Call Now to Set Up Tutoring:

(888) 888-0446

HiSET: Math : Measurement and Geometry

Study concepts, example questions & explanations for HiSET: Math

Create An Account

All HiSET: Math Resources

125 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #11 : Dilations

On the graph of the equation

$x^{2}+ y^{2} = 64$ ,

perform a dilation with center (0,4) and scale factor $\frac{1}{3}$ .

Give the equation of the resulting circle.

Possible Answers:

None of the other choices gives the correct response.

$x^{2}+\left (y- \frac{4}{3} \right )^{2} = \frac{64}{9}$

$x^{2}+\left (y- \frac{4}{3} \right )^{2} = \frac{16}{9}$

$x^{2}+\left (y- \frac{8}{3} \right )^{2} = \frac{16}{9}$

$x^{2}+\left (y- \frac{8}{3} \right )^{2} = \frac{64}{9}$

Correct answer:

$x^{2}+\left (y- \frac{8}{3} \right )^{2} = \frac{64}{9}$

Explanation:

The graph of the equation

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

is a circle with center (h, k) and radius .

$x^{2}+ y^{2} = 64$ ,

$(x-0) ^{2}+ (y-0)^{2} = 8^{2}$

is a circle with center at origin (0,0) and radius 8.

A dilation of a circle with scale factor $\frac{1}{3}$ will result in multiplying that radius by $\frac{1}{3}$ , so the radius of the circle will be $8\cdot \frac{1}{3}= \frac{8}{3}$ .

To find the center of the image, note that the origin is 4 units below (0,4) . The center of the new circle must be $4 \cdot \frac{1}{3} = \frac{4}{3}$ units below (0,4) , so this center will be $\left ( 0,4- \frac{4}{3} \right )$ , or $\left ( 0, \frac{8}{3} \right )$ . See the figure below:

Substituting in the circle formula, this is

$(x-0)^{2}+\left (y- \frac{8}{3} \right )^{2} = \left (\frac{8}{3} \right )^{2}$ ,

$x^{2}+\left (y- \frac{8}{3} \right )^{2} = \frac{64}{9}$ .

Report an Error

Example Question #12 : Dilations

Obtuse triangle

On the above obtuse triangle perform a dilation of scale factor $\frac{1}{2}$ with the center of the dilation at the circumcenter of the triangle. Let the images of , , and be , , and C ' , respectively.

Which of the following correctly shows $\bigtriangleup A'B'C'$ relative to $\bigtriangleup ABC$ ?

Possible Answers:

Obtuse triangle

Correct answer:

Obtuse triangle

Explanation:

The circumcenter of a triangle can be located by finding the intersection of the perpendicular bisectors of the three sides of the triangle. The perpendicular bisectors are shown below, with point of intersection :

Obtuse triangle

Construct $\overline{AO}$ , $\overline{BO}$ , and $\overline{CO}$ . A dilation of scale factor $\frac{1}{2}$ with center can be performed by letting A ' , , and be the midpoints of $\overline{AO}$ , $\overline{BO}$ , and $\overline{CO}$ , respectively:

Obtuse triangle

Removing the perpendicular bisectors and , we see that the correct choice is the figure

Obtuse triangle

Report an Error

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

Find the circumference of a circle with the following diameter:

d=14

Possible Answers:

$16\pi$

$14\pi$

$15\pi$

$8\pi$

$7\pi$

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.

$C=14\pi$

Report an Error

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.

Report an Error

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

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

Report an Error

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.

Report an Error

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...

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

Report an Error

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.

Report an Error

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.

Report an Error

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.

Report an Error

View Tutors

Yucheng
Certified Tutor

The University of Texas at Austin, Bachelor of Science, Mechanical Engineering.

View Tutors

Nandini
Certified Tutor

George Brown College, Master's/Graduate, Project management . George Brown College, Master's/Graduate, Human Resource Managem...

View Tutors

Lilia
Certified Tutor

Pangasinan State University, Bachelor of Science, Elementary School Teaching. Walden University, Doctor of Education, Urban E...

All HiSET: Math Resources

125 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

SSAT Tutors in Boston, English Tutors in Atlanta, SSAT Tutors in Dallas Fort Worth, Biology Tutors in Washington DC, GRE Tutors in Miami, SSAT Tutors in Miami, Statistics Tutors in Seattle, Calculus Tutors in Philadelphia, GRE Tutors in Houston, MCAT Tutors in Boston

Popular Courses & Classes

ISEE Courses & Classes in Miami, LSAT Courses & Classes in Washington DC, ACT Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in Philadelphia, SAT Courses & Classes in Miami, MCAT Courses & Classes in Atlanta, ACT Courses & Classes in Los Angeles, GMAT Courses & Classes in Chicago, GRE Courses & Classes in Washington DC, SSAT Courses & Classes in Dallas Fort Worth

Popular Test Prep

GRE Test Prep in Phoenix, ACT Test Prep in Philadelphia, SSAT Test Prep in Boston, ACT Test Prep in San Francisco-Bay Area, SAT Test Prep in Los Angeles, ACT Test Prep in Los Angeles, SAT Test Prep in Washington DC, GMAT Test Prep in Chicago, GMAT Test Prep in Philadelphia, MCAT Test Prep in Dallas Fort Worth

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

HiSET: Math : Measurement and Geometry

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

Example Question #11 : Dilations

Example Question #12 : Dilations

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

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

All HiSET: Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: