Chapter 3.6, Problem 13PSB

To determine

To Find: How much more the longest side of triangle is than the average of sides of the triangle.

Answer to Problem 13PSB

The difference between the average of sides and longest side of triangle $=\text{ 4}$

Explanation of Solution

Given information:

In the given figure $\overline{AB}=x\text{ + 5 ,}\overline{BC}=\text{ 2}x\text{ + 1, }\overline{AC}=4x\text{ - 6 }$ also the average of the length of the sides of triangle is $14$

Formula used:

  $\text{ Average of numbers = }\frac{sumofallnumbers}{totalnumbers}$

Calculation:

As per the given question we have,

  $\overline{AB}=x\text{ + 5 }\mathrm{.............................................}\text{(i)}$

  $\begin{array}{l}\overline{BC}=\text{ 2}x\text{ + 1}\mathrm{.............................................}\text{(ii)}\\ \overline{AC}=4x\text{ - 6 }\mathrm{..........................................}\text{(iii)}\end{array}$

Also, average is given as $\text{14}\mathrm{..........................}\text{(iv)}$$$ Using the above formula, we can write as:

  $\begin{array}{l}Average=\frac{sumofallnumbers}{totalnumbers}\\ \\ \Rightarrow \text{ 14 = }\frac{(x+5)+(2x+1)+(4x-\text{6)}}{3}\\ \\ \Rightarrow \text{ 14 = }\frac{7x}{3}\\ \\ \Rightarrow 42\text{ =7}x\\ \\ \Rightarrow x\text{ =6}\end{array}$

using this value of $x\text{ =6 }$ in (i), (ii) and (iii)we will get:

  $\begin{array}{l}\overline{AB}=x\text{ + 5 = 6 + 5 = 11}\\ \overline{BC}=\text{ 2}x\text{ + 1 = (2}\times \text{6)+1 =12+1 =13}\\ \overline{AC}=4x\text{ - 6 =(4}\times \text{6) - 6 = 24 - 6 =18}\end{array}$

As the largest side of triangle is $18$ and average is $14$ so their difference = $18-14\text{ = 4}$

Conclusion:

The difference between the average of sides and longest side of triangle $=\text{ 4}$