Two corners of an isosceles triangle are at #(2 ,6 )# and #(7 ,5 )#. If the triangle's area is #64 #, what are the lengths of the triangle's sides?

1 Answer
Harish Chandra Rajpoot
Jul 11, 2018

#5.099#, #25.6155# & #25.6155#

Explanation:

The length of side of isosceles triangle joining the vertices #(2, 6)# & #(7, 5)#
#=\sqrt{(2-7)^2+(6-5)^2}#

#=\sqrt{26}#

If #h# is the length of altitude dropped from third vertex to the side/base joining the vertices #(2, 6)# & #(7, 5)# then the area of isosceles triangle

#\frac{1}{2}(\text{base})\times \text{(altitude)}#

#1/2 \sqrt{26}h=64#

#h=128/\sqrt{26}#

Now, using the Pythagorean theorem in a right triangle with legs #\sqrt26# & #128\/sqrt{26}#, length of each of equal sides of isosceles triangle

#\sqrt{(\sqrt26)^2+(128/\sqrt{26})^2}#

#=25.6155#

hence, the sides of given isosceles triangle are #\sqrt{26}=5.099#, #25.6155# & #25.6155#

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