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Symbol Defintion
Example : Sqr(x) = square root of x
Q01. Diagram and question
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- Draw a triangle ABC and sides are a,b,c
- Let E be a point on AC and CE = b/n
- Let T be a point on BA and TA = c/n
- Let F be a point of BC and BF = a/n
- Join ET, TF and FE so that make four triangles
- 1. Prove that area of triangle EFT = (area ABC)*(1 - 3*(n-1))/(n^2)
- 2. Prove that area FEC = area EAT = area TBF
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Q02. Area of triangle and sine law
Sine law
- a = 2*R*sin(A)
- b = 2*R*sin(B)
- c = 2*R*sin(C)
- = b*c*sin(A)/2
- = c*a*sin(B)/2
- = a*b*sin(B)/2
- = b*c*sin(A)/2 = b*c*(a/(2*R))/2 = a*b*c/(4*R)
- = c*a*sin(B)/2 = c*a*(b/(2*R))/2 = a*b*c/(4*R)
- = a*b*sin(B)/2 = a*b*(c/(2*R))/2 = a*b*c/(4*R)
- Hence area ABC = 2*(R^2)*sin(A)*sin(B)*sin(C)
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Q03. Area EFT = (area ABC)*(1 - 3*(n-1)/(n^2))
Question
- Find the area of triangle ETF = (are ABC)*(1 - 3*(n-1)/(n^2))
- Area of triangle CEF
- Area = (FC*CE)*sin(C)/2
- Area = (a/n)*((n-1)*b/n)*sin(C)/2
- Area = ((n-1)/(n^2))*a*b*sin(C)/2
- Area = ((n-1)/(n^2))*(area of triangle ABC)
- Area of triangle AET
- Area = (AE*AT)*sin(A)/2
- Area = (b/n)*((n-1)*c/n)*sin(A)/2
- Area = ((n-1)/(n^2))*b*c*sin(A)/2
- Area = ((n-1)/(n^2))*(area of triangle ABC)
- Area of triangle BFT
- Area = (BT*BF)*sin(B)/2
- Area = (c/n)*((n-1)*a/n)*sin(B)/2
- Area = ((n-1)/(n^2))*b*c*sin(B)/2
- Area = ((n-1)/(n^2))*(area of triangle ABC)
- Hence area EFT
- Area = ABC - CEF - AET - BFT
- Area = (area ABC)*(1 - 3*(n-1)/(n^2))
- Area = (area ABC)*((n^2 - 3*n + 3)/(n^2)) and n > 1
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Q04. Verify
Exercies : Use above results to find
- Area EFT = (1/4)*(Area ABC) if n = 2
- Area EFT = (3/9)*(Area ABC) if n = 3
- Area EFT = (07/16)*(Area ABC) if n = 4
- Area EFT = (13/25)*(Area ABC) if n = 5
- Area = (1 - 3*(3*(area ABC))/16)
- Area = (1 - 9/16)*(area ABC)
- Area = 7*(area ABC)/16
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