Hello everyone.
In this segment, we are going to estimate the moment of inertia for the isosceles triangle 
The isosceles triangle is a special case of the triangle ABC , for which the two sides AC and CB are equal.
And if we drop a perpendicular to the base, we will find that the dimension a=b/2 
So
actually we are going to use the expressions which we have estimated for the Ixy and readjust
and substitute the value of a as a=b/2
Since we have Ixy=(b*h^2/24)*(2a+b) and our a, which is 
the distance from our perpendicular to the left corner A is =b/2 and our two axes
We can draw, that this is the X axis  ,and this is y axis, and our CG
we have x' and y'.
the Cg
somewhere here, so our Ix y about these two axes X &Y passing by point A, 
will be (b*h^2/24)*(2a+b), where 2a=b , so inside bracket
we have (2b)*(b*h^2/24) at 
the end.
We have (b^2*h^2).
/12)
Remember that X Cg was(a+b)/3 and ycg 
was (h/3) and since a again 
=b/2, we can write that Xcg=(b+b/2)/3
which will (3*b/6) =b/2, and ycg=h/3
So we are going to use.
We have (b^2*h^2/12) as total value minus the area 
Ixy cg=Ixy-A* x*y̅
So our Area (1/2b*h)*(x̅)*( y̅),x̅=b/2 and y̅
is h/3, we move to the next  slide in which we have Ixycg=b^2*h^2/12-(1/2*b*h)
*(h/3), we will find out that
open one bracket (our denominator =(2*2*3)= 12,
we have  bh*b*h= b^2*h^2 /12, so we have 
inside the  bracket (1-1)=0,  which means that at the CG we have major axes  x' and y'
and the product of inertia will be= zero.
Thanks a lot and see you.
Goodbye.
