Chapter 15, Problem 13RP

(a)

To determine

To Find: The measurement of required segment in given triangle.

Answer to Problem 13RP

  $\begin{array}{l}PQ=\sqrt{34}\\ QR=3\sqrt{2}\\ PR=7\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}\Delta PQR\\ P=\left(-1,-2\right)\\ Q=\left(4,1\right)\\ R=\left(6,-2\right)\end{array}$

Formula Used:

  $\text{Distance Formula}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

  $\begin{array}{l}\Delta PQR\\ P=\left(-1,-2\right)\\ Q=\left(4,1\right)\\ R=\left(6,-2\right)\\ PQ=\sqrt{{\left(4-\left(-1\right)\right)}^2+{\left(1-\left(-2\right)\right)}^2}=\sqrt{25+9}\\ \Rightarrow PQ=\sqrt{34}\\ QR=\sqrt{{\left(6-4\right)}^2+{\left(\left(-2\right)-1\right)}^2}=\sqrt{9+9}\\ \Rightarrow QR=3\sqrt{2}\\ PR=\sqrt{\left(6-{\left(-1\right)}^2+{\left(\left(-2\right)-\left(-2\right)\right)}^2\right)}=\sqrt{49}\\ \Rightarrow PR=7\end{array}$

Conclusion:

Thus the measurement of required segment is

  $\begin{array}{l}PQ=\sqrt{34}\\ QR=3\sqrt{2}\\ PR=7\end{array}$

(b)

To determine

To Find: The type triangle is given.

Answer to Problem 13RP

Acute

Explanation of Solution

Given:

  $\begin{array}{l}\Delta PQR\\ P=\left(-1,-2\right)\\ Q=\left(4,1\right)\\ R=\left(6,-2\right)\end{array}$

Formula Used:

  $\text{Distance Formula}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

  $\begin{array}{l}Acute:a^2+b^2>c^2\\ Obtuse:a^2+b^2<c^2\\ Right:a^2+b^2=c^2\end{array}$

Calculation:

  $\begin{array}{l}\Delta PQR\\ P=\left(-1,-2\right)\\ Q=\left(4,1\right)\\ R=\left(6,-2\right)\\ PQ=\sqrt{{\left(4-\left(-1\right)\right)}^2+{\left(1-\left(-2\right)\right)}^2}=\sqrt{25+9}\\ \Rightarrow PQ=\sqrt{34}\\ \Rightarrow P{Q}^2=34\\ QR=\sqrt{{\left(6-4\right)}^2+{\left(\left(-2\right)-1\right)}^2}=\sqrt{9+9}\\ \Rightarrow QR=3\sqrt{2}\\ \Rightarrow Q{R}^2=18\\ PR=\sqrt{\left(6-{\left(-1\right)}^2+{\left(\left(-2\right)-\left(-2\right)\right)}^2\right)}=\sqrt{49}\\ \Rightarrow PR=7\\ \Rightarrow P{R}^2=49\\ P{Q}^2+Q{R}^2=34+18=52>49=P{R}^2\end{array}$

Conclusion:

Thus the given triangle is Acute.

(c)

To determine

To Arrange: The angles of given triangle in ascending order.

Answer to Problem 13RP

  $\angle P<\angle R<\angle Q$

Explanation of Solution

Given:

  $\begin{array}{l}\Delta PQR\\ P=\left(-1,-2\right)\\ Q=\left(4,1\right)\\ R=\left(6,-2\right)\end{array}$

Formula Used:

  $\text{Distance Formula}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

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

Calculation:

  

  $\begin{array}{l}\Delta PQR\\ P=\left(-1,-2\right)\\ Q=\left(4,1\right)\\ R=\left(6,-2\right)\\ PQ=\sqrt{{\left(4-\left(-1\right)\right)}^2+{\left(1-\left(-2\right)\right)}^2}=\sqrt{25+9}\\ \Rightarrow PQ=\sqrt{34}\\ \\ QR=\sqrt{{\left(6-4\right)}^2+{\left(\left(-2\right)-1\right)}^2}=\sqrt{9+9}\\ \Rightarrow QR=3\sqrt{2}\\ \\ PR=\sqrt{\left(6-{\left(-1\right)}^2+{\left(\left(-2\right)-\left(-2\right)\right)}^2\right)}=\sqrt{49}\\ \Rightarrow PR=7\\ \text{Angle opposite to:}\\ PRis\angle Q\\ PQis\angle R\\ QRis\angle P\\ \because PR>PQ>QR\\ \therefore \angle P<\angle R<\angle Q\end{array}$

Conclusion:

Thus the required arrangement is $\angle P<\angle R<\angle Q$ .