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Two altitudes of a triangle have lengths 12 and 14. What is the longest possible length of the third altitude, if it is a positive integer?

 Aug 16, 2017

Best Answer 

 #1
avatar+21869 
+2

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

if it is a positive integer?

 

Let h = the length of the third altitude.

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

 

Let A = Area of the triangle.

Let 2A = a*12
Let 2A = b*14
Let 2A = c*h

 

1.

\(\begin{array}{|rcll|} \hline a &=& \frac{2A}{12} \\ b &=& \frac{2A}{14} \\ c &=& \frac{2A}{h} \\ \hline \end{array}\)

 

2.

A triangle exists with side lengths a, b, and c
if and only if they satisfy the three triangle inequalities:

\(\begin{array}{|lrcll|} \hline (1) & a & < & b + c \\ (2) & b & < & c + a \\ (3) & c & < & a + b \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1) & \frac{2A}{12} & < & \frac{2A}{14} + \frac{2A}{h} \quad & | \quad :2A \\ & \frac{1}{12} & < & \frac{1}{14} + \frac{1}{h} \\\\ (2) & \frac{2A}{14} & < & \frac{2A}{h} + \frac{2A}{12} \quad & | \quad :2A \\ & \frac{1}{14} & < & \frac{1}{h} + \frac{1}{12} \\\\ (3) & \frac{2A}{h} & < & \frac{2A}{12} + \frac{2A}{14} \quad & | \quad :2A \\ & \frac{1}{h} & < & \frac{1}{12} + \frac{1}{14} \\ \hline \end{array}\)

 

(1):

\( \begin{array}{|rcll|} \hline \frac{1}{12} & < & \frac{1}{14} + \frac{1}{h} \\ \frac{1}{12}-\frac{1}{14} & < & \frac{1}{h} \\ \frac{12-14}{12\cdot 14} & < & \frac{1}{h} \\ \frac{2}{168} & < & \frac{1}{h} \\ \frac{1}{84} & < & \frac{1}{h} \quad & | \quad \cdot 84h \\ \mathbf{ h } & \mathbf{ < } & \mathbf{ 84 } \\ \hline \end{array} \)

 

(2):

\(\begin{array}{|rcll|} \hline \frac{1}{14} & < & \frac{1}{h} + \frac{1}{12} \\ \frac{1}{14} - \frac{1}{12} & < & \frac{1}{h} \\ \frac{12-14}{14\cdot 12} & < & \frac{1}{h} \\ -\frac{2}{168} & < & \frac{1}{h} \\ -\frac{1}{84} & < & \frac{1}{h} \quad & | \quad \cdot 84h \\ -h & < & 84 \quad & | \quad \cdot (-1) \qquad switch "<" to ">" \\ \mathbf{ h } & \mathbf{ > } & \mathbf{ -84 } \\ \hline \end{array} \)

 

(3):

\(\begin{array}{|rcll|} \hline \frac{1}{h} & < & \frac{1}{12} + \frac{1}{14} \\ \frac{1}{h} & < & \frac{14+12}{12\cdot 14} \\ \frac{1}{h} & < & \frac{26}{168} \\ \frac{1}{h} & < & \frac{13}{84} \quad & | \quad \cdot \frac{84}{13}\ h \\ \frac{84}{13} & < & h \\ \mathbf{ 6.46153846154 } & \mathbf{ < } & \mathbf{ h } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline 6.46153846154 < h < 84 \\ \hline \end{array}\)

 

The longest possible length of the third altitude, if it is a positive integer is 83

 

laugh

 Aug 17, 2017
 #1
avatar+21869 
+2
Best Answer

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

if it is a positive integer?

 

Let h = the length of the third altitude.

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

 

Let A = Area of the triangle.

Let 2A = a*12
Let 2A = b*14
Let 2A = c*h

 

1.

\(\begin{array}{|rcll|} \hline a &=& \frac{2A}{12} \\ b &=& \frac{2A}{14} \\ c &=& \frac{2A}{h} \\ \hline \end{array}\)

 

2.

A triangle exists with side lengths a, b, and c
if and only if they satisfy the three triangle inequalities:

\(\begin{array}{|lrcll|} \hline (1) & a & < & b + c \\ (2) & b & < & c + a \\ (3) & c & < & a + b \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1) & \frac{2A}{12} & < & \frac{2A}{14} + \frac{2A}{h} \quad & | \quad :2A \\ & \frac{1}{12} & < & \frac{1}{14} + \frac{1}{h} \\\\ (2) & \frac{2A}{14} & < & \frac{2A}{h} + \frac{2A}{12} \quad & | \quad :2A \\ & \frac{1}{14} & < & \frac{1}{h} + \frac{1}{12} \\\\ (3) & \frac{2A}{h} & < & \frac{2A}{12} + \frac{2A}{14} \quad & | \quad :2A \\ & \frac{1}{h} & < & \frac{1}{12} + \frac{1}{14} \\ \hline \end{array}\)

 

(1):

\( \begin{array}{|rcll|} \hline \frac{1}{12} & < & \frac{1}{14} + \frac{1}{h} \\ \frac{1}{12}-\frac{1}{14} & < & \frac{1}{h} \\ \frac{12-14}{12\cdot 14} & < & \frac{1}{h} \\ \frac{2}{168} & < & \frac{1}{h} \\ \frac{1}{84} & < & \frac{1}{h} \quad & | \quad \cdot 84h \\ \mathbf{ h } & \mathbf{ < } & \mathbf{ 84 } \\ \hline \end{array} \)

 

(2):

\(\begin{array}{|rcll|} \hline \frac{1}{14} & < & \frac{1}{h} + \frac{1}{12} \\ \frac{1}{14} - \frac{1}{12} & < & \frac{1}{h} \\ \frac{12-14}{14\cdot 12} & < & \frac{1}{h} \\ -\frac{2}{168} & < & \frac{1}{h} \\ -\frac{1}{84} & < & \frac{1}{h} \quad & | \quad \cdot 84h \\ -h & < & 84 \quad & | \quad \cdot (-1) \qquad switch "<" to ">" \\ \mathbf{ h } & \mathbf{ > } & \mathbf{ -84 } \\ \hline \end{array} \)

 

(3):

\(\begin{array}{|rcll|} \hline \frac{1}{h} & < & \frac{1}{12} + \frac{1}{14} \\ \frac{1}{h} & < & \frac{14+12}{12\cdot 14} \\ \frac{1}{h} & < & \frac{26}{168} \\ \frac{1}{h} & < & \frac{13}{84} \quad & | \quad \cdot \frac{84}{13}\ h \\ \frac{84}{13} & < & h \\ \mathbf{ 6.46153846154 } & \mathbf{ < } & \mathbf{ h } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline 6.46153846154 < h < 84 \\ \hline \end{array}\)

 

The longest possible length of the third altitude, if it is a positive integer is 83

 

laugh

heureka Aug 17, 2017
 #2
avatar+98197 
0

 

Thanks, heureka.....I like your method of solving this one...!!!

 

 

cool cool cool

 Aug 17, 2017

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