NCERT Solutions for Maths Class 9 Chapter 6

EXERCISE 6.1

Question 1 In the given figure, line $AB$ and $CD$ interesect at $O$. If $\angle AOC+\angle BOE=70^0$ and $\angle BOD=40^0$ find reflex $\angle COE$.

Answer: 

Given: The line $AB$ and $CD$ intersect at $O$.

$\angle AOC+\angle BOE=70^0$

$\angle BOD=40^0$

And $AB$ is a straight line: rays n the line are $OC$ and $OE$.

If, $\angle AOC+\angle COE+\angle BOE=180^0$

$⇒(\angle AOC+\angle BOE)+\angle COE=180^0$

$⇒70^0+\angle COE=180^0$

$⇒\angle CO=180^0-70^0=110^0$

Reflex $\angle COE=360^0-110^0=250^0$

∴ Reflex $\angle COE=250^0$

Then $CD$ is a straight line; rays on the ine are $OE$ and $OB$.

And $\angle COE+\angle BOE+\angle BOD=180^0$

$⇒110^0+\angle BOE+40^0=180^0$

$⇒\angle BOE=180^0-150^0=30^0$

Hence, $\angle BOE=30^0$ and Reflex $\angle COE=250^0$

Question 2: In the given figure, lines $XY$ and $MN$ interesect at $O$. If $\angle POy=90^0$ and $a:b=2:3$, find $C$.

Answer: 

Given The line $XY$ and $MN$ intersect at $O$.

Then, $\angle POY=90^0$ and $a:b=2:3$

Let us assume the common ratio between $a$ and $b$ be $x$.

$∴a=2x$, and $b=3x$

And $XY$ is a straight line, rays on the line are $OM$ and $OP$.

Because $\angle XOM+\angle MOP+\angle POY=180^0$

$b+a+\angle POY=180^0$

$3x+2x+90^0=180^0$

$5x=90^0$

$∴x=180^0$

Then,

$a=2x=2\times18=36^0$

$b=3x=3\times18=54^0$

Now, $MN$ is a straight line. Ray on the line is $OX$.

The Linear Pair is,

$b+c=180^0$

$54^0+c=180^0$

$c=180^0-54^0$

$=126^0$

$∴c=126^0$

Question 3: In the given figure, $\angle PQR=\angle PRQ$, then prove that $\angle PQS=\angle PRT$

Anwer In the given figure, $ST$ is a straight line and ray $QP$ stands on it.

The Linear Pair is.

$∴\angle PQS+\angle PQR=180^0$

$\angle PQR=180^0-\angle PQS$    …(1)

$∴\angle PRT+\angle PRQ=180^0$

$\angle PRQ=180^0-\angle PRT$  ..(2)

It is denoted as $\angle PQR=\angle PRQ$

Now, equating equations (1) and (2). We get

$180^0-\angle PQS=180^0-\angle PRT$

$∴\angle PQS=\angle PRT$

Hence proved.

Question 4: In the givn figure, if $x+y=w+z$ then prove that $AOB$ is a line.

Answer:

Given $x+y=w+z$.

A complete angle becomes,

$x+y+z+w=360^0$

The

$x+y+x+y=360^0$

$2(x+y)=360^0$

$∴x+y=180^0$

Since $x$ and $y$ form a linear pair, therefore, $AOB$ is a line.

Hence proved.

Question 5: In the given figure, $POQ$ is a line. Ray $OR$ is perpendicular to line $PQ$. $OS$ is another ray lying between rays $OP$ and $OR$. Prove that $\angle ROS=\frac{1}{2}(\angle QOS-\angle POS)$

Answer: 

Given $OR\bot PQ$

$\angle POR=90^0$

$\angle POS+\angle SOR=90^0$

$\angle ROS=90^0+\angle POS$   …(1)

Because $OR\bot PQ$ Then,

$\angle QOR=90^0$

$\angle QOS-\angle ROS=90^0$

$\angle ROS=\angle QOS-90^0$     …(2)

Add equation (1) and (2), we obtain

$2\angle ROS=\angle QOS-\angle POS$

Hence proved.

Question 6: It is given that $\angle XYZ=64^0$ and $XY$ is produced to point $P$. Draw a figure from the given information. If ray $YQ$ bisects $\angle ZYP$, find $\angle XYQ$ and reflex $angle QYP$.

Answer: 

Given: the line $YQ$ bisects $\angle PYZ$

Hence, $\angle QYP=\angle ZYQ$

It can be observed that $PX$ is a line. Rays on the line are $YQ$ and $YZ$.

Then, $\angle XYZ+\angle ZYQ+\angle QYP=180^0$

$64+2\angle QYP=180^0$

$2\angle QYP=180-64=116$

$\angle QYP=58$

Also, $\angle ZYQ=\angle QYP=58^0$

Reflex $\angle QYP=360^0-58^0=302^0$

$\angle XYQ=\angle XYZ+\angle ZYQ$

$=64+58=122$

$∴\angle XYQ=122$ and Reflex $\angle QYP=302^0$

Exercise 6.2

Question 1: In the given figure, find the values of $x$ and $y$ and then show that $AP\parallel CD$.

Answer: 

It can be observed that,

By linear pair,

$50+x=180^0$

$∴x=130^0$

By vertically opposite angles property.

$y=130$

A $x$ and $y$ are alternate interior angles for line $AB$ and $CD$ and also measures of these angles are equal to each other, therefore, line $AB\parallel CD$.

Hence proved.

Question 2: In the given figure, if $AB\parallel CD,\space CD\parallel EF$ and $y:z=3:7$, find $x$.

Answer: 

Given $y:z=3:7,\space AB\parallel CD$ and $CD\parallel EF$

If, the lines parallel to the same line are parallel to each other. So,

$∴AB\parallel CS\parallel EF$

The alternate interrior angle are equal. Then

$x=z$   … (1)

Now, $y:z=3:7$

Let us assume that the common ratio between $y$ and $z$ be $a$.

$∴y=3a$ and $z=7a$

Also, co-interior angles on the same side of the transversal. Then,

$x+y=180^0$

By equation  (1)

$z+y=180^0$

$7a+3a=180^0$

$10a=180^0$

$∴a=18^0$

We get,

$z=7a$

$=7\times18=126^0$

$∴x=126^0$

Question 3: In the figure, if $AB\parallel CD,\space EF\parallel CD$ and $\angle GED=126^0$, find $\angle AGE,\space \angle GEF$ and $\angle FGE$.

Answer: Give that, $AB\parallel CD$ and $EF\parallel CD$. Then,

$\angle GED=126^0$

$\angle GEF+\angle FED=126^0$

$\angle GEF+90=126^0$

$\angle GEF=36^0$

As $\angle AGE$ and $\angle GED$ are alternate interior angles. We get,

$\angle AGE=\angle GED=126^0$

However, by linear pair to obtain $\angle AGE+\angle FGE=180^0$

$126^0+\angle FGE=180^0$

$\angle FGE=180^0-126^0=54^0$

Hence, $\angle AGE=126^0,\space \angle GEF=26^0$, and $\angle FGe=54^0$

Question 4: In the given figure, if $PQ\parallel ST,\space \angle PQR=110^0$ and $\angle RST=120^0$, find $\angle QRS$. [Hint: Draw a line parallel to $ST$ through point $R$.]

Answer: Given: $PQ\parallel ST,\space \angle PQR=110$ and $\angle RST=130^0$

Now, draw the line $XY$ parallel to $ST$ and passing through point $R$.

Bym Co-interior angles on the same side of transversal $QR$. We get,

$\angle PQR+\angle QRX=180^0$

$110^0+\angle QRX=180^0$

$\angle QRX=70^0$

Also, Co-interior angles on the same side of transversal $SR$ is

$\angle RST+\angle SRY=180^0$

$70+\angle QRS=50=180^0$

$\angle QRS=180^0-120^0=60^0$

$∴\angle QRS=60^0$

Question 5: In the given figure, if $AB\parallel CD,\space \angle APQ=50^0$ and $\angle PRD=127^0$, find $x$ and $y$.

Answer: Alternate interior angle

$\angle APR=\angle PRD$

$50+y=127^0$

$y=127^0-50^0$

$y=77^0$

Also,

$\angle APQ=\angle PQR$

$50^0=x$

$∴x=50^0$ and $y-77^0$

Question 6: In the figure, $PQ$ and $RS$ are two mirrors palced parallel to each other. An incident ray $AB$ strike the mirror $PQ$ at $B$, the reflected ray moved along the path $BC$ and strikes the mirror $RS$ at $C$ and again reflects back along $CD$. Prove tha $AB\parallel Cd$.

Answer: Let us draw $BM\bot PQ$ and $CN\bot RS$.

$∴BM\parallel CN$

Thus, $BM$ and $CN$ are two parallel lines and a transversal line $BC$ cuts them at $B$ and $C$ respectively.

$\angle 2=\angle 3$

By laws of reflection we get, $\angle 1=\angle 2$ and $\angle3=\angle 4$

$\angle 1=\angle2=\angle3=\angle4$

Also,

$\angle1+\angle2=\angle3+\angle4$

$\angle ABC=\angle DCB$

Howevere, these are alternate interior angle. Then,

$∴AB\parallel CD$

Excercise 6.3

Question 1: In the given figure, sides $QP$ and $RQ$ of $\triangle PQR$ are produced to point $S$ and $T$ respectively. If $\angle SPR=135^0$ and $PQT=110^0$, find $\angle PRQ$.

Answer: Given that, $\angle SPR=135^0$ and $\angle PQT=110^0$

By using linear pair angles property, we get,

$\angle SPR+\angle QPR=180^0$

$135^0+\angle QPR=180^0$

$∴\angle QPR=45^0$

Here also use linear pair property.

$\angle PQT+\angle PQR=180^0$

$110+\angle PQR=180^0$

$∴\angle PQR=70^0$

As the sum of all interior angles of a triangles is $180^0$

Therefore, for $\triangle PQR$ we get,

$\angle QPR+\angle PQR+\angle PRQ=180^0$

$45+70+\angle PRQ=180^0$

$\angle PRQ=180^0-115^0$

$∴\angle PRQ=65^0$

Question 2: In the figure, $\angle X=62^0,\space\angle XYZ=54^0$. If $YO$ and $ZO$ are the bisects of $\angle XYZ$ and $\angle XZY$ respectively of $\tringle XYZ$, find $\angle OZY$ and $\angle YOZ$.

Answer: As the sum of all interior angles of a triangle is $180^0$.

Therefore, for $\triangle XYZ$ we get,

$\angle X+\angle XYZ+\angle XZY=180^0$

$62^0+54^0+\angle XZY=180^0$

$\angle XZY=180^0-116$

$∴\angle XZY=64^0$

Since, $OZ$ is the bisector of $\angle XZY$.

$\angle OZY=\frac{64}{2}=32^0$

Similarly, $\angle OYZ=\frac{54}{2}=27^0$

Using angle sum property for $\triangle OYZ$, to obtain

$\angle OYZ+\angle YOZ+\angle OZY=180^0$

$27+\angle YOZ+32=180^0$

$\angle YOZ=180^0-59^0$

$∴\angle YOZ=121^0$

Hence, $\angle OZY=32$ and $\angle YOZ=121^0$

Question 3: In the figure, if $AB\parallel DE,\space \angle BAC=35^0$ and $\angle CDE=53^0$, find $\angle DCE$.

Answer: Given: $AB\parallel DE$

$AE$ is a transversal to $AB\parallel DE$. Then, by alternate interior angles property we get,

$\angle BAC=\angle CED$

$\angle CED=35^0$

In $\triangle CDE$, use angle sum property of a triangle, to obtain

$\angle CDE=35^0$

$53^0+35^0+\angle DCE=180^0$

$\angle DCE=180^0-88^0$

$∴\angle DCE=92^0$

Question 4: In given figure, if lines $PQ$ and $RS$ intersect at point $T$, such that $\angle PRT=40^0,\space\angle RPT=95^0$ and $\angle TSQ=75^0$, find $\angle SQT$.

Anwer: Using angle sum property for $\triangle PRT$, to obtain

$40^0+95^0+\angle PTR=180^0$

$\angle PTR=45^0$

By using vertically opposite property we get,

$\angle STQ=\angle PTR=45^0$

$∴\angle STQ=45^0$

Again, using angle sum property for $\triangle STQ$, to obtain

$\angle STQ+\angle SQT+\angle QST=180^0$

$45+\angle SQT+75^0=180^0$

$\angle SQT=180^0-120^0$

$∴\angle SQT=60^0$

Question 5: In the given figure, if $PQ\bot PS,\space PQ\parallel SR,\space \angle SQR=28^0$ and $\angle QRT=65^0$ then find the values of $x$ and $y$.

Answer: Given that $PQ\parallel SR$ and $QR$ is a transversal line.

By using alternate interior angles property we get,

$\angle PQR=\angle QRT$

$x+28^0=65^0$

$x=65^0-28^0$

$∴x=37^0$

By using the angle sum property for $\triangle SPQ$, to obtain

$\angle SPQ+x+y=180^0$

$90^0+37+y=180^0$

$y=180^0-127^0$

$∴y=53^0$

Hence, $x=37^0$ and $y=53^0$

Question 6: In the figure, the side $QR$ of $\triangle PQR$ is produced to a point $S$. If the busectors of $\angle PQR$ and $PRS$ meet at point $T$, then prove that $\angle QTR=\frac{1}{2}\angle QPR$.

Answer: In $\triangle QTR,\space \angle TRS$ is an exterior angle.

$∴\angle QTR+\angle TQR+\angle TRS$

$\angle QTR+\angle TRS-\angle TQR$     …(1)

For $\triangle PQR,\space \angle PRS$ is an external angle.

$∴\angle QPR+\angle PQR+\angle PRS$

As $QT$ and $RT$ are angle bisectors. Then,

$\angle QPR+2\angle TQR=2\angle TRS$

$\angle QPR=2(\angle TRS-\angle TQR)$

By using equation (1), we get,

$\angle QPR=2\angle QTR$

$∴\angle QTR=\frac{1}{2}\angle QPR$