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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 ## The above figure shows a sector of a circle. What is the are tagged by: Max@Math Revolution ##### This topic has 2 expert replies and 0 member replies ### GMAT/MBA Expert ## The above figure shows a sector of a circle. What is the are ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult [Math Revolution GMAT math practice question]10.11 The above figure shows a sector of a circle. What is the area of the sector? 1) x = 120 2) AB=6âˆš3 _________________ 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]

The above figure shows a sector of a circle. What is the area of the sector?

1) x = 120
2) AB=6âˆš3
(We assume - as part of the definition of a sector a circle - that the "origin" of the angle x shown in the figure is the center of the circle.)

The variable R will denote the radius of the circle. All angles are measured in degrees.
$? = \frac{{x\,}}{{360\,}}\left( {\pi {R^{\,2}}} \right)$

$x = 120\,\,\,\left( {{\text{both}}\,\,{\text{figures}}} \right)$

$AB = 6\sqrt 3 \,\,\,\left( {{\text{both}}\,\,{\text{figures}}} \right)$
$x = 90\,\,\,\mathop {\,\,\, \Rightarrow \,\,\,\,}\limits^{L\,,\,L\,,\,L\sqrt 2 } \,\,\,R\sqrt 2 = 6\sqrt 3 \,\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\,\frac{{\sqrt 2 }}{2}\,} \,\,\,\,\,R = 3\sqrt 6 \,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,?\,\,\, = \,\,\,\frac{1}{4}\,\left( {\pi \, \cdot 9 \cdot 6} \right)$
$x = 120\,\,\,\mathop \Rightarrow \limits_{\left( * \right)}^{30\,,\,60\,,\,90\,\,} \,\,\,\,?\,\,\, = \,\,\,\frac{1}{3}\,\left( {\pi \, \cdot 36} \right)\,\,\, = 12\pi \,\,\, \ne \,\,\,\,\frac{1}{4}\,\left( {\pi \, \cdot 9 \cdot 6} \right)\,\,\,\,\,\,$
$\left( * \right)\,\,30\,,\,60\,,\,90\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered} \,L\sqrt 3 \,\,\, = \,\,\frac{{6\sqrt 3 }}{2} \hfill \\ \,2L = R \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,R = 6$

$\left( {1 + 2} \right)\,\,\,?\,\, = \,\,12\pi \,\,\,\left( {{\text{shown}}\,\,{\text{above}}} \right)$

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

Regards,
Fabio.

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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 area of a sector is (1/2) Ï€r^2*(x/360), where r is the radius of the sector. Since we have two variables, r and x, C is most likely to be the answer and we need to check both conditions together first.

Conditions 1) and 2):
Quadrilateral OACB is a kite, so its diagonals bisect each other at right angles, and bisect the angles at the vertices.
Since AB = 6âˆš3, AD = AB/2 = 3âˆš3. Since x =120, angle AOD has measure 60, the triangle ODA is a right triangle and OD:OA:DA = 1:2:âˆš3. This yields OA:DA = r: 3âˆš3 = 2: âˆš3, which implies that r = 6.
Thus, the area of the sector is (1/2) Ï€6^2*(120/360) = (1/2) (1/3)36Ï€ = 6Ï€.

Both conditions together are sufficient.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

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