What is the area of an equilateral triangle with an inscribed circle of radius #6sqrt3?#

2 Answers
Somebody N. · EZ as pi
May 25, 2018

#color(blue)(324sqrt(3)" units squared")#

Explanation:

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From the diagram:

Since the sides #AB# , #AC# ,#BC# are tangents to the circle, then:

#/_ODA=/_OEA=/_OFC=90^@#

The radius is perpendicular to the tangent at the point of tangency.

Because the triangle is equilateral #BF=FC# and #AF# is the perpendicular bisector, bisecting #/_DAE#

#/_DAE=60^@#, #/_DAO=1/2/_DAE=30^@#

#AO=(6sqrt(3))/sin(30)=12sqrt(3)#

#OF="radius"=6sqrt(3)#

#AF=12sqrt(3)+6sqrt(3)=18sqrt(3)#

#/_FCE=60^@# , #/_FCO=1/2/_FCE=30^@#

#FC=(6sqrt(3))/tan(30)#

Area of triangle:

#1/2 "base"xx"height"=FCxxAF=(6sqrt(3))/tan(30)*18sqrt(3)#

#=(6sqrt(3))/(sqrt(3)/3)*18sqrt(3)=324sqrt(3) # units squared.

Note:

This was only possible to solve given just the radius of the circle, because of the unique properties of the equilateral triangle and its symmetry. This would not have been possible if the triangle had been non-equilateral.

EZ as pi
May 25, 2018

#324sqrt3#

Explanation:

Referring to the same diagram as in Somebody N's answer.

In the diagram: #Delta ABC# is equilateral.

#:. /_A= /_B =/_C= 60° and#

# BE=CD=AF# and are lines of symmetry,

#/_OFC = 90°# (angle between tangent and radius.)

In #Delta OFC: /_C=30°,, /_F=90°, and color(red)(OF = 6sqrt3)#

#(FC)/(6sqrt3)= Tan 60° = sqrt3#

#FC = 6sqrt3 sqrt3 = 6xx3=18.#

#:. color(blue)(BC = 2xx FC =36)#

#Area Delta = 1/2color(blue)(b)color(red)(h)#

For #Delta OBC: A = 1/2 xx36xx6sqrt3 = 108sqrt3#

#Area Delta ABC = 3 xx 108sqrt3 = 324sqrt3#

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