Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 618.
Thursday, June 9, 2011
Problem 618: Triangle Area, 60 Degrees, Incircle, Incenter, Tangent
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∆=(1/2)bc sin 60° = (√3/4)bc
ReplyDelete(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
S=(1/2)bc.sinA=bc.sin(A/2).cos(A/2)
ReplyDeletesin²(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)
Pure Geometry solution
ReplyDeleteLet 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
Join EF, AEF is an equilaterial triangle
ReplyDeleteLet AE=AE=EF=x & radius of circle=r
<EOF=120
By cosine law, x^2=r^2+r^2-2r^2cos120
x=r*sqrt3
r=x/sqrt3
[ABC]=[(AB)(BC)sin60]/2=[(2+x)(5+x)*sqrt3]/4
Also, [ABC]=r*(2+x+5+x+7)/2=r(7+x)---------(*)
r(7+x)=[(2+x)(5+x)*sqrt3]/4----------(2)
Sub (1) in (2)
4x(7+x)=[(x+2)(x+5)*3]
4x^2+28x=3x^2+21x+30
x^2+7x-30=0
(x+10)(x-3)=0
x=3 or -10 (rej)
r=sqrt3
By (*), [ABC]=r(7+x)=10sqrt3