My Math Forum area of triangle given heights

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November 16th, 2018, 08:58 PM   #1
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area of triangle given heights

In the attached, what is the area of triangle ABC?

Obviously they gave 22.5 and 67.5 to tell us that they add to 90 degrees, and then I applied formulas of right angle triangle, plus this makes the triangle ABC isosceles, but then what? And also BE and EC equal I found....

My drawing may be out of scale. We are supposed to solve this only with a pencil and paper, no calculators, rulers.
Attached Images
 04-5.jpg (36.8 KB, 4 views)

Last edited by skipjack; November 17th, 2018 at 12:06 AM.

 November 17th, 2018, 02:01 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 Use the fact that CF/EF = sqrt2 - 1.
November 17th, 2018, 02:31 AM   #3
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Quote:
 Originally Posted by skipjack Use the fact that CF/EF = sqrt2 - 1.
so you used tan 22.5 but we will not have calculators in the exam, so is there another way?

i think we are supposed to know sin cos tan of only 30 60 45 90 ..... and no trigonometry in this exam

 November 17th, 2018, 02:43 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 You can show that CF/EF = sqrt2 - 1 by applying the angle bisector theorem to a triangle with sides 1, 1 and sqrt2 and angles 45 degrees, 45 degrees and 90 degrees.
November 17th, 2018, 05:01 AM   #5
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Quote:
 Originally Posted by skipjack You can show that CF/EF = sqrt2 - 1 by applying the angle bisector theorem to a triangle with sides 1, 1 and sqrt2 and angles 45 degrees, 45 degrees and 90 degrees.
which triangle is 45 45 90 here? can not see it

 November 17th, 2018, 05:30 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 Produce FC to L, such that LF = EF. Triangle EFL now has the required angles and EC bisects angle LEF. The angle bisector theorem can be proved without using trigonometry, so the method I've given doesn't explicitly use trigonometry. A related method (that doesn't use the angle bisector theorem) is to produce EF to G so that EF = FG, and produce BC to H, where H lies on AG extended. Also, draw AH, GC and the perpendicular GK from G to BH. Show that CE = CG = GH and that angle CGH is a right angle. Hence calculate GK and KC (which are equal) in terms of CE and apply Pythagoras to triangle GKE (whose hypotenuse is known). Solve the resulting equation to obtain the length of CE. Now use AE = HE = HC + CE = 2KC + CE to calculate AE and calculate the area of triangle ABC as the product of the lengths of CE and AE. Thanks from ketanco

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