Two corners of an isosceles triangle are at #(4 ,2 )# and #(6 ,1 )#. If the triangle's area is #3 #, what are the lengths of the triangle's sides?

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Answer 1 · 2016-04-27T13:22:47

Length of the sides are # (13sqrt(5))/10#

Using Pythagoras the distance between points is

# sqrt ((x_2-x_1)^2+(y_2-y_1)^2) -> sqrt((6-4)^2+(1-2)^2)#

#=sqrt(5)->"triangle base"#
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Tony B

Area = half the base #xx# hight #->3 = sqrt(5)/2xxh#

#=>h=6/sqrt(5)#

Using Pythagoras #(sqrt(5)/2)^2 +h^2=x^2#

#=>x=sqrt((sqrt(5)/2)^2+(6/sqrt(5))^2)#

#=>x=sqrt(5/4+36/5)#

#=>x=sqrt((25+144)/20) = sqrt(169/20)#

#x=sqrt( 13^3/(2^2xx5))#

#=>x=13/2sqrt(1/5) = 12/(2sqrt(5))#

It is frowned upon to have a root in the denominator

Multiply by 1 but in the form of #1=sqrt(5)/sqrt(5)#

#x=13/(2sqrt(5))xxsqrt(5)/sqrt(5) = (13sqrt(5))/10#