Two corners of an isosceles triangle are at #(2 ,4 )# and #(1 ,8 )#. If the triangle's area is #64 #, what are the lengths of the triangle's sides?

Question

1176 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-20T15:56:51

#color(blue)((5sqrt(44761))/34 , (5sqrt(44761))/34, sqrt(17)#

Let #A=(2,4), and B=(1,8)#

Then side #c=AB#

Length of #AB=sqrt((1-2)^2+(8-4)^2)=sqrt(17)#

Let this be the base of the triangle:

Area is:

#1/2ch=64#

#1/2sqrt(17)(h)=64#

#h=128/sqrt(17)#

For isosceles triangle:

#a=b#

Since the height bisects the base in this triangle:

#a=b=sqrt((c/2)^2+(h^2))#

#a=b=sqrt((sqrt(17)/2)^2+(128/sqrt(17))^2)=(5sqrt(44761))/34~~31.11#

Sides are:

#color(blue)((5sqrt(44761))/34 , (5sqrt(44761))/34, sqrt(17)#