Two corners of an isosceles triangle are at #(8 ,5 )# and #(6 ,2 )#. If the triangle's area is #4 #, what are the lengths of the triangle's sides?

1 Answer
Feb 7, 2017

Lengths of triangle's sides are # 3.61(2dp) , 2.86(dp), 2.86(dp)# unit.

Explanation:

Length of base of isoceles triangle is #b=sqrt((x_1-x_2)^2+(y_1-y_2)^2) = sqrt((8-6)^2+(5-2)^2)= sqrt(4+9)=sqrt 13=3.61(2dp)#

Area of isoceles triangle is # A_t = 1/2 *b*h or 4 = 1/2*sqrt13*h or h = 8/sqrt 13 =2.22(2dp)#. Where #h# be the altitude of triangle.

Legs of isoceles triangle are #l_1=l_2= sqrt( h^2 +(b/2)^2) =sqrt( 2.22^2 +(3.61/2)^2) = 2.86(2dp)#unit

Lengths of triangle's sides are # 3.61(2dp) , 2.86(dp),2.86(dp)# unit. [Ans]