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Two corners of an isosceles triangle are at #(8 ,5 )# and #(1 ,7 )#. If the triangle's area is #15 #, what are the lengths of the triangle's sides?

Answer 1

see below.

Name the points #M(8,5) and N(1,7)#
By Distance formula,
#MN=sqrt((1-8)^2+(7-5)^2)=sqrt53#
Given Area #A=15#,
#MN# can be either one of the equal sides or the base of the isosceles triangle.

Case 1) : #MN# is one of the equal sides of the isosceles triangle.
#A=1/2bh#, where #b and h# are the base and the height of the triangle, respectively.
#=> 15=1/2*MN*h#
#=> h=(2*15)/sqrt53=30/sqrt53#
#=> MP=PN# (the equal side) #= sqrt(((MN)/2)^2+h^2)#
#= sqrt((sqrt53/2)^2+(30/sqrt53)^2)#
#=sqrt(6409/212)#

Therefore, lengths of the triangle's sides are #sqrt(6409/212), sqrt(6409/212), sqrt53#

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Answer 2

Sadie Adams

The lengths of the sides of the isosceles triangle are approximately 8.37 units, 8.37 units, and 7.42 units.

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Answer from HIX Tutor

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