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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?
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 $2:1$ .
$|CO| = (2/3) |CX| = (2/3) √(75)$ =
$(2/3)5√3 = (10/3)√3$.
a/(sqrt(3))Which is equal toOA = OB = OC (= about 5.77). $\endgroup$ – Coto May 12 '17 at 12:46