Call Now to Set Up Tutoring:

(888) 888-0446

HiSET: Math : Properties of polygons and circles

Study concepts, example questions & explanations for HiSET: Math

Create An Account Create Tests & Flashcards

All HiSET: Math Resources

125 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 Next →

Example Question #11 :Measurement And Geometry

Pool shape

A quadrilateral is shown, and the angle measures of 3 interior angles are given. Findx, the missing angle measure.

Possible Answers:

$33^{\circ}$

$52^{\circ}$

$94^{\circ}$

$128^{\circ}$

$228^{\circ}$

Correct answer:

$128^{\circ}$

Explanation:

The sum of the measures of the interior angles of a quadrilateral is 360 degrees. The sum of the measures of the interior angles of any polygon can be determined using the following formula:

$Sum \ of\ Interior \ Angles = 180^{\circ}(n-2)$ , whereis the number of sides.

For example, with a quadrilateral, which has 4 sides, you obtain the following calculation:

$Sum \ of\ Interior \ Angles = 180^{\circ}(4-2)=180^{\circ}\cdot2=360^{\circ}$

Solving forrequires setting up an algebraic equation, adding all 4 angles to equal 360 degrees:

$50^{\circ}+67^{\circ}+115^{\circ}+x^{\circ}=360^{\circ}$

Solving foris straightforward: subtract the values of the 3 known angles from both sides:

$x^{\circ}= 360^{\circ}-50^{\circ}-67^{\circ}-115^{\circ}$

$x^{\circ}=128^{\circ}$

Report an Error

Example Question #12 :Measurement And Geometry

Intercepted 2

is the center of the above circle. Calculate $m \angle AOB$ .

Possible Answers:

$140^{\circ }$

$125^{\circ }$

$135^{\circ }$

$130^{\circ }$

$120^{\circ }$

Correct answer:

$135^{\circ }$

Explanation:

$\angle AOB$ is the central angle that intercepts $\overarc{AB}$ , so

$m \angle AOB = m \overarc{AB}$ .

Therefore, we need to find $m \overarc{AB}$ to obtain our answer.

If the sides of an angle with vertex outside the circle are both tangent to the circle, the angle formed is half the difference of the measures of the arcs. Therefore,

$\frac{m \overarc{ACB} - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{AB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{ACB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for:

$\frac{(360- x )- x }{2} =45$

$\frac{360- 2 x }{2} =45$

Multiply both sides by 2:

$\frac{360- 2 x }{2} \cdot 2 =45 \cdot 2$

360- 2 x = 90

两边同时减去360:

360- 2 x - 360 = 90 - 360

-2x = -270

Divide both sides by:

$-2x \div (-2) = -270 \div (-2)$

x = 135 ,

the measure of $\overarc{AB}$ and, consequently, that of $\angle AOB$ .

Report an Error

Example Question #13 :Measurement And Geometry

Inscribed heptagon

The above figure shows a regular seven-sided polygon, orheptagon, inscribed inside a circle.is the common center of the figures.

Give the measure of $\angle DBO$ .

Possible Answers:

$65 \frac{1}{7}^{\circ }$

$62 \frac{4}{7}^{\circ }$

$64 \frac{2}{7}^{\circ }$

$63\frac{3}{7}^{\circ }$

$61 \frac{5}{7}^{\circ }$

Correct answer:

$64 \frac{2}{7}^{\circ }$

Explanation:

Consider the figure below, which adds some radii of the heptagon (and circle):

Inscribed heptagon

$\overline{OB}$ , as a radius of a regular polygon, bisects $\angle EBD$ . The measure of this angle can be calculated using the formula

$m = \frac{180 (n-2)}{n}$ ,

where n = 7 :

$m \angle EBD = \frac{180^{\circ } (7-2)}{7} = \frac{180^{\circ } (5)}{7} = \frac{900}{7}^{\circ } = 1 28 \frac{4}{7}^{\circ }$

Consequently,

$\angle DBO = \frac{1}{2} \cdot m \angle EBD = \frac{1}{2} \cdot 1 28 \frac{4}{7}^{\circ } = 64 \frac{2}{7}^{\circ }$ ,

the correct response.

Report an Error

Example Question #1 :Properties Of Polygons And Circles

Inscribed heptagon

The above figure shows a regular seven-sided polygon, orheptagon, inscribed inside a circle.is the common center of the figures.

Give the measure of $\angle AOB$ .

Possible Answers:

$155 \frac{1}{7}^{\circ }$

$156 \frac{4}{7}^{\circ }$

$154 \frac{2}{7}^{\circ }$

$153 \frac{5}{7}^{\circ }$

$152 \frac{6}{7}^{\circ }$

Correct answer:

$154 \frac{2}{7}^{\circ }$

Explanation:

Examine the diagram below, which divides $\angle AOB$ into three congruent angles, one of which is $\angle 1$ :

Inscribed heptagon

The measure of a central angle of a regular-sided polygon which intercepts one side of the polygon is $\frac{360 ^{\circ }}{n}$ ; setting n = 7 , the measure of $\angle 1$ is

$m \angle 1 = \frac{360 ^{\circ }}{7} = 51 \frac{3}{7}^{\circ }$ . $\angle AOC$ has measure three times this; that is,

$m \angle AOC = 3 \times51 \frac{3}{7}^{\circ } = 154 \frac{2}{7}^{\circ }$

Report an Error

Example Question #15 :Measurement And Geometry

Inscribed heptagon

The above figure shows a regular seven-sided polygon, orheptagon, inscribed inside a circle.is the common center of the figures.

Give the measure of $\angle AOC$ .

Possible Answers:

$103 \frac{5}{7}^{\circ }$

$101 \frac{3}{7}^{\circ }$

$104 \frac{2}{7}^{\circ }$

$102 \frac{6}{7}^{\circ }$

$105 \frac{1}{7}^{\circ }$

Correct answer:

$102 \frac{6}{7}^{\circ }$

Explanation:

Examine the diagram below, which divides $\angle AOC$ into two congruent angles, one of which is $\angle 1$ :

Inscribed heptagon

The measure of a central angle of a regular-sided polygon which intercepts one side of the polygon is $\frac{360 ^{\circ }}{n}$ ; setting n = 7 , the measure of $\angle 1$ is

$m \angle 1 = \frac{360 ^{\circ }}{7} = 51 \frac{3}{7}^{\circ }$ . $\angle AOC$ has measure twice this; that is,

$m \angle AOC = 2 \times51 \frac{3}{7}^{\circ } = 102 \frac{6}{7}^{\circ }$

Report an Error

Example Question #1 :Angle Measure, Central Angles, And Inscribed Angles

Inscribed nonagon

The above figure shows a regular ten-sided polygon, ordecagon, inscribed inside a circle.is the common center of the figures.

Give the measure of $\angle AOB$ .

Possible Answers:

$72^{\circ }$

$75^{\circ }$

$80^{\circ }$

$76^{\circ }$

$84^{\circ }$

Correct answer:

$72^{\circ }$

Explanation:

Examine the diagram below, which divides $\angle AOB$ into two congruent angles, one of which is $\angle 1$ :

Inscribed nonagon

The measure of a central angle of a regular-sided polygon which intercepts one side of the polygon is $\frac{360 ^{\circ }}{n}$ ; setting n = 10 , the measure of $\angle 1$ is

$m \angle 1 = \frac{360 ^{\circ }}{10} =36^{\circ }$ . $\angle AOB$ has measure twice this; that is,

$m \angle AOB = 2 \times 36 ^{\circ } = 72 ^{\circ }$ .

Report an Error

Example Question #17 :Measurement And Geometry

Inscribed nonagon

The above figure shows a regular ten-sided polygon, ordecagon, inscribed inside a circle.is the common center of the figures.

Give the measure of $\angle OBC$ .

Possible Answers:

$36 ^{\circ }$

$45^{\circ }$

$42^{\circ }$

$40^{\circ }$

$32^{\circ }$

Correct answer:

$36 ^{\circ }$

Explanation:

Consider the triangle $\bigtriangleup OBC$ . Since $\overline{OB}$ and $\overline{OC}$ are radii, they are congruent, and by the Isosceles Triangle Theorem, $m \angle OBC = m \angle OCB$ .

Now, examine the figure below, which divides $\angle BOC$ into three congruent angles, one of which is $\angle 1$ :

Inscribed nonagon

The measure of a central angle of a regular-sided polygon which intercepts one side of the polygon is $\frac{360 ^{\circ }}{n}$ ; setting n = 10 , the measure of $\angle 1$ is

$m \angle 1 = \frac{360 ^{\circ }}{10} =36^{\circ }$ . $\angle BOC$ has measure three times this; that is,

$m \angle BOC = 3 \times 36 ^{\circ } = 108 ^{\circ }$ .

The measures of the interior angles of a triangle total $108^{\circ }$ , so

$m \angle OCB + m \angle OBC+m \angle BOC = 180$

Substituting 108 for $\angle BOC$ and $m \angle OBC$ for $m \angle OCB$ :

$m \angle OBC+ m \angle OBC+108 = 180$

$2 m \angle OBC+108 = 180$

$2 m \angle OBC+108- 108 = 180- 108$

$2 m \angle OBC=72$

$2 m \angle OBC \div 2 =72 \div 2$

$m \angle OBC= 36$

Report an Error

Example Question #11 :Measurement And Geometry

Inscribed nonagon

The above figure shows a regular ten-sided polygon, ordecagon, inscribed inside a circle.is the common center of the figures.

Give the measure of $\angle CBE$ .

Possible Answers:

$42^{\circ }$

$32^{\circ }$

$30^{\circ }$

$40^{\circ }$

$36^{\circ }$

Correct answer:

$36^{\circ }$

Explanation:

Through symmetry, it can be seen that Quadrilateral BCDE is a trapezoid, such that $\overline{BC} || \overline{DE }$ . By the Same-Side Interior Angle Theorem, $\angle CBE$ and $\angle DEB$ are supplementary - that is,

$m \angle CBE +\angle DEB = 180$ .

The measure of $\angle DEB$ can be calculated using the formula

$m = \frac{180 (n-2)}{n}$ ,

where n = 10 :

$m \angle DEB= \frac{180^{\circ } (10-2)}{10} = \frac{180^{\circ } (8)}{10} = \frac{1,440}{10}^{\circ } =144 ^{\circ }$

Substituting:

$m \angle CBE +144= 180$

$m \angle CBE +144 - 144 = 180 - 144$

$m \angle CBE = 36$

Report an Error

Example Question #19 :Measurement And Geometry

If two angles are supplementary and one angle measures $95^{\circ}$ , what is the measurement of the second angle?

Possible Answers:

$85^{\circ}$

$90^{\circ}$

$80^{\circ}$

$95^{\circ}$

Correct answer:

$85^{\circ}$

Explanation:

Step 1: Define supplementary angles. Supplementary angles are two angles whose sum is $180^{\circ}$ .

Step 2: Find the other angle by subtracting the given angle from the maximum sum of the two angles.

So, $180^{\circ}-95^{\circ}=85^\circ$

The missing angle (or second angle) is $85^{\circ}$

Report an Error

Example Question #1 :Properties Of Polygons And Circles

$\angle 1$ and $\angle 2$ are complementary angles.

$\angle 1$ and $\angle 3$ are supplementary angles.

$\angle 2 \cong \angle 4$

$m \angle 3= 112^{\circ }$

Evaluate $m \angle 4$ .

Possible Answers:

$68^{\circ }$

$112^{\circ }$

$90^{\circ }$

$22^{\circ }$

$158^{\circ }$

Correct answer:

$22^{\circ }$

Explanation:

$\angle 1$ and $\angle 3$ are supplementary angles, so, by definition,

$m \angle 1 + m \angle 3 = 180^{\circ }$

$m \angle 3= 112^{\circ }$ , so substitute and solve for $m \angle 1$ :

$m \angle 1 + 112^{\circ } = 180^{\circ }$

$m \angle 1 + 112^{\circ } -112^{\circ } = 180^{\circ }-112^{\circ }$

$m \angle 1 = 68 ^{\circ }$

$\angle 1$ and $\angle 2$ are complementary angles, so, by definition,

$m \angle 1 + m \angle 2 = 90^{\circ }$

Substitute and solve for $m \angle 2$ :

$68 ^{\circ } + m \angle 2 = 90^{\circ }$

$68 ^{\circ } + m \angle 2 - 68 ^{\circ } = 90^{\circ }-68 ^{\circ }$

$m \angle 2 =22^{\circ }$

$\angle 2 \cong \angle 4$ - that is, the angles have the same measure. Therefore,

$m \angle 4 =22^{\circ }$ .

Report an Error

← Previous 1 2 3 Next →

View Tutors

Jose
Certified Tutor

Old Dominion University, Current Undergrad Student, Mathematics.

View Tutors

Jennifer
Certified Tutor

State University of West Georgia, Bachelor in Arts, Mass Communications. Walden University, Doctor of Education, Curriculum a...

View Tutors

Justice
Certified Tutor

All HiSET: Math Resources

125 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Calculus Tutors in New York City,Biology Tutors in Los Angeles,English Tutors in Washington DC,French Tutors in Chicago,Algebra Tutors in New York City,Math Tutors in San Diego,ACT Tutors in Philadelphia,GMAT Tutors in Phoenix,Chemistry Tutors in Philadelphia,Chemistry Tutors in Miami

Popular Courses & Classes

GMAT Courses & Classes in San Diego,SAT Courses & Classes in New York City,SSAT Courses & Classes in Los Angeles,GRE Courses & Classes in Boston,GMAT Courses & Classes in Chicago,MCAT Courses & Classes in Phoenix,SAT Courses & Classes in Washington DC,LSAT Courses & Classes in Chicago,LSAT Courses & Classes in Los Angeles,ISEE Courses & Classes in San Diego

Popular Test Prep

ACT Test Prep in Boston,LSAT Test Prep in Boston,ISEE Test Prep in San Diego,GMAT Test Prep in Washington DC,SAT Test Prep in San Francisco-Bay Area,GRE Test Prep in San Francisco-Bay Area,GMAT Test Prep in San Diego,SAT Test Prep in Denver,ACT Test Prep in Denver,GMAT Test Prep in Atlanta

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:

copyrigh的物理或电子签名t 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 : Properties of polygons and circles

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

Example Question #11 :Measurement And Geometry

Example Question #12 :Measurement And Geometry

Example Question #13 :Measurement And Geometry

Example Question #1 :Properties Of Polygons And Circles

Example Question #15 :Measurement And Geometry

Example Question #1 :Angle Measure, Central Angles, And Inscribed Angles

Example Question #17 :Measurement And Geometry

Example Question #11 :Measurement And Geometry

Example Question #19 :Measurement And Geometry

Example Question #1 :Properties Of Polygons And Circles

All HiSET: Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Find the Best Tutors

Email address:
Your name:

Email address:
Your name:
Feedback: