## RS Aggarwal Class 9 Chapter 4 – Lines And Triangles Ex 4A Solutions Free PDF

The RS Aggarwal Class 9 Solutions Chapter 4 Angles, Lines and Triangles help you strengthen your concepts and revise important topics thoroughly so that you can prepare for the exam in a desirable manner and ensure to score good marks in the exam. It comprises a great number of questions to practice which will easily help you to clear your concepts. Students should refer to these solutions while preparing for their exam as it covers the entire syllabus updated as per the CBSE. The solutions are prepared by our subject experts in every simple language.

The RS Aggarwal Class 9 Solutions will help students to solve the questions with ease and perform well in the exam. All the solutions are explained in a pretty detailed way which promotes easy learning and understanding of the concepts. By solving these solutions students will gain the confidence to solve questions asked from any topic in the exam.

## Download PDF of RS Aggarwal Class 9 Solutions Chapter 4â€“ Angles, Lines and Triangles Ex 4A

**Question 1: In the given figure, AB is a mirror. PQ is the incident ray and QR, the reflected ray. If \(\angle PQR=112^{\circ}\), find \(\angle PQA\).**

**Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **

Ans:

We know that the angle of incidence = angle of reflection.

Hence, let \(\angle PQA = \angle BQR = x^{o}\)

Since, AQB is a straight line, we have

Therefore, \( \angle PQA + \angle PQR + \angle BQR = 180^{o}\)

x + 112 + x = 180^{o}

2x = 68

X = 34^{o}

Therefore, \( \angle PQA = 34^{o}\)

**Question 2: If two straight lines intersect each other than prove that ray opposite to the bisector of one of the angles so formed bisect the vertically opposite angles.**

Ans:

Let AB and CD be the two lines intersecting at a point O and let ray OE bisect \(\angle AOC\) . Now draw a ray OF in the opposite direction of OE, such that EOF is a straight line.

Let \(\angle COE = 1, \angle AOE = 2, \angle BOF = 3\ and\ \angle DOF = 4\)

We know that vertically opposite angles are equal.

Therefore, \( \angle 1 = \angle 4\ and\ \angle 2 = \angle 3\)

But, \(\angle 1 = \angle 2\) [Since OE bisects \(\angle AOC\) ]

Therefore, \( \angle 4 = \angle 3\)

Hence, the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.

**Question 3: Prove that the bisector of two adjacent supplementary angles include a right angle.**

Ans: Let AOB denote a straight line and let \(\angle AOC\ and\ \angle BOC\) be the supplementary angles.

Thus, we have:

\(\angle AOC = x^{o}\) and \(\angle BOC = (180 -x)^{o}\)

Let OE bisects \(\angle AOC\) and OF bisect \(\angle BOC\)

Then, we have:

\(\angle AOE = \angle COE = \frac{1}{2} x^{o}\) and

\(\angle BOF = \angle FOC = \frac{1}{2}(180 – x)^{o}\)

Therefore,

\(\angle COE + \angle FOC = \frac{1}{2} x + \frac{1}{2} (180 – x)^{o}\)

\( = \frac{1}{2}(x + 180 – x)\)

\(= \frac{1}{2}(180^{o})\)

= 90^{o}

**Q4) In the adjoining figure \(AB\left | \right |CD\) are cut by a traversal t at E and F respectively. If L1 = \(70^{\circ}\) , find the measure of each of the remaining marked angles.**

**Ans.**

We have, \(\angle 1=70^{\circ}\) . Then,

\(\angle 1=\angle 5\) [Corresponding angle ]

Therefore, \(\angle 5=70^{\circ}\)

\(\angle 1=\angle 3\) [Vertically-opposite angles]

Therefore, \(\angle 3=70^{\circ}\)

\(\angle 5=\angle 7\) [Vertically-opposite angles]

Therefore, \(\angle 7=70^{\circ}\)

\(\angle 1+\angle 2=180^{\circ}\) [Since AFB is a straight line ]

Therefore, \(70^{\circ}+\angle 2=180^{\circ}\)

Ã \(\angle 2=110^{\circ}\) [Vertically-opposite angles]

Ã \(\angle 4=110^{\circ}\)

\(\angle 2=\angle 6\) [Corresponding angles]

Ã \(\angle 6=110^{\circ}\)

\(\angle 6=\angle 8\) [Vertically-opposite angles]

Ã \(\angle 8=110^{\circ}\)

Therefore, \(\angle 1=70^{\circ}\) , \(\angle 2=110^{\circ}\) , \(\angle 3=70^{\circ}\) , \(\angle 4=110^{\circ}\) , \(\angle 5=70^{\circ}\) , \(\angle 6=110^{\circ}\) , \(\angle 7=70^{\circ}\) and \(\angle 8=110^{\circ}\) .

**Q5) In the adjoining figure, \(AB\left | \right | CD\) are cut by a transversal t at E and F respectively. If L2:L1 = 5:4 , find the measure of each one of the marked angles.**

**Given \(AB\left | \right | CD\) and a line t intersects them at E and F forming angles**

**L1 + L2 + L3 + L4 + L5 + L6 + L7 + L8**

**Ans. **

Given, L2:L1 = 5:4

Let L2 = 5y and L1 = 4y

But L2 + L1 = \(180^{\circ}\) [Linear pair]

Ã 5y + 4y = \(180^{\circ}\)

Ã Y = \(\frac{180^{\circ}}{9}=20^{\circ}\)

Therefore, L2 = 5y = 5 x \(20^{\circ}\) = \(100^{\circ}\)

And L1 = 4y = 4 x \(20^{\circ}\) = \(80^{\circ}\)

But L1 = L3 [vertically opp. Angles]

Therefore, L3 = \(80^{\circ}\)

Similarly, Since L2 = L4 [vertically opp. Angles]

Therefore, L4 = \(100^{\circ}\)

Since, L1 = L5 [corresponding angles]

Therefore, L5 = \(80^{\circ}\)

Since, L4 = L6 [Alternate angles]

Therefore, L6 = \(100^{\circ}\)

Since, L3 = L7 [Corresponding angles]

Therefore, L7 = \(80^{\circ}\)

Since, L4 = L8 [Corresponding angles]

Therefore, L8 = \(100^{\circ}\)

Hence, L3 = \(80^{\circ}\) ,L4 = \(100^{\circ}\) , L5 = \(80^{\circ}\) , L6 = \(100^{\circ}\) , L7 = \(80^{\circ}\) , L8 = \(100^{\circ}\) .

**Q6) In the adjoining fig. ABCD is a quadrilateral in which \(AB\left | \right |DC\) and \(AD\left | \right |BC\) .Prove that \(\angle ADC=\angle ABC\) .**

**Ans. **

Let \(AD\parallel BC\) and CD is the transversal. Then,

\(\angle ADC+\angle DCB=180^{\circ}\) Â â€¦(i) [Consecutive Interior angles]

Also,

\(AB\parallel CD\) and BC is the transversal. Then,

\(\angle DCB+\angle ABC=180^{\circ}\) Â Â â€¦(ii) [Consecutive Interior angles]

From (i) and (ii), we get:

\(\angle ADC+\angle DCB=\angle DCB+\angle ABC\)

Ã \(\angle ADC=\angle ABC\)

**Q7) In each of the figure find the angle \(AB\parallel CD\). Find the value of x in each case.**

**(i)**

**Ans. (i)**

In the fig. \(AB\parallel CD\parallel EF\)

Now, \(AB\parallel EF\) and BE is the transversal. Then,

\(\angle ABE=\angle BEF\) [Alternate interior angles]

Ã \(\angle BEF=35^{\circ}\)

Again, \(EF\parallel CD\) and DE is the transversal.

Then,

\(\angle DEF=\angle FED\)

Ã \(\angle FED=65^{\circ}\)

Therefore, \(x^{\circ}=\angle BEF+\angle FED\)

= \((35+65)^{\circ}\)

= \(100^{\circ}\)

Or, x = 100

**(ii)**

Draw \(EO\parallel AB\parallel CD\)

Then, \(\angle EOB+\angle EOD=x^{\circ}\)

Now, \(EO\parallel AB\) and BO is the transversal.

Therefore, \(\angle EOB+\angle ABO=180^{\circ}\) [Consecutive Interior angles]

Ã \(\angle EOB+55^{\circ}=180^{\circ}\)

Ã \(\angle EOB=155^{\circ}\)

Therefore,

\(x^{\circ}=\angle EOB+\angle EOD\)

= \((125+155)^{\circ}\)

= \(280^{\circ}\)

Or, x = 280

**(iii)**

Draw \(EF\parallel AB\parallel CD\) .

Then, \(\angle AEF+\angle CEF=x^{\circ}\)

Now, \(EF\parallel AB\) and AE is the transversal.

Therefore, \(\angle AEF+\angle BAE=180^{\circ}\) [Consecutive interior angles]

Ã \(\angle AEF+116=180\)

Ã \(\angle AEF=64^{\circ} \)

Again, \(EF\parallel CD\) and CE is the transversal.

\(\angle CEF+\angle ECD=180^{\circ}\) [Consecutive Interior angles]

Ã \(\angle CEF+124=180\)

Ã \(\angle CEF=56^{\circ}\)

Therefore,

\(x^{\circ}=\angle AEF+\angle CEF\)

= \((64+56)^{\circ}\)

= \(120^{\circ}\)

Or, x = 120

**Q8) In the given figure, \(AB\parallel CD\parallel EF\). Find the value of x.**

**Ans. **

Given, \(EF\parallel CD\) and CE is the transversal.

Then,

\(\angle ECD+\angle CEF=180^{\circ}\) [Consecutive Interior angles]

Ã \(\angle ECD+130^{\circ}=180^{\circ}\)

Ã \(\angle ECD=50^{\circ}\)

Again, \(AB\parallel CD\) and BC is the transversal.

Then,

\(\angle ABC=\angle BCD\) [Alternate Interior Angles]

Ã \(70^{\circ}=x+50^{\circ}\;therefore [\angle BCD=\angle BCE+\angle ECD]\)

Ã \(x=20^{\circ}\)

**Q9) In the given figure , \(AB\parallel CD\) . Find the value of x.**

**Ans. **

Draw \(EF\parallel AB\parallel CD\)

\(EF\parallel CD\) and CE is the transversal.

Then,

\(\angle ECD+\angle CEF=180^{\circ}\) [Anfles on the same side of a transversal are supplementary]

Ã \(130^{\circ}+\angle CEF=180^{\circ}\)

Ã \(\angle CEF=50^{\circ}\)

Again, \(EF\parallel AB\) and AE is the transversal.

Then,

\(\angle BAE+\angle AEF=180^{\circ}\) [Angles on the same side of a transversal line are supplementary]

Ã \(x^{\circ}+20^{\circ}+50^{\circ}=180^{\circ}\;[\angle AEF=\angle AEC+\angle CEF]\)

Ã \(x^{\circ}+70^{\circ}=180^{\circ}\)

Ã \(x^{\circ}=110^{\circ}\)

Ã x = 110

**Q10) In the given figure, \(AB\parallel CD\) , Prove that**

**\(\angle BAE-\angle DCE=\angle AEC\)**

**Ans. **

Draw \(EF\parallel AB\parallel CD\) through E.

Now, \(EF\parallel AB\) and AE is the transversal.

Then,

\(\angle BAE+\angle AEF=180^{\circ}\) [Angles on the same side of a transversal line are supplementary]

Again, \(EF\parallel CD\) and CE is the transversal.

Then,

\(\angle DCE+\angle CEF=180^{\circ}\) [Angles on the same side of a transversal line are supplementary]

Ã \(\angle DCE+(\angle AEC+\angle AEF)=180^{\circ}\)

Ã \(\angle DCE+\angle AEC+180^{\circ}-\angle BAE=180^{\circ}\)

Ã \(\angle BAE-\angle DCE=\angle AEC\)

**Q11) In the given figure, \(AB\parallel CD\; and \; CD\parallel EF\). Find the value of x.**

**Ans. **

We have, \(AB\parallel CD\) and \(BC\parallel ED\).

\(BD\parallel ED\) and CD is the transversal.

Then,

\(\angle BCD+\angle CDE=180^{\circ}\) [Angles on the same side of a transversal line are supplementary]

Ã \(\angle BCD+75=180\)

Ã \(\angle BCD=105^{\circ}\)

\(AB\parallel CD\) and BC is the transversal.

\(\angle ABC=\angle BCD\) (alternate angles)

Ã \(x^{\circ}=105^{\circ}\)

Ã x = 105

**Q12) In the given figure , \(AB\parallel CD\) .Prove that P + q â€“ r = Â \(180^{\circ}\) .**

**Ans. **

Draw \(PFQ\parallel AB\parallel CD\)

Now, \(PFQ\parallel AB\) and EF is the transversal.

Then,

\(\angle AEF+\angle EFP=180^{\circ}\) Â Â Â â€¦(1)

[Angles on the same side of a transversal line are supplementary]Also, \(PFQ\parallel CD\)

\(\angle PFQ=\angle FGD=r^{\circ}\) [Alternate angles]

And \(\angle EFP=\angle EFG-\angle PFG=q^{\circ}-r^{\circ}\)

Putting the value of \(\angle EFP\) in equ. (i)

We get,

\(p^{\circ}+q^{\circ}-r^{\circ}=180^{\circ}\)

Ã p + q â€“ r = 180

**Q13) In the given figure, \(AB\parallel PQ\). Find the value of x and y.**

**Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **

**Ans. **

Given \(AB\parallel PQ\)

Let CD be the transversal cutting AB and PQ at E and F, respectively.

Then,

\(\angle CEB+\angle BEG+\angle GEF=180^{\circ}\) [Since CD is a straight line]

Ã \(75^{\circ}+20^{\circ}+\angle GEF=180^{\circ}\)

Ã \(\angle GEF=85^{\circ}\)

We know that the sum of angles of a triangle is \(180^{\circ}\)

therefore, \( \angle GEF+\angle EGF+\angle EFG=180\)

Ã \(85^{\circ}+x+25^{\circ}=180^{\circ}\)

Ã \(110^{\circ}+x=180^{\circ}\)

Ã x = \(70^{\circ}\)

And

\(\angle FEG+\angle BEG=\angle DFQ\) [Corresponding angles]

Ã \(85^{\circ}+20^{\circ}=\angle DFQ\)

Ã \(\angle DFQ=105^{\circ}\)

\(\angle EFG+\angle GFQ+\angle DFQ=180^{\circ}\) [Since CD is a straight line]

Ã \(25^{\circ}+y+105^{\circ}=180^{\circ}\)

Ã \(y=50^{\circ}\)

therefore, \( x=70^{\circ}\) and \(y=50^{\circ}\)

**Q14) In the given figure, \(AB\parallel CD\). Find the value of x.**

**Ans. **

\(AB\parallel CD\) and AC is the transversal.

Then,

\(\angle BAC+\angle ACD=180^{\circ}\) [Consecutive Interior angles]

Ã 75 Â + \(\angle ACD=180\)

Ã \(\angle ACD=105^{\circ}\)

And,

\(\angle ACD=\angle ECF\) [Vertically â€“opposite angles]

Ã \(\angle ECF=105^{\circ}\)

We know that the sum of the angles of a triangle is \(180^{\circ}\)

\(\angle ECF+\angle CFE+\angle CEF=180^{\circ}\)

Ã \(105^{\circ}+30^{\circ}+x=180^{\circ}\)

Ã \(135^{\circ}+x=180^{\circ}\)

Ã \(x=45^{\circ}\)

**Q15) In the given figure, \(AB\parallel CD\). Find the value of x.**

**Ans. **

\(AB\parallel CD\) and PQ is the transversal.

Then,

\(\angle PEF=\angle EGH\) [Corresponding Angles]

Ã \(\angle EGH=85^{\circ}\)

And,

\(\angle EGH+\angle QGH=180^{\circ}\) [Since PQ is a straight line]

Ã \(85^{\circ}+\angle QGH=180^{\circ}\)

Ã \(\angle QGH=95^{\circ}\)

Also,

\(\angle CHQ+\angle GHQ=180^{\circ}\) [Since CD is a straight line]

Ã \(115^{\circ}+\angle GHQ=180^{\circ}\)

Ã \(\angle GHQ=65^{\circ}\)

We know that the sum of angles of a triangle is \(180^{\circ}\)

Ã \(\angle QGH+\angle GHQ+\angle GQH=180^{\circ}\)

Ã \(95^{\circ}+65^{\circ}+x=180^{\circ}\)

Ã x = \(20^{\circ}\)

**Q16) In the given figure, \(AB\parallel CD\). Find the value of x,y and z.**

**Ans. **

\(\angle ADC= \angle DAB\) [Alternate interior angles]

Ã z = \(75^{\circ}\)

\(\angle ABC=\angle BCD\) [Alternate Interior Angles]

Ã x = \(75^{\circ}\)

We know that the sum of the angles of triangle is \(180^{\circ}\)

Ã \(35^{\circ}+y+75^{\circ}=180^{\circ}\)

Ã y = \(70^{\circ}\)

Therefore, \(x=35^{\circ},y=70^{\circ}\;and\;z=75^{\circ}\)

**Q17) In the given figure, \(AB\parallel CD\). Find the value of x,y and z.**

**Ans. **

\(AB\parallel CD\) and let EF and EG be the transversals.

Now, \(AB\parallel CD\) and EF is the transversal.

Then,

\(\angle AEF=\angle EFG\) [Alternate angles]

Ã \(y^{\circ}=75^{\circ}\)

Ã y = 75

Also,

\(\angle EFC+\angle EFD=180^{\circ}\) [Since CFGD is a straight line]

Ã x + y =180

Ã x + 75 =180

Ã x = 105

And,

\(\angle EGF+\angle EGD=180^{\circ}\) [Since CFGD is a straight line]

Ã \(\angle EGF+125=180\)

Ã \(\angle EGF=55^{\circ}\)

We know that the sum of angles of a triangle is \(180^{\circ}\)

\(\angle EFG+\angle GEF+\angle EGF=180^{\circ}\)

Ã \(y+z+55=180\)

Ã 75 + z + 55 = 180

Ã z = 50

Therefore, x = 105, y = 75 and z = 50

**Q18) In the given figure, \(AB\parallel CD\; and \; EF\parallel GH\). Find the value of x,y,z and t.**

**Ans. **

In the given figure,

x = \(60^{\circ}\) [Vertically-opposite Angles]

\(\angle PRQ=\angle SQR\) [Alternate angles]

y = \(60^{\circ}\)

\(\angle APR=\angle PQS\) [Corresponding Angles]

Ã \(110^{\circ}=\angle PQR+60^{\circ}\;because [\angle PQS=\angle PQR+\angle RQS]\)

Ã \(\angle PQR=50^{\circ}\)

\(\angle PQR+\angle RQS+\angle BQS=180^{\circ}\) [Since AB is straight line]

Ã \(50^{\circ}+60^{\circ}+z=180^{\circ}\)

Ã \(110^{\circ}+z=180^{\circ}\)

Ã \(z=70^{\circ}\)

\(\angle DSH=z\) [Corresponding Angles]

Ã \(\angle DSH=70^{\circ}\)

therefore, \( \angle DSH=t\) [Vertically-opposite Angles]

Ã t = \(70^{\circ}\)

Therefore,\(;x=60^{\circ},z=70^{\circ}\;and\;t=70^{\circ}\)

**Q19) For what value of x will the lines l and m be parallel to each other?**

**Â Â Â Â Â Â Â Â Â Â **

**Ans. **

For the lines l and m to be parallel

**(i)**

\(\Leftrightarrow\) 3x â€“ 20 = 2x +10 Â [Corresponding angles]

\(\Leftrightarrow\) x = 30

(ii)

\(\Leftrightarrow\) 3x + 5 + 4x = 180 [Consecutive Interior Angles]

\(\Leftrightarrow\) 7x = 175

\(\Leftrightarrow\) x = 25

**Q20) If two straight lines are perpendicular to the same line, prove that the lines are parallel to each other.**

**Ans:**

**Given: **Two lines m and n are perpendicular to a given line l.

**To Prove : **\(m\parallel n\)

Proof : Since \(m\perp l\)

So, \(\angle 1=90^{\circ}\)

Again, Since \(n\perp l\)

\(\angle 2=90^{\circ}\)

therefore,\( \angle 1=\angle 2=90^{\circ}\)

But \(\angle 1\) and \(\angle 2\) are the corresponding angles made by the transversal l with lines m and n and they are proved to be equal.

Thus, \(m\parallel n\)‘

### Key Features of RS Aggarwal Class 9 Solutions Chapter 4â€“ Angles, Lines And Triangles Ex 4A

- It provides ample techniques to solve difficult and tricky questions.
- It is an extremely vital resource for revision purpose.
- It acts as a guide for the students while preparing for their exam.
- It improves your performance and problem-solving skills.