If the circle has a radius of 4, what is the perimeter of the inscribed equilateral triangle?
Answer: $12\sqrt{3}$
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
|||||||||||||||||
|
|
See here for the image I am referring to:
the angle $\angle ABC$ is $60$, so the angle $\angle AOC$ is $120$ (inscribed angles to a chord in a circle have half the value the angle from the center $O$ to that same chord). In triangle $\triangle ACO$, $OA = OC = \text{radius}$, then $$\angle IAO = \angle ICO = (180 - \angle AOC)/2 = 30$$ ($I$ is midpoint of $AC$, and $OI$ is perpendicular to $AC$) Triangle $\triangle AIO$ is both half an equilateral triangle, and a right triangle at $I$; then $$OI = \frac{1}{2} OA = \frac{1}{2}\cdot 4 = 2$$ $$AI = \sqrt{OA^2 - OI^2} = \sqrt{4^2 - 2^2} = \sqrt{12} = 2\sqrt{3}$$ $$AC = 2 AI\text{ (as $I$ is the midpoint of $AC$)} = 4\sqrt{3}$$ so the perimeter of $\triangle ABC = 3 \cdot AC = \fbox{$12\sqrt{3}\;$}$ |
|||||||||
|
|
Hint: Use the law of sines on an interior triangle:
|
|||||||||||
|
|
Without trigonometry: since $\,\Delta ABC\,$ is equilateral, its circumcenter is the same as its incenter is the same as the intersection point of its medians. Call this point $\,O\,\Longrightarrow \,$ if $\,AM\,$ is the whole median from $\,A\,$ to BC, then $$4=AO=\frac{2}{3}AM\Longrightarrow x = 6$$ (since the intersection point of the medians cuts each of them in a$\,1:2\,$ proportion, the longest side being always on the vertex side). Now draw the triangle $\,\Delta AMB\,$ ,with sides $\,x\,,\,x/2\,,\,6\,\,,\,\,x=\,$ the triangle's side, and use Pythagoras (in an equilateral triangle, each median is its root vertex's angle bisector and also the height to the other side): $$x^2=6^2+\left(\frac{x}{2}\right)^2\Longrightarrow \frac{3}{4}x^2=36\Longrightarrow x=\frac{12}{\sqrt 3}\Longrightarrow P_{\Delta ABC}=3x=12\sqrt 3$$ |
||||
|
|
|
As you said you are preparing for a standardized test, if you are preparing for CAT/GRE/GMAT level test then this is somewhat fast approach: If you observe carefully, we are given the circumradius of the equilateral triangle. If $s$ and $h$ be the side and the height respectively of an equilateral traingle then we know circumradius is given by $\frac 23 \times h$. Thus, $$ \frac 23 \times h = 4\implies h = 6$$ Again, $h=6=\frac{\sqrt{3}}2 \times s \implies s = \frac {12}{\sqrt{3}} \implies \text{ perimeter } = 3s = 12\sqrt{3}$ |
|||
