Chapter 15, Problem 11RP

(a)

To determine

To Find: The longest side in the given triangle.

Answer to Problem 11RP

  $AB$

Explanation of Solution

Given:

  $\begin{array}{l}\Delta ABC\text{ is an isosceles triangle}\text{.}\\ \angle C=obtuse\\ AB=6\end{array}$

Formula Used:

In a triangle side opposite to greater angle is always greater and opposite to smallest angel is always smallest or vice-versa.

Calculation:

  $\begin{array}{l}\Delta ABC\text{ is an isosceles triangle}\text{.}\\ \angle C=obtuse\\ AB=6\\ \because \angle C\text{ is obtuse and other two angles of triangle are equal so }\\ \angle C\text{is the biggest angle}\\ AB\text{ is opposite to }\angle C.\\ \therefore AB\text{is the longest}\text{.}\end{array}$

  

Conclusion:

Thus the required longest side is $AB$

(b)

To determine

To Find: The numbers in which perimeter can lie.

Answer to Problem 11RP

Explanation of Solution

Given:

  $\begin{array}{l}\Delta ABC\text{ is an isosceles triangle}\text{.}\\ \angle C=obtuse\\ AB=6\end{array}$

Formula Used:

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

  $Perimeter=Sumofallsides$

Calculation:

  

  $\begin{array}{l}\Delta ABC\text{ is an isosceles triangle}\text{.}\\ \angle C=obtuse\\ AB=6\\ \because \angle C\text{ is obtuse and other two angles of triangle are equal so }\\ \angle C\text{is the biggest angle}\\ AB\text{ is opposite to }\angle C.\\ \therefore AB\text{is the longest}\text{.}\\ AB=6\\ \because AB\text{is the longest then other sides must be less than it}\\ \text{And should also satisfy the angle-inequality theorem}\text{.}\\ BC\&AC<AB\\ And\\ BC+AC>AB\\ \Rightarrow BC+BC>AB\\ \Rightarrow 2BC>AB\\ \Rightarrow BC>\frac{6}{2}\\ \Rightarrow BC>3\\ \text{for}BC=3\\ Perimeter=3+3+6=12\\ forBC=6\\ Perimeter=6+6+6=18\\ thus,12<perimeter<18\end{array}$

Conclusion:

Thus the perimeter of given triangle must lie between $12<perimeter<18$ .