How do you find the area of a isosceles triangle with base 10 and perimeter 36?

##60## ##square units##
I have two solutions for you using the formulas
##A=1/2*base*height=1/2bh##
-and using –
##A=sqrt(s(s-a)(s-b)(s-c))##
where:
##a b and c =>## sides of the triangles
##s=>## semi-perimeter ##=P/2=(a+b+c)/2##
Since it is an isosceles triangle the measure of the sides given its periemter would be ##13## ##13## and ##10##. **Because two sides of an isosceles triangle is congruent.
Using ##A=1/2bh## :
First is to find the height.
Using this figure:
Know that the height of the triangle bisects the side opposite the vertex.
We can use pythagorean formula to find the height:
##h=sqrt((13^2)-(5^2))##
##h=12##
then we solve for the area:
##A=1/2bh=1/2(10)(12)##
##A=60## square units
Using Heron’s Formula:
Let the sides of the triangle be:
##a=13##
##b=13##
##c=10##
##s=(a+b+c)/2=(13+13+10)/2=18##
##A=sqrt(s(s-a)(s-b)(s-c))##
##A=sqrt(18(18-13)(18-13)(18-10))##
##A=60## square units