Chapter 15, Problem 6RP

(a)

To determine

To Find: The congruent segment.

Answer to Problem 6RP

  $CE\cong DE$

Explanation of Solution

Given:

  $\begin{array}{l}\angle DCE={50}^o\\ \angle CDE={50}^o\end{array}$

  

Formula Used:

In a triangle side opposite to equal angles are always or vice-versa.

Calculation:

  $\begin{array}{l}In\Delta CDE\\ \because \angle DCE\cong \angle CDE={50}^o\\ \therefore CE\cong DE\end{array}$

Conclusion:

Thus the required congruent segments are $CE\cong DE$

(b)

To determine

To Find: The requiredshorter segment in the given figure.

Answer to Problem 6RP

  $\overline{BE}$

Explanation of Solution

Given:

  $\begin{array}{l}\angle BCE={35}^o\\ \angle CEB={90}^o\\ \angle CBE={55}^o\end{array}$

  

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:

  $\begin{array}{l}In\Delta BEC\\ \angle BCE={35}^o\\ \angle CEB={90}^o\\ \angle CBE={55}^o\\ \because \angle BCE<\angle CBE<\angle CEB\\ \because \overline{BE}\text{ is opposite to }\angle BCE\\ \therefore \overline{BE}<\overline{EC}\end{array}$

Conclusion:

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

(c)

To determine

To Define: The name of the required segment.

Answer to Problem 6RP

Hypotenuse

Explanation of Solution

Given:

  $\begin{array}{l}\angle PQS=\angle 2\\ \angle QSR=\angle 1\end{array}$

  

Formula Used:

Pythagorean-Theorem: It states that in right angled triangle the longest segment is termed as Hypotenuse.

Calculation:

  $\begin{array}{l}In\Delta BEC\\ \angle BEC={90}^o\\ \overline{BC}\text{ opposite to}\angle BEC\\ \overline{BC}\text{ is greatest segment so in Right angled triangle the }\\ \text{largest segment is termed as Hypotenuse}\text{. }\end{array}$

Conclusion:

Thus the name of required segment is Hypotenuse.

(d)

To determine

To Find: The required longer segment in the given figure.

Answer to Problem 6RP

  $\overline{DE}$

Explanation of Solution

Given:

  $\begin{array}{l}\angle DAE={35}^o\\ \angle AED={120}^o\\ \angle ADE={27}^o\end{array}$

  

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:

  $\begin{array}{l}In\Delta ADE\\ \angle DAE={35}^o\\ \angle AED={120}^o\\ \angle ADE={27}^o\\ \because \angle ADE<\angle DAE<\angle AED\\ \because \overline{DE}\text{ is opposite to }\angle DAE\\ \therefore \overline{AE}<\overline{DE}\end{array}$

Conclusion:

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

(b)

To determine

To Find: The required shortest segment in the given figure.

Answer to Problem 6RP

  $\overline{AE}$

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:

  $\begin{array}{l}In\text{quadrilateral}ABCDand\Delta ADE\\ \angle DAE={35}^o\\ \angle AED={120}^o\\ \angle ADE={27}^o\\ \because \angle ADE<\angle DAE<\angle AED\\ \because \overline{AE}\text{ is opposite to }\angle ADE\\ \therefore \overline{AE}<\overline{DE}\\ \because \angle ADE={27}^o\text{ is the smallest angle in the given figure}\\ \therefore \overline{AE}\text{ is the smallest segment in the given quadiralteral}\text{.}\end{array}$

Conclusion:

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