This seems to be the same as one of the arguments I give in the post, when I speak of ½bw. Yes, the result is not terribly difficult, as I mention, but I find it nevertheless interesting to discuss various proofs of it. For example, I like the shearing proof best, since it avoids any need to discuss formulas or algebra, and seems totally clear. Meanwhile, the first proof is very clear, again without any algebra, for the non-backtracking case.
+1/2 hbn= 1/2 h x (b1+b2+…+bn) = 1/2h x length of the rectangle = 1/2 the area of the rectangle.
This seems to be the same as one of the arguments I give in the post, when I speak of ½bw. Yes, the result is not terribly difficult, as I mention, but I find it nevertheless interesting to discuss various proofs of it. For example, I like the shearing proof best, since it avoids any need to discuss formulas or algebra, and seems totally clear. Meanwhile, the first proof is very clear, again without any algebra, for the non-backtracking case.
Seems trivial. Let b1, b2, etc. represent the bases of the n triangles so formed. Area = I/2 hb1 + 1/2 hb2 +..