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HiSET: Math : Circumference

Study concepts, example questions & explanations for HiSET: Math

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125 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Equations Of Circles

Find the circumference of a circle with the following diameter:

d=14

Possible Answers:

$16\pi$

$7\pi$

$8\pi$

$15\pi$

$14\pi$

Correct answer:

$14\pi$

Explanation:

A circle's circumference is calculated using the following formula:

$C=2\pi r$

In this equation, the variable, , is the circle's radius.

In our problem we are given the diameter. The diameter is related to the radius in the following manner:

d=2r

Let's rewrite our formula by substituting the diameter for the radius.

$C=d\pi$

Let's substitute and solve.

$C=14\pi$

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Example Question #142 : Hi Set: High School Equivalency Test: Math

Which of the following is closest to the length of a 270-degree arc of a circle with radius 20?

Possible Answers:

105

Correct answer:

Explanation:

A circle with radius 20 has as its circumference $2 \pi$ times this, or

$C = 2 \pi r = 2 \pi (20 )= 40 \pi$

There are 360 degrees in a circle, so a 270-degree arc is

$\frac{270}{360} = \frac{270 \div 90}{360 \div 90} = \frac{3}{4}$

of the circle. Therefore, the length of this arc is

$L= \frac{3}{4} C = \frac{3}{4} (40 \pi)= 30 \pi$

$\pi \approx 3.14159$ ,

so a reasonable estimate of the length of this arc is

$L \approx 30 (3.14159) \approx 94.2477$

Of the five choices, 95 comes closest to the correct length.

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Example Question #3 : Equations Of Circles

A circle on the coordinate plane has a diameter with endpoints (3,4) and (9, 10) . Give its equation.

Possible Answers:

$(x-7)^{2} + (y-6)^{2}= 36$

$(x-6)^{2} + (y-7)^{2}= 36$

$(x-7)^{2} + (y-6)^{2}= 18$

Insufficient information exists to determine its equation.

$(x-6)^{2} + (y-7)^{2}= 18$

Correct answer:

$(x-6)^{2} + (y-7)^{2}= 18$

Explanation:

The equation, in standard form, of a circle on the coordinate plane with center (h, k) and radius is

$(x-h)^{2} + (y-k)^{2}= r^{2}$ .

It is necessary to find the center and the radius of the circle in order to determine its equation.

The center of the circle is the midpoint of a given diameter. The midpoint $\left ( x_{M}, y_{M} \right )$ of a segment with endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ can be found by applying the midpoint formula:

$x_{M}= \frac{x_{1}+ x_{2} }{2}$

$y_{M}= \frac{y_{1}+ y_{2} }{2}$

Setting $x_{1}= 3, x_{2}= 9$ :

$x_{M}= \frac{3+9 }{2}= \frac{12}{2} = 6$

Setting $y_{1}= 4, y_{2}= 10$

$y_{M}= \frac{4+10 }{2} = \frac{14}{2} = 7$

The midpoint, and the center of the circle, is at (6, 7) .

The radius of the circle is the distance from the center to an endpoint, so we can use the distance formula

$r = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }$ .

Substitute $x_{1}= 6, x_{2}= 9 , y_{1}= 7, y_{2}= 10$ :

$r = \sqrt{(9-6)^{2}+(10-7 )^{2} }$

$= \sqrt{3^{2}+3 ^{2} }$

$= \sqrt{9+9 }$

$= \sqrt{18 }$

and

$r^{2} = 18$

Now, set $r^{2} = 18 , h = 6, k= 7$ in the standard form of the equation of a circle, and the result is

$(x-6)^{2} + (y-7)^{2}= 18$

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Example Question #4 : Equations Of Circles

A circle on the coordinate plane has a radius with endpoints (3,4) and (9, 10) . Give its equation.

Possible Answers:

$(x-4)^{2}+ (y-10)^{2} = 18$

$(x-4)^{2}+ (y-10)^{2} = 72$

Insufficient information exists to determine its equation.

$(x-3)^{2}+ (y-9)^{2} = 72$

$(x-3)^{2}+ (y-9)^{2} = 18$

Correct answer:

Insufficient information exists to determine its equation.

Explanation:

The equation, in standard form, of a circle on the coordinate plane with center (h, k) and radius is

$(x-h)^{2} + (y-k)^{2}= r^{2}$ .

It is necessary to find the center and the radius of the circle in order to determine its equation. The radius of the circle can be determined by applying the distance formula to the coordinates of the endpoints of the radius. However, it is not clear from the information given which endpoint is the center of the circle. Therefore, while can be determined, (h, k) cannot. The correct response is that insufficient information exists to determine its equation.

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Example Question #1 : Equations Of Circles

What is the circumference of a circle with an area of $1369\pi$ ?

Possible Answers:

$35\pi$

$70\pi$

$37\pi$

$74\pi$

Correct answer:

$74\pi$

Explanation:

Step 1: Since we are given the area, let's first find the radius of the circle.

If the area of a circle is calculated by: $A=\pi r^2$ , we can plug in the values that are given to us. $A=1369\pi$ .

We have $1369\pi=\pi r^2$ . We will divide by $\pi$ .

1369=r^2 . To find the radius, we will take the square root of both sides.

So, r=37 .

Step 2: Now that we have the radius, we will now find the circumference.

The circumference formula by using the radius is $C=2\pi r$ .

Plug in all the information that we have...

$C=2\pi(37)\Rightarrow C=74\pi$

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Example Question #5 : Equations Of Circles

A circle on the coordinate plane has center (5, -9) and circumference $50 \pi$ .

Give its equation.

Possible Answers:

$(x-5)^{2}+ (y+9)^{2}= 2,500$

$(x-5)^{2}+ (y+9)^{2}= 5$

$(x-5)^{2}+ (y+9)^{2}= 625$

$(x-5)^{2}+ (y+9)^{2}= 50$

$(x-5)^{2}+ (y+9)^{2}= 25$

Correct answer:

$(x-5)^{2}+ (y+9)^{2}= 625$

Explanation:

The standard form of the equation of a circle with center at (h, k) and radius is

$(x-h)^{2}+ (y-k)^{2}= r^{2}$

The circumference of the circle is $50 \pi$ ; to find the radius, divide this circumference by $2 \pi$ :

$r = \frac{C}{2\pi }= \frac{50 \pi }{2\pi } = 25$

Now, set h = 5, k = -9, r = 25 in the equation of the circle:

$(x-5)^{2}+ (y-(-9))^{2}= 25^{2}$

Simplifying:

$(x-5)^{2}+ (y+9)^{2}= 625$ ,

the correct equation.

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Example Question #5 : Equations Of Circles

A circle on the coordinate plane has center (5, -9) and area $50 \pi$ .

Give its equation.

Possible Answers:

$(x-5)^{2}+ (y+9)^{2}= 2,500$

$(x-5)^{2}+ (y+9)^{2}= 625$

$(x-5)^{2}+ (y+9)^{2}= 50$

$(x-5)^{2}+ (y+9)^{2}= 25$

$(x-5)^{2}+ (y+9)^{2}= 5$

Correct answer:

$(x-5)^{2}+ (y+9)^{2}= 50$

Explanation:

The standard form of the equation of a circle with center at (h, k) and radius is

$(x-h)^{2}+ (y-k)^{2}= r^{2}$

The area of the circle is $50 \pi$ ; to find the radius, substitute in the formula:

$\pi r^{2}= A$

$\pi r^{2}= 50 \pi$

$\frac{ \pi r^{2}}{\pi} = \frac{50 \pi}{\pi}$

$r^{2} = 50$

We do not need to find . Set $h = 5, k = -9, r ^{2} = 50$ in the equation of the circle:

$(x-5)^{2}+ (y-(-9))^{2}= 50$

Simplifying:

$(x-5)^{2}+ (y+9)^{2}=50$ ,

the correct equation.

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Example Question #141 : Hi Set: High School Equivalency Test: Math

A circle on the coordinate plane has its center at (2,9 ) and passes through (10, 3) . Give its circumference.

Possible Answers:

The information given is insufficient to answer the question

$100 \pi$

$10 \pi$

$25 \pi$

$20 \pi$

Correct answer:

$20 \pi$

Explanation:

The radius of the circle is the distance between its center and a point that it passes through. This radius can be calculated using the distance formula:

$r = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }$ .

Setting $x_{1} = 2, x_{2} = 10, y_{1}=9, y_{2} =3$ :

$r = \sqrt{(10-2)^{2}+(3-9)^{2} }$

$= \sqrt{8^{2}+(-6)^{2} }$

$= \sqrt{64+36}$

$= \sqrt{100}$

=10

The circumference of a circle is equal to $2 \pi$ multiplied by the radius, so

$C= 2 \pi r = 2 \pi (10) = 20 \pi$ ,

the correct response.

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Example Question #1 : Equations Of Circles

Give the circumference of the circle on the coordinate plane with equation

$x^{2}+y^{2}+ 6x-20y+ 9=0$ .

Possible Answers:

$40 \pi$

$100 \pi$

$10 \pi$

$50 \pi$

$20 \pi$

Correct answer:

$20 \pi$

Explanation:

We must first rewrite the equation of the circle in standard form

$(x-h)^{2} + (y-k)^{2}= r^{2}$ ;

will be the radius.

$x^{2}+y^{2}+ 6x-20y+ 9=0$

Subtract 9 from both sides, and rearrange the terms remaining on the left as follows:

$x^{2}+y^{2}+ 6x-20y+ 9- 9 =0- 9$

$x^{2}+ 6x + \underline{\; \; \; \; \; \; }+y^{2}-20y+ \underline{\; \; \; \; \; \; } = - 9$

Note that blanks have been inserted after the linear terms. In these blanks, complete two perfect square trinomials by dividing linear coefficients by 2 and squaring:

$(6 \div 2)^{2} = 3^{2} = 9$

$(20 \div 2)^{2} = 10^{2} = 100$

Add to both sides, as follows:

$x^{2}+ 6x + 9+y^{2}-20y+ 100 = - 9+9+ 100$

$x^{2}+ 6x + 9+y^{2}-20y+ 100 = 100$

Rewrite the expression on the left as the sum of the squares of two binomials.

$(x+3)^{2} + (y-10)^{2} = 100$

The equation is now in standard form.

$r^{2} = 100$

The radius is the square root of this, or

r = 10

The circumference of a circle is the radius multiplied by $2 \pi$ , so

$C = 2 \pi r = 2 \pi (10)= 20 \pi$ ,

the correct response.

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HiSET: Math : Circumference

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

Example Question #1 : Equations Of Circles

Example Question #142 : Hi Set: High School Equivalency Test: Math

Example Question #3 : Equations Of Circles

Example Question #4 : Equations Of Circles

Example Question #1 : Equations Of Circles

Example Question #5 : Equations Of Circles

Example Question #5 : Equations Of Circles

Example Question #141 : Hi Set: High School Equivalency Test: Math

Example Question #1 : Equations Of Circles

All HiSET: Math Resources

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