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

(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.

=>

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 \((\frac{1}{2}) πr^2*(\frac{x}{360})\), 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 \(AB = 6√3, AD = \frac{AB}{2} = 3√3\). Since \(x =120^o\), angle \(AOD\) has measure \(60^o\), the triangle \(ODA\) is a right triangle and \(OD:OA:DA = 1:2:√3\). This yields \(OA:DA = r: 3√3 = 2: √3,\)which implies that \(r = 6.\)
Thus, the area of the sector is\((\frac{1}{2}) π6^2*(\frac{120}{360}) = (\frac{1}{2}) (\frac{1}{3})36π = 6π.\)

Both conditions together are sufficient.

Therefore, C is the answer.
Answer: C

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.

\(πr^2\) is the area of the circle, and \(\frac{x}{360}\) (\(x\) in degrees) is the fraction of the area we are interested in. (Think about a slice of pizza.)

In short: "1/2" should not be present in the formula quoted.

Regards,
Fabio.

Dear fskilnik, could you please explain a few things ? Why do you take a sector with 90 degrees ? And then what is this notation : ⇒(∗)30,60,90 and where does the 1/3 * (π⋅36) come from ? I don't really understand your explanations.

Thanks a lot for your time ~

Hi, EricD28 !

Thank you for your interest in my solution.

> Question 01: when considering statement (2) ALONE, we must obey the AB given length, but we are allowed to choose different values for x.
I chose first 90 degrees, then 120 degrees. These are particular cases especially easy to deal with the corresponding calculations.
The calculations were necessary to *guarantee* we can find two different numerical values for the question asked. This is what we call a BIFURCATION.

> Question 02: this notation means "using the 30-60-90 triangle shortcut, we may conclude that..."
(This shortcut is carefully explained in our method and, I believe, in almost every GMAT course).

> Question 03: 1/3 comes from the 120/360 degrees ratio, while pi*(6^2) is the area of the circle with radius 6.

Regards,
Fabio.

Thank you very much Fabio, i solved using sin 60 but it is the same as the 30 60 90 (i didn't know this theorem, thx for the tip)

Have a good day

Hi, EricD28.

YES, you may use basic trigonometry to justify the shortcut´s validity.
Another possibility: think about the height of an equilateral triangle, obtained when you "double" your 30-60-90 (along the hypotenuse).

Regards,
Fabio.