(1) Parallel lines: Two lines, having no common points, are parallel lines. [Undefined term: The distance between the two lines are constant] (i) Perpendicular lines: Two lines which form angle of 90° with each other, are perpendicular lines. (iii) Line segment: The line which have two end point, isRead more
(1) Parallel lines: Two lines, having no common points, are parallel lines. [Undefined term: The distance between the two lines are constant]
(i) Perpendicular lines: Two lines which form angle of 90° with each other, are perpendicular lines.
(iii) Line segment: The line which have two end point, is a line segment.
(iv) Radius of a Circle: The line segment from the centre of circle to the circle is radius of circle.
(v) Square: A quadrilateral having all sides equal and all angles right angle.
There are several undefined terms which should be listed. They are consistent, because they deal with two different situations (i) says that Given two points A and B, there is a point C lying on the line in between them; (ii) says that: Given A and B, you can take C not lying on the line through A aRead more
There are several undefined terms which should be listed. They are consistent, because they deal with two different situations
(i) says that
Given two points A and B, there is a point C lying on the line in between them;
(ii) says that:
Given A and B, you can take C not lying on the line through A and B.
These ‘postulates’ do not follow from Euclid’s postulates. However, they follow from Axiom 5.1.
Given: AC = BC ⇒ AC + AC = AC + BC [∵ Equals are added to equals] ⇒ 2AC = AB [∵ AC +BC coincides with AB] ⇒ AC = 1/2 AB [∵ Things which are halves of the same things are equal to one another]
Given: AC = BC
⇒ AC + AC = AC + BC [∵ Equals are added to equals]
⇒ 2AC = AB [∵ AC +BC coincides with AB]
⇒ AC = 1/2 AB [∵ Things which are halves of the same things are equal to one another]
(i) False, as there are infinite number of line that can pass through one point. (ii) False, because one and only one line can be drawn through two distinct points. (iii) True, because a terminated line can be produced both the sides infinitely. (iv) True, if two circles are equal (i.e. their areasRead more
(i) False, as there are infinite number of line that can pass through one point.
(ii) False, because one and only one line can be drawn through two distinct points.
(iii) True, because a terminated line can be produced both the sides infinitely.
(iv) True, if two circles are equal (i.e. their areas πr² are equal), then their radii are also equal.
(v) True, according to Euclid’s axiom, “Things which are equal to the same thing are equal to one another”
(i) To represent the equation 2x + 9 = 0 in one variable, we will use number line. 2x + 9 = 0 ⇒ x = -9/2 (ii) To represent the equation 2x + 9 = 0 in two variable, we will use Cartesian plane. Now the equation: 2x + 0.y = -9 ⇒ x = (-9 - 0.y)/2 Putting y = 1, we have, x = (-9 - 0 × 1)/2 = -9/2 PuttinRead more
(i) To represent the equation 2x + 9 = 0 in one variable, we will use number line.
2x + 9 = 0
⇒ x = -9/2
(ii) To represent the equation 2x + 9 = 0 in two variable, we will use Cartesian plane. Now the equation:
2x + 0.y = -9
⇒ x = (-9 – 0.y)/2
Putting y = 1, we have, x = (-9 – 0 × 1)/2 = -9/2
Putting y = 2, we have, x = (-9 – 0 × 2)/2 = -9/2
Hence, A(-9/2, 1) and B(-9/2, 2) are the two solutions of the given equation.
(i) Taking Celsius on x-axis and Fahrenheit on y-axis, the linear equation is given by: y = (9/5)x + 32 For plotting the graph: Putting x = 0, we have, y = (9/5) × 0 + 32 = 32 putting x = 5, we have, y = (9/5) × 5 + 32 = 41 Putting x = 10, we have, y = (9/5) × 10 + 32 = 50 Hence, A(0, 100), B(5, 41)Read more
(i) Taking Celsius on x-axis and Fahrenheit on y-axis, the linear equation is given by: y = (9/5)x + 32
For plotting the graph:
Putting x = 0, we have, y = (9/5) × 0 + 32 = 32
putting x = 5, we have, y = (9/5) × 5 + 32 = 41
Putting x = 10, we have, y = (9/5) × 10 + 32 = 50
Hence, A(0, 100), B(5, 41) and c(10, 50) are the solutions of the equation.
(ii) If the temperature is 30° C, then
F = (9/5) × 30 + 32 = 54 + 32 = 86
Hence, if the temperature is 30°C, the temperature in Fahrenheit is 86°F.
(iii) If the temperature is 95°F, then
95 = (9/5)C + 32
⇒ 95 – 32 = (9/5)C
⇒ 63 × 5/9 = C
⇒ C = 35°
If the temperature is 95°F, the temperature in Celsius is 35°C.
(iv) If temperature is 0°C, then
F = (9/5) × 0 + 32 = 0 + 32 = 32
If the temperature is 0°F, then
0 = (9/5)C + 32
⇒ -32 = (9/5)C
⇒ -32 × 5/9 = C
⇒ – 160/9 = C
⇒ C = -17.8°
If the temperature is 0°C, the temperature in Fahrenheit is 32°F and if the temperature is 0°F, the temperature in Celsius is -17.8°C.
(v) Let x° be the temperature which is numerically the same in both Fahrenheit and Celsius, then
x = (9/5)x + 32
⇒ x – 32 = (9/5)x
⇒ (x – 32) × 5 = 9x
⇒ 5x – 160 = 9x
⇒ 4x = -160
⇒ x – 40°
Hence, -40° is the temperature which is numerically the same in both Fahrenheit and Celsius.
(i) Equation y = 3 can be represented in one variable on number line. (ii) For two variables representation of y = 3, we will use Cartesian plane. Now the equation: 0.x + y = 3 ⇒ y = 3 - 0.x Putting x = 1, we have, y = 3 - 0.1 = 3 Putting x = 2, we have, y = 3 - 0.2 = 3 Hence, A(1, 3) and B(2, 3) arRead more
(i) Equation y = 3 can be represented in one variable on number line.
(ii) For two variables representation of y = 3, we will use Cartesian plane. Now the equation:
0.x + y = 3
⇒ y = 3 – 0.x
Putting x = 1, we have, y = 3 – 0.1 = 3
Putting x = 2, we have, y = 3 – 0.2 = 3
Hence, A(1, 3) and B(2, 3) are the two solutions of the given equation.
(i) (0,2) Given equation: x - 2y = 4 In x- 2y = 4, putting x = 0 and y = 2, we have, 0 - 2 × 2 = -4 ≠ 4 Therefore, (0, 2) is not a solution of the equation.
(i) (0,2)
Given equation: x – 2y = 4
In x- 2y = 4, putting x = 0 and y = 2, we have, 0 – 2 × 2 = -4 ≠ 4
Therefore, (0, 2) is not a solution of the equation.
(ii) (2, 0) Given equation: x - 2y = 4 Inx-2y = 4, putting x = 2 and y = 0, we have, 2 - 2 × 0 = 2 ≠ 4 Hence, (2, 0) is not a solution of the equation.
(ii) (2, 0)
Given equation: x – 2y = 4
Inx-2y = 4, putting x = 2 and y = 0, we have, 2 – 2 × 0 = 2 ≠ 4
Hence, (2, 0) is not a solution of the equation.
Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them?.
(1) Parallel lines: Two lines, having no common points, are parallel lines. [Undefined term: The distance between the two lines are constant] (i) Perpendicular lines: Two lines which form angle of 90° with each other, are perpendicular lines. (iii) Line segment: The line which have two end point, isRead more
(1) Parallel lines: Two lines, having no common points, are parallel lines. [Undefined term: The distance between the two lines are constant]
See less(i) Perpendicular lines: Two lines which form angle of 90° with each other, are perpendicular lines.
(iii) Line segment: The line which have two end point, is a line segment.
(iv) Radius of a Circle: The line segment from the centre of circle to the circle is radius of circle.
(v) Square: A quadrilateral having all sides equal and all angles right angle.
Consider two ‘postulates’ given below:
There are several undefined terms which should be listed. They are consistent, because they deal with two different situations (i) says that Given two points A and B, there is a point C lying on the line in between them; (ii) says that: Given A and B, you can take C not lying on the line through A aRead more
There are several undefined terms which should be listed. They are consistent, because they deal with two different situations
See less(i) says that
Given two points A and B, there is a point C lying on the line in between them;
(ii) says that:
Given A and B, you can take C not lying on the line through A and B.
These ‘postulates’ do not follow from Euclid’s postulates. However, they follow from Axiom 5.1.
If a point C lies between two points A and B such that AC = BC, then prove that AC = ½ AB. Explain by drawing the figure.
Given: AC = BC ⇒ AC + AC = AC + BC [∵ Equals are added to equals] ⇒ 2AC = AB [∵ AC +BC coincides with AB] ⇒ AC = 1/2 AB [∵ Things which are halves of the same things are equal to one another]
Given: AC = BC
See less⇒ AC + AC = AC + BC [∵ Equals are added to equals]
⇒ 2AC = AB [∵ AC +BC coincides with AB]
⇒ AC = 1/2 AB [∵ Things which are halves of the same things are equal to one another]
Which of the following statements are true and which are false? Give reasons for your answers.
(i) False, as there are infinite number of line that can pass through one point. (ii) False, because one and only one line can be drawn through two distinct points. (iii) True, because a terminated line can be produced both the sides infinitely. (iv) True, if two circles are equal (i.e. their areasRead more
(i) False, as there are infinite number of line that can pass through one point.
See less(ii) False, because one and only one line can be drawn through two distinct points.
(iii) True, because a terminated line can be produced both the sides infinitely.
(iv) True, if two circles are equal (i.e. their areas πr² are equal), then their radii are also equal.
(v) True, according to Euclid’s axiom, “Things which are equal to the same thing are equal to one another”
Give the geometric representations of 2x + 9 = 0 as an equation
(i) To represent the equation 2x + 9 = 0 in one variable, we will use number line. 2x + 9 = 0 ⇒ x = -9/2 (ii) To represent the equation 2x + 9 = 0 in two variable, we will use Cartesian plane. Now the equation: 2x + 0.y = -9 ⇒ x = (-9 - 0.y)/2 Putting y = 1, we have, x = (-9 - 0 × 1)/2 = -9/2 PuttinRead more
(i) To represent the equation 2x + 9 = 0 in one variable, we will use number line.
See less2x + 9 = 0
⇒ x = -9/2
(ii) To represent the equation 2x + 9 = 0 in two variable, we will use Cartesian plane. Now the equation:
2x + 0.y = -9
⇒ x = (-9 – 0.y)/2
Putting y = 1, we have, x = (-9 – 0 × 1)/2 = -9/2
Putting y = 2, we have, x = (-9 – 0 × 2)/2 = -9/2
Hence, A(-9/2, 1) and B(-9/2, 2) are the two solutions of the given equation.
In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius: F:(9/5) C+32
(i) Taking Celsius on x-axis and Fahrenheit on y-axis, the linear equation is given by: y = (9/5)x + 32 For plotting the graph: Putting x = 0, we have, y = (9/5) × 0 + 32 = 32 putting x = 5, we have, y = (9/5) × 5 + 32 = 41 Putting x = 10, we have, y = (9/5) × 10 + 32 = 50 Hence, A(0, 100), B(5, 41)Read more
(i) Taking Celsius on x-axis and Fahrenheit on y-axis, the linear equation is given by: y = (9/5)x + 32
For plotting the graph:
Putting x = 0, we have, y = (9/5) × 0 + 32 = 32
putting x = 5, we have, y = (9/5) × 5 + 32 = 41
Putting x = 10, we have, y = (9/5) × 10 + 32 = 50
Hence, A(0, 100), B(5, 41) and c(10, 50) are the solutions of the equation.
(ii) If the temperature is 30° C, then
F = (9/5) × 30 + 32 = 54 + 32 = 86
Hence, if the temperature is 30°C, the temperature in Fahrenheit is 86°F.
(iii) If the temperature is 95°F, then
95 = (9/5)C + 32
⇒ 95 – 32 = (9/5)C
⇒ 63 × 5/9 = C
⇒ C = 35°
If the temperature is 95°F, the temperature in Celsius is 35°C.
(iv) If temperature is 0°C, then
F = (9/5) × 0 + 32 = 0 + 32 = 32
If the temperature is 0°F, then
0 = (9/5)C + 32
⇒ -32 = (9/5)C
⇒ -32 × 5/9 = C
⇒ – 160/9 = C
⇒ C = -17.8°
If the temperature is 0°C, the temperature in Fahrenheit is 32°F and if the temperature is 0°F, the temperature in Celsius is -17.8°C.
(v) Let x° be the temperature which is numerically the same in both Fahrenheit and Celsius, then
See lessx = (9/5)x + 32
⇒ x – 32 = (9/5)x
⇒ (x – 32) × 5 = 9x
⇒ 5x – 160 = 9x
⇒ 4x = -160
⇒ x – 40°
Hence, -40° is the temperature which is numerically the same in both Fahrenheit and Celsius.
Give the geometric representations of y = 3 as an equation.
(i) Equation y = 3 can be represented in one variable on number line. (ii) For two variables representation of y = 3, we will use Cartesian plane. Now the equation: 0.x + y = 3 ⇒ y = 3 - 0.x Putting x = 1, we have, y = 3 - 0.1 = 3 Putting x = 2, we have, y = 3 - 0.2 = 3 Hence, A(1, 3) and B(2, 3) arRead more
(i) Equation y = 3 can be represented in one variable on number line.
See less(ii) For two variables representation of y = 3, we will use Cartesian plane. Now the equation:
0.x + y = 3
⇒ y = 3 – 0.x
Putting x = 1, we have, y = 3 – 0.1 = 3
Putting x = 2, we have, y = 3 – 0.2 = 3
Hence, A(1, 3) and B(2, 3) are the two solutions of the given equation.
Check which of the following are solution of the equation x – 2y = 4 and which are not: (0, 2)
(i) (0,2) Given equation: x - 2y = 4 In x- 2y = 4, putting x = 0 and y = 2, we have, 0 - 2 × 2 = -4 ≠ 4 Therefore, (0, 2) is not a solution of the equation.
(i) (0,2)
See lessGiven equation: x – 2y = 4
In x- 2y = 4, putting x = 0 and y = 2, we have, 0 – 2 × 2 = -4 ≠ 4
Therefore, (0, 2) is not a solution of the equation.
Check which of the following are solutions of the equation x – 2y = 4 and which are not: (2, 0)
(ii) (2, 0) Given equation: x - 2y = 4 Inx-2y = 4, putting x = 2 and y = 0, we have, 2 - 2 × 0 = 2 ≠ 4 Hence, (2, 0) is not a solution of the equation.
(ii) (2, 0)
See lessGiven equation: x – 2y = 4
Inx-2y = 4, putting x = 2 and y = 0, we have, 2 – 2 × 0 = 2 ≠ 4
Hence, (2, 0) is not a solution of the equation.
Check which of the following are solutions of the equation x – 2y = 4 and which are not: (4, 0)
(iii) (4,0) Given equation: x - 2y = 4 In x-2y = 4, putting x = 4 and y = 0, we have, 4 - 2 × 0 = 4 Hence, (4,0) is a solution of the equation.
(iii) (4,0)
See lessGiven equation: x – 2y = 4
In x-2y = 4, putting x = 4 and y = 0, we have, 4 – 2 × 0 = 4
Hence, (4,0) is a solution of the equation.