Hello everyone, in this segment.
We are going to estimate the product moment of inertia for the triangle abc, the base =b 
and the height is =h and the  two intersecting X-axis and Y-axis, they are intersecting at point
A.
If we drop a perpendicular line from point c  to the base ab, it will be intersecting at point d.
with distance to the left =a, for the Ixy.
We can consider this triangle abc as composed of two right angles triangles, the first one which is triangle
adc.
We call it Triangle #1 while the second one is a triangle bcd, we call the Triangle
#2, from the previous calculations, we have Ixy of the first triangle
axes X and Y.
passing by at the edge, at the intersection of the hypotenuse and the base, the value is
= h^2*a^2/8 and our base =a.
So we have the expression of Ixy= (h^2*a^2/8), while  for 
the other triangle,  triangle #2, we have a base=(b-a),from the general triangle after subtracting a,
and the height =h, and our calculation normally for two intersecting axes X and Y, which
is passing by the base and on  the opposite side of the triangle, which means it we cannot utilize
The Ixy estimated  for that axis because, it will be called  Ycd, anyway  Ixy for the Y passing
by Ycd = (base)^2*h^2/24, in that regard our base 
=(b-a)^2*h^2/24, while
Ixy at cg=-(base)^2*h^2/24, which means that 
= -(b-a)^2*h^2/72, in order to estimate the Ixy  for the axis Y which is
apart by a distance =a, we are going to utilize the Ixy at the CG, then we are going to add the area
multiplied by x̅ * multiplied by y̅ , so our final Ixy will be Ixy for the first triangle.
which is (h^2*a^2/8), (+) ixy for the second triangle at the Cg, which
is= -(b-a)^2*h^2/72(+) the area which is=(1/2(b-a)*(h)
*x *y, which are the distances of the CG , normal distance to x and y axes, the
first, x distance will be (a+1/3(b-a)),, which will give us, (2a+b)*1/3
and the y, the height in the y-direction is = (h/3), since we have the 72 here.
and we have a denominator of (3*3*2), which will give us (18), I'm going to it convert to 72
by multiplying by (4/4), so we have a common 72, the denominator, and we have
here (b-a)* h^2/72, that is a common numerator factor, we open one bracket
we have here (-)(b-a)^2, which means
we can take (b-a) out
We have still (-(b-a), for here since, we introduced ( 4 ) in the numerator.
then we will multiply  4 *(2a ), which is 8a, plus
(4)*(b), which is  (4b), summing  together, we have ( 4b-b )=3 *b, and we have
( 8a+a), which will be (9*a),  we will introduce  again and rearrange.
We have (h^2*a^2/8)+( b-a)*h^2/72
*(3*b+9*a)
We have a common factor of 3
we move to the second slide,  we are getting (3)out, so we have Ixy= h^2*a^2/8
+ (3/72).
h^2  and our new bracket will be (b-a)*(b+3a), we have
(h^2*a^2/8 +
we have (3/72), will give us 1/ (24) , and the
product of two brackets will be (b^2-b*a+3*a*b-3*a^2)
we will find that if we make the denominator as (24), we multiplied here (3)/(3) 
we have (+3a^2+b^2+2*a*b-3a^2), they will cancel each other
we left with Ixy=h^2*(1/24)*( 
(b^2+2a*b), that is  our expression, at the end but we have a common b inside the bracket,  we get it out
so we have (b*h^2/24)*(b+2a), there is another expression utilizing  
the area, so we can re adjust these terms  by considering that (b*h^2/24) composed of 
(1/2*b*h)*
(h*(2a+b) , so we have the first expression Ixy
=B*h^/24*(2a+b), while for the area which is (1/2*b*h),  we can write
that (1/2b*h) as At.
and then we have this area *h/12*(2a+b)
this is Ixy  for the triangle at  two axes intersecting at point a, in order to get the expression of Ixy
at the Cg, we are going to subtract the multiplication of the area by the distance (x cg*ycg) as follows
here we have the Xcg= (a+b)/3 ,  while the  the y̅
(h/3)
So again we have our expression which was (b*h^2/24)*(2a+b)
- the area which is (1/2*b*h) *(a+b)/3 *
by (h/3),  in order to make a common denominator by (72), we multiply first the term (3)/(3)
and for the second term, we will multiply by(4)/(4), let us check
we have here in the denominator of (72) while  we have here for the second term (
(2*3*3*4),(6*3) is 18 and( 18 *4) is72
So we have also in the numerator common factor b*h^2/72, we have here 
(3*2) in which 6*a +3*b 
we have a negative for  (-4a) 
and and negative for (-4b) 
So we are going to sum inside the bracket, we have (6a
-4a) will be(2a) (-4b+3b), give  (-b), that is the expression for  Ixy' at this C.g, Ix'y'
passing by the Cg for the triangle.
Thanks a lot and See you.
