 # Exercise 4.1: Lines and Triangles

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 $\angle PQA$.

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 = 180o

2x = 68

X = 34o

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,

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

$= \frac{1}{2}(x + 180 – x)$

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

= 90o

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, $70^{\circ}+\angle 2=180^{\circ}$

à $\angle 2=110^{\circ}$ [Vertically-opposite angles]

à $\angle 4=110^{\circ}$

$\angle 2=\angle 6$ [Corresponding angles]

à $\angle 6=110^{\circ}$

$\angle 6=\angle 8$ [Vertically-opposite angles]

à $\angle 8=110^{\circ}$

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]

à 5y + 4y = $180^{\circ}$

à 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,

$\angle ADC+\angle DCB=180^{\circ}$   …(i) [Consecutive Interior angles]

Also,

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

$\angle DCB+\angle ABC=180^{\circ}$    …(ii) [Consecutive Interior angles]

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

$\angle ADC+\angle DCB=\angle DCB+\angle ABC$

à $\angle ADC=\angle ABC$

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]

à $\angle BEF=35^{\circ}$

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

Then,

$\angle DEF=\angle FED$

à $\angle FED=65^{\circ}$

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]

à $\angle EOB+55^{\circ}=180^{\circ}$

à $\angle EOB=155^{\circ}$

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]

à $\angle AEF+116=180$

à $\angle AEF=64^{\circ}$

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

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

à $\angle CEF+124=180$

à $\angle CEF=56^{\circ}$

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]

à $\angle ECD+130^{\circ}=180^{\circ}$

à $\angle ECD=50^{\circ}$

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

Then,

$\angle ABC=\angle BCD$ [Alternate Interior Angles]

à $70^{\circ}=x+50^{\circ} ;Therefore, [\angle BCD=\angle BCE+\angle ECD]$

à $x=20^{\circ}$

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]

à $130^{\circ}+\angle CEF=180^{\circ}$

à $\angle CEF=50^{\circ}$

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]

à $x^{\circ}+20^{\circ}+50^{\circ}=180^{\circ}\;[\angle AEF=\angle AEC+\angle CEF]$

à $x^{\circ}+70^{\circ}=180^{\circ}$

à $x^{\circ}=110^{\circ}$

à 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]

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

à $\angle DCE+\angle AEC+180^{\circ}-\angle BAE=180^{\circ}$

à $\angle BAE-\angle DCE=\angle AEC$

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]

à $\angle BCD+75=180$

à $\angle BCD=105^{\circ}$

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

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

à $x^{\circ}=105^{\circ}$

à x = 105

Q12) In the given figure , $AB\parallel CD$ .Prove that P + q – r =  $180^{\circ}$ .

Ans.

Draw $PFQ\parallel AB\parallel CD$

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

Then,

$\angle AEF+\angle EFP=180^{\circ}$     …(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,

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

à p + q – r = 180

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

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]

à $75^{\circ}+20^{\circ}+\angle GEF=180^{\circ}$

à $\angle GEF=85^{\circ}$

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

therefore, $\angle GEF+\angle EGF+\angle EFG=180$

à $85^{\circ}+x+25^{\circ}=180^{\circ}$

à $110^{\circ}+x=180^{\circ}$

à x = $70^{\circ}$

And

$\angle FEG+\angle BEG=\angle DFQ$ [Corresponding angles]

à $85^{\circ}+20^{\circ}=\angle DFQ$

à $\angle DFQ=105^{\circ}$

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

à $25^{\circ}+y+105^{\circ}=180^{\circ}$

à $y=50^{\circ}$

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]

à 75  + $\angle ACD=180$

à $\angle ACD=105^{\circ}$

And,

$\angle ACD=\angle ECF$ [Vertically –opposite angles]

à $\angle ECF=105^{\circ}$

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

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

à $105^{\circ}+30^{\circ}+x=180^{\circ}$

à $135^{\circ}+x=180^{\circ}$

à $x=45^{\circ}$

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]

à $\angle EGH=85^{\circ}$

And,

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

à $85^{\circ}+\angle QGH=180^{\circ}$

à $\angle QGH=95^{\circ}$

Also,

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

à $115^{\circ}+\angle GHQ=180^{\circ}$

à $\angle GHQ=65^{\circ}$

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

à $\angle QGH+\angle GHQ+\angle GQH=180^{\circ}$

à $95^{\circ}+65^{\circ}+x=180^{\circ}$

à 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]

à z = $75^{\circ}$

$\angle ABC=\angle BCD$ [Alternate Interior Angles]

à x = $75^{\circ}$

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

à $35^{\circ}+y+75^{\circ}=180^{\circ}$

à 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]

à $y^{\circ}=75^{\circ}$

à y = 75

Also,

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

à x + y =180

à x + 75 =180

à x = 105

And,

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

à $\angle EGF+125=180$

à $\angle EGF=55^{\circ}$

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

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

à $y+z+55=180$

à 75 + z + 55 = 180

à 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]

à $110^{\circ}=\angle PQR+60^{\circ}\;because [\angle PQS=\angle PQR+\angle RQS]$

à $\angle PQR=50^{\circ}$

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

à $50^{\circ}+60^{\circ}+z=180^{\circ}$

à $110^{\circ}+z=180^{\circ}$

à $z=70^{\circ}$

$\angle DSH=z$ [Corresponding Angles]

à $\angle DSH=70^{\circ}$

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

à 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?

Ans.

For the lines l and m to be parallel

(i)

$\Leftrightarrow$ 3x – 20 = 2x +10  [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, $m\parallel n$