Two corners of an isosceles triangle are at #(7 ,6 )# and #(4 ,9 )#. If the triangle's area is #24 #, what are the lengths of the triangle's sides?

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Answer 1 · 2017-06-13T06:43:52

The length of the other sides is #=11.5#

The length of the base is

#b=sqrt((7-4)^2+(6-9)^2)=sqrt(3^2+3^2)=3sqrt2#

Let the altitude of the triangle be #=h#

Then,

The area is #A=1/2bh#

#1/2*3sqrt2*h=24#

#h=(2*24)/(3sqrt2)=8sqrt2#

The other sides of the triangle are

#a=c=sqrt(h^2+(b/2)^2)#

#=sqrt((8sqrt2)^2+(3/2sqrt2)^2)#

#=sqrt(128+9/2)#

#=sqrt(265/2)#

#=11.5#