Two corners of an isosceles triangle are at #(7 ,2 )# and #(3 ,6 )#. If the triangle's area is #6 #, what are the lengths of the triangle's sides?

Question

Two corners of an isosceles triangle are at #(7  ,2   )# and #(3   ,6   )#.  If the triangle's area is #6   #, what are the lengths of the triangle's sides?

Avery Allen · Accepted Answer

The lengths of sides are: #a=5/2sqrt2=3.5355339# and #b=5/2sqrt2=3.5355339# and #c=4sqrt2=5.6568542#

First we let #C(x, y)# be the unknown 3rd corner of the triangle.

Also Let corners #A(7, 2)# and #B(3, 6)#

We set the equation using sides by distance formula #a=b#

#sqrt((x_c-3)^2+(y_c-6)^2)=sqrt((x_c-7)^2+(y_c-2)^2)#

simplify to obtain #x_c-y_c=1" " "#first equation

Use now the matrix formula for Area:

#Area=1/2((x_a,x_b,x_c,x_a),(y_a,y_b,y_c,y_a))=#

#=1/2(x_ay_b+x_by_c+x_cy_a-x_by_a-x_cy_b-x_ay_c)#

#Area=1/2((7,3,x_c,7),(2,6,y_c,2))=#

#Area=1/2*(42+3y_c+2x_c-6-6x_c-7y_c)#

#Area=6# this is given

We now have the equation

#6=1/2*(42+3y_c+2x_c-6-6x_c-7y_c)#

#12=-4x_c-4y_c+36#

#x_c+y_c=6" " " #second equation

Solving simultaneously the system #x_c-y_c=1# #x_c+y_c=6# #x_c=7/2# and #y_c=5/2#

We can now solve for the lengths of sides #a# and #b# #a=b=sqrt((x_b-x_c)^2+(y_b-y_c)^2)#

#a=b=sqrt((3-7/2)^2+(6-5/2)^2)#

#a=b=5/2sqrt(2)=3.5355339" " "#units

compute side #c#:

#c=sqrt((x_a-x_b)^2+(y_a-y_b)^2)#

#c=sqrt((7-3)^2+(2-6)^2)#

#c=sqrt(2(16))#

#c=4sqrt2=5.6568542#

Violet Aldrich · Answer

undefined The lengths of the triangle's sides are approximately 4.24 units, 4.24 units, and 3.78 units.