Chapter 15, Problem 7RP

To determine

To Find: In the given set that cannot be sides of a triangle.

Answer to Problem 7RP

  $a)20,40,20$

Explanation of Solution

Given:

  $a)20,40,20$

  $b)30,40,20$

  $c)20,20,20$

  $d)30,40,50$

Formula Used:

Triangle Inequality Theorem: It states that sum of two sides of a triangle is always greater than the third side.

Calculation:

  $\begin{array}{l}a)20,40,20\\ 20+20=40\\ \text{Sum of two sides are equal to third}\text{.}\\ \\ b)30,40,20\\ 30+40=70>20\\ \text{Sum of two sides are greater than third}\text{.}\\ \\ c)20,20,20\\ 20+20=40>20\\ \text{Sum of two sides are greater than third}\text{.}\\ \\ d)30,40,50\\ 30+40=70>50\\ \text{Sum of two sides are greater than third}\text{.}\end{array}$

Conclusion:

Thus the required set is $a)20,40,20$