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Creating a ring using two circles in c++ program

Create a class called circle which has a private member radius.Define methods for area and perimeter in this class.Use this class to create a class called ring which is the combination of two circles.Define area and perimeter of ring using area and perimeter of circles,define a constructor which creates rings of specific inner and outer radii.Create and delete objects of the ring class.Program must allow the  user to create a ring of his choice and display the area and perimeter of the ring so created.

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Nice try!

If nobody has helped you since September 2008, when you asked the question, you probably scored a "zero" in your computer-programming course, and probably don't need an answer to your "homework" assignment.

Posted on Dec 05, 2009

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A circle with a radius of 3 kilometers

What is the question?
A circle with a radius of 3 kilometers has a perimeter of about 18.85 kilometers and an area of about 28.27 square kilometers.

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Area of circle

(pi)R squared (Fixya won't let me use the sign for Pi)

Area of a circle=pi*radius^2 or pi times radius times radius

example: suppose the radius of a circle is 10 inches, then the Area of the circle would be pi times 10 times 10 or 100*pi

pi represents the circumference (distance around) of a circle / diameter and for any circle is constantly the same number which is approximately 3.14

so in my example 100*pi=100*3.14 or 314 square inches

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I want to cut a circle where the outside line is 18 inches

To get a circle with a perimeter of 18 inches, you want a diameter of about 5.73 inches (just under 5 3/4 inches) or a radius of 2.86 inches (call it 2 7/8 inches). Get a compass, set its legs 2 and 7/8 inches apart, put one end at the center, and turn.

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The radius and the coordinates of the center will completely define a circle. Some other measurements include the diameter, circumference, and area. Then there are parts of the circle, such as arc and chord.

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How to calculate the sides of a hexagon?

Well it depends. If the hexagon is irregular (sides are not equal) there is no formula to calculate the sides as they can have arbitrary values. You must measure them.
If the hexagon is regular you may be able to relate the measure of a side to the radius of the circle in which it is inscribed. If you have the radius of the circle, the side is equal to the radius. If you have the value of perimeter you divide that value by 6.
There is also a formula that relates the area of the hexagon to the measure of the side s. The formula is Area=(6/4)(s^2)cot(PI/6), where cot is the cotangent function, its angle is in radian. In degrees Pi/6 is 30 degrees.

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If you know about area formulas of class 8 and 9,please solve my question?there is a the between of the circle,there is a rectangle of dimension 4cm and 3cm.if the rectangle will cut out what...

The diameter of the circle is the diagonal of the rectangle use Pythagoras this rectangle is a made from two 3,4,5 triangles, so the diameter is 5cm, the radius will be 2.5cm don't forget the BODMAS ordering rule (Brackets of Division Multiplication Addition and Subtraction) Also rounding to 2 decimal places. This solution is based on all four corners of the rectangle touching the circumference of the circle.

Area of a circle is P (22/7) i x Radius (squared ^2) ====> 22/7 x 2.5cm ^ 2 ====> 22/7 x 6.25cm

TOTAL AREA of CIRCLE = 19.64 cm^2

Area of the rectangle is length x width ====> 4 cm x 3 cm


The remaining portion is the circle area minus the rectangle 19.64-12 = 7.64 cm^2

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What is the radius of a 30'' pipe

30" diameter? Radius is 15" (half of diameter).

30" circumference (distance around) is a little more complicated...

The number pi.gif is the ratio of the circumference of a circle to the diameter. The value of pi.gif is approximately 3.14159265358979323846...The diameter of a circle is twice the radius. Given the diameter or radius of a circle, we can find the circumference. We can also find the diameter (and radius) of a circle given the circumference. The formulas for diameter and circumference of a circle are listed below. We round pi.gif to 3.14 in order to simplify our calculations. diam_formula.gif circum_formula.gif

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A pool with a radius of 2.5 metres surrounded by a path 0.8 metres wide what is the perimetee of the pool .using the value of 3.142

You must have been taught the relation between the circumference (perimeter) of a circle, the radius and pi. Use it to calculate the perimeter. It is true for all circular figure.
Depending on what is really asked (the perimeter of the actual pool where you find the water, or the general pool area where no shoes are allowed)
1st case: use the radius that was given.
2nd case: the total radius is the radius given plus the width of the path.

If you do not have a formula for the perimeter that involves the radius, but have one that involves the diameter you can use it too, knowing that the diameter is twice the radius or the radius is one half of the diameter.

As you might have guessed I was not going to make things too easy for you by providing a ready-made answer, but I gave you all the hints that will help you solve the problem.

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Hi everybody can anyone help me with the project ''Data Hiding in Audio Files'' project in Java and how to run the project with full source code to

The Circle Class Using Data Hiding and Encapsulation package shapes; // Specify a package for the class public class Circle { // The class is still public // This is a generally useful constant, so we keep it public public static final double PI = 3.14159; protected double r; // Radius is hidden, but visible to subclasses // A method to enforce the restriction on the radius // This is an implementation detail that may be of interest to subclasses protected checkRadius(double radius) { if (radius < 0.0) throw new IllegalArgumentException("radius may not be negative."); } // The constructor method public Circle(double r) { checkRadius(r); this.r = r; } // Public data accessor methods public double getRadius() { return r; }; public void setRadius(double r) { checkRadius(r); this.r = r; } // Methods to operate on the instance field public double area() { return PI * r * r; } public double circumference() { return 2 * PI * r; } }

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