Chapter 15, Problem 2RP

(a)

To determine

To Find: The longest segment in the given figure.

Answer to Problem 2RP

  $\overline{WZ}$

Explanation of Solution

Given:

  $\begin{array}{l}XY=XW\\ \angle YXZ={40}^o\\ \angle ZXW={50}^o\end{array}$

  

Formula Used:

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

Angle-Sum Property: In a triangle the sum of all interior angle is equal to 180.

Calculation:

  $\begin{array}{l}In\Delta XWY\\ XY=XW\\ \angle YXZ={40}^o\\ \angle ZXW={50}^o\\ \because XY=XW\Rightarrow \angle Y=\angle W\\ \because \overline{WZ}\text{ is opposite to }\angle {50}^o\\ \therefore \overline{WZ}>\overline{ZY}\end{array}$

Conclusion:

Thus the required segment is $\overline{WZ}$

(b)

To determine

To Find: The longest segment in the given figure.

Answer to Problem 2RP

  $\overline{AB}$

Explanation of Solution

Given:

  $\angle ACB={90}^o$

  

Formula Used:

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

Angle-Sum Property: In a triangle the sum of all interior angle is equal to 180.

Calculation:

  $\begin{array}{l}In\Delta ABC\\ \angle ACB={90}^o\\ \because \overline{AB}\text{ is opposite of }\angle ACB\\ \therefore \overline{AB}>\overline{BC}\end{array}$

Conclusion:

Thus the required segment is $\overline{AB}$

(c)

To determine

To Find: The longest segment in the given figure.

Answer to Problem 2RP

  $\overline{QR}$

Explanation of Solution

Given:

  

Formula Used:

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

Angle-Sum Property: In a triangle the sum of all interior angle is equal to 180.

Calculation:

  $\begin{array}{l}\angle PRQ={60}^o\\ \angle PQR={20}^o+{30}^o={50}^o\\ \angle PQR+\angle PRQ+\angle RPQ={180}^o\\ \Rightarrow 60+50+\angle RPQ=180\\ \Rightarrow \angle RPQ=180-110={70}^o\\ \because \overline{QR}\text{ is opposite to }\angle RPQ\\ \therefore \overline{QR}>\overline{PQ}or\overline{PR}\end{array}$

Conclusion:

Thus the required segment is $\overline{QR}$