DAV Class 8 Maths Chapter 10 Worksheet 2
Parallel Lines Worksheet 2
1. In the given figure, \(l \parallel m\) and \(p \perp l\) and \(q \perp m\). Show that \(p \parallel q\). Give reasons.
Solution
\[ \begin{aligned} \color{magenta}{\textbf{Given:}}& \ l \parallel m \\ &p\perp l \\ & q\perp m\\ \\ \color{magenta}{\textbf{To prove:}}&\ p \parallel q \\ \\ \color{magenta}{\textbf{Proof:}}& \\ l & \parallel m \ \\ \angle 1 & = \angle 3 \ \color{magenta}(\text{corresponding angles})\\[6pt] \text{Since} \ \angle 1 & = 90^\circ \\ \Rightarrow \ \angle 3 & = 90^\circ \\[6pt] \therefore \ & {\color{green} p \perp m } \\ & q\perp m \ \color{magenta}(\text{Given})\\ \\ \therefore \color{green}p \parallel q & \begin{cases} \color{magenta}\text{Two lines in the same plane} \\ \color{magenta}\text{perpendicular to a given line} \\ \color{magenta}\text{are parallel to each other}\\ \end{cases}\\[6pt] \end{aligned} \]2. What can you say about the quadrilateral ABCD in the given figure. Is it a rectangle? Justify your answer.
Solution
\[ \begin{aligned} \color{magenta}{\textbf{Given:}} & \ \angle 1 = 90^\circ \\ & \ \angle 4 = 90^\circ \\ \\ \color{magenta}{\textbf{To prove:}}& \ ABCD \text{ is a rectangle} \\ \\ \color{magenta}{\textbf{Proof:}}& \\ \angle 1 + \angle 5 & = 180^\circ \ \color{magenta}(\text{Linear pair})\\ 90^\circ + \angle 5 & = 180^\circ \\ \angle 5 & = 180^\circ - 90^\circ \\ \angle 5 & = 90^\circ \\ \\ \angle 1 & = \angle 2 \ \color{magenta}(\text{Alternate interior angles})\\ \angle 2 & = 90^\circ \\ \\ \angle 1 & = \angle 3 \ \color{magenta}(\text{corresponding angles})\\ \angle 3 & = 90^\circ \\[6pt]\angle 2 = \angle 3 = \angle 4 & = \angle 5 = 90^\circ \\[6pt] \therefore ABCD & \text{ is a rectangle or square.} \end{aligned} \]3. In the given figure, \( \triangle ABC\) and \(AD\) is an altitude. Show that: (i) \(BP \parallel AD\), (ii) \(CQ \parallel AD\), (iii) \(BP \parallel CQ\).
Solution
\[ \begin{aligned} \color{magenta}{\textbf{To prove:}}& \ (i)\ BP\parallel AD \\ &(ii)\ CQ\parallel AD \\ & (iii)\ BP\parallel CQ \\[6pt] \color{magenta}{\textbf{Proof:}}& \\[2pt] \textbf{(i)}\ \angle PBA &=\angle DAB =40^\circ \ \color{magenta}\text{(Alternate interior angles)}\\ \Rightarrow \ & \color{green}{BP\parallel AD} \\[10pt]\textbf{(ii)}\ \angle ADC &=\angle DCQ =90^\circ \ \color{magenta}\text{(Alternate interior angles)}\\ \Rightarrow \ & \color{green}{CQ\parallel AD}\\[10pt] \textbf{(iii)} \ BP & \parallel AD\ \text{(from (i))} \\ AD & \parallel CQ\ \text{(from (ii))}\\ \Rightarrow \ & \color{green}{BP\parallel CQ} \end{aligned} \]4. Draw a line segment \(AB\) of length \(5\) cm. At \(A\) and \(B\), construct lines perpendicular to \(AB\). Also, draw the perpendicular bisector of \(AB\). Are these three lines parallel to each other? Justify your answer.
Solution
\[ \begin{aligned} AD & \perp AB \ \color{magenta}\text{(Given)} \\ CF & \perp AB \ \color{magenta}\text{(Given)} \\ BE & \perp AB \ \color{magenta}\text{(Given)} \\ \\ \therefore \color{green} AD \parallel CF \parallel BE & \begin{cases} \color{magenta}\text{Lines in the same plane} \\ \color{magenta}\text{perpendicular to a given line} \\ \color{magenta}\text{are parallel to each other}\\ \end{cases}\\[6pt] \end{aligned} \]5. If \( \angle DAC = 30^\circ\), find the angles of \( \triangle ABC\) in the given figure.
Solution
\[ \begin{aligned} \color{magenta}{\textbf{To find:}} & \ \angle ABC,\ \angle ACB,\ \angle A \\[6pt] \color{magenta}{\textbf{Proof:}} & \\ \\ In \ \triangle ADC \\ \angle DAC + \angle ADC + \angle ACD &= 180^\circ \ \color{magenta}(\text{angle sum property})\\ 30^\circ + 90^\circ + \angle ACD &= 180^\circ \\ 120^\circ + \angle ACD &= 180^\circ \\ \angle ACD &= 180^\circ - 120^\circ \\ \color{green}\angle ACD &= \color{green} 60^\circ \\ \\\angle A & = \angle BAD + \angle CAD \\ \angle A & = 40^\circ + 30^\circ \\ \color{green} \angle A & = \color{green} 70^\circ \\ \\ In \ \triangle ABC \\ \angle A + \angle ABC + \angle ACB &= 180^\circ \ \color{magenta}(\text{angle sum property})\\ 70^\circ + \angle ABC + 60^\circ &= 180^\circ \\ 130^\circ + \angle ABC &= 180^\circ \\ \angle ABC &= 180^\circ - 130^\circ \\ \color{green}\angle ABC &= \color{green} 50^\circ \\ \\ \end{aligned} \]Answer \( \angle A = {\color{red}{70^\circ}} ,\angle ABC = {\color{red}{50^\circ}} , \angle ACB = {\color{red}{60^\circ}}\)