Chapter 15.3, Problem 13PSB

To determine

To Find: The type of given triangle.

Answer to Problem 13PSB

  $\Delta PQR\text{ is right }\Delta$

Explanation of Solution

Given:

  $\begin{array}{l}\to PQ=1\frac{1}{2}\\ \to QR=2\frac{1}{2}\\ \to PR=2\end{array}$

Formula used:

Pythagorean-Theorem:

  $\begin{array}{l}\text{ for acute }\Delta \\ a^2+b^2>c^2\end{array}$

  $\begin{array}{l}\text{for obtuse triangle}\\ a^2+b^2<c^2\end{array}$

  $\begin{array}{l}\text{For right angle triangles}\\ {\text{a}}^2+b^2=c^2\end{array}$

Calculation:

  $\begin{array}{l}\text{⇒}P{Q}^2={\left(\frac{3}{2}\right)}^2=\frac{9}{4}\\ \text{⇒}P{R}^2={\left(\frac{5}{2}\right)}^2=\frac{25}{4}\\ \text{⇒}Q{R}^2\text{=}{\left(2\right)}^2=4\\ \text{⇒}P{Q}^2+Q{R}^2=\frac{9}{4}+4=\frac{9+16}{4}=\frac{25}{4}=P{R}^2\\ \text{⇒}P{Q}^2+Q{R}^2=P{R}^2\\ \text{So, it is right angled triangle}\text{.}\end{array}$

Conclusion:

Thus $\Delta PQR$ is right triangle