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April 9th, 2019, 11:32 PM   #1
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Square in a right triangle.

Hi,

Please solve the question attached in the image file.

Thx.
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File Type: jpg maths_q_trig.jpg (16.1 KB, 16 views)
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April 10th, 2019, 03:15 AM   #2
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$\displaystyle \dfrac{1092}{\sqrt{3145}} \approx 19.472$
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April 10th, 2019, 03:16 AM   #3
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Are you required to use trigonometry? This can be solved using Pythagoras and the properties of similar triangles, though the working is a bit tedious. What have you tried?
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April 10th, 2019, 03:10 PM   #4
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mrtwhs gave you the solution.
WHAT will you do if your teacher asks HOW YOU got the solution?

Label the triangle ABC and the square DEFG.
So we have 4 similar triangles: ABC, ADG, BEF and CFG.
Can you "see" that?
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April 10th, 2019, 03:59 PM   #5
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The correct solution is (1092/4237)sqrt(3145) m.
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April 10th, 2019, 07:03 PM   #6
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Hmmm...mrtwhs gets ~19.472, skip gets ~14.454, and I get ~15.273

I used 56 as the hypotenuse (should be ~56.08); shouldn't matter much...

I'll have another look tomorrow....musta goofed somewhere...
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April 11th, 2019, 01:19 AM   #7
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Below is a diagram (to scale) that accurately confirms my answer.
SquareInTriangle.PNG
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April 11th, 2019, 03:06 AM   #8
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Quote:
Originally Posted by mrtwhs View Post
$\displaystyle \dfrac{1092}{\sqrt{3145}} \approx 19.472$
Please do explain.
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April 11th, 2019, 03:07 AM   #9
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Quote:
Originally Posted by Denis View Post
mrtwhs gave you the solution.
WHAT will you do if your teacher asks HOW YOU got the solution?

Label the triangle ABC and the square DEFG.
So we have 4 similar triangles: ABC, ADG, BEF and CFG.
Can you "see" that?
True and thanks. But plz elaborate two three steps.
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April 11th, 2019, 03:13 AM   #10
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Quote:
Originally Posted by Denis View Post
Hmmm...mrtwhs gets ~19.472, skip gets ~14.454, and I get ~15.273
So just take the average!
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