with four copies of the same triangle. Three of these have been rotated 90°, 180°, and 270°, respectively.
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Proof of Pythagoras Theorem (III) (contd) Each has area ab/2. Lets put them together without additional rotations so that they form a square with side c.
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Proof of Pythagoras Theorem (III) (contd) The square has a square hole with the side (a - b). Summing up its area (a - b)² and 2ab, the area of the four triangles (4·ab/2), we get C²=(a-b)²+2ab C²= a²+b² -2ab+2ab C²=a²+b² Hence Proved.
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Pythagoras Theorem Proof (Through Similarity) (IV)Theorem: In a right triangle, the square of the hypotenuse isequal to the sum of the squares of the other two sides.Given: A right-angled triangle with angle A = 90*
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Pythagoras Theorem Proof (Through Similarity) (IV) (contd.)To Prove: (Hypotenuse)2 = (Base) 2 + (Perpendicular) 2Construction: From A draw AD perpendicular to BCProof: In triangles ADC and BAC, (i) angle ADC = angle BAC [both90*] (ii) angle C = angle C[common] By AA similarity criterion, Triangle ADC is similar to BAC. Since corresponding sides are proportional in similar triangles, CD/AC = AC/BC AC2 = CD X BC(a) In triangles ADB and BAC, (i) angle BDA = angle BAC [both 90*]
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Pythagoras Theorem Proof (Through Similarity) (IV) (contd.)So, By AA similarity criterion, Triangle ADB is similar to BAC. BD/AB = AB/BC AB 2 = BC X BD (b) Adding (a) and (b), AB 2 + AC 2 = CD X BC + BC X BD AB 2 + AC 2 = BC( CD + BD) AB 2 + AC 2 = BC(BC) AB 2 + AC 2 = BC 2 Hence Proved
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PYTHAGORAS THEOREM PROOF (VI) "The square on the hypotenuse of a right triangle is equal to the sum of the squares on the two legs" (Eves 80-81). This theorem is talking about the area of the squares that are built on each side of the right triangleAccordingly, we obtain the following areas for the squares, where thegreen and blue squares are on the legs of the right triangle and the redsquare is on the hypotenuse. area of the green square is area of the blue square is area of the red square isFrom our theorem, we have the followingrelationship:area of green square + area of blue square = area of red square orAs I stated earlier, this theorem was named after Pythagoras because he was the first toprove it. He probably used a dissection type of proof similar to the following in provingthis theorem.
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Some real life uses of Pythagoras TheoremArchitectureand ConstructionNavigationEarthquake LocationCrime Scene InvestigationArrow or Missile Trajectory
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Pythagoras theorem is used in Coordinate Geometry. It is used in finding the Euclidean distance formula d = (x₂ - x₁)² + (y₂ - y₁)² (x₁, y₁)d² = a² + b² = (x₂ - x₁)² + (y₂ - y₁)² d = distanced = (x₂ - x₁)² + (y₂ - y₁)² b= y₂ - y₁ (x₂, y₂) a = x₂ - x₁
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One of the Pythagorean triplet is a multiple of 3 One of the Pythagorean triplet is a multiple of 4 One of the Pythagorean triplet is a multiple of 5Some examples:(3,4,5) (5,12,13) (7,24,25)(8,15,17) (9,40,41) (11,60,61)(12,35,37) (13,84,85) (16,63,65) If you multiply each member of the Pythagorean triplet by n, where n isa positive real number then, the resulting set is another PythagoreantripletFor example, (3,4,5) and (6,8,10) are Pythagorean triplets.The only fundamental Pythagorean triangle whose area is twice itsperimeter is (9, 40, 41).
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APPLICATIONS OF CONVERSE OF PYTHAGORAS THEOREMThe converse of Pythagoras theorem can be used tocategorize trianglesIf a² + b² = c² , then triangle ABC is a right angled triangleIf a² + b² < c² , then triangle ABC is an obtuse angled triangleIf a² + b² > c² , then triangle ABC is an acute angled triangle
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