Question

Triangle $ABC$ has altitudes $\overline{AD},$ $\overline{BE},$ and $\overline{CF}.$ If $AD = 12,$ $BE = 14,$ and $CF$ is a positive integer, then find the largest possible value of $CF.$
Hint(s):
What formula involves the length of an altitude? Use this formula to write inequalities.

Answer 1

Answer:

CF = 7

Step-by-step explanation:

Given: Δ ABC

Altitudes ⇒ AD = 12 , BE = 14  and  CF = ?

The area of the triangle = 0.5 AD * BC ⇒ (1)

OR area                          = 0.5 BE * AC  ⇒ (2)

OR area                          = 0.5 CF * AB  ⇒ (3)

By equating (1) and (3)

∴ 0.5 CF * AB = 0.5 AD * BC

∴ $CF = \frac{AD*BC}{AB} =\frac{12BC}{AB}$ ⇒ (4)

By equating (1) and (2)

0.5 BE * AC = 0.5 AD * BC

∴ $\frac{AC}{BC} = \frac{AD}{BE} =\frac{12}{14} =\frac{6}{7}$ (5)

We should know that about the relation between the sides of the triangle:

The sum of two sides will be greater than the third side

So, AB < BC + AC    ⇒ divide both sides by BC

∴ $\frac{AB}{BC} < 1+\frac{AC}{BC}$

By substitution from (5) with AC/BC

∴ $\frac{AB}{BC} <1+\frac{6}{7}$

∴ $\frac{AB}{BC} <\frac{13}{7}$

∴ $\frac{BC}{AB} > \frac{7}{13}$ ⇒ (6)

By substitution from (6) at (4)

∴ CF > 12 * 7/13

CF > 6.46

But CF is a positive integer

∴ CF = 7