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Mathematics Dictionary
Dr. K. G. Shih
Area of Triangle


  • Q01 | - Area of triangle = b*c*sin(A)/2
  • Q02 | - Area of triangle = Sqr(s(s-a)*(s-b)*(s-c)/(b*c))
  • Q03 | - Area of triangle = a*b*c/(4*R)
  • Q04 | - Area of triangle = 2*(R^2)*sin(A)*sin(B)*sin(C)
  • Q05 | - Area of triangle = r*s
  • Q06 | - Area of triangle = (x1*y2 + x3*y1 + x2*y3 - x3*y2 - x2*y1 - x1*y3)
  • Q07 | - Colinear and blanking a polygon

    • Q01. Area of triangle = b*c*sin(A)/2

    Proof
    • Draw triangle ABC : Let BC = a, CA = b and AB = c
    • Draw CF perpendicular to AB
    • Hence CF = AC*sin(A)
    • Area of triangle = AB*CF/2
    • Hence area of triangle = b*c*sin(A)/2
    Formula
    • 1. Area of triangle = b*c*sin(A)/2
    • 2. Area of triangle = c*a*sin(B)/2
    • 3. Area of triangle = a*b*sin(C)/2

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    Q02. Area of triangle = Sqr(s(s-a)*(s-b)*(s-c)/(b*c))

    Hint : It requires     Proof
    • Since area of triangle = b*c*sin(A)/2 .............. (1)
    • Since sin(A) = 2*Sqr(s*(s-a)*(s-b)*(s-c))/(b*c) .... (2)
    • Substitute (2) into (1)
    • Hence area of triangle = Sqr(s*(s-a)*(s-b)*(s-c)/(b*c))
    • This is called Heron formula (MD 2002 program 20 13 or 20 16).

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    Q03. Area of triangle = a*b*c/4

    Hint : It requires
    • 1. Sine law : a = 2*R*sin(A), b = 2*R*sin(B) and c = 2*R*sin(C)
    • 2. Area of triangle = a*b*sin(C)/2
    Proof
    • Area of triangle = a*b*sin(C)/2
    • sin(C) = c/(2*R)
    • Hence area of triangle = a*b*c/(4*R)
    • Where R is radius of circum-circle

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    Q04. Area of triangle = 2*(R^2)*sin(A)*sin(B)*sin(C)

    Hint : It requires
    • 1. Sine law : a = 2*R*sin(A), b = 2*R*sin(B) and c = 2*R*sin(C)
    • 2. Area of triangle = a*b*sin(C)/2
    Proof
    • Area of triangle = a*b*sin(C)/2
    • a = 2*R*sin(A)
    • b = 2*R*sin(B)
    • Hence Area of triangle = 2*(R^2)*sin(A)*sin(B)*sin(C)

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    Q05. Area of triangle = r*s

    Definition
    • Radius of in-circle = r
    • s = (a + b + c)/2 for triangle ABC
    Construction
    • Draw triangle ABC
    • Draw bisectors of angles
    • The bisectors of angles meet at I
    • From I draw lines perpendicular to each side
    • The distance from I to side is r
    Proof
    • Area of triangle AIB = r*c/2
    • Area of triangle BIC = r*a/2
    • Area of triangle CIA = r*b/2
    • Area of triangle ABC = AIB + BIC + CIA
      • = r*(a + b + c)/2
      • = r*s

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    Q06. Area of triangle = (x1*y2 + x3*y1 + x2*y3 - x1*y2 - x2*y1 - x1*y3)/2

    Definition
    • A(x1, y1), B(x2, y2), C(x3, y3) are three points
    • Find area of triangle ABC
    Formula
    • | x1 y1 1 |
    • | x2 y2 1 | = 0.5*(Area of triangle ABC)
    • | x3 y3 1 |
    Proof
    • Area of Square APQR
      • = (x2 - x1)*(y3 - y1)
      • = x2*y3 - x2*y1 - x1*y3 + x1*y1 ........... (1)
    • Area of triangle APB
      • = (x2 - x1)*(y2 - y1)/2
      • = (x2*y2 - x2*y1 - x1*y2 + x1*y1)/2 ....... (2)
    • Area of triangle BQC
      • = -(x2 - x3)*(y2 - y3)/2
      • = -(x2*y2 - x2*y3 - x3*y2 + x3*y3)/2 ...... (3)
    • Area of triangle ACR
      • = (x3 - x1)*(y3 - y1)/2
      • = (x3*y3 - x3*y1 - x1*y3 + x1*y1)/2 ....... (4)
    • Area ABC = (1) - (2) - (3) - (4)
      • = x2*y3 - x2*y1 - x1*y3 + x1*y1
      • - 0.5*(x2*y2 - x2*y1 - x1*y2 + x1*y1)
      • - 0.5*(-(x2*y2 - x2*y3 - x3*y2 + x3*y3)
      • - 0.5*(x3*y3 - x3*y1 - x1*y3 + x1*y1 )
      • = x2*y3 - x2*y1 - x1*y3 - 0.5*(-x2*y1 - x1*y2)
      • - 0.5*(x2*y3 + x3*y2) - 0.5*(-x3*y1 - x1*y3)
      • = (x2*y3 - 0.5*x2*y3) - x2*y1 + 0.5*x2*y1 - x1*y3 + 0.5*x1*y3
      • + 0.5*x1*y2 - 0.5*x3*y2 + 0.5*x1*y3
      • = 0.5*(x1*y2 + x3*y1 + x2*y3 - x3*y2 - x2*y1 - x1*y3)

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    Q07. Colinear and blanking a polygon Colinear : 3 points (x1, y1), (x2, y2), (x3, y3) on a line
    • | x1 y1 1 |
    • | x2 y2 1 | = 0
    • | x3 y3 1 |
    Point (xp, yp) inside a polygon : A,B,P in clockwise direction
    • | x1 y1 1 |
    • | x2 y2 1 | = +
    • | xp yp 1 |
    Point (xp, yp) inside a polygon : B,C,P in clockwise direction
    • | x2 y2 1 |
    • | x3 y3 1 | = +
    • | xp yp 1 |
    Do this check for point C,D,P; D,E,P; ....
    • If all results in same sign, then point P is inside polygon ABCDEF
    • This method is used to balnk a polygon on the computer

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