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A triangle has sides measuring 10, 17 and 21 units

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A triangle has sides measuring 10, 17 and 21 units  [#permalink]

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New post 30 Apr 2017, 23:40
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67% (01:33) correct 33% (00:38) wrong based on 3 sessions

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A triangle has sides measuring 10, 17 and 21 units. A square is inscribed in the triangle such that one side of the square lies on the longest side of the triangle. The other two vertices of the square touch the two shorter sides of the triangle. What is the length (in units) of the side of the square?

A. 168/29
B. 188/29
C. 178/59
D. 153/59
E. 64/11

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Re: A triangle has sides measuring 10, 17 and 21 units  [#permalink]

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New post 30 Apr 2017, 23:45
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As this is a good question, I am posting the solution below:

By Heron's Formula, the area, A, of a triangle with sides a, b, c is given by A = square root[s(s − a)(s − b)(s − c)], where s = ½(a + b + c) is the semi-perimeter of the triangle.

Then s = ½(10 + 17 + 21) = 24, and A = 84.

View the diagram here: https://ibb.co/chNhVk

Now drop a perpendicular of length h onto the side of length 21.
We also have A = ½ × base × perpendicular height.
Hence A = 21h/2 = 84, from which h = 8.

Notice that the triangle above the square is similar to the whole triangle. (This follows because its base, the top of the square, is parallel to the base of the whole triangle.)

Let the square have side of length d.

Considering the ratio of altitude to base in each triangle, we have 8/21 = (8 − d)/d = 8/d − 1.

Therefore the length of the side of the square is 168/29.
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Re: A triangle has sides measuring 10, 17 and 21 units  [#permalink]

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New post 01 May 2017, 01:33
14101992 wrote:
As this is a good question, I am posting the solution below:

By Heron's Formula, the area, A, of a triangle with sides a, b, c is given by A = square root[s(s − a)(s − b)(s − c)], where s = ½(a + b + c) is the semi-perimeter of the triangle.

Then s = ½(10 + 17 + 21) = 24, and A = 84.

View the diagram here: https://ibb.co/chNhVk

Now drop a perpendicular of length h onto the side of length 21.
We also have A = ½ × base × perpendicular height.
Hence A = 21h/2 = 84, from which h = 8.

Notice that the triangle above the square is similar to the whole triangle. (This follows because its base, the top of the square, is parallel to the base of the whole triangle.)

Let the square have side of length d.

Considering the ratio of altitude to base in each triangle, we have 8/21 = (8 − d)/d = 8/d − 1.

Therefore the length of the side of the square is 168/29.



With due respect to the efforts of post this question and solution, I want readers to know that

In Short, This question is IRRELEVANT for GMAT as GMAT doesn't expect the use of Heron's Formula that has been used to explain. Also, The question can't be solved using limited properties needed for GMAT so good for Math geeks/lovers but not good question for GMAT strictly aspirants.

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

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Re: A triangle has sides measuring 10, 17 and 21 units &nbs [#permalink] 01 May 2017, 01:33
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