Thursday, June 9, 2011

Problem 618: Triangle Area, 60 Degrees, Incircle, Incenter, Tangent

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 618.

 Problem 618: Triangle Area, 60 Degrees, Incircle, Incenter, Tangent

3 comments:

  1. ∆=(1/2)bc sin 60° = (√3/4)bc
    (1/2)=cos 60°=(b² + c² -a²)/2bc= (b² + c²-49)/2bc
    b²+c²=bc+49
    s-b=2, s-c=5 and so b-c=3
    2bc=(b²+c²)-(b-c)²=bc+49-9=bc+40
    bc=40
    ∴∆=(√3/4)bc=10√3 sq units

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  2. S=(1/2)bc.sinA=bc.sin(A/2).cos(A/2)
    sin²(A/2)=(s-b).(s-c)/bc
    bc=(s-b).(s-c)/sin²(A/2)
    S=(s-b).(s-c).cotan(A/2)
    S=2*5*sqr(3)

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  3. Pure Geometry solution

    Let h be the length of the altitude from B and let AE = AF = p

    So h = (p+2)√3/2

    And h^2 + (p/2 + 4)^2 = 49 from which p = 3 and h = 5√3/2

    So the Area(ABC) = ½ h(p+5) = 10√3


    Sumith Peiris
    Moratuwa
    Sri Lanka

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