Two corners of an isosceles triangle are at #(3 ,9 )# and #(2 ,7 )#. If the triangle's area is #4 #, what are the lengths of the triangle's sides?

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Answer 1 · 2016-05-04T14:48:56

#color(brown)("As a simplified exact value:")#

#color(blue)(s=sqrt(549)/(2sqrt(17))=(3sqrt(1037))/34)#

#color(brown)("As an approximate decimal")#

#color(blue)(s~~2.831" to 3 decimal places")#

Tony B

Let the vertices be A,B and C
Let the corresponding sides be a, b, and c.

Let the width be w
Let the vertical height be h
Let the length of sides a and c be s

Given: Area = 4
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the value of w")#

Using Pythagoras #" "w=sqrt( (9-7)^2+(3-2)^2)#

#color(blue)(=> w= sqrt(16+1) = sqrt(17))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the value of h")#

Given area# = 4 =1/2wh#

#color(blue)(h=8/w= 8/sqrt(17))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Using Pythagoras

#s^2=(w/2)^2+h^2#

#s^2=(sqrt(17)/2)^2+(8/sqrt(17))^2#

#s=sqrt(17/4+64/17)#

#s=sqrt(545/68)#

#color(brown)("As a simplified exact value this:")#

#color(blue)(s=sqrt(549)/(2sqrt(17))=(3sqrt(1037))/34)#

#color(brown)("As an approximate decimal")#

#color(blue)(s~~2.831" to 3 decimal places")#