How to find the length of a radius - Basic Geometry

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Question

A circle has an area of 36π inches. What is the radius of the circle, in inches?

Answer

We know that the formula for the area of a circle is π_r_2. Therefore, we must set 36π equal to this formula to solve for the radius of the circle.

36π = π_r_2

36 = _r_2

6 = r

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Question

Circle X is divided into 3 sections: A, B, and C. The 3 sections are equal in area. If the area of section C is 12π, what is the radius of the circle?

Circle X

Answer

Find the total area of the circle, then use the area formula to find the radius.

Area of section A = section B = section C

Area of circle X = A + B + C = 12π+ 12π + 12π = 36π

Area of circle = where r is the radius of the circle

36π = πr2

36 = r2

√36 = r

6 = r

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Question

The specifications of an official NBA basketball are that it must be 29.5 inches in circumference and weigh 22 ounces. What is the approximate radius of the basketball?

Answer

To Find your answer, we would use the formula: C=2πr. We are given that C = 29.5. Thus we can plug in to get \[29.5\]=2πr and then multiply 2π to get 29.5=(6.28)r. Lastly, we divide both sides by 6.28 to get 4.70=r. (The information given of 22 ounces is useless)

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Question

A circle with center (8, –5) is tangent to the y-axis in the standard (x,y) coordinate plane. What is the radius of this circle?

Answer

For the circle to be tangent to the y-axis, it must have its outer edge on the axis. The center is 8 units from the edge.

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Question

A circle has an area of $49\pi \ in^{2}$ . What is the radius of the circle, in inches?

Answer

We know that the formula for the area of a circle is πr_2. Therefore, we must set 49_π equal to this formula to solve for the radius of the circle.

49_π_ = _πr_2

49 = _r_2

7 = r

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Question

The area of a circle is one square yard. Give its radius in inches, to the nearest tenth of an inch.

Answer

The area of a circle is

$A = \pi r^{2}$

Substitute 1 for :

$1 = \pi r^{2}$

$r = \sqrt{\frac{1}{\pi }}$

This is the radius in yards. The radius in inches is 36 times this.

$36\sqrt{\frac{1}{\pi }} \approx 20.3$

20.3 inches is the radius.

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Question

The circle shown below has an area equal to $49\pi$ . What is the length of the radius, , of this circle?

Answer

The formula for the area of a circle is $A=\pi R^{2}$ . We can fill in what we know, the area, and then solve for the radius, .

$49\pi =\pi R^{2}$

Divide each side of the equation by $\pi$ :

$49=R^{2}$

Take the square root of each side:

$\sqrt{49}=\sqrt{R^{2}}$

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Question

A circle has a circumference of $32 \pi$ inches. What is the radius of the circle?

Answer

The circumference of a circle is given by $c = 2\pi * r$ , where is the circumference and is the radius.

Plug in the given circumference for and solve for :

$32 \pi = 2\pi * r$

$\frac{(32 \pi)}{2\pi} = \frac{(2\pi * r)}{2\pi}$

$r= 16\ in$

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Question

What is the name of the segment in red?

Answer

The radius is the distance from the center of a circle to any point on it's perimeter.

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Question

What is the name of the segment in brown?

Answer

A chord is a line segment which joins two points on a curve. A chord does not go through the center of a circle.

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Question

The diameter of a circle is 16 centimeters. What is the circle's radius in centimeters?

Answer

The radius is half of the diameter. To find the radius, simply divide the diameter by 2.

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Question

Find the radius of a circle inscribed in a square that has a diagonal of $12\sqrt2$ .

Answer

Notice that the diagonal of the square is also the hypotenuse of a right isosceles triangle whose legs are also the sides of the square. You should also notice that the diameter of the circle has the same length as that of a side of the square.

In order to find the radius of the circle, we need to first use the Pythagorean theorem to find the length of the side of the square.

$\text{Diagonal}^2=\text{side}^2+\text{side}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{12\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=12$

Now keep in mind the following relationship between the diameter and the side of the square:

$\text{Diameter}=\text{side}=12$

Recall the relationship between the diameter and the radius.

$\text{Radius}=\frac{1}{2}(\text{Diameter})$

Substitute in the value of the radius by plugging in the value of the diameter.

$\text{Radius}=\frac{1}{2}(12)$

Solve.

$\text{Radius}=6$

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Question

Find the radius of a circle inscribed in a square that has a diagonal of $2\sqrt2$ .

Answer

Notice that the diagonal of the square is also the hypotenuse of a right isosceles triangle whose legs are also the sides of the square. You should also notice that the diameter of the circle has the same length as that of a side of the square.

In order to find the radius of the circle, we need to first use the Pythagorean theorem to find the length of the side of the square.

$\text{Diagonal}^2=\text{side}^2+\text{side}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{2\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=2$

Now keep in mind the following relationship between the diameter and the side of the square:

$\text{Diameter}=\text{side}=2$

Recall the relationship between the diameter and the radius.

$\text{Radius}=\frac{1}{2}(\text{Diameter})$

Substitute in the value of the radius by plugging in the value of the diameter.

$\text{Radius}=\frac{1}{2}(2)$

Solve.

$\text{Radius}=1$

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Question

Find the radius of a circle inscribed in a square that has a diagonal of $3\sqrt2$ .

Answer

Notice that the diagonal of the square is also the hypotenuse of a right isosceles triangle whose legs are also the sides of the square. You should also notice that the diameter of the circle has the same length as that of a side of the square.

In order to find the radius of the circle, we need to first use the Pythagorean theorem to find the length of the side of the square.

$\text{Diagonal}^2=\text{side}^2+\text{side}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{3\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=3$

Now keep in mind the following relationship between the diameter and the side of the square:

$\text{Diameter}=\text{side}=3$

Recall the relationship between the diameter and the radius.

$\text{Radius}=\frac{1}{2}(\text{Diameter})$

Substitute in the value of the radius by plugging in the value of the diameter.

$\text{Radius}=\frac{1}{2}(3)$

Solve.

$\text{Radius}=\frac{3}{2}$

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Question

Find the radius of a circle inscribed in a square with a diagonal of $6\sqrt2$ .

Answer

Notice that the diagonal of the square is also the hypotenuse of a right isosceles triangle whose legs are also the sides of the square. You should also notice that the diameter of the circle has the same length as that of a side of the square.

In order to find the radius of the circle, we need to first use the Pythagorean theorem to find the length of the side of the square.

$\text{Diagonal}^2=\text{side}^2+\text{side}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{6\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=6$

Now keep in mind the following relationship between the diameter and the side of the square:

$\text{Diameter}=\text{side}=6$

Recall the relationship between the diameter and the radius.

$\text{Radius}=\frac{1}{2}(\text{Diameter})$

Substitute in the value of the radius by plugging in the value of the diameter.

$\text{Radius}=\frac{1}{2}(6)$

Solve.

$\text{Radius}=3$

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Question

Find the length of the radius of a circle inscribed in a square with a diagonal of $5\sqrt2$ .

Answer

Notice that the diagonal of the square is also the hypotenuse of a right isosceles triangle whose legs are also the sides of the square. You should also notice that the diameter of the circle has the same length as that of a side of the square.

In order to find the radius of the circle, we need to first use the Pythagorean theorem to find the length of the side of the square.

$\text{Diagonal}^2=\text{side}^2+\text{side}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{5\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=5$

Now keep in mind the following relationship between the diameter and the side of the square:

$\text{Diameter}=\text{side}=5$

Recall the relationship between the diameter and the radius.

$\text{Radius}=\frac{1}{2}(\text{Diameter})$

Substitute in the value of the radius by plugging in the value of the diameter.

$\text{Radius}=\frac{1}{2}(5)$

Solve.

$\text{Radius}=\frac{5}{2}$

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Question

Find the length of the radius of a circle inscribed in a square with a diagonal of $4\sqrt2$ .

Answer

Notice that the diagonal of the square is also the hypotenuse of a right isosceles triangle whose legs are also the sides of the square. You should also notice that the diameter of the circle has the same length as that of a side of the square.

In order to find the radius of the circle, we need to first use the Pythagorean theorem to find the length of the side of the square.

$\text{Diagonal}^2=\text{side}^2+\text{side}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{4\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=4$

Now keep in mind the following relationship between the diameter and the side of the square:

$\text{Diameter}=\text{side}=4$

Recall the relationship between the diameter and the radius.

$\text{Radius}=\frac{1}{2}(\text{Diameter})$

Substitute in the value of the radius by plugging in the value of the diameter.

$\text{Radius}=\frac{1}{2}(4)$

Solve.

$\text{Radius}=2$

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Question

Find the length of the radius of a circle inscribed in a square that has a diagonal of $7\sqrt2$ .

Answer

Notice that the diagonal of the square is also the hypotenuse of a right isosceles triangle whose legs are also the sides of the square. You should also notice that the diameter of the circle has the same length as that of a side of the square.

In order to find the radius of the circle, we need to first use the Pythagorean theorem to find the length of the side of the square.

$\text{Diagonal}^2=\text{side}^2+\text{side}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{7\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=7$

Now keep in mind the following relationship between the diameter and the side of the square:

$\text{Diameter}=\text{side}=7$

Recall the relationship between the diameter and the radius.

$\text{Radius}=\frac{1}{2}(\text{Diameter})$

Substitute in the value of the radius by plugging in the value of the diameter.

$\text{Radius}=\frac{1}{2}(7)$

Solve.

$\text{Radius}=\frac{7}{2}$

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Question

Find the length of the radius of a circle inscribed in a square that has a diagonal of $8\sqrt2$ .

Answer

Notice that the diagonal of the square is also the hypotenuse of a right isosceles triangle whose legs are also the sides of the square. You should also notice that the diameter of the circle has the same length as that of a side of the square.

In order to find the radius of the circle, we need to first use the Pythagorean theorem to find the length of the side of the square.

$\text{Diagonal}^2=\text{side}^2+\text{side}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{8\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=8$

Now keep in mind the following relationship between the diameter and the side of the square:

$\text{Diameter}=\text{side}=8$

Recall the relationship between the diameter and the radius.

$\text{Radius}=\frac{1}{2}(\text{Diameter})$

Substitute in the value of the radius by plugging in the value of the diameter.

$\text{Radius}=\frac{1}{2}(8)$

Solve.

$\text{Radius}=4$

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Question

Find the length of the radius of a circle inscribed in a square that has a diagonal of $\sqrt2$ .

Answer

Notice that the diagonal of the square is also the hypotenuse of a right isosceles triangle whose legs are also the sides of the square. You should also notice that the diameter of the circle has the same length as that of a side of the square.

In order to find the radius of the circle, we need to first use the Pythagorean theorem to find the length of the side of the square.

$\text{Diagonal}^2=\text{side}^2+\text{side}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=1$

Now keep in mind the following relationship between the diameter and the side of the square:

$\text{Diameter}=\text{side}=1$

Recall the relationship between the diameter and the radius.

$\text{Radius}=\frac{1}{2}(\text{Diameter})$

Substitute in the value of the radius by plugging in the value of the diameter.

$\text{Radius}=\frac{1}{2}(1)$

Solve.

$\text{Radius}=\frac{1}{2}$

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