RS AGGARWAL CLASS 9 CHAPTER 7 LINES AND ANGLES EXERCISE 7C

  EXERCISE 7C 

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Question 1:

In the given figure, l || m and a transversal t cuts them. If ∠1 = 120°, find the measure of each of the remaining marked angles.

Answer 1:

We have, $\angle 1=120°$. Then,
$$\begin{gathered} \angle 1=\angle 5 \left[\text{Corresponding angles}\right] \\ \Rightarrow \angle 5=120° \\ \angle 1=\angle 3 \left[\text{Vertically-opposite angles}\right] \\ \Rightarrow \angle 3=120° \end{gathered}$$

$$\begin{gathered} \angle 5=\angle 7 \left[\text{Vertically-opposite angles}\right] \\ \Rightarrow \angle 7=120° \\ \angle 1+\angle 2=180° \left[\text{Since AFB is a straight line}\right] \\ \Rightarrow 120°+\angle 2=180° \end{gathered}$$

$$\begin{gathered} \Rightarrow \angle 2=60° \\ \angle 2=\angle 4 \left[\text{Vertically-opposite angles}\right] \\ \Rightarrow \angle 4=60° \\ \angle 2=\angle 6 \left[\text{Corresponding angles}\right] \end{gathered}$$

$$\begin{gathered} \Rightarrow \angle 6 =60° \\ \angle 6=\angle 8 \left[\text{Vertically-opposite angles}\right] \\ \Rightarrow \angle 8=60° \\ \therefore \angle 1=120°\mathbf{,}\mathbf{ }\angle 2=60°, \angle 3=120°, \angle 4=60°, \angle 5=120°, \\ \angle 6 =60°, \angle 7=120° \text{and} \angle 8=60° \end{gathered}$$

Question 2:

In the given figure, l || m and a transversal t cuts them. If ∠7 = 80°, find the measure of each of the remaining marked angles.

Answer 2:

In the given figure, ∠7 and ∠8 form a linear pair.

∴ ∠7 + ∠8 = 180º

⇒ 80º + ∠8 = 180º

⇒ ∠8 = 180º − 80º = 100º

Now, 

∠6 = ∠8 = 100º       (Vertically opposite angles)

∠5 = ∠7 = 80º         (Vertically opposite angles)

It is given that, l || m and t is a transversal.

∴ ∠1 = ∠5 = 80º      (Pair of corresponding angles)

∠2 = ∠6 = 100º         (Pair of corresponding angles)

∠3 = ∠7 = 80º           (Pair of corresponding angles)

∠4 = ∠8 = 100º         (Pair of corresponding angles)

Question 3:

In the given figure, l || m and a transversal t cuts them. If ∠1 : ∠2 = 2 : 3, find the measure of each of the marked angles.

Answer 3:

Let ∠1 = 2k and ∠2 = 3k, where k is some constant.

Now, ∠1 and ∠2 form a linear pair.

∴ ∠1 + ∠2 = 180º

⇒ 2k + 3k = 180º

⇒ 5k = 180º

⇒ k = 36º

∴ ∠1 = 2k = 2 × 36º = 72º

∠2 = 3k = 3 × 36º = 108º

Now, 

∠3 = ∠1 = 72º         (Vertically opposite angles)

∠4 = ∠2 = 108º       (Vertically opposite angles)

It is given that, l || m and t is a transversal.

∴ ∠5 = ∠1 = 72º       (Pair of corresponding angles)

∠6 = ∠2 = 108º         (Pair of corresponding angles)

∠7 = ∠1 = 72º           (Pair of alternate exterior angles)

∠8 = ∠2 = 108º         (Pair of alternate exterior angles)

Question 4:

For what value of x will the line l and m be parallel to each other?

Answer 4:

For the lines l and m to be parallel
$$\begin{gathered} \iff 3x-20=2x+10 \left[\text{Corresponding Angles}\right] \\ \iff x=\mathbf{30} \end{gathered}$$
 

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Question 5:

For what value of x will the lines l and m be parallel to each other?

Answer 5:

$$\begin{gathered} \iff 3x+5+4x=180 \left[\text{Consecutive Interior Angles}\right] \\ \iff 7x=175 \\ \iff x=25 \end{gathered}$$

Question 6:

In the given figure, AB || CD and BC || ED. Find the value of x.

Answer 6:

$BC\parallel ED$ and CD is the transversal.
Then,

$$\begin{gathered} \angle BCD+\angle CDE=180° \left[\text{Angles on the same side of a transversal line are supplementary}\right] \\ \Rightarrow \angle BCD+75=180 \\ \Rightarrow \angle BCD=105° \end{gathered}$$

$AB\parallel CD$ and BC is the transversal.

$$\begin{gathered} \angle ABC=\angle BCD (\mathrm{alternate} \mathrm{angles}) \\ \Rightarrow x°=105° \\ \Rightarrow x=105 \end{gathered}$$

Question 7:

In the given figure, AB || CD || EF. Find the value of x.

Answer 7:

$EF\parallel CD$ and CE is the transversal.
Then,
$$\begin{gathered} \angle ECD+\angle CEF=180° \left[\text{Consecutive Interior Angles}\right] \\ \Rightarrow \angle ECD+130°=180° \\ \Rightarrow \angle ECD=50° \end{gathered}$$
Again, $AB\parallel CD$ and BC is the transversal.
Then,
$$\begin{gathered} \angle ABC=\angle BCD \left[\text{Alternate Interior Angles}\right] \\ \Rightarrow 70°=x+50° \left[\because \angle BCD=\angle BCE+\angle ECD\right] \\ \Rightarrow x=\mathbf{20}\mathbf{°} \end{gathered}$$

Question 8:

In the given figure, AB || CD. Find the values of x, y and z.

Answer 8:

$AB\parallel CD$ and let EF and EG be the transversals.
Now, $AB\parallel CD$  and EF is the transversal.
Then,
$$\begin{gathered} \angle AEF=\angle EFG \left[\text{Alternate Angles}\right] \\ \Rightarrow y°=75° \\ \Rightarrow y=75 \end{gathered}$$
Also,
$$\begin{gathered} \angle EFC+\angle EFD=180° \left[\text{Since CFD is a straight line}\right] \\ \Rightarrow x+y=180 \\ \Rightarrow x+75=180 \\ \Rightarrow x=105 \end{gathered}$$
And,
$$\begin{gathered} \angle EGF+\angle EGD=180° \left[\text{Since CFGD is a straight line}\right] \\ \Rightarrow \angle EGF+125=180 \\ \Rightarrow \angle EGF=55° \end{gathered}$$
We know that the sum of angles of a triangle is $180°$
$$\begin{gathered} \angle EFG+\angle GEF+\angle EGF=180° \\ \Rightarrow y+z+55=180 \\ \Rightarrow 75+z+55=180 \\ \Rightarrow z=50 \\ \therefore x=\mathbf{105}, y=\mathbf{75} \text{and }z=\mathbf{50} \end{gathered}$$

Question 9:

In each of the figures given below, AB || CD. Find the value of x in each case.

Answer 9:

(i)

Draw $EF\parallel AB\parallel CD$.
Now, $AB\parallel EF$ and BE is the transversal.
Then,
$$\begin{gathered} \angle ABE=\angle BEF \left[\text{Alternate Interior Angles}\right] \\ \Rightarrow \angle BEF=35° \end{gathered}$$
Again, $EF\parallel CD$ and DE is the transversal.
Then,
$$\begin{gathered} \angle DEF=\angle FED \\ \Rightarrow \angle FED=65° \\ \therefore x°=\angle BEF+\angle FED \\ =\left(35+65\right)° \\ =100° \\ \mathrm{or}, x=100 \end{gathered}$$

(ii)


Draw $EO\parallel AB\parallel CD$.
Then, $\angle EOB+\angle EOD=x°$
Now, $EO\parallel AB$ and BO is the transversal.
$$\begin{gathered} \therefore \angle EOB+\angle ABO=180° \left[\text{Consecutive Interior Angles}\right] \\ \Rightarrow \angle EOB+55°=180° \\ \Rightarrow \angle EOB=125° \end{gathered}$$
Again, $EO\parallel CD$ and DO is the transversal.
$$\begin{gathered} \therefore \angle EOD+\angle CDO=180° \left[\text{Consecutive Interior Angles}\right] \\ \Rightarrow \angle EOD+25°=180° \\ \Rightarrow \angle EOD=155° \end{gathered}$$
Therefore,
$$\begin{gathered} x°=\angle EOB+\angle EOD \\ =\left(125+155\right)° \\ =280° \\ \mathrm{or}, x=280 \end{gathered}$$

(iii)

Draw $EF\parallel AB\parallel CD$.
Then, $\angle AEF+\angle CEF=x°$
Now, $EF\parallel AB$ and AE is the transversal.
$$\begin{gathered} \therefore \angle AEF+\angle BAE=180° \left[\text{Consecutive Interior Angles}\right] \\ \Rightarrow \angle AEF+116=180 \\ \Rightarrow \angle AEF=64° \end{gathered}$$
Again, $EF\parallel CD$ and CE is the transversal.
$$\begin{gathered} \angle CEF+\angle ECD=180° \left[\text{Consecutive Interior Angles}\right] \\ \Rightarrow \angle CEF+124=180 \\ \Rightarrow \angle CEF=56° \end{gathered}$$

Therefore,
$$\begin{gathered} x°=\angle AEF+\angle CEF \\ =\left(64+56\right)° \\ =120° \\ \mathrm{or}, x=120 \end{gathered}$$

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Question 10:

In the given figures, AB || CD. Find the value of x.

Answer 10:

Draw $EF\parallel AB\parallel CD$.
$EF\parallel CD$ and CE is the transversal.
Then,
$$\begin{gathered} \angle ECD+\angle CEF=180° \left[\text{Angles on the same side of a transversal line are supplementary}\right] \\ \Rightarrow 130°+\angle CEF=180° \\ \Rightarrow \angle CEF=50° \end{gathered}$$
Again, $EF\parallel AB$ and AE is the transversal.
Then,
$$\begin{gathered} \angle BAE+\angle AEF=180° \left[\text{Angles on the same side of a transversal line are supplementary}\right] \\ \Rightarrow x°+20°+50°=180° \left[\angle AEF=\angle AEC+\angle CEF\right] \\ \Rightarrow x°+70°=180° \\ \Rightarrow x°=110° \\ \Rightarrow x=110 \end{gathered}$$

Question 11:

In the given figure, AB || PQ. Find the values of x and y.

Answer 11:

Given, $AB\parallel PQ$.
Let CD be the transversal cutting AB and PQ at E and F, respectively.
Then,
$$\begin{gathered} \angle CEB+\angle BEG+\angle GEF=180° \left[\text{Since CD is a straight line}\right] \\ \Rightarrow 75°+20°+\angle GEF=180° \\ \Rightarrow \angle GEF=85° \end{gathered}$$
We know that the sum of angles of a triangle is $180°$.
$$\begin{gathered} \therefore \angle GEF+\angle EGF+\angle EFG=180 \\ \Rightarrow 85°+x+25°=180° \\ \Rightarrow 110°+x=180° \\ \Rightarrow x=70° \end{gathered}$$
And
$$\begin{gathered} \angle FEG+\angle BEG=\angle DFQ \left[\text{Corresponding Angles}\right] \\ \Rightarrow 85°+20°=\angle DFQ \\ \Rightarrow \angle DFQ=105° \\ \angle EFG+\angle GFQ+\angle DFQ=180° \left[\text{Since CD is a straight line}\right] \\ \Rightarrow 25°+y+105°=180° \\ \Rightarrow y=50° \\ \therefore x=70° \text{and} y=50° \end{gathered}$$

Question 12:

In the given figure, AB || CD. Find the value of x.

Answer 12:

$AB\parallel CD$ and AC is the transversal.
Then,
$$\begin{gathered} \angle BAC+\angle ACD=180° \left[\text{Consecutive Interior Angles}\right] \\ \Rightarrow 75+\angle ACD=180 \\ \Rightarrow \angle ACD=105° \end{gathered}$$
And,
$$\begin{gathered} \angle ACD=\angle ECF \left[\text{Vertically-Opposite Angles}\right] \\ \Rightarrow \angle ECF=105° \end{gathered}$$
We know that the sum of the angles of a triangle is $180°$.
$$\begin{gathered} \angle ECF+\angle CFE+\angle CEF=180° \\ \Rightarrow 105°+30°+x=180° \\ \Rightarrow 135°+x=180° \\ \Rightarrow x=45° \end{gathered}$$

Question 13:

In the given figure, AB || CD. Find the value of x.

Answer 13:

$AB\parallel CD$ and PQ is the transversal.
Then,
$$\begin{gathered} \angle PEF=\angle EGH \left[\text{Corresponding Angles}\right] \\ \Rightarrow \angle EGH=85° \end{gathered}$$
And,
$$\begin{gathered} \angle EGH+\angle QGH=180° \left[\text{Since PQ is a straight line}\right] \\ \Rightarrow 85°+\angle QGH=180° \\ \Rightarrow \angle QGH=95° \end{gathered}$$
Also,
$$\begin{gathered} \angle CHQ+\angle GHQ=180° \left[\text{Since CD is a straight line}\right] \\ \Rightarrow 115°+\angle GHQ=180° \\ \Rightarrow \angle GHQ=65° \end{gathered}$$
We know that the sum of angles of a triangle is $180°$.
$$\begin{gathered} \Rightarrow \angle QGH+\angle GHQ+\angle GQH=180° \\ \Rightarrow 95°+65°+x=180° \\ \Rightarrow x=20° \\ \therefore x=20° \end{gathered}$$

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Question 14:

In the given figure, AB || CD. Find the values of x, y and z.

Answer 14:

$$\begin{gathered} \angle ADC=\angle DAB \left[\text{Alternate Interior Angles}\right] \\ \Rightarrow z=75° \\ \angle ABC=\angle BCD \left[\text{Alternate Interior Angles}\right] \\ \Rightarrow x=35° \end{gathered}$$
We know that the sum of the angles of a triangle is $180°$.
$$\begin{gathered} \Rightarrow 35°+y+75°=180° \\ \Rightarrow y=70° \\ \therefore x=35°, y=70° \text{and }z=75°. \end{gathered}$$

Question 15:

In the given figure, AB || CD. Prove that ∠BAE − ∠DCE = ∠AEC.

Answer 15:

Draw $EF\parallel AB\parallel CD$ through E.
Now, $EF\parallel AB$ and AE is the transversal.
Then, $\angle BAE+\angle AEF=180° \left[\text{Angles on the same side of a transversal line are supplementary}\right]$
Again, $EF\parallel CD$ and CE is the transversal.
Then,
$$\begin{gathered} \angle DCE+\angle CEF=180° \left[\text{Angles on the same side of a transversal line are supplementary}\right] \\ \Rightarrow \angle DCE+\left(\angle AEC+\angle AEF\right)=180° \\ \Rightarrow \angle DCE+\angle AEC+180°-\angle BAE=180° \\ \Rightarrow \angle BAE-\angle DCE=\angle AEC \end{gathered}$$

Question 16:

In the given figure, AB || CD. Prove that p + q − r = 180.

Answer 16:

Draw $PFQ\parallel AB\parallel CD$.
Now, $PFQ\parallel AB$ and EF is the transversal.
Then,
$$\begin{gathered} \angle AEF+\angle EFP=180°.....\left(1\right) \\ \left[\text{Angles on the same side of a transversal line are supplementary}\right] \end{gathered}$$
Also, $PFQ\parallel CD$.

$$\begin{gathered} \angle PFG=\angle FGD=r°\left[\text{Alternate Angles}\right] \\ \mathrm{and} \angle EFP=\angle EFG-\angle PFG=q°-r° \\ \mathrm{putting} \mathrm{the} \mathrm{value} \mathrm{of} \angle EFP \mathrm{in} \mathrm{eqn}. \left(\mathrm{i}\right) \\ \mathrm{we} \mathrm{get}, \\ p°+q°-r°=180° \\ \Rightarrow p+q-r=180 \end{gathered}$$

Question 17:

In the given figure, AB || CD and EF || GH. Find the values of x, y, z and t.

Answer 17:

In the given figure,
$$\begin{gathered} x=60° \left[\text{Vertically-Opposite Angles}\right] \\ \angle PRQ=\angle SQR \left[\text{Alternate Angles}\right] \\ y=60° \\ \angle APR=\angle PQS \left[\text{Corresponding Angles}\right] \\ \Rightarrow 110°=\angle PQR+60° \left[\because \angle PQS=\angle PQR+\angle RQS\right] \\ \Rightarrow \angle PQR=50° \end{gathered}$$
$$\begin{gathered} \angle PQR+\angle RQS+\angle BQS=180° \left[\text{Since AB is a straight line}\right] \\ \Rightarrow 50°+60°+z=180° \\ \Rightarrow 110°+z=180° \\ \Rightarrow z=70° \end{gathered}$$
$$\begin{gathered} \angle DSH=z \left[\text{Corresponding Angles}\right] \\ \Rightarrow \angle DSH=70° \\ \therefore \angle DSH=t \left[\text{Vertically-Opposite Angles}\right] \\ \Rightarrow t=70° \\ \therefore x=60°, y=60°, z=70° \text{and} t=70°. \end{gathered}$$

Question 18:

In the given figure, AB || CD and a transversal t cuts them at E and F respectively. If EG and FG are the bisectors of ∠BEF and ∠EFD respectively, prove that ∠EGF = 90°.

Answer 18:

It is given that, AB || CD and t is a transversal.

∴ ∠BEF + ∠EFD = 180°     .....(1)     (Sum of the interior angles on the same side of a transversal is supplementary)

EG is the bisector of ∠BEF.    (Given)

∴ ∠BEG = ∠GEF = $\frac{1}{2}$∠BEF

⇒ ∠BEF = 2∠GEF                .....(2)

Also, FG is the bisector of ∠EFD.    (Given)

∴ ∠EFG = ∠GFD = $\frac{1}{2}$∠EFD

⇒ ∠EFD = 2∠EFG                .....(3)

From (1), (2) and (3), we have

2∠GEF + 2∠EFG = 180°

⇒ 2(∠GEF + ∠EFG) = 180°

⇒ ∠GEF + ∠EFG = 90°            .....(4)

In ∆EFG,

∠GEF + ∠EFG + ∠EGF = 180°          (Angle sum property)

⇒ 90° + ∠EGF = 180°                         [Using (4)]

⇒ ∠EGF = 180° − 90° = 90°

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Question 19:

In the given figure, AB || CD and a transversal t cuts them at E and F respectively. If EP and FQ are the bisectors of ∠AEF and ∠EFD respectively, prove that EP || FQ .



 

Answer 19:

It is given that, AB || CD and t is a transversal.

∴ ∠AEF = ∠EFD           .....(1)         (Pair of alternate interior angles)

EP is the bisectors of ∠AEF.        (Given)

∴ ∠AEP = ∠FEP = $\frac{1}{2}$∠AEF

⇒ ∠AEF = 2∠FEP          .....(2)

Also, FQ is the bisectors of ∠EFD.

∴ ∠EFQ = ∠QFD = $\frac{1}{2}$∠EFD

⇒ ∠EFD = 2∠EFQ         .....(3)

From (1), (2) and (3), we have

2∠FEP = 2∠EFQ

⇒ ∠FEP = ∠EFQ

Thus, the lines EP and FQ are intersected by a transversal EF such that the pair of alternate interior angles formed are equal. 

∴ EP || FQ        (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel)

Question 20:

In the given figure, BA || ED and BC || EF. Show that ∠ABC = ∠DEF.

Answer 20:

It is given that, BA || ED and BC || EF.

Construction: Extend DE such that it intersects BC at J. Also, extend FE such that it intersects AB at H.



Now, BA || JD and BC is a transversal.

∴ ∠ABC = ∠DJC      .....(1)       (Pair of corresponding angles)

Also, BC || HF and DJ is a transversal.

∴ ∠DJC = ∠DEF      .....(2)       (Pair of corresponding angles)

From (1) and (2), we have

∠ABC = ∠DEF

Question 21:

In the given figure, BA || ED and BC || EF. Show that ∠ABC + ∠DEF = 180°.

Answer 21:

It is given that, BA || ED and BC || EF.

Construction: Extend ED such that it intersects BC at G. 



Now, BA || GE and BC is a transversal.

∴ ∠ABC = ∠EGC      .....(1)       (Pair of corresponding angles)

Also, BC || EF and EG is a transversal.

∴ ∠EGC + ∠GEF = 180°      .....(2)       (Interior angles on the same side of the transversal are supplementary)

From (1) and (2), we have

​∠ABC + ∠GEF = 180°        

Or ∠ABC + ∠DEF = 180°          

Question 22:

In the given figure, m and n are two plane mirrors perpendicular to each other. Show that the incident ray CA is parallel to the reflected ray BD.

Answer 22:

AP is normal to the plane mirror OA and BP is normal to the plane mirror OB.

It is given that the two plane mirrors are perpendicular to each other.

Therefore, BP || OA and AP || OB.

So, BP ⊥ AP       (OA ⊥ OB)

⇒ ∠APB = 90°    .....(1)

In ∆APB,

​∠2 + ∠3 + ∠APB = 180°      (Angle sum property)

∴ ∠2 + ∠3 + 90° = 180°       [Using (1)]

⇒ ∠2 + ∠3 = 180° − 90° = 90°

⇒ 2∠2 + 2∠3 = 2 × 90° = 180°        .....(2)

By law of reflection, we have

∠1 = ∠2  and ∠3 = ∠4                     .....(3)       (Angle of incidence = Angle of reflection)   

From (2) and (3), we have

∠1 + ∠2 + ∠3 + ∠4 = 180°

⇒ ∠BAC + ∠ABD = 180°            (∠1 + ∠2 = ∠BAC and ∠3 + ∠4 = ∠ABD)

Thus, the lines CA and BD are intersected by a transversal AB such that the interior angles on the same side of the transversal are supplementary.

∴ CA || BD    

Question 23:

In the figure given below, state which lines are parallel and why?

Answer 23:

Here, ∠BAC = ∠ACD = 110°

Thus, lines AB aand CD are intersected by a transversal AC such that the pair of alternate angles are equal.

∴ AB || CD     (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel)

Thus, line AB is parallel to line CD.

Also, ∠ACD + ∠CDE = 110° + 80° = 190° ≠ 180°

If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal are supplementary, then the two lines are parallel.

Therefore, line AC is not parallel to line DE.      

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Question 24:

Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.

Answer 24:

Let the two lines m and n be respectively perpendicular to the two parallel lines p and q.
To prove: m is parallel to n.
Proof: Since, m is perpendicular to p
$\therefore$$\angle 1=90°$
Also, ​n is perpendicular to q 
$\therefore$$\angle 3=90°$
Since p and q are parallel and m is a transversal line 
$\therefore$ $\angle 2=\angle 1=90°$              [Corresponding angles]
Also, $\angle 2=\angle 3=90°$
We know that if two corresponding angles are equal then the two lines containing them must be parallel.
Therefore, the lines m and n are parallel to each other.