Bunuel wrote:

The height of an equilateral triangle is the side of a smaller equilateral triangle, as shown above. If the side of the large equilateral triangle is 1, what is AB?

A. 1 - √3/2

B. 0.25

C. 2 - √3

D. 1/3

E. 1 - √3/4

(The formula below we suggest you memorize. It will be used here twice.)

\({h_{eq}} = {{L\sqrt 3 } \over 2}\,\,\,\,\,\left( * \right)\,\,\,\,\,\,\,\,\left( {{\rm{height}}\,\,{\rm{of}}\,\,{\rm{an}}\,\,{\rm{equilateral}}\,\,{\rm{triangle}}\,\,{\rm{with}}\,\,{\rm{side}}\,\,L} \right)\)

\(? = AB = {L_{\,{\rm{large}}}} - {h_{{\rm{eq}}\,{\rm{small}}}}\,\, = \,\,\,1 - \,\,{?_{{\rm{temporary}}}}\)

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

\(? = 1 - {3 \over 4} = {1 \over 4}\)

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

Regards,

Fabio.

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Fabio Skilnik :: GMATH method creator (Math for the GMAT)

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