NMTC Practice Problem-10

Geometry Level 3

$ABC$ is an equilateral triangle. Its centroid is $G$ and $AG=2 cm$ . $S$ is the circumcircle of the triangle. Find the area of the shaded part if in the circum circle $S$ everything except the triangular(ABC) is shaded. If this can be written in the form $k\pi-\sqrt b$ find $k+b$

The answer is 31.

2 solutions

Trevor Arashiro
Aug 21, 2014

Because the triangle is equilateral, it's area can be represented by $\dfrac{3\sqrt3g^2}{4}$ where g is the length from the centroid to a vertex. Also, because the triangle is equilateral, g is equal to the circumradius.

Thus we have $(2)^2\pi-\dfrac{3\sqrt3(2)^2}{4}\Rightarrow\boxed{4\pi-\sqrt{27}}$

Excellent.

Krishna Ar - 6 years, 9 months ago

Thank you.

Trevor Arashiro - 6 years, 9 months ago

Can you tell me as to why you tag all your problems with Easy money? What does that mean?

Krishna Ar - 6 years, 9 months ago

@Krishna Ar – Two reasons, on my school math team, "easy money" is a phrase that is thrown around everyday. That along with "OOOHHHHHH!!!!!!, That's WEAK, girly man, and a bunch of others. Second reason, it's so that incase I want to find only the problems I created, I don't have to go to my profile and sift through all of my posts, because a ton are reshares. So I just search up easy money and all my problems are there. Also, I'm getting my "irl" friends to tag their problems with easymoney so we can easily find each others posts as well. Unfortunately, they don't have as many posts as I do because I joined brilliant way before them.

Trevor Arashiro - 6 years, 9 months ago

@Trevor Arashiro – :D. Nice explanation

Krishna Ar - 6 years, 9 months ago

Rifath Rahman
Aug 31, 2014

Nice problem,after solving all that you get 4 * pi-3 * sqrt3,and start to wonder how could that be sqrt b,then you see 4 * pi-sqrt 3^2 * sqrt 3=4 * pi-sqrt 27,so ans is 4+27=31

NMTC Practice Problem-10

The answer is 31.

2 solutions

0 pending reports