+0

# Geometry

0
270
2
+75

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?

eileenthecoolbean  Aug 16, 2017

### Best Answer

#1
+18827
+1

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

heureka  Aug 17, 2017
Sort:

### 2+0 Answers

#1
+18827
+1
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

heureka  Aug 17, 2017
#2
+80941
0

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

CPhill  Aug 17, 2017

### 36 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details