| From: ROLL::USENET "USENET Newsgroup Distributor" 18-JUL-1984 22:07
To: HARE::STAN
Subj: USENET net.math newsgroup articles
Newsgroups: net.math
Path: decwrl!dec-rhea!dec-tonto!luong
Subject: Solution to Geometry V
Posted: Tue Jul 17 09:02:48 1984
>Subject: Geometry V
>Once again, I have another problem, although could be one of the last. Here
>goes: In the obtuse triangle ABC (angle C is obtuse), point M is on AB such
>that AM = MB, point D is on BC such that MD is prpndclr to BC, and point E
>is a point on MA such that EC is prpndclr to BC. Given the area of triangle
>ABC is 24, then find area of triangle BED. Good luck!
>Billy Pizer
Denoting by (XYZ) the area of a triangle XYZ:
(MCB) = 0.5 BC.MD and (ECB) = 0.5 BC.EC
therefore:
(MCB)/(ECB) = MD/EC
= DB/CB (since MD and EC are parallel) (1)
Draw perpendiculars CC' and DD' to EB.
Now:
(DEB) = 0.5 EB.DD' and (ECB) = 0.5 EB.CC'
Therefore:
(DEB)/(ECB) = DD'/CC'
= DB/CB (since DD' and CC' are parallel) (2)
Combining (1) and (2) shows that:
(DEB) = (MCB) (3)
But (MCB) = 0.5 CC'.MB
(MCB) = 0.5 CC'.(AB/2) = (0.5 CC'.AB)/2 = (ABC)/2 = 12 (4)
Finally, combining (3) and (4):
(DEB) = 12
==========
Van Luong Nguyen,
Digital Equipment Corporation.
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| Newsgroups: net.math
Path: decwrl!amd!dual!zehntel!hplabs!oliveb!tymix!kanner
Subject: queer identity
Posted: Tue Jul 24 16:09:49 1984
Newsgroups: net.puzzle,net.math
Path: decwrl!decvax!ittvax!dcdwest!sdcsvax!bmcg!asgb!gupta
Subject: Geometry Puzzle 5
Posted: Thu Jul 26 12:22:14 1984
>Once again, I have another problem, although could be one of the last. Here
>goes: In the obtuse triangle ABC (angle C is obtuse), point M is on AB such
>that AM = MB, point D is on BC such that MD is prpndclr to BC, and point E
>is a point on MA such that EC is prpndclr to BC. Given the area of triangle
>ABC is 24, then find area of triangle BED. Good luck!
>Billy Pizer
Solution>
As M is the mid-point of AB,
area(MCB) = area(MCA) = 1/2 area(ABC) = 12 ... (1)
Also,
area(DMC) = area(DME) ... (2)
because they are on the same base (DM) and
have the same height (DM is parallel to CE).
Now,
area(MCB) = area(MDB) + area(DMC)
= area(MDB) + area(DME) ... (from 2)
= area(BED)
.
. . area(BED) = 12 ... (from 1)
Yogesh Gupta
sdcrdcf --- bmcg!asgb!gupta
sdcsvax ---/
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