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Area Ascertainment II (Posted on 2014-10-24) Difficulty: 3 of 5
In triangle ABC, it is known that AB = 9 and BC/AC = 40/41.

Find the maximum possible area of the triangle ABC.

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Some Thoughts re(2): Nice problem and nice decimal point Comment 8 of 8 |
(In reply to re: Nice problem and nice solutions. by Steve Herman)

If a decimal point was missing, than it is there now.

As  to  41^2 = 40^2 + 9^2

....if k=1 then the sides are 40 and 41, the area =  .5*40*41 *sin C.

The maximum is achieved at C=90deg, so ABC is a right-angled triangle.

Makes sense.


  Posted by Ady TZIDON on 2014-10-27 05:33:23
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