Two corners of an isosceles triangle are at #(7 ,2 )# and #(3 ,9 )#. If the triangle's area is #24 #, what are the lengths of the triangle's sides?

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Answer 1 · 2017-07-28T08:04:03

The lengths of the sides of the isoceles triangle are #8.1u#, #7.2u# and #7.2u#

The length of the base is

#b=sqrt((3-7)^2+(9-2)^2)=sqrt(16+49)=sqrt65=8.1u#

The area of the isoceles triangle is

#area=a=1/2*b*h#

#a=24#

Therefore,

#h=(2a)/b=(2*24)/sqrt65=48/sqrt65#

Let the length of the sides be #=l#

Then, by Pythagoras

#l^2=(b/2)^2+h^2#

#l^2=(sqrt65/2)^2+(48/sqrt65)^2#

#=65/4+48^2/65#

#=51.7#

#l=sqrt51.7=7.2u#