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Firlakuza [10]
3 years ago
6

How do i measure the diameter of a circle?

Mathematics
2 answers:
Reil [10]3 years ago
6 0
A circle diameter is the measurement of a straight line that goes across the middle from one point on the edge of the circle and to another point on the opposite edge of the circle. Depending on the dimensions you do know, you can figure the diameter through a number of methods. You will need to use the value pi to calculate this. Pi, typically expressed as 3.141593, is a mathematical constant that is an odd number.

Measure the diameter of a circle using a ruler. Unlike the imaginary ones in math problems, if you are trying to find the diameter of an actual circle, use a ruler to measure one end of the circle to the other, making sure that your ruler reaches the exact center. If you're not sure where the center is you should do your best to measure the results a couple of times and then average them.

When you know the radius, multiply by 2. The radius is the exact center of the circle to the external line. It is just half the diameter, and so the calculation of the radius actually needs to be multiplied. If the radius is 5 cm, for instance, the diameter is 10 cm.
trapecia [35]3 years ago
3 0

Answer:

Step-by-step explanation:

If you know the radius of the circle, double it to get the diameter. The radius is the distance from the center of the circle to its edge. If the radius of the circle is 4 cm, then the diameter of the circle is 4 cm x 2, or 8 cm. If you know the circumference of the circle, divide it by π to get the diameter

hope that helps!

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Answer:

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The restriction is that the perimeter have to be 12 ft at most:

P=(a+2b)+\dfrac{\pi a}{2}=2b+a+(\dfrac{\pi}{2}) a=2b+(1+\dfrac{\pi}{2})a=12

We can express b in function of a as:

2b+(1+\dfrac{\pi}{2})a=12\\\\\\2b=12-(1+\dfrac{\pi}{2})a\\\\\\b=6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a

Then, the area become:

A=ab+\dfrac{\pi a^2}{8}=a(6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a)+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a^2+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}-\dfrac{\pi}{8}\right)a^2\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{8}\right)a^2

To maximize the area, we derive and equal to zero:

\dfrac{dA}{da}=6-2\left(\dfrac{1}{2}+\dfrac{\pi}{8}\right )a=0\\\\\\6-(1-\pi/4)a=0\\\\a=\dfrac{6}{(1+\pi/4)}\approx6/1.78\approx 3.36

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b=6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a\\\\\\b=6-0.393*3.36=6-1.32\\\\b=4.68

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Try this solution:

Step-by-step explanation:

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2) for the vertex of the function: x₀= -b/2a, where a;b;c - numbers from ax²+bx+c=0 (from x²-6x-7=0).

x₀=6/2=3;

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It means, the coordinates of the vertex are (3;-16).

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