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A New way, The Area of Trapezium
A New way, The Area of Trapezium by Piyush Goel Lot of mathematicians have proved Pythagoras theorem in their own ways. If you google it you will indeed found hundred of ways. Meanwhile I was also sure that maybe one day I could find something new out of this incredible Pythagoras theorem and Recently I got something which I would like to share with you. To Prove: Deriving the equation of area of trapezium using Arcs Proof: There is a triangle ABC with sides a b and c as shown in the figure. Now, Area of ∆ BCEG = Area of ∆ BDC +Area of ⌂ DCEF + Area of ∆ EFG c^2=ac/2+ Area of ⌂ DCEF + (c-b) c/2 (2c^2– ac –c^2+ bc )/2=Area of ⌂ DCEF (c^2– ac+ bc )/2=Area of ⌂ DCEF c(c– a+ b)/2=Area of ⌂ DCEF Area of ⌂ DCEF=BC(DE+CF)/2 Copyrighted©PiyushGoel
Himesh Bhai
How the Factorial Function came into Existence.
A note on the Factorial Function Write down 0,1,2,3,4,5 and put parallel square of each number like 0,1,4,9,16,25, then start to subtract the bigger one to the lower one (1–0),(4–1),(9–4),(16–9) and (25–16) to get 1,3,5,7,9 and again subtract the bigger one to the lower one (3–1),(5–3),(7–5) and (9–7) to get (2,2,2,2). Again we squared each number, at the same time we cubed each number (0,1,8,27,64,125,216) and the same procedure follows, subtract the bigger one to the lower one (1–0),(8–1),(27–8),(64–27) ,(125–64) and(216–125) to get (1,7,19,37,61,91) and again (7–1),(19–7),(37–19),(61–37) and (91–61), to get (6,12,18,24,30) same again(12–6),(18–12),24–18) and (30–24) till the result come out here we get (6,6,6,6). At the same time if we do it again for 4 and 5 (power).When we get 2 for 2 ,6 for 3 ,24 for 4 and 120 for 5. The result is the factorial function. https://piyushtheorem.wordpress.com/2017/02/08/a-note-on-the-factorial-function/
Piyush Theorem
PiyushTheorem PiyushTheorem: In a Right-Angled Triangle with sides in A.P. Series, the distance between the point of intersection of median & altitude at the base is 1/10 Th the sum of other two sides. This Theorem applies in Two Conditions: 1.The Triangle must be Right-Angled. 2.Its Sides are in A.P. Series. 1.Proof with Trigonometry Tan α =AD/DC AD= DC Tan α —————–1 Tan α = AD/DE AD= DE Tan2 α —————-2 DC Tan α = DE Tan 2 α (DE+EC) Tan α = DE Tan 2 α DE Tan α + EC Tan α = DE Tan 2 α DE Tan α + EC Tan α = 2 DE Tan α / (1- Tan2 α ) DE Tan α – DE Tan3 α + EC Tan α –EC Tan3 α = 2DE Tan α EC Tan α –EC Tan3 α – DE Tan3 α = 2DE Tan α – DE Tan α Tan α (EC – EC Tan2 α – DE T an2 α )= DE Tan α DE Tan2 α – DE = EC Tan2 α – EC -DE ( Tan2 α + 1) = -EC (1 – Tan2 α ) DE (sin2 α /cos2 α + 1) = EC (1- sin2 α /cos2 α ) DE (sin2 α + cos2 α /cos2 α ) = EC (cos2 α – sin2 α /cos2 α ) DE (sin2 α + cos2 α ) = EC(cos2 α –sin2 α ) DE (sin2 α + cos2 α ) = EC (cos2 α –sin2 α ) where (sin2 α + cos2 α =1) & (cos2 α –sin2 α = cos2 α ) DE= EC cos2 α cos α =a/a+d & sin α = (a-d)/ (a +d) cos2 α = a2/ (a +b) 2 sin2 α = (a-d) 2/ (a+ d) 2 DE= EC (cos2 α – sin2 α ) = EC (a2 / (a +b) 2 – (a-d) 2/ (a +d) 2 = EC (a2 – (a-d) 2/ (a +d) 2 = EC (a –a +d) (a+ a-d)/ (a+ d) 2 = EC (d) (2a -d)/ (a+ d) 2 = (a +d)/2(d) (2a -d)/ (a +d) 2 ————- where EC= (a +d)/2 = (d) (2a -d)/2(a +d) = (d) (8d -d)/2(4d+d) ——————where a= 4d (as per the Theorem) = 7d2 /2(5d) = 7d /10 = (3d+4d)/10= (AB+AC)/10 2.Proof with Obtuse Triangle Theorem AC2=EC2 +AE2 +2CE.DE where EC = ( a +d) /2,AE=( a +d)/2 a2 = (a +d/2)2 + (a+ d/2)2 + 2(a +d)/2DE = (a +d/2) (a+d+2DE) = (a +d/2) (a+d+2DE) where a=4d 16d2 = (5d/2) (5d+2DE) 32d/5 = 5d + 2DE 32d/5 – 5d = 2DE 32d -25d/5 = 2DE DE =7d/10 = (3d+4d)/10 = (AB+AC)/10 3.Proof with Acute Triangle Theorem AB2= AC2+BC2 – 2BC.DC (a-d) 2= a2 + (a+ d) 2 -2(a+ d) (DE+EC) where AB= (a-d), AC=a, BC =( a +d) & EC= (a +d)/2 (a-d) 2 – (a +d)2 = a2 -2(a +d)(DE+EC) (a- d –a-d) (a -d +a +d) = a2 -2(a+ d) (2DE+a+d)/2 2(-2d) (2a) = 2a2 -2(a +d) (2DE+a+d) -8ad – 2a2 = -2(a +d) (2DE+a+d) -2a (4d +a) = -2(a +d) (2DE+a+d) a (4d + a) = (a +d)(2DE+a+d) 4d (4 d + 4d) = (4d+d) (2DE+4d+d) 4d (8d) = (5d) (2DE+5d) 32d2/5d = (2DE+5d) 32d/5 = (2DE+5d) 32d/5 – 5d = 2DE (32d – 25d)/5 = 2 DE DE = 7d/10 = (3d+4d)/10 = (AB+AC)/10 4. Proof with Co-ordinates Geometry Equation of BE Y – 0 =b-0/0-a(X – a) Y = -b/a(X) + b——————- (1) M1 = -b/a For perpendicular M1M2= -1 So M2=a/b Equation of AC Y – 0 = a/b(X-0) Y=a/b(X) —————— (2) Put Y value in equation (1) a/b(X) + b/a(X) =b X (a2+b2/a b) = b X = ab2/ (a2 + b2) To get Value of Y, put X value in equation (2) Y = a/b (ab2/ (a2+b2) Y = a2b/ (a2+b2) Here we got co-ordinates of Point C – ab2/ (a2 + b2), a2b/ (a2+b2) and co-ordinates of point d is (a/2, b/2) because d is midpoint. As per the “Theorem” a=z-d, b=z, c = z+ d (z +d) 2= (z-d) 2+z2 from here z=4d so a=3d and b=4d Put value of a & b ab2/ (a2 + b2), a2b/ (a2+b2) & (a/2, b/2) ab2/ (a2 + b2) = 48d/25 a2b/ (a2+b2) = 36d/25 a/ 2=3d/2 b/ 2 =4d/2 CD2= (48d/25 -3d/2)2-(36d/25-4d/2)2 = (96d-75d/50)2 + (72d-100d/50)2 = (21d/50)2 + (-28d/50)2 = (441d2/2500) + (784d2/2500) = (1225d2/2500) CD= 35d/50 = 7d/10 = 7d/10 = (3d+4d)/10 = (AB+AE)/10 https://piyushtheorem.wordpress.com/2017/02/08/a-theorem-on-right-angled-triangles/