GMAT Prep
What is the greatest possible area of a triangular region with one vertex at the center of a circle with radius one and the other two vertices on the circle?
$$A.\ \frac{ \sqrt{3}}{4}$$
$$B.\ \frac{1}{2}$$
$$C.\ \frac{\pi}{4}$$
$$D.\ 1$$
$$E.\ \sqrt{2}$$
OA B
What is the greatest possible area of a triangular region
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- fskilnik@GMATH
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AAPL wrote:GMAT Prep
What is the greatest possible area of a triangular region with one vertex at the center of a circle with radius one and the other two vertices on the circle?
$$A.\ \frac{ \sqrt{3}}{4} \,\,\,\,\,\,B.\ \frac{1}{2}\,\,\,\,\,\,C.\ \frac{\pi}{4}\,\,\,\,\,\,D.\ 1\,\,\,\,\,\,E.\ \sqrt{2}$$
$$?\,\,\, = \,\,{S_{\Delta ABC}}\,\,\max $$
Let C be the center of the circle with unitary radius.
Without loss of generality, we may (and will) assume point A is one unit at the right of point C (as shown in the figure on the left).
For the last vertex (B), without loss of generality (in terms of exploring possible areas) there are only two possibilities:
(1) B is in the arc AD (figure in the middle) or (2) B is in the arc DE (figure on the right)
In BOTH cases we have:
$${S_{\Delta ABC}} = {{AC \cdot h} \over 2} = {h \over 2}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = {1 \over 2}\,\,\,\,\,\,\left( {{\rm{when}}\,\,h = CD\,,\,\,{\rm{i}}{\rm{.e}}{\rm{.}},\,\,B = D} \right)$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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Given two sides of a triangle, the greatest possible area will be achieved if the two sides form a RIGHT ANGLE and serve as the triangle's base and height.AAPL wrote:GMAT Prep
What is the greatest possible area of a triangular region with one vertex at the center of a circle with radius one and the other two vertices on the circle?
$$A.\ \frac{ \sqrt{3}}{4}$$
$$B.\ \frac{1}{2}$$
$$C.\ \frac{\pi}{4}$$
$$D.\ 1$$
$$E.\ \sqrt{2}$$
OA B
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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The triangle with the largest possible area, given the above constraints, is an isosceles right triangle. Thus, since the base and height of the triangle is 1, the maximum area is 1/2 x 1 x 1 = 1/2.AAPL wrote:GMAT Prep
What is the greatest possible area of a triangular region with one vertex at the center of a circle with radius one and the other two vertices on the circle?
$$A.\ \frac{ \sqrt{3}}{4}$$
$$B.\ \frac{1}{2}$$
$$C.\ \frac{\pi}{4}$$
$$D.\ 1$$
$$E.\ \sqrt{2}$$
OA B
Answer: B
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