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

Labels:
60 degrees,
area,
incenter,
incircle,
tangency point,
tangent,
triangle

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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