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Two corners of an isosceles triangle are at #(1 ,7 )# and #(5 ,3 )#. If the triangle's area is #6 #, what are the lengths of the triangle's sides?

Answer 1

Let the coordinates of the third corner of the isosceles triangle be #(x,y)#. This point is equidistant from other two corners.

#(x-1)^2+(y-7)^2=(x-5)^2+(y-3)^2#

#=>x^2-2x+1+y^2-14y+49=x^2-10x+25+y^2-6y+9#

#=>8x-8y=-16#

#=>x-y=-2#

#=>y=x+2#

Now the perpendicular drawn from #(x,y)# on the line segment joining two given corners of triangle will bisect the side and the coordinates of this mid point will be #(3,5)#.

So height of the triangle

#H=sqrt((x-3)^2+(y-5)^2)#

And base of the triangle

#B=sqrt((1-5)^2+(7-3)^2)=4sqrt2#

Area of the triangle

#1/2xxBxxH=6#

#=>H=12/B=12/(4sqrt2)#

#=>H^2=9/2#

#=>(x-3)^2+(y-5)^2=9/2#

#=>(x-3)^2+(x+2-5)^2=9/2#

#=>2(x-3)^2=9/2#

#=>(x-3)^2=9/4#

#=>x=3/2+3=9/2=4.5#

So #y=x+2=4.5+2=6.5#

Hence length of each equal sides

#=sqrt((5-4.5)^2+(3-6.5)^2)#

#=sqrt(0.25+12.25)=sqrt12.5=2.5sqrt2#

Hence lengths of three sides are #2.5sqrt2,2.5sqrt2,4sqrt2#

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Answer 2

Savannah Aguilar

The lengths of the triangle's sides are 4, 4, and 4√2.

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Answer from HIX Tutor

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