Two corners of an isosceles triangle are at #(2 ,5 )# and #(9 ,8 )#. If the triangle's area is #12 #, what are the lengths of the triangle's sides?

1 Answer
Rafael
Oct 11, 2016

#sqrt(1851/76)#

Explanation:

The two corners of the isosceles triangle are at (2,5) and (9,8). To find the length of the line segment between these two points, we will use the distance formula (a formula derived from the Pythagorean theorem).

Distance Formula for points #(x_1,y_1)# and #(x_2,y_2)#:
#D=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

So given the points #(2,5)# and #(9,8)#, we have:
#D=sqrt((9-2)^2+(8-5)^2)#
#D=sqrt(7^2+3^2)#
#D=sqrt(49+9)#
#D=sqrt(57)#

So we know that the base has a length #sqrt(57)#.

Now we know that the area of the triangle is #A=(bh)/2#, where b is the base and h is the height. Since we know that #A=12# and #b=sqrt(57)#, we can compute for #h#.

#A=(bh)/2#
#12=(sqrt(57)h)/2#
#24=(sqrt(57)h)#
#h=24/sqrt(57)#

Finally to find the length of a side, we will use Pythagorean theorem (#a^2+b^2=c^2#). From the image, you can see that we can divide an isosceles triangle into two right triangles. So to find the length of one side, we can take one of the two right triangles then use the height #24/sqrt(57)# and the base #sqrt(57)/2#. Take note that we divided the base by two.
enter image source here

#a^2+b^2=c^2#
#(24/sqrt(57))^2+(sqrt(57)/2)^2=c^2#
#576/57+57/4=c^2#
#192/19+57/4=c^2#
#(768+1083)/76=c^2#
#1851/76=c^2#
#c=sqrt(1851/76)#

So the length of its sides is #sqrt(1851/76)#

Related questions
See all questions in Perimeter and Area of Triangle
Impact of this question
1587 views around the world
You can reuse this answer
Creative Commons License