Using the Ratio to Find x and y

Oct 13, 2025 by wbrum.org 84 views

Introduction

In geometry, a circle is a set of points that are all equidistant from a central point called the centre. The diameter of a circle is a line segment that passes through the centre and connects two points on the circle. In this problem, we are given a circle with centre 0, and a diameter AD. We are also given a point B on the circle, and a ratio DAB = 50. Our goal is to find the coordinates of point B, denoted by x and y.

Understanding the Problem

To solve this problem, we need to understand the properties of circles and diameters. A circle is a set of points that are all equidistant from a central point. The diameter of a circle is a line segment that passes through the centre and connects two points on the circle. In this case, the diameter AD passes through the centre 0 and connects points A and D.

Drawing a Diagram

Let's draw a diagram to visualize the problem. We can draw a circle with centre 0 and diameter AD. We can also draw a point B on the circle. Since DAB = 50, we can draw a line segment AB that is 50 times the length of BD.

  A
  / \
 /   \
/_____\
|  0  |
|_____/
  D
  \ /
   \/
    B

Using the Ratio to Find x and y

Since DAB = 50, we can write an equation to represent the ratio of the lengths of AB and BD. Let's say the length of BD is x. Then, the length of AB is 50x. We can use the Pythagorean theorem to find the length of AD.

Applying the Pythagorean Theorem

The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, the right-angled triangle is ABD, and the hypotenuse is AD.

AD^2 = AB^2 + BD^2
AD^2 = (50x)^2 + x^2
AD^2 = 2500x^2 + x^2
AD^2 = 2501x^2

Finding the Length of AD

Since AD is the diameter of the circle, we know that AD is equal to twice the radius of the circle. Let's call the radius of the circle r. Then, AD = 2r.

AD = 2r
AD^2 = (2r)^2
AD^2 = 4r^2

Equating the Two Expressions for AD^2

We can equate the two expressions for AD^2 to find the value of x.

2501x^2 = 4r^2
x^2 = 4r^2 / 2501
x^2 = 4r^2 * (1/2501)
x^2 = 4r^2 * (1/2501)
x = sqrt(4r^2 * (1/2501))
x = 2r * sqrt(1/2501)
x = 2r * (1/sqrt(2501))
x = 2r * (1/50.01)
x = 0.04r

Finding the Value of r

Since the diameter AD is equal to twice the radius r, we can write an equation to represent the relationship between AD and r.

AD = 2r
AD^2 = (2r)^2
AD^2 = 4r^2

Using the Given Information to Find r

We are given that DAB = 50, which means that the length of AB is 50 times the length of BD. We can use this information to find the value of r.

AB = 50BD
AB^2 = (50BD)^2
AB^2 = 2500BD^2
AB^2 = 2500x^2

Equating the Two Expressions for AB^2

We can equate the two expressions for AB^2 to find the value of r.

2500x^2 = 4r^2
x^2 = 4r^2 / 2500
x^2 = 4r^2 * (1/2500)
x = sqrt(4r^2 * (1/2500))
x = 2r * sqrt(1/2500)
x = 2r * (1/sqrt(2500))
x = 2r * (1/50)
x = 0.02r

Finding the Value of x

We can now find the value of x by substituting the value of r into the expression for x.

x = 0.02r
x = 0.02 * (AD/2)
x = 0.02 * (AD/2)
x = 0.01AD

Finding the Value of AD

We are given that DAB = 50, which means that the length of AB is 50 times the length of BD. We can use this information to find the value of AD.

AB = 50BD
AB^2 = (50BD)^2
AB^2 = 2500BD^2
AB^2 = 2500x^2

Equating the Two Expressions for AB^2

We can equate the two expressions for AB^2 to find the value of AD.

2500x^2 = 4r^2
x^2 = 4r^2 / 2500
x^2 = 4r^2 * (1/2500)
x = sqrt(4r^2 * (1/2500))
x = 2r * sqrt(1/2500)
x = 2r * (1/sqrt(2500))
x = 2r * (1/50)
x = 0.02r

Finding the Value of AD

We can now find the value of AD by substituting the value of r into the expression for AD.

AD = 2r
AD = 2 * (0.02AD)
AD = 0.04AD
1 = 0.04
AD = 25

Finding the Value of x

We can now find the value of x by substituting the value of AD into the expression for x.

x = 0.01AD
x = 0.01 * 25
x = 0.25

Finding the Value of y

We can now find the value of y by using the Pythagorean theorem.

y^2 = AD^2 - x^2
y^2 = 25^2 - 0.25^2
y^2 = 625 - 0.0625
y^2 = 624.9375
y = sqrt(624.9375)
y = 25

Conclusion

In this problem, we were given a circle with centre 0, and a diameter AD. We were also given a point B on the circle, and a ratio DAB = 50. Our goal was to find the coordinates of point B, denoted by x and y. We used the Pythagorean theorem and the given ratio to find the values of x and y. The final answer is x = 0.25 and y = 25.

Introduction

In our previous article, we solved a problem involving a circle with centre 0, and a diameter AD. We were given a point B on the circle, and a ratio DAB = 50. Our goal was to find the coordinates of point B, denoted by x and y. In this article, we will answer some frequently asked questions about the problem.

Q: What is the formula for finding the value of x?

A: The formula for finding the value of x is x = 0.01AD.

Q: How do I find the value of AD?

A: To find the value of AD, you can use the formula AD = 2r, where r is the radius of the circle. Alternatively, you can use the given ratio DAB = 50 to find the value of AD.

Q: What is the relationship between the values of x and y?

A: The values of x and y are related by the Pythagorean theorem, which states that y^2 = AD^2 - x^2.

Q: How do I find the value of y?

A: To find the value of y, you can use the Pythagorean theorem, which states that y^2 = AD^2 - x^2. Then, take the square root of both sides to find the value of y.

Q: What is the significance of the ratio DAB = 50?

A: The ratio DAB = 50 is used to find the value of AD, which is then used to find the values of x and y.

Q: Can I use this method to find the values of x and y for any circle?

A: Yes, you can use this method to find the values of x and y for any circle, as long as you know the ratio DAB and the diameter AD.

Q: What are the limitations of this method?

A: The limitations of this method are that it assumes a circle with a diameter AD and a point B on the circle. It also assumes that the ratio DAB = 50 is known.

Q: Can I use this method to find the values of x and y for a circle with a different ratio?

A: Yes, you can use this method to find the values of x and y for a circle with a different ratio, as long as you know the new ratio and the diameter AD.

Q: How do I apply this method to a real-world problem?

A: To apply this method to a real-world problem, you would need to know the ratio DAB and the diameter AD of the circle. You would then use the formulas x = 0.01AD and y^2 = AD^2 - x^2 to find the values of x and y.

Q: What are some common applications of this method?

A: Some common applications of this method include finding the coordinates of points on a circle, calculating the area of a circle, and determining the circumference of a circle.

Q: Can I use this method to find the values of x and y for a circle with a different shape?

A: No, this method is only applicable to circles with a diameter AD and a point B on the circle. It is not applicable to circles with a different shape.

Q: How do I know if this method is applicable to my problem?

A: To determine if this method is applicable to your problem, you would need to know the ratio DAB and the diameter AD of the circle. If you have this information, you can use the formulas x = 0.01AD and y^2 = AD^2 - x^2 to find the values of x and y.

Q: What are some common mistakes to avoid when using this method?

A: Some common mistakes to avoid when using this method include:

Not knowing the ratio DAB and the diameter AD of the circle
Not using the correct formulas to find the values of x and y
Not checking the validity of the results

Q: How do I troubleshoot common issues with this method?

A: To troubleshoot common issues with this method, you would need to check the following:

Make sure you have the correct ratio DAB and diameter AD of the circle
Check that you are using the correct formulas to find the values of x and y
Verify that the results are valid and make sense in the context of the problem

Conclusion

In this article, we have answered some frequently asked questions about the problem of finding the coordinates of point B on a circle with centre 0 and diameter AD, given a ratio DAB = 50. We have also discussed some common applications of this method and some common mistakes to avoid when using it.