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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

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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 $5\frac 5 {21} < c < 110$.

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

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  • $\begingroup$ 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. $\endgroup$ – John Omielan Jan 3 at 1:52
  • $\begingroup$ thanks @JohnOmielan for pointing that out.it is fixed now. $\endgroup$ – Rakibul Islam Prince Jan 3 at 2:34

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