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Komok [63]
3 years ago
7

I dont get this question.

Mathematics
1 answer:
SashulF [63]3 years ago
8 0
It is 12 - 2/3 = 18, then solve :)
hope that helps :D
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Two altitudes of a triangle have lengths $12$ and $14$. What is the longest possible integer length of the third altitude
nataly862011 [7]

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

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A = 1/2*c*h

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Now, we will use the triangle inequalities:

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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.

Learn more about TRIANGLE here brainly.com/question/2217700

#SPJ4

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