Question

Two altitudes of a triangle have lengths $12$ and $14$. What is the longest possible integer length of the third altitude

Answer 1

The longest possible altitude of the third altitude (if it is a positive integer) is 83.

According to statement

Let h is the length of third altitude

Let a, b, and c be the sides corresponding to the altitudes of length 12, 14, and h.

From Area of triangle

A = 1/2*B*H

Substitute the values in it

A = 1/2*a*12

a = 2A / 12 -(1)

Then

A = 1/2*b*14

b = 2A / 14 -(2)

Then

A = 1/2*c*h

c = 2A / h -(3)

Now, we will use the triangle inequalities:

• a < b+c

2A/12 < 2A/14 + 2A/h

Solve it and get

h<84

• b < a+c

2A/14 < 2A/12 + 2A/h

Solve it and get

h > -84

• c < a+b

2A/h < 2A/12 + 2A/14

Solve it and get

h > 6.46

From all the three inequalities we get:

6.46<h<84

So, the longest possible altitude of the third altitude (if it is a positive integer) is 83.

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