Class 9 Maths RD Sharma Solutions – Page 16

RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles VSAQS

January 5, 2022June 22, 2018 by Prasanna

RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles VSAQS

Other Exercises

Question 1.
How many least number of distinct points determine a unique line?
Solution:
At least two distinct points determine a unique line.

Question 2.
How many lines can be drawn through both of the given points?
Solution:
Through two given points, one line can be drawn.

Question 3.
How many lines can be drawn through a given point?
Solution:
Through a given point, infinitely many lines can be drawn.

Question 4.
in how many points two distinct lines can intersect?
Solution:
Two distinct lines can intersect at the most one point.

Question 5.
In how many points a line, not in a plane, can intersect the plane?
Solution:
A line not in a plane, can intersect the plane at one point.

Question 6.
In how many points two distinct planes can intersect?
Solution:
Two distinct planes can intersect each other at infinite number of points.

Question 7.
In how many lines two distinct planes can intersect?
Solution:
Two distinct planes intersect each other in one line.

Question 8.
How many least number of distinct points determine a unique plane?
Solution:
Three non-collinear points can determine a unique plane.

Question 9.
Given three distinct points in a plane, how many lines can be drawn by joining them?
Solution:
Through three given points, one line can be drawn of they are collinear and three if they are non-collinear.

Question 10.
How many planes can be made to pass through a line and a point not on the line?
Solution:
Only one plane can be made to pass through a line and a point not on the line.

Question 11.
How many planes can be made to pass through two points?
Solution:
Infinite number of planes can be made to pass through two points.

Question 12.
How many planes can be made to pass through three distinct points?
Solution:
Infinite number of planes can be made of they are collinear and only one plane, if they are non-collinear.

Hope given RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles VSAQS are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles Ex 9.1

January 5, 2022June 22, 2018 by Prasanna

RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles Ex 9.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles Ex 9.1

Other Exercises

Question 1.
Define the following tenns :
(i) Line segment
(ii) Collinear points
(iii) Parallel lines
(iv) Intersecting lines
(v) Concurrent lines
(vi) Ray
(vii) Half-line.
Solution:
(i) A line segment is a part of a line which lies between two points on it and it is denoted as \(\overline { AB }\) or only by AB. It has two end points and is measureable.

(ii) Three or more points which lie on the same straight line, are called collinear points.
(iii) Two lines which do not intersect each other at any point are called parallel lines.
(iv) If two lines have one point in common, are called intersecting lines.
(v) If two or more lines which pass through a common point are called concurrent lines.
(vi) Ray : A part of a line which has one end point.
(vii) Half line : If A, B, C, be the points on a line l, such that A lies between B and C and we delete the point from line l, two parts of l that remain are each called a half-line.
RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles Ex 9.1 Q1.2

Question 2.
(i) How many lines can pass through a given point?
(ii) In how many points can two distinct lines at the most intersect?
Solution:
(i) Infinitely many lines can pass through a given point.
(ii) Two distinct lines at the most intersect at one point.

Question 3.
(i) Given two points P and Q, find how many line segments do they determine?
(ii) Name the line segments determined by the three collinear points P, Q and R.
Solution:
(i) Only one line segment can be drawn through two given points P and Q.
(ii) Three collinear points P, Q and R, three lines segments determine : \(\overline { PQ }\) , \(\overline { QR }\) and \(\overline { PR }\) .

Question 4.
Write the truth value (T/F) of each of the following statements:
(i) Two lines intersect in a point.
(ii) Two lines may intersect in two points.
(iii) A segment has no length.
(iv) Two distinct points always determine a line.
(v) Every ray has a finite length.
(vi) A ray has one end-point only.
(vii) A segment has one end-point only.
(viii) The ray AB is same as ray BA.
(ix) Only a single line may pass through a given point.
(x) Two lines are coincident if they have only one point in common.
Solution:
(i) False : As two lines do not intersect also any a point.
(ii) False : Two lines intersect at the most one point.
(iii) False : A line segment has definitely length.
(iv) True.
(v) False : Every ray has no definite length.
(vi) True.
(vii) False : A segment has two end point.
(viii)False : Rays AB and BA are two different rays.
(ix) False : Through a given point, infinitely many lines can pass.
(x) False : Two lines are coincident of each and every points coincide each other.

Question 5.
In the figure, name the following:
RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles Ex 9.1 Q5.1

(i) Five line segments.
(ii) Five rays.
(iii) Four collinear points.
(iv) Two pairs of non-intersecting line segments.
Solution:
From the given figure,
(i) Five line segments are AC, PQ, PR, RS, QS.
(ii) Five rays : \(\xrightarrow { PA }\) , \(\xrightarrow { RB }\) , \(\xrightarrow { PB }\) , \(\xrightarrow { CS }\) , \(\xrightarrow { DS }\) .
(iii) Four collinear points are : CDQS, APR, PQL, PRB.
(iv) Two pairs of non-intersecting line segments an AB and CD, AP and CD, AR and CS, PR and QS.

Question 6.
Fill in the blanks so as to make the following statements true:
(i) Two distinct points in a plane determine a _____ line.
(ii) Two distinct_____ in a plane cannot have more than one point in common.
(iii) Given a line and a point, not on the line, there is one and only _____ line which passes through the given point and is_____ to the given line.
(iv) A line separates a plane into ____ parts namely the____ and the____ itself.
Solution:
(i) Two distinct points in a plane determine a unique line.
(ii) Two distinct lines in a plane cannot have more than one point in common.
(iii) Given a line and a point, not on the line, there is one and only perpendicular line which passes through the given point and is perpendicular to the given line.
(iv) A line separates a plane into three parts namely the two half planes, and the one line itself.

Hope given RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles Ex 9.1 are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 8 Lines and Angles MCQS

January 5, 2022June 22, 2018 by Prasanna

RD Sharma Class 9 Solutions Chapter 8 Lines and Angles MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 8 Lines and Angles MCQS

Other Exercises

Mark the correct alternative in each of the following:
Question 1.
The point of intersect of the co-ordiante axes is
(a) ordinate
(b) abscissa
(c) quadrant
(d) origin
Solution:
Origin (d)

Question 2.
The abscissa and ordinate of the origin are
(a) (0, 0)
(b) (1, 0)
(c) (0, 1)
(d) (1, 1)
Solution:
The abscissa and ordinate of the origin are (0, 0). (a)

Question 3.
The measure of the angle between the co-ordinate axes is
(a) 0°
(b)   90°
(c) 180°
(d)   360°
Solution:
The measure of the angle between the coordinates of axes is 90°. (b)

Question 4.
A point whose abscissa and ordinate are 2 and -5 respectively, lies in
(a) First quadrant
(b) Second quadrant
(c) Third quadrant
(d) Fourth quadrant
Solution:
The point whose abscissa is 2 and ordinate -5 will lies in fourth quadrant. (d)

Question 5.
Points (-4, 0) and (7, 0) lie
(a) on x-axis
(b)   y-axis
(c) in first quadrant
(d) in second quadrant
Solution:
∵ The ordinates of both the points are 0
∴ They lie on x-axis            (a)

Question 6.
The ordinate of any point on x-axis is
(a) 0
(b) 1
(c) -1
(d) any number
Solution:
The ordinate of any point lying on x-axis is 0.          (a)

Question 7.
The abscissa of any point on y-axis is
(a) 0
(b) 1
(c) -1
(d) any number
Solution:
The abscissa of any point on y-axis is 0. (a)

Question 8.
The abscissa of a point is positive in the
(a) First and Second quadrant
(b) Second and Third quadrant
(c) Third and Fourth quadrant
(d) Fourth and First quadrant
Solution:
The abscissa of a point is positive in the fourth and First quadrant. (d)

Question 9.
A point whose abscissa is -3 and ordinate 2 lies in
(a) First quadrant
(b) Second quadrant
(c) Third quadrant
(d) Fourth quadrant
Solution:
A point (-3, 2) will lies in second quadrant. (b)

Question 10.
Two points having same abscissae but different ordinates lie on
(a) x-axis
(b) y-axis
(c) a line parallel to y-axis
(d) a line parallel to x-axis
Solution:
Two points having same abscissae but different ordinates is a line parallel to y- axis. (c)

Question 11.
The perpendicular distance of the point P (4, 3) from x-axis is
(a) 4
(b) 3
(c) 5
(d) none   of these
Solution:
The perpendiculat distance of the point P (4, 3) from x-axis is 3.         (b)

Question 12.
The perpendicular distance of the point P (4, 3) from y-axis is
(a) 4
(b) 3
(c) 5
(d) none of these
Solution:
perpendicular distance of the point P (4, 3) from y-axis is 4.         (a)

Question 13.
The distance of the point P (4, 3) from the origin is
(a) 4
(b) 3
(c) 5
(d) 7
Solution:
The distance of the point P (4, 3) from origin is \(\sqrt { { 4 }^{ 2 }+{ 3 }^{ 3 } }\)
= \(\sqrt { 16+9 }\)
= \(\sqrt { 25 }\) = 5 (c)

Question 14.
The area of the triangle formed by the points A (2, 0), B (6, 0) and C (4, 6) is
(a) 24 sq. units
(b) 12 sq. units
(c) 10 sq. units
(d) none of these
Solution:
RD Sharma Class 9 Solutions Chapter 8 Lines and Angles MCQS Q14.1

Question 15.
The area of the triangle formed by the points P (0, 1), Q (0, 5) and R (3, 4) is
(a) 16 sq. units
(b) 8 sq. units
(c) 4 sq. units
(d) 6 sq. units
Solution:
RD Sharma Class 9 Solutions Chapter 8 Lines and Angles MCQS Q15.1

Hope given RD Sharma Class 9 Solutions Chapter 8 Lines and Angles MCQS are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry MCQS

January 5, 2022June 22, 2018 by Prasanna

RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry MCQS

Other Exercises

Mark the correct alternative in each of the following:
Question 1.
If (4, 19) is a solution of the equation y = ax + 3, then a =
(a) 3
(b) 4
(c) 5
(d) 6
Solution:
∵ (4, 19) is a solution of equation
y = ax + 3
∴ x = 4, y= 19 will satisfy the equation
∴ 19 = a x 4 + 3 = 4a + 3
4a = 19-3 = 16 ⇒ a= \(\frac { 16 }{ 4 }\) = 4
∴ a = 4 (b)

Question 2.
If (a, 4) lies on the graph of 3x + y = 10, then the value of a is
(a) 3
(b) 1
(c) 2
(d) 4
Solution:
∵ (a, 4) is the solution of the equation 3x + y = 10
∴ x = a, y = 4 will satisfy the equation
∴ Substituting the value of x and y in the equation
3 xa + 4= 10 ⇒ 3a =10- 4 = 6
⇒ a = \(\frac { 6 }{ 3 }\) = 2
∴ a = 2 (c)

Question 3.
The graph of the linear equation 2x – y= 4 cuts x-axis at
(a) (2, 0)
(b) (-2, 0)
(c) (0, -4)
(d) (0, 4)
Solution:
∵ graph of the equation,
2x – y = 4 cuts x-axis
∴ y = 0
∴ 2x – 0 = 4 ⇒ 2x = 4
⇒ x = \(\frac { 4 }{ 2 }\) = 2
∴ The line cuts x-axis at (2, 0) (a)

Question 4.
How many linear equations are satisfied by x = 2 and y = -3 ?
(a) Only one
(b)   Two
(c) Three
(d)    Infinitely many
Solution:
∵ From a point, infinitely number of lines can pass.
∴ The solution x = 2, y = -3 is the solution of infinitely many linear equations.       (d)

Question 5.
The equation x – 2 = 0 on number line is represented by
(a) aline
(b) a point
(c) infinitely many lines
(d) two lines
Solution:
The equation x – 2 = 0
⇒ x = 2
∴ It is representing by a point on a number line. (b)

Question 6.
x = 2, y = -1 is a solution of the linear equation
(a) x   + 2y = 0
(b) x + 2y = 4
(c) 2x + y = 0
(d) 2x + y = 5
Solution:
x = 2, y = -1
Substituting the values of x and y in the equations one by one, we get (a) x + 2y = 0
⇒ 2 + 2(-1) = 0
⇒ 2 – 2 = 0
⇒ 0 = 0 which is true                             (a)

Question 7.
If (2k – 1, k) is a solution of the equation 10x – 9y = 12, then k =
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
∵ (2k – 1, k) is a solution of the equation 10x – 9y = 12
Substituting the value of x and y in the equation
10(2k – 1) – 9k = 12
⇒ 20k – 10-9k= 12
⇒ 20k – 9k = 12 + 10
⇒ 11k = 22
⇒ k =\(\frac { 22 }{ 11 }\) = 2∴ k = 2                                                 (b)

Question 8.
The distance between the graph of the equation x = – 3    and x   = 2      is
(a) 1
(b) 2
(c) 3
(d) 5
Solution:
The distance between the graphs of the equation
x = -3 and x = 2 will be
2(-3) = 2+ 3 = 5 (b)

Question 9.
The distance   between the graphs of the equations y = -1 and y = 3    is
(a) 2
(b) 4
(c) 3
(d) 1
Solution:
The distance between the graphs of the equation
y = -1 and y = 3
is 3 – (-1) = 3 + 1 = 4                            (b)

Question 10.
If the graph of the equation 4x + 3y = 12 cuts the co-ordinate axes at A and B, then hypotenuse of right triangle AOB is of length
(a) 4 units
(b) 3 units
(c) 5 units
(d) none of these
Solution:
Equation is 4x + 3y = 12
If it cuts the x-axis, then y = 0
∴ 4x x 3 x 0 = 12
⇒ 4x = 12 ⇒ x = \(\frac { 12 }{ 4 }\) = 3
OA = 3 units
∴ The point of intersection of x-axis is (3, 0)
Again if it cuts the y-axis, then x = 0 , Y= 0
∴ 4x x 3 x 0 = 12
⇒ 4x = 12 ⇒ x = \(\frac { 12 }{ 3 }\) = 4
⇒ OB = 4 units
∴ The point of intersection is (0, 4)
∴ In right ΔAOB,
AB² = AO² + OB²= (3)² + (4)²
= 9 + 16 = 25
= (5)²∴ AB = 5 units (c)

Hope given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS

January 5, 2022June 21, 2018 by Prasanna

RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS

Other Exercises

Question 1.
Write the equation representing x-axis.
Solution:
The equation of x-axis is, y = 0.

Question 2.
Write the equation representing y-axis.
Solution:
The equation of y-axis is, x = 0.

Question 3.
Write the equation of a line passing through the point (0, 4) and parallel to x-axis.
Solution:
The equation of the line passing through the point (0,4) and parallel to x-axis will be y = 4.

Question 4.
Write the equation of a line passing through the point (3, 5) and parallel to x-axis.
Solution:
The equation of the line passing through the point (3, 5) and parallel to x-axis will be y = 5.

Question 5.
Write the equation of a line parallel toy-axis and passing through the point (-3, -7).
Solution:
The equations of the line passing through the point (-3, -7) and parallel to y-axis will be x = -3.

Question 6.
A line passes through the point (-4, 6) and is parallel to x-axis. Find its equation. A line passes through the point (-4, 6) and is parallel to x-axis. Find its equation.
Solution:
A line parallel to x-axis and passing through the point (-4, 6) will be y = 6.

Question 7.
Solve the equation 3x – 2 = 2x + 3 and represent the solution on the number line.
Solution:
3x – 2 = 2x + 3
⇒ 3x – 2x = 3 + 2 (By terms formation)
⇒ x = 5
∴ x = 5
Solution on the number line is
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS Q7.1

Question 8.
Solve the equation 2y – 1 = y + 1 and represent it graphically on the coordinate plane.

Solution:
2y – 1 = y + 1
⇒ 2y – y = 1 +1
⇒ y = 2
∴ It is a line parallel to x-axis at a distance of 2 units above the x-axis is y = 2.

Question 9.
If the point (a, 2) lies on the graph of the linear equation 2x – 3y + 8 = 0, find the value of a.
Solution:
∵ Points (a, 2) lies on the equation
2x – 3y + 8 = 0
∴ It will satisfy the equation,
Now substituting the value of x = a, y = 2 in the equation
⇒ 2a – 3 x 2+ 8= 0
⇒ 2a + 2= 0
⇒ 2a = -2
⇒ a = \(\frac { -2 }{ 2 }\) = -1
∴ a = -1

Question 10.
Find the value of k for which the point (1, -2) lies on the graph of the linear equation, x – 2y + k = 0.
Solution:
∵ Point (1, -2) lies on the graph of the equation x – 2y + k = 0
∴ x = 1, y = -2 will satisfy the equation
Now substituting the value of x = 1, y = -2 in it
1-2 (-2) + k = 0
⇒ 1 + 4 + k = 0
⇒ 5+ k = 0 ⇒ k =-5
∴ k = -5

Hope given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.