## Knowledge in New Proof

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 ∆&nbsp; BCEG = Area of ∆&nbsp; BDC +Area of ⌂ DCEF + Area of ∆ EFG c^2=ac/2+ Area of ⌂ DCEF + (c-b) &nbsp;c/2 (2c^2&ndash; ac &ndash;c^2+ bc )/2=Area of ⌂ DCEF (c^2&ndash; ac+ bc )/2=Area of ⌂ DCEF c(c&ndash; a+ b)/2=Area of ⌂ DCEF Area of ⌂ DCEF=BC(DE+CF)/2 Copyrighted&copy;PiyushGoel

Piyush Theorem

&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;PiyushTheorem PiyushTheorem: In a Right-Angled Triangle with sides in A.P. Series, the distance between the point of intersection of median &amp; altitude at the base is 1/10 Th&nbsp;the sum of other two sides. This Theorem applies in Two Conditions: 1.The&nbsp;Triangle&nbsp;must be&nbsp;Right-Angled. 2.Its&nbsp;Sides&nbsp;are in&nbsp;A.P. Series. 1.Proof with Trigonometry Tan&nbsp;&nbsp;&alpha;&nbsp;&nbsp;&nbsp;=AD/DC AD= DC&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;&nbsp;&nbsp;&mdash;&mdash;&mdash;&mdash;&mdash;&ndash;1 Tan&nbsp;&nbsp;&alpha;&nbsp;= AD/DE AD= DE&nbsp;Tan2&nbsp;&alpha;&nbsp;&nbsp;&nbsp;&mdash;&mdash;&mdash;&mdash;&mdash;-2 DC&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;= DE&nbsp;Tan 2&nbsp;&alpha; (DE+EC)&nbsp;&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;= DE&nbsp;Tan 2&nbsp;&alpha; DE&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;&nbsp;+ EC&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;= DE&nbsp;Tan 2&nbsp;&alpha; DE&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;&nbsp;+ EC&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;= 2 DE&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;/ (1-&nbsp;Tan2&nbsp;&nbsp;&alpha;&nbsp;&nbsp;&nbsp;) DE&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;&nbsp;&nbsp;&ndash; DE&nbsp;Tan3&nbsp;&nbsp;&alpha;&nbsp;+ EC&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;&ndash;EC&nbsp;Tan3&nbsp;&nbsp;&alpha;&nbsp;&nbsp;= 2DE&nbsp;Tan&nbsp;&nbsp;&alpha; EC&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;&ndash;EC&nbsp;Tan3&nbsp;&nbsp;&alpha;&nbsp;&ndash; DE&nbsp;Tan3&nbsp;&nbsp;&alpha;&nbsp;= 2DE&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;&ndash; DE&nbsp;Tan&nbsp;&nbsp;&alpha; Tan&nbsp;&nbsp;&alpha;&nbsp;(EC &ndash; EC&nbsp;Tan2&nbsp;&nbsp;&alpha;&nbsp;&ndash; DE&nbsp;T an2&nbsp;&nbsp;&alpha;&nbsp;)= DE&nbsp;Tan&nbsp;&nbsp;&alpha; DE&nbsp;Tan2&nbsp;&nbsp;&alpha;&nbsp;&nbsp;&ndash; DE = EC&nbsp;Tan2&nbsp;&nbsp;&alpha;&nbsp;&nbsp;&ndash; EC -DE (&nbsp;Tan2&nbsp;&nbsp;&alpha;&nbsp;+ 1) = -EC (1 &ndash;&nbsp;Tan2&nbsp;&nbsp;&alpha;&nbsp;) DE (sin2&nbsp;&alpha;&nbsp;&nbsp;/cos2&nbsp;&alpha;&nbsp;+ 1) = EC (1-&nbsp;sin2&nbsp;&alpha;&nbsp;&nbsp;/cos2&nbsp;&alpha;&nbsp;) DE (sin2&nbsp;&alpha;&nbsp;+&nbsp;cos2&nbsp;&alpha;&nbsp;/cos2&nbsp;&alpha;&nbsp;) = EC (cos2&nbsp;&alpha;&nbsp;&ndash;&nbsp;sin2&nbsp;&alpha;&nbsp;/cos2&nbsp;&alpha;&nbsp;) DE (sin2&nbsp;&alpha;&nbsp;&nbsp;+&nbsp;cos2&nbsp;&alpha;&nbsp;) = EC(cos2&nbsp;&alpha;&nbsp;&nbsp;&ndash;sin2&nbsp;&alpha;&nbsp;) DE (sin2&nbsp;&alpha;&nbsp;&nbsp;+&nbsp;cos2&nbsp;&alpha;&nbsp;) = EC (cos2&nbsp;&alpha;&nbsp;&nbsp;&ndash;sin2&nbsp;&alpha;&nbsp;) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;where (sin2&nbsp;&alpha;&nbsp;&nbsp;+&nbsp;cos2&nbsp;&alpha;&nbsp;=1) &amp; (cos2&nbsp;&alpha;&nbsp;&nbsp;&ndash;sin2&nbsp;&alpha;&nbsp;=&nbsp;cos2&nbsp;&alpha;&nbsp;&nbsp;) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;DE= EC&nbsp;cos2&nbsp;&alpha;&nbsp;&nbsp; cos&nbsp;&alpha;&nbsp;&nbsp;&nbsp;=a/a+d &nbsp;&nbsp;&amp;&nbsp;sin&nbsp;&alpha;&nbsp;= (a-d)/ (a +d) cos2&nbsp;&alpha;&nbsp;&nbsp;= a2/ (a +b)&nbsp;2 sin2&nbsp;&alpha;&nbsp;&nbsp;= (a-d)&nbsp;2/ (a+ d)&nbsp;2 DE= EC (cos2&nbsp;&alpha;&nbsp;&nbsp;&nbsp;&nbsp;&ndash;&nbsp;sin2&nbsp;&alpha;&nbsp;) = EC (a2&nbsp;/ (a +b)&nbsp;2&nbsp;&ndash; (a-d)&nbsp;2/ (a +d)&nbsp;2 = EC (a2&nbsp;&ndash; (a-d)&nbsp;2/ (a +d)&nbsp;2 = EC (a &ndash;a +d) (a+ a-d)/ (a+ d)&nbsp;2 = EC (d) (2a -d)/ (a+ d)&nbsp;2 = (a +d)/2(d) (2a -d)/ (a +d)&nbsp;2&nbsp;&mdash;&mdash;&mdash;&mdash;- where EC= (a +d)/2 = (d) (2a -d)/2(a +d) = (d) (8d -d)/2(4d+d) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;where a= 4d (as per the Theorem) = 7d2&nbsp;/2(5d) = 7d /10 = (3d+4d)/10= (AB+AC)/10 2.Proof with Obtuse Triangle Theorem AC2=EC2&nbsp;+AE2&nbsp;+2CE.DE&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;where EC = ( &nbsp;a +d) /2,AE=( a +d)/2 a2&nbsp;= (a +d/2)2&nbsp;+ (a+ d/2)2&nbsp;+ 2(a +d)/2DE = (a +d/2) (a+d+2DE) = (a +d/2) (a+d+2DE) &nbsp;&nbsp;where a=4d 16d2&nbsp;= (5d/2) (5d+2DE) 32d/5 = 5d + 2DE 32d/5 &ndash; 5d = 2DE 32d -25d/5 = 2DE DE =7d/10 = (3d+4d)/10 = (AB+AC)/10 3.Proof with Acute Triangle Theorem AB2= AC2+BC2&nbsp;&ndash; 2BC.DC (a-d) 2= a2&nbsp;+ (a+ d)&nbsp;2&nbsp;-2(a+ d) (DE+EC) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;where AB= (a-d), AC=a, BC =( a +d) &amp; EC= (a +d)/2 (a-d)&nbsp;2&nbsp;&ndash; (a +d)2&nbsp;= a2&nbsp;&nbsp;-2(a +d)(DE+EC) (a- d &ndash;a-d) (a -d +a +d)&nbsp;&nbsp;= a2&nbsp;-2(a+ d) (2DE+a+d)/2 2(-2d) (2a)&nbsp;=&nbsp;2a2&nbsp;-2(a +d) (2DE+a+d) -8ad&nbsp;&ndash;&nbsp;2a2&nbsp;= -2(a +d) (2DE+a+d) -2a (4d&nbsp;&nbsp;&nbsp;+a) = -2(a +d) (2DE+a+d) a (4d&nbsp;&nbsp;+ a) = (a +d)(2DE+a+d) 4d (4 d&nbsp;+&nbsp;4d) = (4d+d) (2DE+4d+d) 4d (8d)&nbsp;= (5d) (2DE+5d) 32d2/5d = &nbsp;&nbsp;(2DE+5d) 32d/5 = &nbsp;&nbsp;(2DE+5d) 32d/5 &ndash; 5d = &nbsp;&nbsp;2DE (32d &ndash; 25d)/5 = &nbsp;&nbsp;2 DE DE = 7d/10 = (3d+4d)/10 = (AB+AC)/10 4. Proof with Co-ordinates Geometry Equation of BE Y &ndash; 0 =b-0/0-a(X &ndash; a) Y = -b/a(X) + b&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;- (1) M1&nbsp;= -b/a For perpendicular M1M2= -1 So M2=a/b Equation of AC Y &ndash; 0 = a/b(X-0) Y=a/b(X) &mdash;&mdash;&mdash;&mdash;&mdash;&mdash; (2) Put Y value in equation (1) a/b(X) + b/a(X) =b X (a2+b2/a b) = b X = ab2/ (a2&nbsp;+ 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 &ndash; ab2/ (a2&nbsp;+ b2), a2b/ (a2+b2) and co-ordinates of point d is (a/2, b/2) because d is midpoint. As per the &ldquo;Theorem&rdquo; a=z-d, b=z, c = z+ d (z +d)&nbsp;2= (z-d)&nbsp;2+z2&nbsp;from here z=4d so a=3d and b=4d Put value of a &amp; b ab2/ (a2&nbsp;+ b2), a2b/ (a2+b2) &amp; (a/2, b/2) ab2/ (a2&nbsp;+ 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&nbsp;+ (72d-100d/50)2 = (21d/50)2&nbsp;+ (-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/

Dams and purpose of having dams

Dams and purpose of having dams

Problem Based on Engineering Questions

Probability based questions is important to solve because it come in both compatative and Board also please solve as many questions in sysmatic

PROBABILITY AND STATISTICS

THIS PDF CONTAINS NOTES OF PROBABILITY AND STATISTICS

PROBABILITY AND STATISTICS

THIS PDF CONTAINS NOTES OF PROBABILITY AND STATISTICS