Important Questions for CBSE Class 9 Maths Chapter 6 Lines and Angles : FREE PDF Download
Chapter 6 of CBSE Class 9 Maths, Lines and Angles, is a crucial part of the geometry section. It introduces students to fundamental concepts like the types of angles, the Angle Sum Property of a Triangle, and the properties of parallel lines and transversals. In this chapter, students also learn how to prove various geometric theorems and apply these principles to real-life problems, making it a key area of focus for both the Class 9 exams and further studies in mathematics.
To help students better prepare for their exams, solving important questions from this chapter is essential. Accessing FREE important questions for Class 9 Maths can greatly enhance understanding and exam readiness. These sample papers, available as free PDFs, provide a wide range of practice problems, allowing students to test their knowledge, improve their problem-solving skills, and gain confidence. With regular practice, students can better grasp the Class 9 Maths syllabus and perform well in their exams.
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Study Important Questions for Class 9 Maths Chapter 6 - Lines and Angles
1-Marks:
1. Measurement of reflex angle is
(i) ${90^\circ }$
(ii) between ${0^\circ }$ and ${90^\circ }$
(iii) between ${90^\circ }$ and ${180^\circ }$
(iv) between ${180^\circ }$ and ${360^\circ }$
Ans: (iv) between ${180^\circ }$ and ${360^\circ }$
2. The sum of angle of a triangle is
(i) ${0^\circ }$
(ii) ${90^\circ }$
(iii) ${180^\circ }$
(iv) none of these
Ans: (iii) ${180^\circ }$
3. In fig if ${\text{ x}} = {30^\circ }{\text{ }}$ then y =
(i) ${90^\circ }$
(ii) ${180^\circ }$
(iii) ${150^\circ }$
(iv) ${210^\circ }$
Ans: (iii) ${150^\circ }$
4. If two lines intersect each other then
(i) Vertically opposite angles are equal
(ii) Corresponding angle are equal
(iii) Alternate interior angle are equal
(iv) None of these
Ans: (i) Vertically opposite angles are equal
5. The measure of Complementary angle of ${63^\circ }$ is
(a) ${30^\circ }$
(b) ${36^\circ }$
(c) ${27^\circ }$
(d) None of there
Ans: (c) ${27^\circ }$
6. If two angles of a triangle is ${30^\circ }$ and ${45^\circ }$ what is measure of third angle
(a) ${95^\circ }$
(b) ${90^\circ }$
(c) ${60^\circ }$
(d) ${105^\circ }$
Ans: (d) ${105^\circ }$
7. The measurement of Complete angle is
(a) ${0^\circ }$
(b) ${90^\circ }$
(c) ${180^\circ }$
(d) ${360^\circ }$
Ans: (d) ${360^\circ }$
8. The measurement of sum of linear pair is
(a) ${180^\circ }$
(b) ${90^\circ }$
(c) ${270^\circ }$
(d) ${360^\circ }$
Ans: (a) ${180^\circ }$
9. The difference of two complementary angles is ${40^\circ }$. The angles are
(a) ${65^\circ },{35^\circ }$
(b) ${70^\circ },{30^\circ }$
(c) ${25^\circ },{65^\circ }$
(d) ${70^\circ },{110^\circ }$
Ans: (c) ${25^\circ },{65^\circ }$
10. Given two distinct points ${\text{P}}$ and ${\text{Q}}$ in the interior of $\angle ABC$, then $\overrightarrow {AB} $ will be
(a) In the interior of $\angle ABC$
(b) In the interior of $\angle ABC$
(c) On the $\angle ABC$
(d) On the both sides of $\overrightarrow {BA} $
Ans: (c) On the $\angle ABC$
11. The complement of ${(90 - a)^0}$ is
(a) $ - {a^0}$
(b) ${(90 + 2a)^0}$
(c) ${(90 - a)^0}$
(d) ${a^0}$
Ans: (d) ${a^0}$
12. The number of angles formed by a transversal with a pair of lines is
(a) 6
(b) 3
(c) 8
(d) 4
Ans: (c) 8
13. In fig \[{L_1}\parallel {L_2}\] and $\angle 1\, = \,{52^ \circ }$ the measure of $\angle 2$ is.
(A) ${38^\circ }$
(B) ${128^\circ }$
(C) ${52^\circ }$
(D) ${48^\circ }$
Ans: (B) ${128^\circ }$
14. In $fig{\text{ x}} = {30^\circ }$ the value of Y is
(A) ${10^\circ }$
(B) ${40^\circ }$
(C) ${36^\circ }$
(D) ${45^\circ }$
Ans: (B) ${40^\circ }$
15. Which of the following pairs of angles are complementary angle?
(A) ${25^\circ },{65^\circ }$
(B) ${70^\circ },{110^\circ }$
(C) ${30^\circ },{70^\circ }$
(D) ${32.1^\circ },{47.9^\circ }$
Ans: (A) ${25^\circ },{65^\circ }$
16. In fig the measures of $\angle 1$ is.
(A) ${158^\circ }$
(B) ${138^\circ }$
(C) ${42^\circ }$
(D) ${48^\circ }$
Ans: (C) ${42^\circ }$
17. In figure the measure of $\angle a$ is
(a) ${30^\circ }$
(b) ${150^0}$
(c) ${15^\circ }$
(d) ${50^\circ }$
Ans: (a) ${30^\circ }$
18. The correct statement is-
F point in common.
Ans: (c) Three points are collinear if all of them lie on a line.
19. One angle is five times its supplement. The angles are-
(a) ${15^\circ },{75^\circ }$
(b) ${30^\circ },{150^\circ }$
(c) ${36^\circ },{144^0}$
(d) ${160^\circ },{40^\circ }$
Ans: (b) ${30^\circ },{150^\circ }$
20. In figure if and $\angle 1:\angle 2 = 1:2.$ the measure of $\angle 8$ is
(a) ${120^\circ }$
(b) ${60^\circ }$
(c) ${30^\circ }$
(d) ${45^\circ }$
Ans: (b) ${60^\circ }$
2-Marks:
1. In Fig. 6.13, lines ${\text{AB}}$ and ${\text{CD}}$ intersect at $O.$ If $\angle {\text{AOC}} + \angle {\text{BOE}} = {70^\circ }$ and $\angle {\text{BOD}} = {40^\circ }$, find $\angle {\text{BOE}}$ and reflex $\angle {\text{COE}}.$
Ans: According to the question given that, $\angle AOC + \angle BOE = {70^\circ }$ and $\angle BOD = {40^\circ }$.
We need to find $\angle BOE$ and reflex $\angle COE$.
According to the given figure, we can conclude that $\angle COB$ and $\angle AOC$ form a linear pair.
As we also know that sum of the angles of a linear pair is ${180^\circ }$.
So, $\angle COB + \angle AOC = {180^\circ }$
Because,$\angle COB = \angle COE + \angle BOE$, or
So, $\angle AOC + \angle BOE + \angle COE = {180^\circ }$
$ \Rightarrow {70^\circ } + \angle COE = {180^\circ }$
$ \Rightarrow \angle COE = {180^\circ } - {70^\circ }$
$ = {110^\circ }.$
Reflex $\angle COE = {360^\circ } - \angle COE$
$ = {360^\circ } - {110^\circ }$
$ = {250^\circ }.$
$\angle AOC = \angle BOD$ (Vertically opposite angles), or
$\angle BOD + \angle BOE = {70^\circ }$
But, according to the question given that $\angle BOD = {40^\circ }$.
${40^\circ } + \angle BOE = {70^\circ }$
$\angle BOE = {70^\circ } - {40^\circ }$
$ = {30^\circ }$.
Hence, we can conclude that Reflex $\angle COE = {250^\circ }$ and $\angle BOE = {30^\circ }$.
2. In the given figure, $\angle PQR = \angle PRQ$, then prove that $\angle PQS = \angle PRT$.
(Image will be uploaded soon)
Ans: According to the question we need to prove that $\angle PQS = \angle PRT$.
According to the question given that $\angle PQR = \angle PRQ$.
According to the given figure, we can conclude that $\angle PQS$ and $\angle PQR$, and $\angle PRS$ and $\angle PRT$ form a linear pair.
As we also know that sum of the angles of a linear pair is ${180^\circ }$.
So, $\angle PQS + \angle PQR = {180^\circ }$, and(i)
$\angle PRQ + \angle PRT = {180^\circ }$..(ii)
According to the equations (i) and (ii), we can conclude that
$\angle PQS + \angle PQR = \angle PRQ + \angle PRT$
But, $\angle PQR = \angle PRQ.$
So, $\angle PQS = \angle PRT$
Hence, the desired result is proved.
3. In the given figure, find the values of ${\text{x}}$ and ${\text{y}}$ and then show that .
(Image will be uploaded soon)
Ans: According to the question we need to find the value of $x$ and $y$ in the figure given below and then prove that
According to the figure, we can conclude that $y = {130^\circ }$ (Vertically opposite angles), and
$x$ and ${50^\circ }$ form a pair of linear pair.
As we also know that the sum of linear pair of angles is ${180^\circ }$.
$x + {50^\circ } = {180^\circ }$
$x = {130^\circ }$
$x = y = {130^\circ }$
According to the given figure, we can conclude that $x$ and $y$ form a pair of alternate interior angles parallel to the lines AB and CD.
Hence, we can conclude that $x = {130^\circ },y = {130^\circ }$ and.
4. In the given figure, if ${\text{AB}}||{\text{CD}},{\text{CD}}||{\text{EF}}$ and ${\text{y}}:{\text{z}} = 3:7$, find ${\text{x}}$.
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Ans: According to the question given that, and $y:z = 3:7$.
We need to find the value of $x$ in the figure given below.
As we also know that the lines parallel to the same line are also parallel to each other.
We can determine that .
Assume that, $y = 3a$ and $z = 7a$.
We know that angles on same side of a transversal are supplementary.
So, $x + y = {180^\circ }.$
$x = z$ (Alternate interior angles)
$z + y = {180^\circ }$, or $7a + 3a = {180^\circ }$
$ \Rightarrow 10a = {180^\circ }$
$a = {18^\circ }$.
$z = 7a = {126^\circ }$
$y = 3a = {54^\circ }$
Now, $x + {54^\circ } = {180^\circ }$
$x = {126^\circ }$
Hence, we can determine that $x = {126^\circ }$.
5. In the given figure, if and $\angle PRD = {127^\circ }$, find ${\text{x}}$ and ${\text{y}}$.
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Ans: According to the question given that, and $\angle PRD = {127^\circ }$.
As we need to find the value of $x$ and $y$ in the figure.
$\angle APQ = x = {50^\circ }$. (Alternate interior angles)
$\angle PRD = \angle APR = {127^\circ }$. (Alternate interior angles)
$\angle APR = \angle QPR + \angle APQ$
${127^\circ } = y + {50^\circ }$
$ \Rightarrow y = {77^\circ }$
Hence, we can determine that $x = {50^\circ }$ and $y = {77^\circ }$.
6. In the given figure, sides QP and RQ of $\Delta {\text{PQR}}$ are produced to points ${\text{S}}$ and ${\text{T}}$ respectively. If $\angle {\text{SPR}} = {135^\circ }$ and $\angle {\text{PQT}} = {110^\circ }$, find $\angle {\text{PRQ}}$.
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Ans: According to the question given that, $\angle SPR = {135^\circ }$ and $\angle PQT = {110^\circ }$.
As we need to find the value of $\angle PRQ$ in the figure given below.
According to the given figure, we can determine that $\angle SPR$ and $\angle RPQ$, and $\angle SPR$ and $\angle RPQ$ form a linear pair.
As we also know that the sum of angles of a linear pair is ${180^\circ }$.
$\angle SPR + \angle RPQ = {180^\circ }$, and
$\angle PQT + \angle PQR = {180^\circ }$
${135^\circ } + \angle RPQ = {180^\circ }$, and
${110^\circ } + \angle PQR = {180^\circ }$, or
$\angle RPQ = {45^\circ }$, and
$\angle PQR = {70^\circ }.$
According to the figure, we can determine that
$\angle PQR + \angle RPQ + \angle PRQ = {180^\circ }$. (Angle sum property)
$ \Rightarrow {70^\circ } + {45^\circ } + \angle PRQ = {180^\circ }$
$ \Rightarrow {115^\circ } + \angle PRQ = {180^\circ }$
$ \Rightarrow {115^\circ } + \angle PRQ = {180^\circ }$
$ \Rightarrow \angle PRQ = {65^\circ }$.
Hence, we can determine that $\angle PRQ = {65^\circ }$.
7. In the given figure, $\angle {\text{X}} = {62^\circ },\angle {\text{XYZ}} = {54^\circ }.$ If ${\text{YO}}$ and ${\text{ZO}}$ are the bisectors of $\angle {\text{XYZ}}$ and $\angle {\text{XZY}}$ respectively of $\Delta {\mathbf{XYZ}}$, find $\angle {\text{OZY}}$ and $\angle {\text{YOZ}}$.
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Ans: According to the question given that, $\angle X = {62^\circ },\angle XYZ = {54^\circ }$ and YO and ZO are bisectors of $\angle XYZ$ and $\angle XZY$, respectively.
As we need to find the value of$\angle OZY$ and $\angle YOZ$ in the figure.
According to the given figure, we can determine that in $\Delta XYZ$
$\angle X + \angle XYZ + \angle XZY = {180^\circ }$ (Angle sum property)
$ \Rightarrow {62^\circ } + {54^\circ } + \angle XZY = {180^\circ }$
$ \Rightarrow {116^\circ } + \angle XZY = {180^\circ }$
$ \Rightarrow \angle XZY = {64^\circ }$.
According to the question given that, OY and OZ are the bisectors of $\angle XYZ$ and $\angle XZY$, respectively.
$\angle OYZ = \angle XYO = \dfrac{{{{54}^\circ }}}{2} = {27^\circ}$, and
$\angle OZY = \angle XZO = \dfrac{{{{64}^\circ }}}{2} = {32^\circ }$
According to the figure, we can determine that in $\Delta OYZ$
$\angle OYZ + \angle OZY + \angle YOZ = {180^\circ }$. (Angle sum property)
${27^\circ } + {32^\circ } + \angle YOZ = {180^\circ }$
$ \Rightarrow {59^\circ } + \angle YOZ = {180^\circ }$
$ \Rightarrow \angle YOZ = {121^\circ }$.
Hence, we can determine that $\angle YOZ = {121^\circ }$ and $\angle OZY = {32^\circ }{\text{. }}$
8. In the given figure, if ${\text{AB}}||{\text{DE}},\angle {\text{BAC}} = {35^\circ }$ and $\angle {\text{CDE}} = {53^\circ }$, find $\angle {\text{DCE}}$.
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Ans: According to the question given that, and $\angle CDE = {53^\circ }$.
As we need to find the value of $\angle DCE$ in the figure given below.
According to the given figure, we can determine that
$\angle BAC = \angle CED = {35^\circ }$ (Alternate interior angles)
According to the figure, we can determine that in $\Delta DCE$
$\angle DCE + \angle CED + \angle CDE = {180^\circ }$. (Angle sum property)
$\angle DCE + {35^\circ } + {53^\circ } = {180^\circ }$
$ \Rightarrow \angle DCE + {88^\circ } = {180^\circ }$
$ \Rightarrow \angle DCE = {92^\circ }$.
Hence, we can determine that $\angle DCE = {92^\circ }$.
9. In the given figure, if lines PQ and RS intersect at point ${\text{T}}$, such that $\angle {\text{PRT}} = {40^\circ }$, $\angle {\text{RPT}} = {95^\circ }$ and $\angle {\text{TSQ}} = {75^\circ }$, find $\angle {\text{SQT}}$.
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Ans: According to the question given that, $\angle PRT = {40^\circ },\angle RPT = {95^\circ }$ and $\angle TSQ = {75^\circ }$.
As we need to find the value of $\angle SQT$ in the figure.
According to the given figure, we can determine that in $\Delta RTP$
$\angle PRT + \angle RTP + \angle RPT = {180^\circ }$ (Angle sum property)
${40^\circ } + \angle RTP + {95^\circ } = {180^\circ }$
$ \Rightarrow \angle RTP + {135^\circ } = {180^\circ }$
$ \Rightarrow \angle RTP = {45^\circ }$.
According to the figure, we can determine that
$\angle RTP = \angle STQ = {45^\circ }.$ (Vertically opposite angles)
According to the figure, we can determine that in $\Delta STQ$
$\angle SQT + \angle STQ + \angle TSQ = {180^\circ }$. (Angle sum property)
$\angle SQT + {45^\circ } + {75^\circ } = {180^\circ }$
$ \Rightarrow \angle SQT + {120^\circ } = {180^\circ }$
$ \Rightarrow \angle SQT = {60^\circ }$.
Hence, we can determine that $\angle SQT = {60^\circ }$.
10. In fig lines ${\text{XY}}$ and ${\text{MN}}$ intersect at O If $\angle $ POY $ = {90^\circ }$ and ${\text{ab}} = 2:3$ find ${\text{c}}$
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Ans: According to the given figure $\angle {\text{POY}} = {90^\circ }$
a: b: 2: 3
Assume that, $a = 2x$ and $b = 3x$
${\text{a}} + {\text{b}} + \angle {\text{POY}} = {180^\circ }(\because {\text{XOY}}$ is a line $)$
$\Rightarrow$ $2{\text{x}} + 3{\text{x}} + {90^\circ } = {180^\circ }$
$\Rightarrow$ $5x = {180^\circ } - {90^\circ }$
$\Rightarrow$ $5{\text{x}} = {90^\circ }$
$\Rightarrow$ $x = \dfrac{{{{90}^\circ }}}{5} = {18^\circ }$
So, $a = {36^\circ },\quad b = {54^\circ }$
MON is a line.
${\text{b}} + {\text{c}} = {180^\circ }$
$\Rightarrow$ ${54^\circ } + {{\text{c}}^\circ } = {180^\circ }$
$\Rightarrow$ $c = {180^\circ } - {54^\circ } = {126^\circ }$
Hence, the value of $c = {126^\circ }$.
11. In fig find the volume of ${\text{x}}$ and ${\text{y}}$ then Show that $\mathrm{AB} \| \mathrm{CD}$
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Ans: Accordign to the given figure, ${50^\circ } + x = {180^\circ }$
(by linear pair)
$x = {180^\circ } - {50^\circ }$
So, $x = {130^\circ }$
$y = {130^\circ }$ (Because vertically opposite angles are equal)
${\text{x}} = {\text{y}}$ as they are corresponding angles.
So, AB || CD
Hence proved.
12. What value of ${\mathbf{y}}$ would make AOB a line if $\angle {\text{AOC}} = 4{\text{y}}$ and $\angle {\text{BOC}} = 6{\text{y}} + {30^\circ }$
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Ans: According to the question given that, $\angle {\text{AOC}} = 4{\text{y}}$ and $\angle {\text{BOC}} = 6{\text{y}} + {30^\circ }$
$\angle {\text{AOC}} + \angle {\text{BOC}} = {180^\circ }$ (By linear pair)
$4y + 6y + {30^\circ } = {180^\circ }$
$\Rightarrow$ $10y = {180^\circ } - {30^\circ }$
$\Rightarrow$ $10y = {150^\circ } $
$\Rightarrow$ $y = {15^\circ }$
13. In fig ${\text{POQ}}$ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that $\angle {\text{ROS}} = \dfrac{1}{2}(\angle {\text{QOS}} - \angle {\text{POS}})$
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Ans: According to the question,
$R.H.S = \dfrac{1}{2}(\angle QOS - \angle POS)$
$ = \dfrac{1}{2}(\angle {\text{ROS}} + \angle {\text{QOR}} - \angle {\text{POS}})$
$ = \dfrac{1}{2}\left( {\angle {\text{ROS}} + {{90}^\circ } - \angle {\text{POS}}} \right) \ldots \ldots ..$ (1)
Because,$\angle {\text{POS}} + \angle {\text{ROS}} = {90^\circ }$
So, by equation 1
$ = \dfrac{1}{2}(ROS + \angle POS + \angle ROS - \angle POS)$ (by equation 1)
$ = \dfrac{1}{2} \times 2\angle {\text{ROS}} = \angle {\text{ROS}}$
= L.H.S
Hence proved.
14. In fig lines l1and l2 intersected at O , if ${\text{x}} = {45^\circ }$ find ${\text{x}},{\text{y}}$ and ${\text{u}}$
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Ans: According to the question given that,
$x = {45^\circ }$
So, ${\text{z}} = {45^\circ }$ (Because vertically opposite angles are equal)
${\text{x}} + {\text{y}} = {180^\circ }$
${45^\circ } + y = {180^\circ }$ (By linear pair)
$y = {180^\circ } - 45$
$y = {135^\circ }$
${\text{y}} = {\text{u}}$
Hence, the value of ${\text{u}} = {135^\circ }$ (Vertically opposite angles)
15. The exterior angle of a triangle is ${110^\circ }$ and one of the interior opposite angle is ${35^\circ }$. Find the other two angles of the triangle.
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Ans: As we all know that the exterior angle of a triangle is equal to the sum of interior opposite angles.
So, $\angle {\text{ACD}} = \angle {\text{A}} + \angle {\text{B}}$
${110 ^\circ}= \angle {\text{A}} + {35^\circ }$
$\Rightarrow$ $\angle {\text{A}} = {110^\circ } - {35^\circ }$
$\Rightarrow$ $\angle {\text{A}} = {75^\circ }$
$\Rightarrow$ $\angle {\text{C}} = 180 - (\angle {\text{A}} + \angle {\text{B}})$
$\Rightarrow$ $\angle {\text{C}} = 180 - \left( {{{75}^\circ } + {{35}^\circ }} \right)$
$\angle {\text{C}} = {70^\circ }$
16. Of the three angles of a triangle, one is twice the smallest and another is three times the smallest. Find the angles.
Ans: Assume that the smallest angle be ${x^\circ }$
Then the other two angles are $2{x^\circ }$ and $3{x^\circ }$
${x^\circ } + 2{x^\circ } + 3{x^\circ } = {180^\circ }$ As we know that the sum of three angle of a triangle is $\left. {{{180}^\circ }} \right$
$6{x^\circ } = {180^\circ }$
${\text{x}} = \dfrac{{180}}{6}$
$ = {30^\circ }$
Hence, angles are ${30^\circ },{60^\circ }$ and ${90^\circ }$.
17. Prove that if one angle of a triangle is equal to the sum of other two angles, the triangle is right angled.
Ans: According to the question given that in $\Delta ABC,\,\,\angle {\text{B}} = \angle {\text{A}} + \angle {\text{C}}$
To prove: $\Delta ABC$ is right angled.
Proof: $\angle A + \angle B + \angle C = {180^\circ } \ldots ..$ (1) (As we know that the sum of three angles of a $\Delta {\text{ABC}}$ is $\left. {{{180}^\circ }} \right)$
$\angle A + \angle C = \angle B \ldots ..$ (2)
From equations (1) and (2),
$\angle B + \angle B = {180^\circ }$
$2\angle {\text{B}} = {180^\circ }$
$\angle {\text{B}} = {90^\circ }$
18. In fig. sides QP and RQ of $\Delta PQR$ are produced to points ${\text{S}}$ and ${\text{T}}$ respectively. If $\angle SPR = {135^\circ }$ and $\angle {\text{PQT}} = {110^\circ }$, find $\angle {\text{PRQ}}$.
(Image will be uploaded soon)
Ans: According to the given figure,
$\angle PQT + \angle PQR = {180^\circ }$
${110^\circ } + \angle PQR = {180^\circ }$
$\angle PQR = {180^\circ } - {110^\circ }$
$\angle {\text{PQR}} = {70^\circ }$
Also, $\angle {\text{SPR}} = \angle {\text{PQR}} + \angle {\text{PRQ}}$ (According to the Interior angle theorem)
${135^\circ } = {70^\circ } + \angle PRQ$
$\angle {\text{PRQ}} = {135^\circ } - {70^\circ }$
Hence, the value of $\angle {\text{PRQ}} = {65^\circ }$.
19. In fig the bisector of $\angle ABC$ and $\angle {\text{BCA}}$ intersect each other at point O prove that $\angle BOC = {90^\circ } + \dfrac{1}{2}\angle A$
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Ans: According to the question given that in \(\vartriangle ABC\) such that the bisectors of $\angle ABC$ and $\angle {\text{BCA}}$ meet at a point O.
To Prove $\angle BOC = {90^\circ } + \dfrac{1}{2}\angle A$
Proof: In $\vartriangle BOC$
$\angle 1 + \angle 2 + \angle BOC = {180^\circ }$ (1)
In $\vartriangle ABC$
$\angle A + \angle B + \angle C = {180^\circ }$
$\angle A + 2\angle 1 + 2\angle 2 = {180^\circ }$
(BO and CO bisects $\angle B$ and $\angle {\text{C}}$ )
$ \Rightarrow \dfrac{{\angle A}}{2} + \angle 1 + \angle 2 = {90^\circ }$
$\angle 1 + \angle 2 = {90^\circ } - \dfrac{{\angle A}}{2}$
(Divide forth side by 2)
$\angle 1 + \angle 2 = {90^\circ } - \dfrac{{\angle A}}{2}$ in (i)
Substituting, ${90^\circ } - \dfrac{{\angle A}}{2} + \angle BOC = {180^\circ }$
$ \Rightarrow \angle BOC = {90^\circ } + \dfrac{{\angle A}}{2}$
Hence proved.
20. In the given figure $\angle POR$ and $\angle QOR$ form a linear pair if ${\mathbf{a}} - {\mathbf{b}} = {80^\circ }$. Find the
value of 'a' and 'b'.
(Image will be uploaded soon)
Ans: $a + b = {180^\circ } \to (1)$ (By line as pair)
$a - b = {80^0} \to (2)$
$2{\text{a}} = {260^\circ }$ (Adding equations (1) and (2))
${\text{a}} = {130^\circ }$
Put ${\text{a}} = {130^\circ }$ in equation (1)
${130^\circ } + b = {180^\circ }$
${\text{b}} = {180^\circ } - {130^\circ } = {50^\circ }$
Hence the value of ${\text{a}} = {130^\circ }$ and ${\text{b}} = {50^\circ }$.
21. If ray ${\text{OC}}$ stands on a line ${\text{AB}}$ such that $\angle AOC = \angle BOC$, then show that $\angle AOC = {90^\circ }$
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Ans: According to the question given that,
$\angle AOC = \angle BOC$
$\angle {\text{AOC}} + \angle {\text{BOC}} = {180^\circ }$ (By lines pair)
$\angle {\text{AOC}} + \angle {\text{AOC}} = {180^\circ }$
$2\angle {\text{AOC}} = {180^\circ }$
$\angle {\text{AOC}} = {90^\circ } = \angle B{\text{OC}}$
22. In the given figure show that $\mathrm{AB} \| \mathrm{EF}$
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Ans: $\angle {\text{BCD}} = \angle {\text{BCE}} + \angle {\text{ECD}}$
$ = {36^\circ } + {30^\circ } = {66^\circ } = \angle ABC$
So, (Alternate interior angles are equal)
Again, $\angle {\text{ECD}} = {30^\circ }$ and $\angle {\text{FEC}} = {150^\circ }$
So, $\angle {\text{ECD}} + \angle {\text{FEC}} = {30^\circ } + {150^\circ } = {180^\circ }$
Therefore, (We know that the sum of consecutive interior angle is $\left. {{{180}^\circ }} \right)$
$A B \| C D$ and $\mathrm{CD} \| \mathrm{EF}$
Then $\mathrm{AB} \| \mathrm{EF}$
Hence proved.
23. In figure if and $\angle PRD = {127^\circ }$ Find ${\text{x}}$ and ${\text{y}}$.
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Ans: $A B \| C D$ and PQ is a transversal
$\angle {\text{APQ}} = \angle {\text{PQD}}$ (Pair of alternate angles)
${50^\circ } = {\text{x}}$
Also and ${\text{PR}}$ is a transversal
$\angle {\text{APR}} = \angle {\text{PRD}}$
${50^\circ } + Y = {127^\circ }$
${\text{Y}} = {127^\circ } - {50^\circ } = {77^\circ }$
Hence the value of ${\text{x}}\, = \,{50^\circ }$ and ${\text{Y}} = {77^\circ }$.
24. Prove that if two lines intersect each other then vertically opposite angler are equal.
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Ans: According to the given figure: AB and CD are two lines intersect each other at $O$.
To prove: (i) $\angle 1 = \angle 2$ and (ii) $\angle 3 = \angle 4$
Proof:
$\angle 1 + \angle 4 = {180^\circ } \to (i)$ (By linear pair)
$\angle 4 + \angle 2 = {180^\circ }\quad \to (ii)$
$\angle 1 + \angle 4 = \angle 4 + \angle 2$ (By equations (i) and (ii))
$\angle 1 = \angle 2$
Similarly,
$\angle 3 = \angle 4$
Hence proved.
25. The measure of an angle is twice the measure of the supplementary angle. Find measure of angles.
Ans: Assume that the measure be ${x^\circ}$.
Then its supplement is ${180^\circ } - {x^\circ}$.
According to question
${x^\circ} = 2\left( {{{180}^\circ } - {x^\circ}} \right)$
$\Rightarrow$ ${x^\circ} = {360^\circ } - 2{x^\circ}$
$\Rightarrow$ $3x = {360^\circ }$
$\Rightarrow$ $x = {120^\circ }$
The measure of the angles are ${120^\circ }$ and ${60^\circ }$.
26. In fig $\angle PQR = \angle PRQ$. Then prove that $\angle PQS = \angle PRT$.
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Ans: $\angle PQS + \angle PQR = \angle PRQ + \angle PRT$ (By linear pair)
But,
$\angle PQR = \angle PRQ$ (Accordign to the question)
So, $\angle PQS = \angle PRT$
Hence proved.
27. In the given fig $\angle {\text{AOC}} = \angle {\text{ACO}}$ and $\angle {\text{BOD}} = \angle {\text{BDO}}$ prove that AC || DB.
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Ans: According to the question given that,
$\angle AOC = \angle ACO$ and $\angle BOD = \angle BDO$
But,
$\angle AOC = \angle BOD$ (Vertically opposite angles)
$\angle AOC = \angle BOD$ and $\angle BOD = \angle BDO$
$ \Rightarrow \angle ACO = \angle BDO$
So, (By alternate interior angle property)
Hence AC || DB proved.
28. In figure if lines ${\text{PQ}}$ and ${\text{RS}}$ intersect at point ${\text{T}}$. Such that $\angle PRT = {40^\circ }$, $\angle RPT = {95^\circ }$ and $\angle TSQ = {75^\circ }$, find $\angle SQT$.
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Ans: According to the $\Delta $ PRT
$\angle {\text{P}} + \angle {\text{R}} + \angle 1 = {180^\circ }$ (By angle sum property)
${95^\circ } + {40^\circ } + \angle 1 = {180^\circ }$
$\angle 1 = {180^\circ } - {135^\circ }$
$\angle 1 = {45^\circ }$
$\angle 1 = \angle 2$ (Vertically opposite angle)
$\angle 2 = \angle {45^\circ }$
According to the $\Delta {\text{TQS}}\quad \angle 2 + \angle {\text{Q}} + \angle {\text{S}} = {180^\circ }$
${45^\circ } + \angle Q + {75^\circ } = {180^\circ }$
$\angle {\text{Q}} + {120^\circ } = {180^\circ }$
$\angle {\text{Q}} = {180^\circ } - {120^\circ }$
$\angle {\text{Q}} = {60^\circ }$
Hence, the value of $\angle {\text{SQT}} = {60^\circ }$.
29. In figure, if $QT \bot PR,\angle TQR = {40^\circ }$ and $\angle SPR = {50^\circ }$ find $x$ and $y$.
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Ans: According to the $\Delta {\text{TQR}}$
${90^\circ } + {40^\circ } + x = {180^\circ }$ (Angle sum property of triangle)
So,$x = {50^\circ }$
Now, $y = \angle {\text{SPR}} + x$
So, $y = {30^\circ } + {50^\circ } = {80^\circ }$.
Hence, the value of $x = {50^\circ }$ and $y = {80^\circ }$.
30. In figure sides QP and RQ of $\Delta PQR$ are produced to points ${\text{S}}$ and ${\text{T}}$ respectively if $\angle SPR = {135^\circ }$ and $\angle PQT = {110^\circ }$, find $\angle PRQ$.
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Ans: According to the given figure,
${110^\circ } + \angle 2 = {180^\circ }$ (By linear pair)
$\angle 2 = {180^\circ } - {110^\circ }$
$\angle 2 = {70^\circ }$
$\angle 1 + {135^\circ } = {180^\circ }$
$\angle 1 = {180^\circ } - {135^\circ }$
$\angle 1 = {45^\circ }$
$\angle 1 + \angle 2 + \angle {\text{R}} = {180^\circ }$ (By angle sum property)
${45^\circ } + {70^\circ } + \angle R = {180^\circ }$
$\angle {\text{R}} = {180^\circ } - {115^\circ }$
$\angle {\text{R}} = {65^\circ }$
Hence, the value of $\angle {\text{PRQ}} = {65^\circ }$.
31. In figure lines ${\text{PQ}}$ and RS intersect each other at point O. If $\angle POR:\angle ROQ = 5:7$. Find all the angles.
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Ans: $\angle POR + \angle ROQ = {180^\circ }$ (Linear pair of angle)
But, $\angle {\text{POR}}:\angle {\text{ROQ}} = 5:7$ (According to the question)
So, $\angle {\text{POR}} = \dfrac{5}{{12}} \times {180^\circ } = {75^\circ }$
Similarly, $\angle {\text{ROQ}} = \dfrac{7}{{12}} \times {180^\circ } = {105^\circ }$
Now, $\angle {\text{POS}} = \angle {\text{ROQ}} = {105^\circ }$ (Vertically opposite angle)
And $\angle {\text{SOQ}} = \angle {\text{POR}} = {75^\circ }$ (Vertically app angle)
3-Marks:
1. In Fig. 6.16, if $x + y = w + z$, then prove that AOB is a line.
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Ans: As we need to prove that AOB is a line.
According to the question, given that $x + y = w + z$.
As we know that the sum of all the angles around a fixed point is ${360^\circ }$.
Hence, we can determine that $\angle AOC + \angle BOC + \angle AOD + \angle BOD = {360^\circ }$, or
$y + x + z + w = {360^\circ }$
But, $x + y = w + z$ (According to the question).
$2(y + x) = {360^\circ }$
$y + x = {180^\circ }$
According to the given figure, we can determine that $y$ and $x$ form a linear pair.
As we also know that if a ray stands on a straight line, then the sum of the angles of linear pair formed by the ray with respect to the line is ${180^\circ }$.
$y + x = {180^\circ }.$
Hence, we can determine that AOB is a line.
2. It is given that $\angle XYZ = {64^\circ }$ and XY is produced to point ${\mathbf{P}}$. Draw a figure from the given information. If ray YQ bisects $\angle ZYP$, find $\angle XYQ$ and reflex $\angle QYP$.
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Ans: According to the question, given that $\angle XYZ = {64^\circ },XY$ is produced to $P$ and YQ bisects $\angle ZYP$.
As we can determine the given below figure for the given situation:
As we need to find $\angle XYQ$ and reflex $\angle QYP$.
According to the given figure, we can determine that $\angle XYZ$ and $\angle ZYP$ form a linear pair.
As we also know that sum of the angles of a linear pair is ${180^\circ }$.
$\angle XYZ + \angle ZYP = {180^\circ }$
But, $\angle XYZ = {64^\circ }$.
$ \Rightarrow {64^\circ } + \angle ZYP = {180^\circ }$
$ \Rightarrow \angle ZYP = {116^\circ }$.
Ray YQ bisects $\angle ZYP$, or
$\angle QYZ = \angle QYP = \dfrac{{{{116}^\circ }}}{2} = {58^\circ }$
$\angle XYQ = \angle QYZ + \angle XYZ$
$ = {58^\circ } + {64^\circ } = {122^\circ }.$
Reflex $\angle QYP = {360^\circ } - \angle QYP$
$ = {360^\circ } - {58^\circ }$
$ = {302^\circ }$
Hence, we can determine that $\angle XYQ = {122^\circ }$ and reflex $\angle QYP = {302^\circ }$.
3. In the given figure, If ${\text{AB}}\parallel {\text{CD}}$ ,$EF \bot CD$ and $\angle GED = {126^\circ }$, find $\angle AGE,\angle GEF$ and $\angle FGE$.
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Ans: According to the question, given that and $\angle GED = {126^\circ }$.
As we need to find the value of $\angle AGE,\angle GEF$ and $\angle FGE$ in the figure given below.
$\angle GED = {126^\circ }$
$\angle GED = \angle FED + \angle GEF$
But, $\angle FED = {90^\circ }$.
${126^\circ } = {90^\circ } + \angle GEF \Rightarrow \angle GEF = {36^\circ }$
Because, $\angle AGE = \angle GED$ (Alternate angles)
$\angle AGE = {126^\circ }.$
According to the given figure, we can determine that $\angle FED$ and $\angle FEC$ form a linear pair.
As we know that sum of the angles of a linear pair is ${180^\circ }$.
$\angle FED + \angle FEC = 180$
$ \Rightarrow {90^\circ } + \angle FEC = {180^\circ }$
$ \Rightarrow \angle FEC = {90^\circ }$
But $\angle FEC = \angle GEF + \angle GEC$
So, ${90^\circ } = {36^\circ } + \angle GEC$
$ \Rightarrow \angle GEC = {54^\circ }$.
$\angle GEC = \angle FGE = {54^\circ }$ (Alternate interior angles)
Hence, we can determine that $\angle AGE = {126^\circ }$,$\angle GEF = {36^\circ }$ and $\angle FGE = {54^\circ }.$
4. In the given figure, ${\text{PQ}}$ and ${\text{RS}}$ are two mirrors placed parallel to each other. An incident ray ${\text{AB}}$ strikes the mirror ${\text{PQ}}$ at ${\text{B}}$, the reflected ray moves along the path ${\text{BC}}$ and strikes the mirror ${\text{RS}}$ at ${\text{C}}$ and again reflects back along ${\text{CD}}.$ Prove that ${\text{AB}}||{\text{CD}}$.
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Ans: According to the question, given that PQ and RS are two mirrors that are parallel to each other.
As we need to prove that in the given figure.
Now we draw lines BX and CY that are parallel to each other, to get
As we also know that according to the laws of reflection
$\angle ABX = \angle CBX$ and $\angle BCY = \angle DCY.$
$\angle BCY = \angle CBX$ (Alternate interior angles)
As we can determine that $\angle ABX = \angle CBX = \angle BCY = \angle DCY$.
According to the figure, we can determine that
$\angle ABC = \angle ABX + \angle CBX$, and
$\angle DCB = \angle BCY + \angle DCY.$
Hence, we can determine that $\angle ABC = \angle DCB$.
According to the figure, we can determine that $\angle ABC$ and $\angle DCB$ form a pair of alternate interior angles corresponding to the lines AB and CD, and transversal BC.
Hence, we can determine that $\angle A B C=\angle D C B$.
5. In the given figure, if SR, $\angle SQR = {28^\circ }$ and $\angle QRT = {65^\circ }$, then find the values of ${\text{x}}$ and ${\text{y}}$.
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Ans: According to the question, given that and $\angle QRT = {65^\circ }$.
As we need to find the values of $x$ and $y$ in the figure.
As we know that "If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles."
According to the figure, we can determine that
$\angle SQR + \angle QSR = \angle QRT$, or
${28^\circ } + \angle QSR = {65^\circ }$
$ \Rightarrow \angle QSR = {37^\circ }$
According to the figure, we can determine that
$x = \angle QSR = {37^\circ }$ (Alternate interior angles)
According to the figure, we can determine that $\Delta PQS$
$\angle PQS + \angle QSP + \angle QPS = {180^\circ }$. (Angle sum property)
$\angle QPS = {90^\circ }\quad (PQ \bot PS)$
$x + y + {90^\circ } = {180^\circ }$
$ \Rightarrow x + {37^\circ } + {90^\circ } = {180^\circ }$
$ \Rightarrow x + {127^\circ } = {180^\circ }$
$ \Rightarrow x = {53^\circ }$
Hence, we can determine that $x = {53^\circ }$ and $y = {37^\circ }$.
6. In the given figure, the side $QR$ of $\Delta $ PQR is produced to a point $S$. If the bisectors of $\angle PQR$ and $\angle PRS$ meet at point ${\text{T}}$, then prove that $\angle QTR = \dfrac{1}{2}\angle QPR$.
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Ans: As we need to prove that $\angle QTR = \dfrac{1}{2}\angle QPR$ in the figure given below.
As we also know that "If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles."
According to the figure, we can determine that in $\Delta QTR,\angle TRS$ is an exterior angle
$\angle QTR + \angle TQR = \angle TRS$, or
$\angle QTR = \angle TRS - \angle TQR$ ……….(i)
According to the figure, we can determine that in $\Delta QTR,\angle TRS$ is an exterior angle
$\angle QPR + \angle PQR = \angle PRS$
According to the question given that QT and RT are angle bisectors of $\angle PQR$ and $\angle PRS$.
$\angle QPR + 2\angle TQR = 2\angle TRS$
$\angle QPR = 2(\angle TRS - \angle TQR)$
As we need to substitute the value of equation (i) in the above equation, to get
$\angle QPR = 2\angle QTR$, or
$\angle QTR = \dfrac{1}{2}\angle QPR$
Hence, we can determine that the desired result is proved.
7. Prove that sum of three angles of a triangle is ${180^\circ }$
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Ans: According to the question given that, $\vartriangle {\text{ABC}}$
To prove that, $\angle A + \angle B + \angle C = {180^\circ }$
Now we draw through point A.
Proof: Because,
So, $\angle 2 = \angle 4 \to (1)$
Because, Altemate interior angle
And $\angle 3 = \angle 5 \to (2)$
Now we adding the equation (1) and equation (2)
$\angle 2 + \angle 3 = \angle 4 + \angle 5$
Adding both sides $\angle 1$,
$\angle 1 + \angle 2 + \angle 3 = \angle 1 + \angle 4 + \angle 5$
$\angle 1 + \angle 2 + \angle 3 = {180^\circ }\,(Because,\,\,\angle 1,\angle 4$, and $\angle 5$ forms a line)
$\angle {\text{A}} + \angle {\text{B}} + \angle {\text{C}} = {180^\circ }$
8. It is given that $\angle XYZ = {64^\circ }$ and ${\text{XY}}$ is produced to point ${\text{P}}$. Draw a figure from the given information. If ray YQ bisects $\angle ZYP$. Find $\angle XYQ$ and reflex $\angle QYP$.
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Ans: According to the figure,
YQ bisects $\angle ZYP$
So, $\angle 1 = \angle 2$
$\angle 1 + \angle 2 + \angle {64^\circ } = {180^\circ }({\text{YX}}$ is a line)
$\angle 1 + \angle 1 + {64^\circ } = {180^\circ }$
$2\angle 1 = {180^\circ } - {64^\circ }$
$2\angle 1 = {116^\circ }$
$\angle 1 = {58^\circ }$
So, $\angle {\text{XYQ}} = {64^\circ } + {58^\circ } = {122^\circ }$
$\angle 2 + \angle {\text{XYQ}} = {180^\circ }$
$\angle 1 = \angle 2 = \angle QYP = {58^\circ }$
$\angle 2 + {122^\circ } = {180^\circ }$
$\angle 2 = {180^\circ } - {122^\circ }$
$\angle QYP = \angle 2 = {58^\circ }$
Reflex $\angle Q{\text{YP}} = {360^\circ } - \angle QYP$
$ = {360^\circ } - {58^\circ }$
$ = {302^\circ }$
Hence, the value of $\angle {\text{XYQ}}\,{\text{ = }}\,{\text{12}}{{\text{2}}^ \circ }$ and reflex $\angle {\text{QYP}} = \,{302^ \circ }$.
9. In fig if and $\angle {\text{RST}} = {130^\circ }$ find $\angle {\text{QRS}}$.
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Ans: Through point R Draw line XY
Because,$\text{PQ}\|\text{ST}$
$\text{ST}\|\text{KL,}\quad So,\,\text{PQ}\|\text{KL}$
Because, $\text{PQ}\|\text{KL}$
So, $\angle {\text{PQR}} + \angle 1 = {180^\circ }$
(As we know that the sum of interior angle on the same side of transversal is ${180^\circ }$ ) ${110^\circ } + \angle 1 = {180^\circ }$
$\angle 1 = {70^\circ }$
Similarly $\angle 2 + \angle {\text{RST}} = {180^\circ }$
$\angle 2 + {130^\circ } = {180^\circ }$
$\angle 2 = {50^\circ }$
$\angle 1 + \angle 2 + \angle 3 = {180^\circ }$
${70^\circ } + {50^\circ } + \angle 3 = {180^\circ }$
$\angle 3 = {180^\circ } - {120^\circ }$
$\angle 3 = {60^\circ }$
Hence, the value of $\angle {\text{QRS}} = {60^\circ }$.
10. The side BC of $\vartriangle ABC$ is produced from ray $BD$. $CE$ is drawn parallel to $AB$, show that $\angle ACD = \angle A + \angle B$. Also prove that $\angle A + \angle B + \angle C = {180^\circ }$.
Ans: As we can see, $\text{AB}\|\text{CE}$ and ${\text{AC}}$ intersect them
$\angle 1 = \angle 4$ ………. (1) (Alternate interior angles)
Also and BD intersect them
$\angle 2 = \angle 5$ …………. (2) (Corresponding angles)
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Now adding equation (1) and equation (2)
$\angle 1 + \angle 2 = \angle 4 + \angle 5$
$\angle A + \angle B = \angle ACD$
Adding $\angle C$ on both sides, we get
$\angle A + \angle B + \angle C = \angle C + \angle ACD$
$\angle A + \angle B + \angle C = {180^\circ }$
Hence, proved.
11. Prove that if a transversal intersect two parallel lines, then each pair of alternate interior angles is equal.
Ans: According to the question given that, line intersected by transversal ${\text{PQ}}$
To Prove: (i) $\angle 2 = \angle 5$ (ii) $\angle 3 = \angle 4$
Proof:
$\angle 1 = \angle 2$ ………… (i) (Vertically Opposite angle)
$\angle 1 = \angle 5$ ………….. (ii) (Corresponding angles)
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By equations (i) and (ii)
$\angle 2 = \angle 5$
Similarly, $\angle 3 = \angle 4$
Hence Proved.
12. In the given figure $\Delta {\text{ABC}}$ is right angled at $A$. $AD$ is drawn perpendicular to $BC$. Prove that $\angle BAD = \angle ACB$
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Ans: According to the figure,
${\text{AD}} \bot BC$
So, $\angle ADB = \angle ADC = {90^\circ }$
From $\vartriangle {\text{ABD}}$
$\angle {\text{ABD}} + \angle {\text{BAD}} + \angle {\text{ADB}} = {180^\circ }$
$\angle {\text{ABD}} + \angle {\text{BAD}} + {90^\circ } = {180^\circ }$
$\angle {\text{ABD}} + \angle {\text{BAD}} = {90^\circ }$
$\angle {\text{BAD}} = {90^\circ } - \angle ABD \to (1)$
But $\angle {\text{A}} + \angle {\text{B}} + \angle {\text{C}} = {180^\circ }$ in $\vartriangle {\text{ABC}}$
$\angle {\text{B}} + \angle {\text{C}} = {90^\circ },\quad Because,\,\,\angle {\text{A}} = {90^\circ }$
$\angle {\text{C}} = {90^\circ } - \angle B \to \,(2)$
From equations (1) and (2)
$\angle {\text{BAD}} = \angle {\text{C}}$
$\angle {\text{BAD}} = \angle {\text{ACB}}$
Hence proved.
13. In $\Delta {\text{ABC}}\angle B = {45^\circ },\angle C = {55^\circ }$ and bisector $\angle A$ meets ${\text{BC}}$ at a point ${\text{D}}$. Find
$\angle ADB$ and $\angle ADC$
Ans: In $\vartriangle {\text{ABC}}$
$\angle {\text{A}} + \angle {\text{B}} + \angle {\text{C}} = {180^\circ }$ (As we know that the sum of three angle of a $\Delta $ is $\left. {{{180}^\circ }} \right)$
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$ \Rightarrow \angle {\text{A}} + {45^\circ } + {55^\circ } = {180^\circ }$
$\angle {\text{A}} = {180^\circ } - {100^\circ } = {80^\circ }$
${\text{AD}}$ bisects $\angle {\text{A}}$
$\angle 1 = \angle 2 = \dfrac{1}{2}\angle A = \dfrac{1}{2} \times {80^\circ } = {40^\circ }$
Now in $\Delta {\text{ADB}}$,
We have, $\angle 1 + \angle {\text{B}} + \angle {\text{ADB}} = {180^\circ }$
$ \Rightarrow {40^\circ } + {45^\circ } + \angle ADB = {180^\circ }$
$ \Rightarrow \angle {\text{ADB}} = {180^\circ } - {85^\circ } = {95^\circ }$
$\angle {\text{ADB}} + \angle {\text{ADC}} = {180^\circ }$
Also ${95^\circ } + \angle ADC = {180^\circ }$
$\angle {\text{ADC}} = {180^\circ } - {95^\circ } = {85^\circ }$
Hence, the value of $\angle {\text{ADB}} = {95^\circ }$ and $\angle {\text{ADC}} = {85^\circ }$
14. In figure two straight lines $AB$ and $CD$ intersect at a point 0 . If $\angle BOD = {x^\circ }$ and $\angle AOD = {(4x - 5)^\circ }$.
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Find the value of ${\mathbf{x}}$ hence find
(a) $\angle AOD$
Ans: $\angle AOB = \angle AOD + \angle DOB$ By linear pair
${180^\circ } = 4x - 5 + x$
${180^\circ } + 5 = 5x$
$5{\text{x}} = 185$
${\text{x}} = \dfrac{{185}}{5} = {37^\circ }$
So, $\angle {\text{AOD}} = 4{\text{x}} - 5$
$ = 4 \times 37 - 5 = 148 - 5$
$ = {143^\circ }$
(b) $\angle BOC$
$\angle {\text{BOC}} = {143^\circ }$
Because, $\angle {\text{AOD}}$ and $\angle {\text{BOC}}$
$\angle {\text{BOD}} = {\text{x}} = {37^\circ }$ (Vertically opposite angles)
(c) $\angle BOC$
$\angle BOD = {37^\circ }$
(d) $\angle AOC$
$\angle AOC = {37^\circ }$
15. The side ${\text{BC}}$ of a $\Delta {\text{ABC}}$ is produced to ${\text{D}}$. The bisector of $\angle {\text{A}}$ meets ${\text{BC}}$ at ${\text{L}}$ as shown if fig. prove that $\angle {\text{ABC}} + \angle {\text{ACD}} = 2\angle {\text{ALC}}$
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Ans: In $\Delta {\text{ABC}}$ we have
$\angle {\text{ACD}} = \angle {\text{B}} + \angle {\text{A}} \to (1)$ (Exterior angle property)
$ \Rightarrow \angle {\text{ACD}} = \angle {\text{B}} + 2\;{\text{L}}1$
$(So,\angle {\text{A}}$ is the bisector of $\angle {\text{A}} = 2\;{\text{L}}1)$
In $\Delta {\text{AB}}L$
$\angle {\text{ALC}} = \angle {\text{B}} + \angle {\text{BA}}L$ (Exterior angle property)
$\angle {\text{A}}L{\text{C}} = \angle {\text{B}} + \angle 1$
$ \Rightarrow 2\angle {\text{ALC}} = 2\angle {\text{B}} + 2\angle 1 \ldots (2)$
Subtracting equation (1) from equation (2)
$2\angle {\text{ALC}} - \angle {\text{ACD}} = \angle {\text{B}}$
$2\angle {\text{ALC}} = \angle {\text{B}} + \angle {\text{ACD}}$
$\angle {\text{ACD}} + \angle {\text{ABC}} = 2\angle {\text{ALC}}$
Hence proved.
16. In fig PT is the bisector of $\angle QPR$ in $\Delta PQR$ and $PS \bot QR$, find the value of $x$
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Ans: Sum of $\angle QPR + \angle Q + \angle R = {180^\circ }$ (According to the angle sum property of triangle)
$\angle QPR = {180^\circ } - {50^\circ } - {30^\circ } = {100^\circ }$
$\angle QPT = \dfrac{1}{2}\angle QPR$
$ = \dfrac{1}{2} \times {100^\circ } = {50^\circ }$
$\angle Q + \angle QPS = \angle PST$ (Exterior angle theorem)
$\angle QPS = {90^\circ } - \angle Q$
$ = {90^\circ } - {50^\circ } = {40^\circ }$
$x = \angle QPT - \angle QPS$
$ = {50^\circ } - {40^\circ } = {10^\circ }$
Hence, the value of $x\, = \,{10^ \circ }$.
17. The sides $BA$ and $DC$ of a quadrilateral $ABCD$ are produced as shown in fig show that $\angle X + \angle Y = \angle a + \angle b$
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Ans: In given fugure join $BD$
$\operatorname{In} \Delta ABD$
$\angle b = \angle ABD + \angle BDA$ (Exterior angle theorem)
$\operatorname{In} \Delta CBD$
$\angle a = \angle CBD + \angle BDC$
$\angle a + \angle b = \angle CBD + \angle BDC + \angle ABD + \angle BDA$
$ = (\angle CBD + \angle ABD) + (\angle BDC + \angle BDA)$
$ = \angle x + \angle y$
$\angle a + \angle b = \angle x + \angle y$
Hence proved.
18. In the ${\text{BO}}$ and ${\text{CO}}$ are Bisectors of $\angle {\text{B}}$ and $\angle {\text{C}}$ of $\Delta {\text{ABC}}$, show that $\angle {\text{BOC}} = {90^\circ } + \dfrac{1}{2}$$\angle {\text{A}}$
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Ans: According to the given figure,
$\angle 1 = \dfrac{1}{2}\angle ABC$
And $\angle 2 = \dfrac{1}{2}\angle ACB$
So, $\angle 1 + \angle 2 = \dfrac{1}{2}(\angle ABC + \angle ACB)\,\,\,\,\,\,\,\,... \ldots (1)$
But,
$\angle ABC + ACB + \angle A = {180^\circ }$
So, $\angle ABC + ACB = {180^\circ } - \angle A$
But,
$\dfrac{1}{2}(\angle ABC + ACB) = {90^\circ } - \dfrac{1}{2}\angle A\,\,\,\,\,\,\,\,\,... \ldots .(2)$
From equation (1) and equation (2) we get
$\angle 1 + \angle 2 = {90^\circ } - \dfrac{1}{2}\angle A\,\,\,\,\,\,\,\,... \ldots ..(3)$
But,
$\angle BOC + \angle 1 + \angle 2 = {180^\circ }$ (Angle of a)
Put the value of $\angle 1\, + \,\angle 2$ in the above equation,
$ = {180^\circ } - \left( {{{90}^\circ } - \dfrac{1}{2}\angle A} \right)$
$ = {90^\circ } + \dfrac{1}{2}\angle A$
Hence proved.
19. In fig two straight lines PQ and RS intersect each other at o, if $\angle {\text{POT}} = {75^\circ }$ Find the values of a, b and c
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Ans: PQ intersect RS at ${\text{O}}$
So,$\angle QOS = \angle POR$ (vertically opposite angles)
${\text{A}} = 4\;{\text{b }}... \ldots .(1)$
Also,
$a + b + {75^\circ } = {180^\circ }\,(Because,\,\,POQ$ is a straight lines)
So, $a + b = {180^\circ } - {75^\circ }$
$ = {105^\circ }$
Using, equation (1) $4b + b = {105^\circ }$
$5b = {105^\circ }$
Or
$b = \dfrac{{105}}{5} = {21^\circ }$
So, $a = 4b$
$a = 4 \times 21$
$a = 84$
Again, $\angle QOR$ and $\angle QOS$ form a linear pair
So, $a + 2c = {180^\circ }$
Using, equation (2)
${84^\circ } + 2c = {180^\circ }$
$2c = {180^\circ } - {84^\circ }$
$2c = {96^\circ }$
$c = \dfrac{{{{96}^\circ }}}{2} = {48^\circ }$
Hence, $a = {84^\circ },b = {21^\circ }$ and $c = {48^\circ }$
20. In figure ray OS stands on a line POQ, ray OR and ray OT are angle bisector of
$\angle POS$ and $\angle SOQ$ respectively. If $\angle POS = x$, find $\angle ROT$.
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Ans: Ray OS stands on the line POQ
So, $\angle {\text{POS}} + \angle {\text{SOQ}} = {180^\circ }$
But $\angle {\text{POS}} = {\text{x}}$
So, ${\text{x}} + \angle {\text{SOQ}} = {180^\circ }$
$\angle {\text{SOQ}} = {180^\circ } - x$
Now ray OR bisects $\angle {\text{POS}}$,
Hence, $\angle {\text{ROS}} = \dfrac{1}{2} \times \angle POS = \dfrac{1}{2} \times x = \dfrac{x}{2}$
Similarly, $\angle {\text{SOT}} = \dfrac{1}{2} \times \angle SOQ = \dfrac{1}{2} \times \left( {{{180}^\circ } - X} \right) = {90^\circ } - \dfrac{x}{2}$
$\angle ROT = \angle ROS + \angle SOT = \dfrac{x}{2} + {90^\circ } - \dfrac{x}{2} = {90^\circ }$
Hence, the value of $\angle ROT\, = \,{90^ \circ }$.
21. If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel.
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Ans: According to the question and figure given that, AD is transversal intersect two lines PQ and RS
We have to prove PQ||RS
Proof: BE bisects $\angle {\text{ABQ}}$
$\angle \mathrm{EBQ}= \dfrac{1}{2}\angle ABQ \to (1)$
Similarity CG bisects $\angle {\text{BCS}}$
So, $\angle 2 = \dfrac{1}{2}\angle BCS \to (2)$
But and ${\text{AD}}$ is the transversal
So,$\quad \angle 1 = \angle 2$
So, $\dfrac{1}{2}\angle ABQ = \dfrac{1}{2}\angle BCS$ (By equations $(1)$ and $(2))$
$ \Rightarrow \angle {\text{ABQ}} = \angle {\text{BCS}}$ (Because corresponding angles are equal)
So, PQ||RS
Hence proved.
22. In figure the sides ${\text{QR}}$ of $\Delta PQR$ is produced to a point ${\text{S}}$. If the bisectors of $\angle PQR$ and $\angle PRS$ meet at point ${\text{T}}$. Then prove that $\angle QRT = \dfrac{1}{2}\angle QPR$
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Ans: Solution, In $\Delta {\text{PQR}}$
$\angle {\text{PRS}} = \angle {\text{Q}} + \angle {\text{P}}$ (By exterior angle theorem)
$\angle 4 + \angle 3 = \angle 2 + \angle 1 + \angle {\text{P}}$
$2\angle 3 = 2\angle 1 + \angle {\text{P}} \to (1)$
So, ${\text{QT}}$ and ${\text{RT}}$ are bisectors of $\angle {\text{Q}}$ and $\angle {\text{PRS}}$
In $\vartriangle {\text{QTR}}$,
$\angle 3 = \angle 1 + \angle {\text{T}} \to $ (2) (By exterior angle theorem)
By equations (1) and (2) we get
$2(\angle 1 + \angle T) = 2\angle 1 + \angle {\text{P}}$
$2\angle 1 + 2\angle {\text{T}} = 2\angle 1 + \angle {\text{P}}$
$\angle {\text{T}} = \dfrac{1}{2}\angle P$
$\angle {\text{QTR}} = \dfrac{1}{2}\angle QPR$
Hence proved.
23. In figure PQ and RS are two mirror placed parallel to each other. An incident ray AB striker the mirror PQ at $B$, the reflected ray moves along the path BC and strike the mirror ${\text{RS}}$ at ${\text{C}}$ and again reflects back along ${\text{CD}}$. Prove that $\mathrm{AB} \| \mathrm{CD}$.
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Ans: Solution, Draw $MB \bot PQ$ and $NC \bot RS$
$\angle 1 = \angle 2 \to (1)$ (Angle of incident)
And $\angle 3 = \angle 4 \to (2)$ (is equal to angle of reflection)
Because, $\angle {\text{MBQ}} = \angle {\text{NCS}} = {90^\circ }$
So, (By corresponding angle property)
Because, $\angle 2 = \angle 3 \to (3)$ (Alternate interior angle)
By equations $(1),(2)$ and (3)
$\angle 1 = \angle 4$
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$\angle 1 + \angle 2 = \angle 4 + \angle 3$
$ \Rightarrow \angle {\text{ABC}} = \angle {\text{BCD}}$
So, (By alternate interior angles)
Hence proved.
4-Marks
1. In fig the side AB and AC of $\vartriangle ABC$ are produced to point E and D respectively. If bisector BO And CO of $\angle {\text{CBE}}$ and $\angle {\text{BCD}}$ respectively meet at point ${\text{O}}$, then prove that $\angle {\text{BOC}} = {90^\circ } - \dfrac{1}{2}\angle {\text{BAC}}$
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Ans: Ray BO bisects $\angle {\text{CBE}}$
So, $\angle {\text{CBO}} = \dfrac{1}{2}\angle {\text{CBE}}$
$ = \dfrac{1}{2}\left( {{{180}^\circ } - y} \right)\,\,\,\left( {Because,\,\,\angle {\text{CBE}} + {\text{y}} = {{180}^\circ }} \right)$
$ = {90^\circ } - \dfrac{y}{2} \ldots \ldots ..$(1)
Similarly, ray CO bisects $\angle BCD$
$\angle {\text{BCO}} = \dfrac{1}{2}\angle BCD$
$ = \dfrac{1}{2}\left( {{{180}^\circ } - Z} \right)$
$ = {90^\circ } - \dfrac{Z}{2} \ldots \ldots \ldots ..$(2)
In $\vartriangle {\text{BOC}}$
$\angle {\text{BOC}} + \angle {\text{BCO}} + \angle {\text{CBO}} = {180^\circ }$
$\angle {\text{BOC}} = \dfrac{1}{2}(y + z)$
But $x + y + z = {180^\circ }$
$y + z = {180^\circ } - x$
$\angle {\text{BOC}} = \dfrac{1}{2}\left( {{{180}^\circ } - x} \right) = {90^\circ } - \dfrac{x}{2}$
$\angle {\text{BOC}} = {90^\circ } - \dfrac{1}{2}\angle {\text{BAC}}$
Hence proved.
2. In given fig. AB CD. Determine $\angle a$.
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Ans: Through O draw a line $l$ parallel to both ${\text{AB}}$ and ${\text{CD}}$
Clearly
$\angle a = \angle 1 + \angle 2$
$\angle 1 = {38^\circ }$
$\angle 2 = {55^\circ }$ (Alternate interior angles)
$\angle a = {55^\circ } + {38^\circ }$
Hence, the value of $\angle a = {93^\circ }$.
3. In fig ${\text{M}}$ and ${\text{N}}$ are two plane mirrors perpendicular to each other. Prove that the incident ray CA is parallel to reflected ray BD.
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Ans: From the figure, it can be seen that ${\text{AP}} \bot {\text{M}}$ and ${\text{BQ}} \bot {\text{N}}$
So, $BQ \bot N$ and $AP \bot M$ and ${\text{M}} \bot {\text{N}}$
So, $\angle {\text{BOA}} = {90^\circ }$
$ \Rightarrow {\text{BQ}} \bot {\text{AP}}$
In $\vartriangle {\text{BOA}}\angle 2 + \angle 3 + \angle {\text{BOA}} = {180^\circ }$ (By angle sum property)
$ \Rightarrow \angle 2 + \angle 3 + {90^\circ } = {180^\circ }$
So, $\angle 2 + \angle 3 = {90^\circ }$
Also $\angle 1 = \angle 2$ and $\angle 4 = \angle 3$
$ \Rightarrow \angle 1 + \angle 4 = \angle 2 + \angle 3 = {90^\circ }$
So, $(\angle 1 + \angle 4) + (\angle 2 + \angle 3) = {90^\circ } + {90^\circ } = {180^\circ }$
$ \Rightarrow (\angle 1 + \angle 2) + (\angle 3 + \angle 4) = {180^\circ }$
or $\angle {\text{CAB}} + \angle {\text{DBA}} = {180^\circ }$
So, (By sum of interior angles of same side of transversal)
Hence proved.
CBSE Class 9 Maths Chapter-6 Important Questions - Free PDF Download
Maths is considered a tough subject but it can be an extremely high-scoring subject just like other subjects. All it needs is a practice of the concepts learned in the chapter. Unless students understand the practical application of the theoretical knowledge, they will not understand the applicability of the mathematical concepts. This can be achieved by practising as many questions based on a particular concept as possible.
By a careful analysis of the past years’ papers, the updated curriculum, and the CBSE guidelines, experts at Vedantu have curated a list of all the important questions for class 9 maths lines and angles. By practising these questions students can gain confidence about their knowledge in the topic.
Topics Covered by Lines and Angles Class 9 Important Questions
The chapter covers basic terms and definitions like a line segment is a part of a line with two endpoints, part of a line with one endpoint is called ray, three or more points on the same line are called collinear points, when two rays emerge from the same point it is called an angle, a vertex is a point from which the rays emerge and the rays that make the angle are called the arms of the angle.
When students practice these questions, they understand the format of answering the different types of questions. They get familiar with the exam pattern as several times questions in the exams are based on the same pattern as these questions. The questions include both short answer type and long answer type questions.
Other Important Topics Covered are:
Types of angles- Acute angle, obtuse angle, right angle, straight angle, reflex angle, complementary angles, supplementary angles, adjacent angles
Intersecting lines and non-interesting lines- If two lines intersect each other then the angle formed vertically opposite to each other are equal
Pair of angles
Parallel lines and a transversal- 4 theorems
Lines parallel to the same line- If two lines are parallel to the same line then they are also parallel to each other
Angle sum property of an angle
Students must try to solve all the questions from the lines and angles class 9 important questions as they cover all the important topics that can be missed out by them during their revision. The solutions have been prepared in an easy to understand format so that students can understand the concept well. It helps the students in solving questions of all difficulty levels. When the basics of these concepts are clear to the students they can understand the detailed concepts of geometry in higher classes with ease.
Class 9 Maths Chapter 6 Lines and Angles Extra Questions
If line AB and CD intersect at a point X such that ACX = 40°, CAX = 95°, and XDB = 75°. Find DBX .
Prove that the sum of the angle of a triangle is 180°.
Prove that two lines are parallel if a transversal line intersects two lines such that the bisectors of a pair of corresponding angles are parallel.
Can all the angles of a triangle be less than 60°? Support your answer with a reason.
Can a triangle have two obtuse angles? Support your answer with a reason.
The sum of the two angles of a triangle is 120° and their difference is 20°. Find the value of all the three angles of a triangle.
If the exterior angle of a triangle is 110 and the interior angle of a triangle is 35°. Find the values of the other two angles of a triangle.
Prove that lines which are parallel to the same line are parallel to each other.
Prove that when two lines intersect each other, then the vertically opposite angles that are formed are equal.
What is the measure of an acute angle, an obtuse angle, a right angle, a reflex angle, and a complementary angle?
Class 9 Maths Chapter 6 Lines and Angles important Topics
Sl.No | Topic | Description |
1 | Basic Terminology | Introduction to terms like lines, angles, rays, and the different types of angles (acute, obtuse, right, etc.). |
2 | Angle Sum Property of a Triangle | The sum of the interior angles of a triangle is always 180°. This is a key theorem used in many geometry problems. |
3 | Types of Angles | Understanding different types of angles such as complementary, supplementary, adjacent, and vertically opposite angles. |
4 | Parallel Lines and Transversals | The study of the relationship between angles when two parallel lines are cut by a transversal (alternate interior, corresponding, and co-interior angles). |
5 | Properties of Parallel Lines | Key properties of parallel lines cut by a transversal, such as the Corresponding Angles Theorem and Alternate Interior Angles Theorem. |
6 | Angle Relationships | Relations between different angles like vertically opposite angles, alternate angles, corresponding angles, and co-interior angles. |
7 | Proofs of Theorems | Proving theorems related to angles in triangles and parallel lines, such as the Angle Sum Property of a triangle and angles formed by transversals. |
8 | Construction of Angles | Drawing angles and constructing parallel lines using a protractor and compass. |
9 | Angle Bisectors | The concept of bisecting angles and constructing the bisector of an angle, along with its properties. |
10 | Application of Lines and Angles | Using the properties of angles in practical problems, including word problems involving parallel lines, transversals, and triangles. |
Benefits of Class 9 Maths Chapter 6 Lines and Angles Important Questions
Math is a subject that requires regular practice. Lines and angles class 9 important questions provide a question bank to the students on class 9 maths chapter 6
The important questions have been compiled along with their step by step detailed answers to help the students know their mistakes if any
These questions cover all the essential topics from the chapter lines and angles and help the students in their revision
Students can easily download important questions for class 9 maths lines and angles in a PDF format for referring to it any time.
With the help of the questions, students will get an idea of the pattern of the question they might get in the exams.
Revision can be done not only for the exams but also for the class test
Students can also take the help of these important questions for completing their assignments
These important questions provide 100 percent accuracy in the solutions and help the students in deriving their answers with ease.
We hope students have found this information on CBSE Important Questions for Class 9 Maths Chapter 6 important questions useful for their studies. Along with important questions, students can also access CBSE Class 9 Maths Chapter 6 NCERT Solutions, CBSE Class 9 Maths Chapter 6 Revision notes, and other related CBSE Class 9 Maths study material. Keep learning and stay tuned with Vedantu for further updates on CBSE Class 9 exams.
Conclusion
CBSE Class 9 Maths Important Questions for Chapter 6 - Lines and Angles serve as a valuable tool for students seeking to excel in their mathematics examinations. These questions are strategically curated to encapsulate the key concepts and theorems presented in the chapter. By working through them, students can reinforce their understanding of lines, angles, and their properties. These questions not only aid in exam preparation but also foster critical thinking and problem-solving skills. As students tackle these questions, they gain confidence in their geometric knowledge, ultimately paving the way for success in their Class 9 mathematics examinations.
FAQs on CBSE Class 9 Maths Important Questions - Chapter 6 Lines and Angles
Q1. Why is Practising Extra Questions for CBSE Class 9 Maths Chapter 6 Lines and Angles important for Students?
Ans: Many reputed online learning platforms prepare a repository of important questions for exam preparation. At Vedantu. Students can find important questions for Class 9 Maths Chapter 6 Lines and Angles. Practising these questions help students in scoring well in the subject. Solving extra questions for any chapter is a great way to boost students’ confidence. By solving important questions for Class 9 Maths Chapter 6 Lines and Angles, students will be able to practice the chapter properly during exam time. These questions will also help in revision.
Q2. Where can I find Important Questions for Maths Chapter 6 Lines and Angles?
Ans: Vedantu is India’s leading online learning platform which provides a well-prepared set of Important Questions for Class 9 Maths Chapter 6. Students can find extra questions for other chapters also on Vedantu. The site offers exceptional exam materials and relevant questions that can be asked in the exams. It is available as a free PDF of Important Questions for Class 9 Maths Chapter 6 Lines and Angles with solutions. The solutions to these questions are also provided by subject experts. These are beneficial in exam preparation and revision during exams.
Q3. What is the angle Sum Property?
Ans: As per the Angle Sum Property, the sum of interior angles of a triangle is equal to 180°. It is an extremely important theorem of maths that helps in many calculations related to a triangle. Students must know how to derive this theorem as it might be asked in the exam.
Q4. Does Vedantu Offer Solutions to the important Questions for Class 9 Maths Chapter 6?
Ans: Yes, of course! At Vedantu, students can find complete solutions for all the questions included in the PDF of important questions for Class 9 Maths Chapter 6. These solutions are prepared by subject matter experts who are well versed in the syllabus and exam guidelines. 100 percent accuracy is maintained in the solutions.
Q5. Where can I find these important questions?
Ans: CBSE Class 9 Maths Important Questions for Chapter 6 can often be found in study guides, online educational platforms, or by performing a quick internet search. They are designed to aid your revision and test your understanding of the chapter.
Q6. Are these questions sufficient for exam preparation, or should I also study the entire chapter thoroughly?
Ans: While important questions are valuable for focused revision, it's essential to have a strong grasp of the entire chapter's concepts. These questions should complement your overall study plan, not replace it.
Q7. Do these important questions include solutions or answers?
Ans: The availability of solutions may vary depending on the source. Some sets of important questions provide solutions or answers, while others may not. It's a good idea to check if solutions are included when using these questions for practice.
8. How can I apply the properties of angles in parallel lines?
In Chapter 6, students learn about properties like alternate interior angles, corresponding angles, and co-interior angles when two parallel lines are cut by a transversal. Practising these important questions will help students understand how to use these angle properties to solve geometrical problems effectively.
9. What are the types of angles covered in Class 9 Maths Chapter 6 - Lines and Angles?
Chapter 6 focuses on various types of angles, including adjacent angles, complementary angles, supplementary angles, and vertically opposite angles. Understanding their properties and relationships is key to solving problems in this chapter.
10. How do I prove the Angle Sum Property of a Triangle?
The Angle Sum Property states that the sum of the interior angles of a triangle is always 180°. In the important questions, students can practice proving this theorem using different methods such as drawing parallel lines or using alternate interior angles.
11. Can I expect questions on real-life applications of Lines and Angles in exams?
Yes, real-life applications such as construction of angles, use of parallel lines, and angle relationships in architecture and design are often asked in the form of word problems or diagrams. These practical applications help students connect theoretical concepts to real-world scenarios.
12. How should I prepare for proofs and theorems in Class 9 Maths Chapter 6 - Lines and Angles?
For proofs, especially theorems like the Angle Sum Property of a Triangle and properties related to parallel lines and transversals, it's important to practice the steps in a structured manner. Focus on the reasoning behind each step and use diagrams wherever necessary. Repeated practice through important questions will help you master these proofs.
13. Are the important questions of Class 9 Maths Chapter 6 - Lines and Angles based on the latest CBSE syllabus?
Yes, the important questions for Chapter 6 - Lines and Angles, provided by Vedantu, are based on the latest CBSE syllabus and exam pattern. They ensure that you are preparing according to the most recent guidelines.
14. What kind of problems should I expect in the exam from Class 9 Maths Chapter 6 - Lines and Angles?
In the exam, you can expect a mix of theoretical questions, problems on properties of angles, proofs, and practical application questions. Topics such as the Angle Sum Property, angles formed by transversals, and angle relationships in parallel lines are frequently tested.
15. Can these important questions of Class 9 Maths Chapter 6 - Lines and Angles help me in solving multiple-choice questions (MCQs)?
Yes, practising important questions helps you to understand the core concepts better, which is very useful for answering MCQs in exams. The questions test your conceptual clarity and problem-solving ability, both of which are essential for excelling in MCQs.