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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote x, y and z are the sides of the triangle shown and h is its tagged by: Max@Math Revolution This topic has 2 expert replies and 0 member replies GMAT/MBA Expert x, y and z are the sides of the triangle shown and h is its Timer 00:00 Your Answer A B C D E Global Stats Difficult [Math Revolution GMAT math practice question] x, y and z are the sides of the triangle shown and h is its height. Is the perimeter, x + y + z of the triangle greater than 1? 1) h = 1/2 2) x = y = 1/3 _________________ Math Revolution Finish GMAT Quant Section with 10 minutes to spare. The one-and-only Worldâ€™s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Only$99 for 3 month Online Course
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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

x, y and z are the sides of the triangle shown and h is its height. Is the perimeter, x + y + z of the triangle greater than 1?

1) h = 1/2
2) x = y = 1/3

$$x + y + z\,\,\mathop > \limits^? \,\,\,1$$
$$\left( 1 \right)\,\,\,\left\{ \matrix{ \,x\mathop > \limits^{\left( * \right)} \,\,\,h = {1 \over 2} \hfill \cr \,z\mathop > \limits^{\left( * \right)} \,\,\,h = {1 \over 2} \hfill \cr y > 0 \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,x + y + z > 1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle$$
$$\left( * \right)\,\,{\rm{in}}\,\,{\rm{any}}\,\,{\rm{given}}\,\,{\rm{right}}\,\,\Delta \,\,{\rm{,}}\,\,\,{\rm{hypotenuse}}\,\,\,{\rm{ > }}\,\,\,\,{\rm{any}}\,\,{\rm{leg}}$$

$$\left( 2 \right)\,\,\left\{ \matrix{ \,{\rm{figure}}\,\,{\rm{on}}\,\,{\rm{the}}\,\,{\rm{left}}\,\,\, \Rightarrow \,\,\,\,\,x + y + z\,\,\,\, < \,\,\,\,3\left( {{1 \over 3}} \right)\,\, = \,\,1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr \,{\rm{figure}}\,\,{\rm{on}}\,\,{\rm{the}}\,\,{\rm{right}}\,\,\, \Rightarrow \,\,\,\,\,x + y + z\,\,\,\,\mathop < \limits^{{\rm{as}}\,{\rm{near}}\,\,{\rm{as}}\,\,{\rm{desired}}!} \,\,\,\,{2 \over 3} + {1 \over 3}\left( {\sqrt 2 } \right)\,\, = \,\,{{\sqrt 2 + 2} \over 3}\,\,\,\,\,\left[ { > \,\,\,1} \right]\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr} \right.$$

The correct answer is therefore (A).

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

Regards,
Fabio.

P.S.: it is interesting to study the case in which D does not belong to the side BC!

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Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

In a triangle, the sum of the lengths of two sides is always greater than the length of the other side.
Thus, from triangle ABD, we must have AB + BD > AD = h, and from triangle ABD, we must have AC + CD > AD = h. If h = 1/2, then AB + BC + CA = AB + BD + DC + CA > 1/2 + 1/2 = 1.
Condition 1) is sufficient.

Condition 2):
If AC = 1/2, then AB + BC + CA = 1/3 + 1/3 + 1/2 = 7/6 > 1 and the answer is â€˜yesâ€™.
If AC = 1/4, then AB + BC + CA = 1/3 + 1/3 + 1/4 = 11/12 < 1 and the answer is â€˜noâ€™.
Since it does not yield a unique solution, condition 2) is not sufficient.

Therefore, the correct answer is A.

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