DAV Class 8 Maths Chapter 10 HOTS
Parallel Lines HOTS
1. In the diagram given below, find the angles \(x,\ y\) and \(z\).
Solution
\[ \begin{aligned} \angle 1 + 110^\circ & = 180^\circ \ \color{magenta}(\text{Linear Pair}) \\ \angle 1 & = 180^\circ - 110^\circ \\ \angle 1 & = 70^\circ \\[6pt] \angle 1 & = y \ \color{magenta}(\text{Alternate interior angles}) \\ \color{green} y & = \color{green}70^\circ \\ \\ \text{In } \triangle ABC & \ \color{magenta}(\text{Angle sum property}) \\ x + y + 50^\circ & = 180^\circ \\ x + 70^\circ + 50^\circ & = 180^\circ \\ x + 120^\circ & = 180^\circ \\ x & = 180^\circ - 120^\circ \\ \color{green} x & = \color{green} 60^\circ \\ \\ x & = z \ \color{magenta}(\text{Corresponding angles}) \\ \color{green} z & = \color{green} 60^\circ \end{aligned} \]Answer \(x=\color{red}{60^\circ}\), \(y=\color{red}{70^\circ}\), \(z=\color{red}{60^\circ}\)
2. In the figure, \(l \parallel m\) and \(t\) is a transversal. The angle bisectors of the interior angles on each side meet at \(P\) and \(Q\). Find the values of angles \(x\) and \(y\).
Solution
\[ \begin{aligned} \color{magenta} \textbf{Given:} & \ l \parallel m \\ & t \text{ is a transversal} \\ \\ \color{magenta} \textbf{To find:} & \ x \text{ and } y \\ \\ \color{magenta} \textbf{Proof:} & \\ \angle 1 + \angle 2 + \angle 7 + \angle 8 & = \ 180^\circ \color{magenta}(\text{interior angles are supplementary})\\[6pt] \angle 2 + \angle 7 & = \ 90^\circ \color{magenta}(\text{interior angles are angle bisectors})\\[6pt] \angle 2 + \angle 7 + \angle 9 & = \ 180^\circ \color{magenta}(\text{Angle sum property})\\ 90^\circ + \angle 9 & = \ 180^\circ \\ \angle 9 & = \ 180^\circ - 90^\circ \\ \angle 9 & = 90^\circ \\[6pt] \angle 9 & = \angle y \ \color{magenta}(\text{Vertically opposite angles}) \\ \color{green} \angle y & = \color{green} 90^\circ \\ \\\angle 3 + \angle 4 + \angle 5 + \angle 6 & = \ 180^\circ \color{magenta}(\text{interior angles are supplementary})\\[6pt] \angle 3 + \angle 6 & = \ 90^\circ \color{magenta}(\text{interior angles are angle bisectors})\\[6pt] \angle x + \angle 3 + \angle 6 & = \ 180^\circ \color{magenta}(\text{Angle sum property})\\ 90^\circ + \angle 9 & = \ 180^\circ \\ \angle x & = \ 180^\circ - 90^\circ \\ \color{green} \angle x & = \color{green} 90^\circ \end{aligned} \]Answer \(x=\color{red}{90^\circ}\), \(y=\color{red}{90^\circ}\)