Given: Lines AB and CD intersects at O such that ∠BOD = 40° and ∠AOC + ∠BOE = 70° ...(1) ∠AOC = ∠BOD [∵ Vertically Opposite Angles] Hence, ∠AOC = 40° [∵ ∠BOD = 40°] Therefore, from the equation (1), we have, 40° + ∠BOE = 70° ⇒ ∠BOE = 70° - 40° = 30° Here, ∠AOC + ∠BOE + ∠COE = 180° [∵ AOB is a straigRead more

Given: Lines AB and CD intersects at O such that ∠BOD = 40° and
∠AOC + ∠BOE = 70° …(1)
∠AOC = ∠BOD [∵ Vertically Opposite Angles]
Hence, ∠AOC = 40° [∵ ∠BOD = 40°]
Therefore, from the equation (1), we have,
40° + ∠BOE = 70°
⇒ ∠BOE = 70° – 40° = 30°
Here, ∠AOC + ∠BOE + ∠COE = 180° [∵ AOB is a straight line]
⇒ 70° + ∠COE = 180° [From the equation (1)]
⇒ ∠COE = 180° – 70° = 110° and
Relex ∠COE = 360° – ∠COE = 360° – 110° = 250°

(iv) 3 = 2x +y ⇒ y = 3 - 2x Putting x = 0, we have, y = 3 - 2 × 0 = 3 Putting x = 1, we have, y = 3 - 2 × 1 = 1 Hence, G(0,3) and H(1,1) are the solutions of the equation.

(iv) 3 = 2x +y
⇒ y = 3 – 2x
Putting x = 0, we have, y = 3 – 2 × 0 = 3
Putting x = 1, we have, y = 3 – 2 × 1 = 1
Hence, G(0,3) and H(1,1) are the solutions of the equation.

Equation of two lines passing through (2, 14) are given by: x + y = 16 and 8x - y = 2. There are infinite number of lines that can pass through (2, 4) as infinite number of lines passes through a point.

Equation of two lines passing through (2, 14) are given by: x + y = 16 and 8x – y = 2.
There are infinite number of lines that can pass through (2, 4) as infinite number of lines passes through a point.

Given that: Distance travelled = x km and total fare = ₹ y Total fare = Fare for first km + Fare for remaining distance Therefore, the equation: y = 8 + 5 × (x - 1) ⇒ y = 5x +3 For the graph: Putting x = 1, we have, y = 5 × 1 + 3 = 8 Putting x = 2, we have, y = 5 × 2 + 3 = 13 Putting x = 3, we have,Read more

Given that: Distance travelled = x km and total fare = ₹ y
Total fare = Fare for first km + Fare for remaining distance
Therefore, the equation: y = 8 + 5 × (x – 1) ⇒ y = 5x +3
For the graph:
Putting x = 1, we have, y = 5 × 1 + 3 = 8
Putting x = 2, we have, y = 5 × 2 + 3 = 13
Putting x = 3, we have, y = 5 × 3 + 3 = 18
Hence, A(1, 8), B(2, 13) and C[3, 18) are solutions of the equation.

## In Fig. 6.13, lines AB and CD intersect at O.

Given: Lines AB and CD intersects at O such that ∠BOD = 40° and ∠AOC + ∠BOE = 70° ...(1) ∠AOC = ∠BOD [∵ Vertically Opposite Angles] Hence, ∠AOC = 40° [∵ ∠BOD = 40°] Therefore, from the equation (1), we have, 40° + ∠BOE = 70° ⇒ ∠BOE = 70° - 40° = 30° Here, ∠AOC + ∠BOE + ∠COE = 180° [∵ AOB is a straigRead more

Given: Lines AB and CD intersects at O such that ∠BOD = 40° and

See less∠AOC + ∠BOE = 70° …(1)

∠AOC = ∠BOD [∵ Vertically Opposite Angles]

Hence, ∠AOC = 40° [∵ ∠BOD = 40°]

Therefore, from the equation (1), we have,

40° + ∠BOE = 70°

⇒ ∠BOE = 70° – 40° = 30°

Here, ∠AOC + ∠BOE + ∠COE = 180° [∵ AOB is a straight line]

⇒ 70° + ∠COE = 180° [From the equation (1)]

⇒ ∠COE = 180° – 70° = 110° and

Relex ∠COE = 360° – ∠COE = 360° – 110° = 250°

## Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the fifth postulate.)

Axiom 5: The whole is greater than the part. Since this is true for anything in any part of the world, this is a universal truth.

Axiom 5:

See lessThe whole is greater than the part.

Since this is true for anything in any part of the world, this is a universal truth.

## Draw the graph of each of the following linear equations in two variables: 3 = 2x + y

(iv) 3 = 2x +y ⇒ y = 3 - 2x Putting x = 0, we have, y = 3 - 2 × 0 = 3 Putting x = 1, we have, y = 3 - 2 × 1 = 1 Hence, G(0,3) and H(1,1) are the solutions of the equation.

(iv) 3 = 2x +y

See less⇒ y = 3 – 2x

Putting x = 0, we have, y = 3 – 2 × 0 = 3

Putting x = 1, we have, y = 3 – 2 × 1 = 1

Hence, G(0,3) and H(1,1) are the solutions of the equation.

## Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Equation of two lines passing through (2, 14) are given by: x + y = 16 and 8x - y = 2. There are infinite number of lines that can pass through (2, 4) as infinite number of lines passes through a point.

Equation of two lines passing through (2, 14) are given by: x + y = 16 and 8x – y = 2.

See lessThere are infinite number of lines that can pass through (2, 4) as infinite number of lines passes through a point.

## If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.

Given equation of line: 3y = ax + 7. Putting x = 3 and y = 4, we have, 3 x 4 = a x3 +7 ⇒ 12 = 3a + 7 ⇒ 12 - 7 = 3a ⇒ a = 5/3

Given equation of line: 3y = ax + 7.

See lessPutting x = 3 and y = 4, we have, 3 x 4 = a x3 +7

⇒ 12 = 3a + 7 ⇒ 12 – 7 = 3a

⇒ a = 5/3

## The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph.

Given that: Distance travelled = x km and total fare = ₹ y Total fare = Fare for first km + Fare for remaining distance Therefore, the equation: y = 8 + 5 × (x - 1) ⇒ y = 5x +3 For the graph: Putting x = 1, we have, y = 5 × 1 + 3 = 8 Putting x = 2, we have, y = 5 × 2 + 3 = 13 Putting x = 3, we have,Read more

Given that: Distance travelled = x km and total fare = ₹ y

See lessTotal fare = Fare for first km + Fare for remaining distance

Therefore, the equation: y = 8 + 5 × (x – 1) ⇒ y = 5x +3

For the graph:

Putting x = 1, we have, y = 5 × 1 + 3 = 8

Putting x = 2, we have, y = 5 × 2 + 3 = 13

Putting x = 3, we have, y = 5 × 3 + 3 = 18

Hence, A(1, 8), B(2, 13) and C[3, 18) are solutions of the equation.

## The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph.

## Use suitable identities to find the following products: (x + 4) (x + 10)

(i) (x + 4)(x + 10) = x² + (4 + 10)x + 4 × 10 [∵ (x + a)(x + b) = x² + (a + b)x + ab ] = x² + 14x + 40

(i) (x + 4)(x + 10)

See less= x² + (4 + 10)x + 4 × 10 [∵ (x + a)(x + b) = x² + (a + b)x + ab ]

= x² + 14x + 40

## Use suitable identities to find the following products: (x + 8) (x – 10)

(ii) (x + 8)(x + -10) = x² + (8 - 10)x + 8 × (-10) [∵ (x + a)(x + b) = x² + (a + b)x + ab ] = x² - 2x - 80

(ii) (x + 8)(x + -10)

See less= x² + (8 – 10)x + 8 × (-10) [∵ (x + a)(x + b) = x² + (a + b)x + ab ]

= x² – 2x – 80

## Use suitable identities to find the following products: (3x + 4) (3x – 5)

(iii) (3x + 4)(3x - 5) = (3x)² + (4 - 5)3x + 4 × (-5) [∵ (x + a)(x + b) = x² + (a + b)x + ab ] = 9x² - 3x - 20

(iii) (3x + 4)(3x – 5)

See less= (3x)² + (4 – 5)3x + 4 × (-5) [∵ (x + a)(x + b) = x² + (a + b)x + ab ]

= 9x² – 3x – 20