Is the area of the triangle drawn in circle A greater than π?

Question

https://gmatclub.com/forum/download/file.php?id=43619 Is the area of the triangle drawn in circle A greater than π? (Assume A to be the center of the circle) (1) x = 36 (2) The radius of the circle is 1. image005.jpg

AvidDreamer09 · Answer

Is the area of the triangle drawn in circle A greater than π? (Assume A to be the center of the circle)

(1) x = 36°
you cant calculate area of a triangle with just the angle, each side could be any unit with angle 36....

(2) The radius of the circle is 1.
Area of triangle formula = (1/2)b*h
base = height = radius = 1
area of the triangle = 0.5
Sufficient

Do let me know if my reasoning is right...... thanks

ShankSouljaBoi · Answer

Base = Height = radius = 1. Wrong assumption please check

ShankSouljaBoi · Answer

1 Insuff
2 Area of entire circle is pi*1^2= pi
Now if the ares of entire circle is pi, the area of triangle in all cases considered will be less than the area of circle (= pi)
So, the area of traingle must be less than Pi
Sufficient.

Kritika2008poddar · Answer

1. Neither radius nor any side measure is given in the question. Hence angle alone is not sufficient

2. If the area of the entire circle= pi. 1^2 = pi. Therefore area of the triangle will be less than pi for sure . Hence B is sufficient

Posted from my mobile device

AvidDreamer09 · Answer

oooh Thanks... that makes more sense... i dunno what i was doing

fskilnik · Answer

{S_\Delta }\,\,\mathop > \limits^? \,\,\pi

(1) In the figure on the left presented below, all circles have center A and "same angle x" (shown in purple).

https://preview.ibb.co/fyk6wp/27_Set18_9v.gif

Please note that choosing convenient radii, we can create "circular sectors" with areas that are arbitrarily small, and arbitrarily large.
Consequence: the corresponding triangles (inside these circular sectors) can have areas that are also arbitrarily small, and arbitrarily large.

This is what we call a "GEOMETRIC BIFURCATION". This argument PROVES mathematically that (1) is insufficient!

(2) For any value of x (*) we have:

\left\{ \matrix{
 \Delta \subset {m{circle}} \hfill \cr 
 \Delta 
e {m{circle}} \hfill \cr} ight.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{S_\Delta } < {S_{{m{circle}}}} = \pi {\left( 1 ight)^2}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{m{NO}}} ightangle

(*) We may implicitly assume that x is positive and less than 180 degrees. (Think about that!)

The correct answer is (B), indeed.

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

Regards, 
Fabio.

bumpbot · Answer

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Is the area of the triangle drawn in circle A greater than π?

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Is the area of the triangle drawn in circle A greater than π?

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