
#1
November 10th, 2016,
07:55
Any thoughts on the sketch?
Many Thanks
Last edited by greg1313; November 10th, 2016 at 08:25.
Reason:
Improve title

November 10th, 2016 07:55
# ADS
Circuit advertisement

#2
November 10th, 2016,
13:24
I presume that the "right triangle" referred to is the one at the right of the parallelogram. It should be clear to you that the right triangle at the left is identical to the first one and has the same area. So in order that the area of either of those two right triangles be "exactly half" the area of the parallelogram, there must be nothing but those two right triangles. That is, there is no "middle section" the "two" verticals must be one that vertical goes from vertex D to vertex B. What does that look like?

#3
November 11th, 2016,
03:33
Thread Author

MHB Seeker
#4
November 11th, 2016,
13:53
Last edited by Klaas van Aarsen; December 10th, 2016 at 20:10.
Reason:
Updated preamble

#5
November 13th, 2016,
08:48
Originally Posted by
mathlearn
Hello
HallsofIvy,
This should be that identical right angled triangle at the left
Then the triangle which has half the area of the parallelogram would be,
Correct ?
Many THanks
No, those are not right triangles. And it is always true that the two triangles you get by drawing a diagonal have area half the parallelogram.
 
   Updated   
Originally Posted by
I like Serena
Hey
mathlearn!
I think the right triangle is fixed.
We have something like this:
And if we make $x$ smaller, we get:
Hmm... let's make $x$ negative:
I think we need $x=0$:
Now the right triangle takes up half of the parallellogram.
Yes, that was what I meant. Nice drawings!