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The heignt of an equilateral triangle is the side of a

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The heignt of an equilateral triangle is the side of a

by AAPL » Mon Oct 15, 2018 5:36 am

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Economist GMAT

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The height of an equilateral triangle is the side of a smaller equilateral triangle, as shown above. If the side of the large triangle is 1, what is AB?
$$A.\ 1-\frac{\sqrt{3}}{2}$$
$$B.\ 0.25$$
$$C.\ 2-\sqrt{3}$$
$$D.\ \frac{1}{3}$$
$$E.\ 1-\frac{\sqrt{3}}{4}$$
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The height of an equilateral triangle is the side of a small

by fskilnik@GMATH » Mon Oct 15, 2018 9:54 am
AAPL wrote:Economist GMAT
The height of an equilateral triangle is the side of a smaller equilateral triangle, as shown above. If the side of the large triangle is 1, what is AB?
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$$A.\ 1-\frac{\sqrt{3}}{2} \,\,\,\,\,\,\, B.\ 0.25 \,\,\,\,\,\,\, C.\ 2-\sqrt{3} \,\,\,\,\,\,\, D.\ \frac{1}{3} \,\,\,\,\,\,\, E.\ 1-\frac{\sqrt{3}}{4}$$
(The formula below we suggest you memorize. It will be used here twice.)
$${h_{eq}} = {{L\sqrt 3 } \over 2}\,\,\,\,\,\left( * \right)\,\,\,\,\,\,\,\,\left( {{\rm{height}}\,\,{\rm{of}}\,\,{\rm{an}}\,\,{\rm{equilateral}}\,\,{\rm{triangle}}\,\,{\rm{with}}\,\,{\rm{side}}\,\,L} \right)$$

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$$? = AB = {L_{\,{\rm{large}}}} - {h_{{\rm{eq}}\,{\rm{small}}}}\,\, = \,\,\,1 - \,\,{?_{{\rm{temporary}}}}$$
$$\Delta \,{\rm{small}}\,\,:\,\,\,\,\,{L_{\,{\rm{small}}}} = {h_{\,{\rm{eq}}\,{\rm{large}}}}\,\,\mathop = \limits^{\left( * \right)} \,\,\,\,{{1 \cdot \sqrt 3 } \over 2}\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{?_{{\rm{temporary}}}} = {h_{{\rm{eq}}\,{\rm{small}}}}\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,{{\sqrt 3 } \over 2} \cdot {{\sqrt 3 } \over 2} = {3 \over 4}$$
$$? = 1 - {3 \over 4} = {1 \over 4}$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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by ceilidh.erickson » Tue Oct 16, 2018 10:57 am
Here's an easier way to look at it: we've created another 30-60-90 triangle between A, B, and the midpoint of the base of the larger triangle.

If the base = 1, then half the base = 1/2. This is the hypotenuse of the right triangle. Thus, AB must be half that length: 1/4.

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by fskilnik@GMATH » Tue Oct 16, 2018 12:37 pm
ceilidh.erickson wrote:Here's an easier way to look at it: we've created another 30-60-90 triangle between A, B, and the midpoint of the base of the larger triangle.

If the base = 1, then half the base = 1/2. This is the hypotenuse of the right triangle. Thus, AB must be half that length: 1/4.
Very nice, ceilidh.erickson!

To validate your argument, it is important to justify the 90-degrees angle.

The figure presented in my post (above) does that.

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Fabio.
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by ceilidh.erickson » Tue Oct 16, 2018 1:15 pm
fskilnik@GMATH wrote:
ceilidh.erickson wrote:Here's an easier way to look at it: we've created another 30-60-90 triangle between A, B, and the midpoint of the base of the larger triangle.

If the base = 1, then half the base = 1/2. This is the hypotenuse of the right triangle. Thus, AB must be half that length: 1/4.
Very nice, ceilidh.erickson!

To validate your argument, it is important to justify the 90-degrees angle.

The figure presented in my post (above) does that.

Regards,
Fabio.
Yes, good point! I didn't include it in my diagram, but we can't just assume it's 90 - we have to justify it with the reasoning Fabio outlined.
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by fskilnik@GMATH » Tue Oct 16, 2018 2:05 pm
ceilidh.erickson wrote: Yes, good point! I didn't include it in my diagram, but we can't just assume it's 90 - we have to justify it with the reasoning Fabio outlined.
I would like to make this PUBLIC compliment to ceilidh.
I explain: in approximately 2 months of mine here (after inactive in this forum since approximately 2012) I realized something as simple as "you are right" is very hard to obtain.
More than that: in one or two occasions, after explaining why some "absolute truths" are not officially sealed, these same "truths" are repeated as if nothing was mentioned.
In short: you made my day, ceilidh!
Kind regards,
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by ceilidh.erickson » Tue Oct 16, 2018 2:13 pm
Well thanks, you made my day, too! We're all here to learn from each other & share what we know, right? I've always found it weird and transparent when people try to "talk over" each other in forums like this. Much better to act like collaborators!
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by fskilnik@GMATH » Tue Oct 16, 2018 2:33 pm
EXACTLY!
The ONLY reason to look at other experts as "competitors" is (in my opinion) stupid. I explain: if someone loves YOUR way of presenting things, this person will NOT be MY "customer", for sure.
In other words, we are just presenting our "dishes", but our restaurant will have a new client only if we offer the kind of food/seasoning he/she prefers!
You have a new friend here, Ceilidh. No ego, just the search for the truth and for beautiful solutions.
(In the posts above, yours is far more beautiful than mine, by the way.)
Kind Regards,
Fabio.
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by ceilidh.erickson » Tue Oct 16, 2018 3:14 pm
Very well said! And thank you ;)
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