Chapter 15.2, Problem 2PSA

To determine

To Find: The sets that could be the lengths of a triangle.

Answer to Problem 2PSA

b

Explanation of Solution

Given:

  $\begin{array}{l}a)3,6,9\\ b)4,5,8\\ c)2,3,8\\ d)\sqrt{2},\sqrt{3},\sqrt{6}\end{array}$

Formula Used:

Sum of two sides should be always greater than the third side.

Calculation:

  $\begin{array}{l}6+9>3\\ 3+9>6\\ 3+6=9\\ \text{These sides can't make a triangle}\text{.}\end{array}$

  $\begin{array}{l}4+5>8\\ 5+8>4\\ 4+8>5\\ \text{all condition satisfied}\text{.}\\ \text{These sides can make a triangle}\text{.}\end{array}$

  $\begin{array}{l}2+3<8\\ 3+8>2\\ 2+8>3\\ \text{all the points not satisfied}\text{.}\\ \text{these sides can't make a triangle}\text{.}\end{array}$

Conclusion:

Thus, the required set is $4,5,8$