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February 6th, 2019, 01:39 AM   #1
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Error in entrance exam?

Hi everyone.

I was going through some entrance exam problems and found the following possible error in one of the questions: https://imgur.com/TMkZ3EV

The size of the shaded area is being asked. The triangles are (I think) supposed to be similar, and this should be a very simple calculation.

The unknown side of the small triangle seems to be 16 (45:36 -> 20:16), and the correct answer is 406.

However, 10:16:20 differs from 27:36:45 and doesn't satisfy the Pythagorean theorem; the short side should be 12, right?

Is there an error or am I just missing something?

Last edited by skipjack; February 7th, 2019 at 04:33 AM.
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February 6th, 2019, 04:22 AM   #2
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The small triangle is obtuse (just), but your diagram isn't precisely to scale, making it look right-angled.
The area of half the rectangle is (27 cm × 36 cm)/2 = 486 cm².
The small triangle's area = (10cm × 20cm × 36/45)/2 = 80 cm² exactly.
Hence the area of the shaded region = 406 cm².
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Last edited by skipjack; February 6th, 2019 at 07:21 PM.
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February 6th, 2019, 05:15 AM   #3
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Thanks. It's been ages since I did math.

Last edited by skipjack; February 6th, 2019 at 07:27 PM.
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February 6th, 2019, 06:48 AM   #4
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There is an equal angle in the smallest and largest right triangle - so they are similar. So

x/20=36/45

The area of the small triangle is (1/2)10x

The area of the shaded triangle is (1/2)(36)(27)-(1/2)10x

Last edited by skipjack; February 6th, 2019 at 07:23 PM.
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February 6th, 2019, 07:38 AM   #5
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Just to make it clear, I thought I had solved this problem before posting, but was left wondering about the side lengths of the small triangle.

Last edited by skipjack; February 6th, 2019 at 07:31 PM.
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February 6th, 2019, 11:32 AM   #6
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The small triangle was perhaps intentionally drawn misleadingly.

Its sides are 10 cm, 20 cm and √260 cm = 16.1245... cm. Note that zylo's answer is flawed; the triangles aren't similar.

Last edited by skipjack; February 6th, 2019 at 07:34 PM.
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February 6th, 2019, 10:58 PM   #7
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Quote:
Originally Posted by skipjack View Post
The small triangle was perhaps intentionally drawn misleadingly.

Its sides are 10 cm, 20 cm and √260 cm = 16.1245... cm. Note that zylo's answer is flawed; the triangles aren't similar.
Misleadingly but "helpfully". The result is the same, after all.

Last edited by r3venans; February 6th, 2019 at 11:09 PM.
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February 7th, 2019, 05:21 AM   #8
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I doubt that it was intended to be helpful (given that providing the rectangle's perimeter isn't helpful), but it's possible that the diagram was redrawn at some stage (prior to publication) by someone who didn't realize that the small triangle is obtuse-angled.

If B is the "top right" vertex of the rectangle, D is the vertex of the small triangle where its longest sides meet, and DE is a perpendicular constructed from D to the diagonal of the rectangle, DE is the altitude of the small triangle corresponding to its base of length 10 cm and triangle DBE is right-angled.

By the use of similar triangles, DE has length 16 cm, and so the small triangle's area is (10 cm × 16 cm)/2 = 80 cm². Triangle DBE's sides have lengths 12 cm, 16 cm and 20 cm. Hence your original thinking can work fine, but needs to involve the construction I've described.

As (12 - 10)² + 16² = 260, you can understand why the length √260 cm that I gave earlier is correct.
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