# Find possible areas of triangle given radius of circumscribed circle

Question: In triangle ABC, AB=4, AC=5, and the radius R of the circumscribed circle is equal to √7. Find all possible values of the area of triangle ABC.

Through using the sine rule, i found the angle of B to be approximately $$70.89339$$, the angle of C to be approximately $$49.10661$$ and the angle of A to be $$60$$. Using the cosine law, I found the missing side length BC to be $$√21$$. Using Heron's law, I then found the area to be $$5√3$$.

However, this question requires more than one area. I am confused as to how to obtain the second area?

## 3 Answers

$AB$ and $AC$ are two chords of a circle of radius $\sqrt7$. Picture this circle together with its diameter through $A$. The two chords can either be on the same side or opposite sides of this radius. You’ve computed the area of the resulting triangle for only one of these cases. Remember that the sine rule cannot tell the difference between $\theta$ and $180-\theta$. Hence you have to check for the possibility of having an obtuse angle. If the angle at $B$ is $\approx 180-71 = 109$ it still leaves room for the angle at $C$ to be $\approx 50$. This will give a different area.

You should also check what happens when you replace $\approx 50$ by $\approx 180-50$. You're right, there is only one possible answer. Since $M$ is the circumcenter of $\triangle ABC$, $MD$ is the perpendicular bisector of $\overline{AB}$. Applying the sine law in $\triangle ABC$ then gives $$|\measuredangle AMD|=\arcsin (\frac {2}{\sqrt {7}}).$$ Analogously, we obtain $$|\measuredangle EMD|=\arcsin (\frac {2.5}{\sqrt {7}}).$$ Overall this leads to $|\measuredangle EMD|= \frac {2\pi}{3}$ and therefore we get $|\measuredangle BAC|=|\measuredangle DAE|=\frac {\pi}{3 }$. Hence, the area of our triangle comes out to be $\frac {1}{2} \cdot 4 \cdot 5 \cdot \sin (\frac {\pi}{3 }) =5\sqrt {3}$, which is exactly your result.

• There is another possibility: the point $C$ can be on the same side of the diameter through $A$ and $B$ instead of on the opposite side as you’ve drawn it. – amd Apr 12 '17 at 0:28
• @amd This cannot be the case since BAC would be acute then but also be equal to $arcsin (\frac {2}{\sqrt{7}})-arcsin (\frac {2.5}{\sqrt{7}})$. But this is roundabout $-22$ degrees or about $158$ degrees which cannot be right in both cases. Hence, what I've drawn is the only case that can occur. – mxian Apr 12 '17 at 0:37
• Draw a circle of radius $5$ centered at $A$. It intersects the circle that you’ve drawn at two points, both of which are perfectly valid choices for $C$. Or, simply reflect the point $C$ that you’ve drawn in the diameter through $A$ to see that there’s another possible triangle. Whoever put together the problem that the OP is trying to solve certainly believed that there’s more than one non-congruent possibility for the inscribed triangle. – amd Apr 12 '17 at 1:11