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?
$$A.\ 1-\frac{\sqrt{3}}{2}$$
$$B.\ 0.25$$
$$C.\ 2-\sqrt{3}$$
$$D.\ \frac{1}{3}$$
$$E.\ 1-\frac{\sqrt{3}}{4}$$
OA B
The heignt of an equilateral triangle is the side of a
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- fskilnik@GMATH
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(The formula below we suggest you memorize. It will be used here twice.)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?
$$A.\ 1-\frac{\sqrt{3}}{2} \,\,\,\,\,\,\, B.\ 0.25 \,\,\,\,\,\,\, C.\ 2-\sqrt{3} \,\,\,\,\,\,\, D.\ \frac{1}{3} \,\,\,\,\,\,\, E.\ 1-\frac{\sqrt{3}}{4}$$
$${h_{eq}} = {{L\sqrt 3 } \over 2}\,\,\,\,\,\left( * \right)\,\,\,\,\,\,\,\,\left( {{\rm{height}}\,\,{\rm{of}}\,\,{\rm{an}}\,\,{\rm{equilateral}}\,\,{\rm{triangle}}\,\,{\rm{with}}\,\,{\rm{side}}\,\,L} \right)$$
$$? = AB = {L_{\,{\rm{large}}}} - {h_{{\rm{eq}}\,{\rm{small}}}}\,\, = \,\,\,1 - \,\,{?_{{\rm{temporary}}}}$$
$$\Delta \,{\rm{small}}\,\,:\,\,\,\,\,{L_{\,{\rm{small}}}} = {h_{\,{\rm{eq}}\,{\rm{large}}}}\,\,\mathop = \limits^{\left( * \right)} \,\,\,\,{{1 \cdot \sqrt 3 } \over 2}\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{?_{{\rm{temporary}}}} = {h_{{\rm{eq}}\,{\rm{small}}}}\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,{{\sqrt 3 } \over 2} \cdot {{\sqrt 3 } \over 2} = {3 \over 4}$$
$$? = 1 - {3 \over 4} = {1 \over 4}$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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- ceilidh.erickson
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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.
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.
Ceilidh Erickson
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- fskilnik@GMATH
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Very nice, ceilidh.erickson!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.
To validate your argument, it is important to justify the 90-degrees angle.
The figure presented in my post (above) does that.
Regards,
Fabio.
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- ceilidh.erickson
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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.fskilnik@GMATH wrote:Very nice, ceilidh.erickson!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.
To validate your argument, it is important to justify the 90-degrees angle.
The figure presented in my post (above) does that.
Regards,
Fabio.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
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- fskilnik@GMATH
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I would like to make this PUBLIC compliment to ceilidh.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 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,
Fabio.
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- ceilidh.erickson
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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!
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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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.
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.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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Portuguese-speakers :: https://www.gmath.com.br
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- ceilidh.erickson
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Very well said! And thank you
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education