- #1
Nguyen Thanh Nam
- 14
- 0
Hello! see if you can get the results same as me ):
Triangle ABC with AB=AC=a, BAC=90 deg. SA is perpendicular to plate (ABC) @ A. SA=a, also! In SB: ES=2EB. H is in plate (SBC) so that AH is perpendicular to plate (SBC). Plate alpha consists of AE and perpendicular ro plate SBC. Figure out the Square of the intersection surface between alpha and Pyramid SABC.
I got it as 2 a^2 / 3 root3. I'm quite doubtful, about you?
The drawing: http://img.photobucket.com/albums/v...orum s/334.jpg
Thanks :-)
Triangle ABC with AB=AC=a, BAC=90 deg. SA is perpendicular to plate (ABC) @ A. SA=a, also! In SB: ES=2EB. H is in plate (SBC) so that AH is perpendicular to plate (SBC). Plate alpha consists of AE and perpendicular ro plate SBC. Figure out the Square of the intersection surface between alpha and Pyramid SABC.
I got it as 2 a^2 / 3 root3. I'm quite doubtful, about you?
The drawing: http://img.photobucket.com/albums/v...orum s/334.jpg
Thanks :-)
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