# Equilateral triangle height problem

I've been stuck with this problem: I've found the Area of the triangle is about: 43.3

And I've found that the height (all heights are equal) is equal to about: 8.66.

Can anyone help me find $OA = OB = OC$ please? If it's not possible, what other information would I need?

• What work have you done? Can you see any similar triangles in your diagram? If so, what would that imply? – Arby May 12 '17 at 12:34
• Never mind, I found a formula for the radius of the circumcircle of an equilateral triangle: a/(sqrt(3)) Which is equal to OA = OB = OC (= about 5.77). – Coto May 12 '17 at 12:46
• Your answer is and must be $2/3$ of the altitude, since the altitudes are medians in this case. – Ethan Bolker May 13 '17 at 0:42

The key to this is looking at all of the different size triangles and the angles they make. Then you'll see a triangle with a 30 degree angle in it. You will also recognise that using the formula $\cos(30)=\frac{adjacent=5}{hypotenuse}$

Re-arrange this to determine the value you require $=5.7735$ To me this problem almost asks to be solved using the most basic methods, namely the Pythagorean theorem along with either similar triangles having similar proportions, or the basic area formula for triangles namely $A=\frac 1 2bh$.

One way to do the similar triangle method: the midpoint of BC (the base in the picture) I will name M. Triangle AMB is congruent to triangle AMC so they are right triangles giving side length of $AM=\sqrt {10^2-5^2}=\sqrt {75}=5\sqrt 3$ by the Pythagorean formula. Triangle AMB can be shown to be similar to triangle OMB, so $\frac {AB} {AM}=\frac {OB} {BM}$ giving $\frac {10} {5\sqrt 3}=\frac {OB} 5$, giving $OB=\frac {10} {\sqrt 3}\approx 5.77$.

Find $OA = OB = OC$.

Let $X$ be the point where line $CO$ intersects line $AB$.

Pythagoras : $|CX|^2 + 5^2 = 10^2$.

$|CX| = √(75)$.

In this equilateral triangle perp. bisectors, altitudes and medians are identical. $O$ is the point of intersection of the medians, and divides every median in the ratio $2:1$ .

$|CO| = (2/3) |CX| = (2/3) √(75)$ =

$(2/3)5√3 = (10/3)√3$.