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Distance to Diagonal (Posted on 2013-05-14) Difficulty: 2 of 5


Let ABCD be a parallelogram with ∠BAC = 45° and ∠ABD = 30°.

What is the distance from B to diagonal AC in terms of the
lengths of sides AB and AD?

See The Solution Submitted by Bractals    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Typo in problem statement | Comment 4 of 6 |
(In reply to Typo in problem statement by Bractals)

It's still pretty easy.
I'll call the distance sought h.
Area of triangle ABC = .5*AB*AD*sin(45º)
also
Area of triangle ABC = .5*AC*h

So .5*AB*AD*sin(45º) = .5*AC*h
h = √(.5)*AB*AD/AC

By law of cosines
AC²=AB² + BC² - 2*AB*BC*cos(135º)

If you substitute this for AC as well as AD for BC the final expression for h is rather messy:

h = √(.5)*AB*AD/√(AB² + AD² + √(2)*AB*AD)

(Note: h can probably be written in terms of just AB, since the triangle ABD is uniquely determined by the ASA congruence theorem.)






  Posted by Jer on 2013-05-16 00:38:14

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