Possible lengths of an altitude of a triangle

I've been preparing for a competition and there is this problem that I cannot solve. Can you please help me and also tell me how to do similar problems if they appear in the future? Problem:
Two of the altitudes of a triangle are 11 units and 10 units. Which of the following can't be the length of an altitude:
(A) 5 units (B) 6 units (c) 7 units (D) 10 units (E) 100 units

Suppose the three sides of the triangle are $$a$$, $$b$$ and $$c$$, and the corresponding altitudes of this triangle are $$h_a=11$$, $$h_b=10$$ and $$h_c$$. Then $$ah_a=bh_b=ch_c = 2A$$, where $$A$$ is the area of the triangle. So, $$11a=10b$$. Write $$a = 10k$$. From the triangular inequality we have $$(11-10)k, i.e. $$k. Plus, we know that $$2A=ah_a=110k$$. Thus, $$h_c$$ must satisfy $$\frac {110k} {21k} < h_c < \frac {110k} k$$, which is $$5\frac 5 {21} < c < 110$$.
The answer is $$a)5$$.
Let,the area be $$k$$ and the unknown altitude is "p" and it's corresponding base is $$c$$,the other two sides are $$a$$ and $$b$$. Now,we can say, $$\frac{1}{2}\times a \times 11 =\frac{1}{2}\times b \times 10 =\frac{1}{2}\times c \times p =k$$ so,$$\frac{1}{2}\times a \times 11=k\implies a=\frac{2k}{11}$$ similarly,$$b=\frac{2k}{10}~~and~~c=\frac{2k}{p}$$ now,according to inequality of sides law, $$a+b\gt c$$ $$\implies \frac{2k}{11} + \frac{2k}{10} \gt \frac{2k}{p}$$ $$p\gt \frac{110}{21}=5.5$$
• Actually, the $10$ and $11$ are the altitudes, not the side lengths, with the $100$ in the answer options also being an altitude. However, you are correct about using the triangle inequality, with the solution using the triangle area formula of $\frac{1}{2}$ the product of base length & altitude length, that the area must obviously be the same for all cases, and then doing the appropriate calculations. – John Omielan Jan 3 at 1:52