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In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that CD = 2 AB? Give reasons for your answer.
No, we cannot conclude that CD = 2AB. The distance of chord AB from the centre being twice the distance of chord CD does not mean that CD is twice AB. According to Theorem 8, the longer chord is closer to the centre, but the relationship between chord length and distance is not directly proportionalRead more
No, we cannot conclude that CD = 2AB. The distance of chord AB from the centre being twice the distance of chord CD does not mean that CD is twice AB. According to Theorem 8, the longer chord is closer to the centre, but the relationship between chord length and distance is not directly proportional. Chord length also depends on the radius of the circle. Therefore, the given distances alone are not sufficient to establish CD = 2AB.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/
See lessFind the product without multiplying directly: (18 x 29)
To find this product using an identity approach, we can break down both numbers relative to base tens, writing them as the binomial product (20 - 2)(30 - 1). Applying the distributive expansion gives 20 times 30 minus 20 times 1 minus 2 times 30 plus 2 times 1. This evaluates to 600 minus 20 minus 6Read more
To find this product using an identity approach, we can break down both numbers relative to base tens, writing them as the binomial product (20 – 2)(30 – 1). Applying the distributive expansion gives 20 times 30 minus 20 times 1 minus 2 times 30 plus 2 times 1. This evaluates to 600 minus 20 minus 60 plus 2. Solving this basic arithmetic sequence yields 522.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessA telecom company charges rupees 600 for a certain recharge scheme. This prepaid balance is reduced by rupees 15 each day after the recharge. (i) Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay. (ii) After how many days will the balance run out? (iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x), reduces with time.
(i) (The prepaid balance is represented by: b(x) = 600 - 15x Here, x shows the number of days and b(x) shows the remaining balance. This is a linear decay because the balance decreases regularly by 15 rupees every day. The reduction is constant, so the graph forms a straight line with a negative sloRead more
(i) (The prepaid balance is represented by:
b(x) = 600 – 15x
Here, x shows the number of days and b(x) shows the remaining balance. This is a linear decay because the balance decreases regularly by 15 rupees every day. The reduction is constant, so the graph forms a straight line with a negative slope.
(ii) To find when the balance becomes zero:
600 – 15x = 0
Adding 15x:
15x = 600
Dividing by 15:
x = 40
Therefore, the prepaid balance will completely run out after 40 days.
(iii) To show how the prepaid balance reduces over time, we calculate the values of b(x) = 600 – 15x for x ranging from 1 to 10 days.
On Day 1, the balance is 600 – 15(1) = 585 rupees.
On Day 2, it is 600 – 15(2) = 570 rupees.
On Day 3, it is 600 – 15(3) = 555 rupees.
On Day 4, it is 600 – 15(4) = 540 rupees.
On Day 5, it is 600 – 15(5) = 525 rupees.
On Day 6, it is 600 – 15(6) = 510 rupees.
On Day 7, it is 600 – 15(7) = 495 rupees.
On Day 8, it is 600 – 15(8) = 480 rupees.
On Day 9, it is 600 – 15(9) = 465 rupees.
On Day 10, it is 600 – 15(10) = 450 rupees.
This structured breakdown clearly illustrates the steady, uniform daily loss of 15 rupees.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-2/
See lessCan you identify the terms, variables and coefficients of this algebraic expression?
The algebraic expression for the total cost is 200l + 160w + 50lw. In this expression, the terms are 200l, 160w and 50lw. The variables used are l and w, which represent the length and width of the garden. The numbers multiplying the variables are called coefficients. Therefore, 200 is the coefficieRead more
The algebraic expression for the total cost is 200l + 160w + 50lw. In this expression, the terms are 200l, 160w and 50lw. The variables used are l and w, which represent the length and width of the garden. The numbers multiplying the variables are called coefficients. Therefore, 200 is the coefficient of l, 160 is the coefficient of w and 50 is the coefficient in the term 50lw. This expression does not contain any constant term.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-2/
See lessHow much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies the brakes?
A sufficient safe distance should be maintained from the truck so that the following vehicle has enough time and space to react and stop safely if the truck suddenly applies its brakes. The required distance depends on factors such as the vehicle’s speed, driver’s reaction time, road condition and bRead more
A sufficient safe distance should be maintained from the truck so that the following vehicle has enough time and space to react and stop safely if the truck suddenly applies its brakes. The required distance depends on factors such as the vehicle’s speed, driver’s reaction time, road condition and braking capacity. A vehicle travelling at higher speed generally requires a greater safe distance to avoid collision.
For more NCERT Solutions of Class 9 Science Exploration Chapter 4 Describing Motion Around Us Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-4/
See less