DAV Class 8 Maths Chapter 10 Worksheet 1
Parallel Lines Worksheet 1
1. In the given diagram, \( l \parallel m \parallel n \) and \( p \) is a transversal. If \( \angle 1 = 110^\circ \), find the angles \(x, y, z \text{ and } w\).
Solution
\[ \begin{aligned} \color{magenta} \text{Given} & : l \parallel m \parallel n \text{ and } p \text{ is transversal}\\ & \angle 1 = 110^\circ \\ \\ \color{magenta} \text{To find} & : x,y,z ,w \\ \\ \angle 1 + \angle x &= 180^\circ \quad \color{magenta} (\text{ Linear pair })\\ 110^\circ + \angle x &= 180^\circ \\ \angle x &= 180^\circ - 110^\circ \\ \color{green} \angle x &= \color{green}{70^\circ} \\[6pt] \angle x &= \angle z \quad \color{magenta} ( \text{ Corresponding angles} ) \\ 70^\circ &= \angle z \\ \implies \color{green} \angle z & = \color{green} 70^\circ \\[6pt] \angle y &= \angle z \quad \color{magenta} ( \text{ Alternate interior angles} ) \\ \color{green} \angle y &= \color{green} 70^\circ \\[6pt] \angle z + \angle w &= 180^\circ \quad \color{magenta} (\text{ Linear pair }) \\ 70^\circ + \angle w &= 180^\circ \\ \angle w &= 180^\circ - 70^\circ \\ \color{green} \angle w &= \color{green} 110^\circ \\\end{aligned} \]Answer \( \angle x={\color{red}{70^\circ}},\ \angle y={\color{red}{70^\circ}},\ \angle z={\color{red}{70^\circ}},\ \angle w={\color{red}{110^\circ}}\)
2. In the quadrilateral \(ABCD\) shown, \(\angle 1=\angle 2=90^\circ\). Is \(AD \parallel BC\)? Justify your answer.
Solution
\[ \begin{aligned} \color{magenta}{\text{Given:}}&\ \angle 1=90^\circ,\ \angle 2=90^\circ\\[6pt] \color{magenta} \text{To prove} & : AD \parallel BC \\ \\ \angle 1 + \angle 2 & = 90^\circ + 90^\circ \\ & = 180^\circ \ \color{magenta} (\text{Co-interior angles are supplementary}) \\[6pt] \therefore & \ AD \parallel BC\end{aligned} \]Answer Yes, \(AD \parallel BC\).
3. Draw a line segment \(\overline{AB}=6\text{ cm}\). Mark two points \(P\) and \(Q\) on it. Draw lines perpendicular to \(AB\) through \(P\) and \(Q\) (i.e., \(PR\) and \(QS\)). What can you say about \(PR\) and \(QS\)? Are these parallel? Justify your answer.
Solution
\[ \begin{aligned} &\angle 1=\angle 2=90^\circ \\[4pt] \angle 1 + \angle 2 & = 90^\circ + 90^\circ \\ & = 180^\circ \ \color{magenta} (\text{Co-interior angles are supplementary}) \\[6pt] \therefore & \ PR \parallel QS\end{aligned} \]Answer Yes, \(PR \parallel QS\).
4. In the given figure, \(l \parallel m \parallel n\). Find \(\angle x\) and \(\angle y\).
Solution
\[ \begin{aligned} \color{magenta}{\text{Given:}}&\ l \parallel m \parallel n \\[4pt] & \angle 1 = 50^\circ \\ & \angle 2 = 30^\circ \\ \\ \color{magenta} \text{To find} & : x,y \\ \\ \angle 1 + \angle x & = 180^\circ \ \color{magenta} (\text{Co-interior angles are supplementary}) \\ 50^\circ + \angle x & = 180^\circ \\ \angle x & = 180^\circ - 50^\circ \\ \therefore \color{green} \angle x & = \color{green} 130^\circ \\ \\\angle 2 & = \angle y \ \color{magenta} (\text{Alternate interior angles}) \\ 30^\circ & = \angle y \\ \therefore \color{green} \angle y & = \color{green}{30^\circ} \end{aligned} \]Answer \(\angle x={\color{red}{130^\circ}}, \ \angle y={\color{red}{30^\circ}}\)
5. \(ABCD\) is a quadrilateral in which all the four angles are equal. Show that \(\overline{AB}\parallel\overline{CD}\) and \(\overline{AD}\parallel\overline{BC}\).
Solution
\[ \begin{aligned} \color{magenta}{\text{Given:}}\;& \angle A=\angle B=\angle C=\angle D = 90^\circ \\ \color{magenta} \text{To prove} & : AB \parallel DC \text{ and } AD \parallel BC \\ \\\left. \begin{aligned} \angle 1 + \angle 4 &= 180^\circ\\ \angle 2 + \angle 3 &= 180^\circ \end{aligned} \right\}\ & \color{magenta}{(\text{co-interior angles are supplementary})} \\\therefore & \ \overline{AB}\parallel\overline{CD} \\ \\ \left. \begin{aligned} \angle 1 + \angle 2 &= 180^\circ\\ \angle 3 + \angle 4 &= 180^\circ \end{aligned} \right\}\ & \color{magenta}{(\text{co-interior angles are supplementary})} \\\therefore & \ \overline{AD}\parallel\overline{BC} \\ \\\end{aligned} \]Answer \( \overline{AB} \parallel \overline{CD} \text{ and } \overline{AD} \parallel \overline{BC} \)
6. In the given figure, if \(\angle BAO=\angle DCO\) and \(OC=OD\), show that \(\overline{AB}\parallel\overline{CD}\).
Solution
\[ \begin{aligned} \color{magenta}{\text{Given: }} & \angle BAO=\angle DCO \\ OC & = OD \\ \\ \color{magenta}{\text{To Prove: }} & \overline{AB}\parallel\overline{CD} \\ \\\text{Consider } & \triangle OCD, \text{ Its an isosceles} \triangle \\\angle DCO & = \angle CDO \ \color{magenta}(\text{Base angles are equal})\\ \\\Rightarrow \angle BAO &= \angle CDO \ \color{magenta}(\text{Since alternate interior angles are equal}) \\[6pt] \therefore & \ {AB} \parallel {CD}\end{aligned} \]Answer \( \overline{AB}\parallel\overline{CD} \)
7. In the given figure, \(\angle A=75^\circ\) and \(CE \parallel AB\). If \(\angle ECD=40^\circ\), find the other two angles of \(\triangle ABC\).
Solution
\[ \begin{aligned} \color{magenta}{\text{Given:}} & \ \angle A=75^\circ \\ CE & \parallel AB \\ \angle ECD & = 40^\circ\\[4pt] \color{magenta}{\text{To find:}} & \ \angle B \text{ and } \angle ACB \\ \\\angle ECD &= \angle ABC \ \color{magenta}{(\text{corresponding angles})}\\ \Rightarrow\ \angle B & = \color{green}{40^\circ}\\ \\ \angle A + \angle B+ \angle ACB &= 180^\circ \ \color{magenta}{(\text{Angle sum property})}\\ 75^\circ + 40^\circ + \angle ACB &= 180^\circ \\ 115^\circ + \angle ACB &= 180^\circ \\ \angle ACB &= 180^\circ - 115^\circ \\ \angle ACB &= \color{green} 65^\circ \\ \end{aligned} \]Answer \( \angle B= {\color{red}{40^\circ}},\ \angle ACB =\color{red}{65^\circ} \)
8. In the given figure, \(l \parallel m\). Find \(\angle x,\ \angle y\) and \(\angle z\).
Solution
\[ \begin{aligned} \color{magenta} \text{Given :} & \ {l \parallel m} \\[4pt] \color{magenta} \text{To find :} & \ \angle x , \angle y , \angle z \\ \\ \angle x &=\color{green}{110^\circ} \ \color{magenta}{(\text{corresponding angles})}\\[8pt] \angle x + \angle z &= 180^\circ \ \color{magenta}(\text{linear pair})\\ \angle z &= 180^\circ - 110^\circ \\ \angle z &=\color{green}{70^\circ}\\[8pt] 40^\circ + \angle 1 &= 180^\circ \ \color{magenta}(\text{linear pair})\\ \angle 1 &= 180^\circ - 40^\circ \\ \angle 1 &= 140^\circ \\ \\ \angle 1 &= \angle y \ \color{magenta}{(\text{corresponding angles})}\\[8pt] \angle y &=\color{green}{140^\circ}\\ \end{aligned} \]Answer \(\angle x= {\color{red}{110^\circ}},\ \angle y={\color{red}{140^\circ}},\ \angle z={\color{red}{70^\circ}}\)
9. In the given figure, show that: (i) \(\overline{AB}\parallel\overline{CD}\), (ii) \(\overline{CD}\parallel\overline{EF}\), (iii) \(\overline{AB}\parallel\overline{EF}\). Justify your answers.
Solution
\[ \begin{aligned} \textbf{(i) Show that: } & \ AB\parallel CD \\ \\ \angle ABC &= 40^\circ \\ \angle BCD &= 20^\circ + 20^\circ \\ \implies & \ 40^\circ \\ \\\angle ABC & = \angle BCD \ \color{magenta} ( \text{ Alternate interior angles} ) \\ \therefore \ \color{green} AB & \parallel \color{green} CD \\ \\\textbf{(ii) Show that: } & \ CD\parallel EF \\ \\ \angle CEF + \angle ECD & = 160^\circ + 20^\circ \\ & = 180^\circ \ \color{magenta}{(\text{co-interior angles are supplementary})} \\[6pt] \therefore \ \color{green} CD & \parallel \color{green} EF \\ \\\textbf{(iii) Show that: } & AB\parallel EF \\ \\ AB & \parallel CD \\ CD &\parallel EF \\ \\ \therefore \ \color{green} AB & \parallel \color{green} EF \\ \end{aligned} \]