Possible lengths of an altitude of a triangle

Question

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

Answer 1

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 $(11-10)k<c<(11+10)k$, i.e. $k<c<21k$. Plus, we know that $2A=ah_a=110k$. Thus, $h_c$ must satisfy $\frac {110k} {21k} < h_c < \frac {110k} k$, which is $5\frac 5 {21} < c < 110$.

Answer 2

Sorry for my previous misinterpretation.

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, $$a+b\gt c$$ $$\implies \frac{2k}{11} + \frac{2k}{10} \gt \frac{2k}{p}$$ $$p\gt \frac{110}{21}=5.5$$