Why can't I use the area to solve for CZ?

Question

Question

Medians $\overline{AX}$ and $\overline{BY}$ of triangle $\triangle ABC$ intersect each other at $O$ perpendicularly. It is given $AX=12$ and $BC=10$. Find the length of median $\overline{CZ}$.

My Answer

We first find AO. $AO=\frac23\cdot12= 8$ We then find OX:$12-8=4$ So $\triangle BOX$ is a 3-4-5 right triangle. We then find $\overline{AB}$, which is $\sqrt{3^2+8^2}=\sqrt{73}$ Since the area of $\triangle ABO=\triangle ABO$, then $\frac{\overline{ZO}\cdot\sqrt{73}}{2}=\frac{3\cdot8}{2}=12$. After simplifying a bit, we find that $\overline{ZO}=\frac{24\sqrt{73}}{73}$. AS $\overline{CZ}=3\overline{ZO}$, $$CZ=\boxed{\frac{72\sqrt{73}}{73}}$$

My Question

After checking the solution manual, I find that my answer is incorrect. The answer is $\overline{CZ}=\frac{3\sqrt{73}}{2}$, which was found using the fact that $\overline{ZO}$ is the median of hypotenuse $\overline{AB}$ of $\triangle AOB$. Why didn't my answer work?

Answer 1

We are given that $AX$ and $BY$ intersect at right angles. This does not imply that $CZ\perp AB$, so $OZ$ is not the altitude of $\triangle AOB$.

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Observe, $Z$ is the midpoint of right triangle $AOB$. Hence, $$ZO=ZA=ZB=\frac{\sqrt{73}}{2}$$ $$\therefore CZ=\frac{3\sqrt{73}}{{2}}$$

Answer 2

You can obtain the median length $ \ \overline{CZ} \ $ from the area of $ \ \triangle ABC \ \ , \ $ but it is a bit more laborious than Sathvik's more direct approach. You already have found correctly the length of the altitude $ \ \overline{CP} \ $ to side $ \ \overline{AB} \ \ . \ $ [Another way to find that is as follows. Having established that the length of $ \ \overline{OB} \ $ is $ \ 3 \ \ , \ $ the length of median $ \ \overline{BY} \ $ is $ \ \frac32·3 \ = \ \frac92 \ \ . \ $ As these medians are the perpendicular diagonals of quadrilateral $ \ BCYX \ \ , \ $ it is straightforward to show that the area of $ \ BCYX \ $ is half the product of the length of its diagonals, $ \ \frac12·12·\frac92 \ = \ 27 \ \ . \ $ Since $ \ X \ $ and $ \ Y \ $ are the midpoints of sides $ \ \overline{BC} \ $ and $ \ \overline{CA} \ \ , \ \triangle YXC \ $ is similar to $ \ \triangle ABC \ \ $ and one-quarter of its area. Hence, the area of $ \ \triangle ABC \ $ is $ \ \frac43 · 27 \ = \ 36 \ \ ; \ $ having determined that the "base" $ \ \overline{AB} \ $ has length $ \ \sqrt{73} \ \ , \ $ we calculate the length of altitude $ \ \overline{CP} \ $ to be $ \ \frac{36}{\frac12·\sqrt{73} } \ = \ \frac{72}{\sqrt{73} } \ \approx \ 8.43 \ . \ ] $

With the diagram rendered more nearly to scale, we find that $ \ \overline{CP} \ $ is exterior to $ \ \triangle ABC \ \ $ (we may also infer this by comparison of its length to the known side lengths). We can now make use of either of the right triangles $ \ \triangle CPB \ $ or $ \ \triangle CPA \ \ $ to find

$$ CP^2 \ + \ PB^2 \ \ = \ \ CB^2 \ \ \Rightarrow \ \ PB^2 \ \ = \ \ 10^2 \ - \ \frac{72^2}{73} \ \ = \ \ \frac{46^2}{73} \ \ ; $$ $$ CZ^2 \ \ = \ \ CP^2 \ + \ PZ^2 \ \ = \ \ \frac{72^2}{73} \ + \ \left( \ \frac{46 }{\sqrt{73} } \ + \ \frac{\sqrt{73}}{2} \ \right)^2 \ \ = \ \ \frac{72^2}{73} \ + \ \left( \frac{165 }{2 · \sqrt{73} } \right)^2 $$ $$ = \ \ \frac{47961}{4·73} \ \ = \ \ \frac{657 }{4 } \ \ \Rightarrow \ \ CZ \ \ = \ \ \frac32·\sqrt{73} $$

or

$$ CP^2 \ + \ PA^2 \ \ = \ \ CA^2 \ \ \Rightarrow \ \ PA^2 \ \ = \ \ 265 \ - \ \frac{72^2}{73} \ \ = \ \ \frac{119^2}{73} \ \ ; $$

$$ CZ^2 \ \ = \ \ CP^2 \ + \ PZ^2 \ \ = \ \ \frac{72^2}{73} \ + \ \left( \ \frac{119}{\sqrt{73} } \ - \ \frac{\sqrt{73}}{2} \ \right)^2 \ \ = \ \ \frac{72^2}{73} \ + \ \left( \frac{165 }{2 · \sqrt{73} } \right)^2 $$ $$ \Rightarrow \ \ CZ \ \ = \ \ \frac32·\sqrt{73} \ \ , $$ as before.