Call Now to Set Up Tutoring:

(888) 888-0446

SAT II Math II : Finding Sides

Study concepts, example questions & explanations for SAT II Math II

Create An Account

All SAT II Math II Resources

2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #41 : Geometry

Triangle

Note: figure NOT drawn to scale.

Refer to the triangle in the above diagram.

AB = 15, AC = 10, a = 60

Evaluate . Round to the nearest tenth, if applicable.

Possible Answers:

$BC \approx 8.1$

$BC \approx 13.2$

$BC \approx 10.6$

$BC \approx 21.8$

$BC \approx 23.2$

Correct answer:

$BC \approx 13.2$

Explanation:

By the Law of Cosines,

$\left (BC \right )^{2} =\left (AB \right )^{2}+ \left (AC \right )^{2} - 2 (AB)(AC) \cos a^{\circ }$

Substitute AB = 15, AC = 10, a = 60 :

$\left (BC \right )^{2} =15^{2}+10^{2} - 2 \cdot15 \cdot 10 \cos 60^{\circ }$

$\left (BC \right )^{2} =225+100 - 300 \cdot \frac{1}{2}$

$\left (BC \right )^{2} =225+100 - 150$

$\left (BC \right )^{2} = 175$

$BC = \sqrt{175} \approx 13.2$

Report an Error

Example Question #2 : Finding Sides

In triangle $\Delta GHJ$ , $m\angle G = 40^{\circ }$ and $m \angle H = 80^{\circ }$ .

Which of the following statements is true about the lengths of the sides of $\Delta GHJ$ ?

Possible Answers:

HJ< GJ<GH

GJ < GH <HJ

GJ<HJ<GH

HJ< GH < GJ

GH<GJ< HJ

Correct answer:

HJ< GH < GJ

Explanation:

In a triangle, the shortest side is opposite the angle of least measure; the longest side is opposite the angle of greatest measure. Therefore, if we order the angles, we can order their opposite sides similarly.

Since the measures of the three interior angles of a triangle must total $180^{\circ }$ ,

$m \angle J + m \angle G + m \angle H = 180^{\circ }$

$m \angle J + 40 ^{\circ }+ 80^{\circ } = 180^{\circ }$

$m \angle J + 120^{\circ } = 180^{\circ }$

$m \angle J =60^{\circ }$

Since

$m \angle G < m \angle J< m \angle H$ ,

we can order the lengths of their opposite sides the same way:

HJ < GH < GJ .

Report an Error

Example Question #1 : Finding Sides

Triangle

Note: figure NOT drawn to scale.

Refer to the above diagram.

AB = 50, a = 63, b = 43 .

Which of the following expressions is equal to ?

Possible Answers:

$AC= \frac{50 \cos 43^{\circ }}{\cos 63^{\circ } }$

$AC= \frac{50 \cos 74^{\circ }}{\cos 43^{\circ } }$

$AC= \frac{50 \cos 43^{\circ }}{\cos 74^{\circ } }$

$AC= \frac{50 \sin 43^{\circ }}{\sin 63^{\circ } }$

$AC= \frac{50 \sin 43^{\circ }}{\sin 74^{\circ } }$

Correct answer:

$AC= \frac{50 \sin 43^{\circ }}{\sin 74^{\circ } }$

Explanation:

By the Law of Sines,

$\frac{AC}{\sin b^{\circ }} = \frac{AB}{\sin c^{\circ } }$ .

Substitute AB = 50 , b = 43 , and c = 180 - (63+ 43) = 180 - 106 = 74 :

$\frac{AC}{\sin 43^{\circ }} = \frac{50}{\sin 74^{\circ } }$

Solve for :

$\frac{AC}{\sin 43^{\circ }}\cdot \sin 43^{\circ }= \frac{50}{\sin 73^{\circ } } \cdot \sin 43^{\circ }$

$AC= \frac{50 \sin 43^{\circ }}{\sin 74^{\circ } }$

Report an Error

Example Question #1 : Finding Sides

Which of the following describes a triangle with sides of length 9 feet, 3 yards, and 90 inches?

Possible Answers:

The triangle is obtuse and isosceles, but not equilateral.

The triangle is acute and scalene.

The triangle is obtuse and scalene.

The triangle is acute and isosceles, but not equilateral.

The triangle is acute and equilateral.

Correct answer:

The triangle is acute and isosceles, but not equilateral.

Explanation:

One yard is equal to three feet; One foot is equal to twelve inches. Therefore, 9 feet is equal to $9 \times 12 = 108$ inches, and 3 yards is equal to $3 \times 36 = 108$ inches. The triangle has sides of measure 90, 108, 108.

We compare the squares of the sides.

$90^{2}+ 108^{2} = 8,100 + 11,664 = 19,764 > 11,664 = 108^{2}$

The sum of the squares of the two smaller sidelengths exceeds that of the third, so the triangle is acute.

The correct response is acute and isosceles.

Report an Error

Example Question #3 : Finding Sides

Triangle

Note: figure NOT drawn to scale.

Refer to the above diagram.

AB = 40, a = 62, b = 44 .

Which of the following expressions is equal to ?

Possible Answers:

$AC= \frac{40 \cos 74^{\circ }}{\cos 44^{\circ } }$

$AC= \frac{40 \cos 44^{\circ }}{\cos 62^{\circ } }$

$AC= \frac{40 \cos 44^{\circ }}{\cos 74^{\circ } }$

$AC= \frac{40 \sin 44^{\circ }}{\sin 74^{\circ } }$

$AC= \frac{40 \sin 44^{\circ }}{\sin 62^{\circ } }$

Correct answer:

$AC= \frac{40 \sin 44^{\circ }}{\sin 74^{\circ } }$

Explanation:

By the Law of Sines,

$\frac{AC}{\sin b^{\circ }} = \frac{AB}{\sin c^{\circ } }$ .

Substitute AB = 40 , b = 44 , and c = 180 - (62+ 44) = 180 - 106 = 74 :

$\frac{AC}{\sin 44^{\circ }} = \frac{40}{\sin 74^{\circ } }$

We can solve for :

$\frac{AC}{\sin 44^{\circ }} \cdot \sin 44^{\circ }= \frac{40}{\sin 74^{\circ } } \cdot \sin 44^{\circ }$

$AC= \frac{40 \sin 44^{\circ }}{\sin 74^{\circ } }$

Report an Error

Example Question #6 : Finding Sides

Triangle

Note: figure NOT drawn to scale.

Refer to the triangle in the above diagram.

BC = 25, a= 60, b = 45 .

Evaluate .

Possible Answers:

$AC= \frac{50\sqrt{6}}{ 3 }$

$AC= \frac{50\sqrt{3}}{ 3 }$

$AC= \frac{25\sqrt{6}}{ 3 }$

$AC= 25\sqrt{3}$

$AC= \frac{25\sqrt{3}}{ 3 }$

Correct answer:

$AC= \frac{25\sqrt{6}}{ 3 }$

Explanation:

By the Law of Sines,

$\frac{AC}{\sin b^{\circ }} = \frac{BC}{\sin a^{\circ } }$

Substitute BC = 25, a= 60, b = 45 and solve for :

$\frac{AC}{\sin 45^{\circ } } = \frac{25}{\sin 60^{\circ }}$

$\frac{AC}{\sin 45^{\circ } } \cdot \sin 45^{\circ }\sin 60^{\circ } = \frac{25}{\sin 60^{\circ }} \cdot \sin 45^{\circ }\sin 60^{\circ }$

$AC\sin 60^{\circ } =25 \sin 45^{\circ }$

$AC\cdot { \frac{\sqrt{3}}{2}}= 25 \cdot \frac{\sqrt{2}}{2}$

$AC\cdot { \frac{\sqrt{3}}{2}}= \frac{25 \sqrt{2}}{2}$

$AC\cdot { \frac{\sqrt{3}}{2}} \cdot \frac{2 \sqrt{3}}{3}= \frac{25 \sqrt{2}}{2} \cdot \frac{2 \sqrt{3}}{3}$

$AC \cdot \frac{6}{6}= \frac{50 \sqrt{2}\sqrt{3}}{6}$

$AC= \frac{25\sqrt{6}}{ 3 }$

Report an Error

Example Question #2 : Finding Sides

Decagon

The above figure is a regular decagon. Evaluate to the nearest tenth.

Possible Answers:

$x \approx 11.4$

$x \approx 9.7$

$x \approx 10.4$

$x \approx 10.8$

$x \approx 11.6$

Correct answer:

$x \approx 11.4$

Explanation:

Two sides of the triangle formed measure 6 each; the included angle is one angle of the regular decagon, which measures

$m = \left [\frac{180(10-2)}{10} \right ] ^{\circ }= 144^{\circ }$ .

Since we know two sides and the included angle of the triangle in the diagram, we can apply the Law of Cosines,

$c^{2}= a^{2}+ b^{2} - 2ab \cos C^{\circ }$

with a = b = 6 and C = 144 :

$x ^{2}= 6^{2} + 6^{2} - 2 \cdot 6 \cdot 6 \cos 144^{\circ }$

$x ^{2}= 36+ 36- 72 \cos 144^{\circ }$

$x ^{2} \approx 72 - 72 \left ( -0.8090\right )$

$x \approx 11.4$

Report an Error

Example Question #352 : Sat Subject Test In Math Ii

Regular Pentagon ABCDE has perimeter 35. $\overline{AB }$ has as its midpoint; segment $\overline{ME}$ is drawn. To the nearest tenth, give the length of $\overline{ME}$ .

Possible Answers:

14.4

11.3

8.7

9.2

8.2

Correct answer:

8.7

Explanation:

The perimeter of the regular pentagon is 35, so each side measures one fifth of this, or 7. Also, since is the midpoint of $\overline{AB}$ , $AM = BM = \frac{1}{2}AB = \frac{1}{2} \cdot 7= 3.5$ .

Also, each interior angle of a regular pentagon measures $108^{\circ }$ .

Below is the pentagon in question, with indicated and $\overline{ME}$ constructed; all relevant measures are marked.

Pentagon 1

A triangle $\bigtriangleup MBC$ is formed with AM = 3.5 , AE= 7 , and included angle measure $\angle A= 108^{\circ }$ . The length of the remaining side can be calculated using the law of cosines:

$c^{2} = a^{2} + b^{2} -2 ab \cos \gamma$

where and are the lengths of two sides, $\gamma$ is the measure of their included angle, and is the length of the third side.

Setting $a = AM = 3.5, b=AE= 7, \gamma= m \angle A = 108^{\circ }$ , and c = ME , substitute and evaluate :

$(ME)^{2} = 3.5 ^{2} + 7^{2} -2 (3.5) (7) \cos 108^{\circ }$

$\approx 12.25 + 49 -49(-0.3090)$

$\approx 61.25 + 15.14$

$(ME)^{2} \approx 76.39$ ;

Taking the square root of both sides:

$ME \approx\sqrt{ 76.39 } \approx 8.7$ ,

the correct choice.

Report an Error

Example Question #51 : Geometry

Regular Hexagon ABCDEF has perimeter 360. $\overline{AF}$ and $\overline{DE}$ have and as midpoints, respectively; segment $\overline{MN}$ is drawn. To the nearest tenth, give the length of $\overline{MN}$ .

Possible Answers:

79.5

94.6

72.4

120.0

90.0

Correct answer:

90.0

Explanation:

The perimeter of the regular hexagon is 360, so each side measures one sixth of this, or 60. Since is the midpoint of $\overline{AB}$ , $AM = MF = \frac{1}{2}AF= \frac{1}{2} \cdot 60 = 30$ .

Similarly, DN = NE = 30 .

Also, each interior angle of a regular hexagon measures $120 ^{\circ }$ .

Below is the hexagon with the midpoints and , and with $\overline{MN}$ constructed. Note that perpendiculars have been drawn to $\overline{MN}$ from and , with feet at points and respectively.

Hexagon

EFXY is a rectangle, so XY = EF= 60 .

$m \angle MBX = m \angle MBC- m \angle XBC = 120 ^{\circ } - 90 ^{\circ } = 30 ^{\circ }$ .

This makes $\overline{MX}$ and $\overline{MF}$ the short leg and hypotenuse of a 30-60-90 triangle; as a consequence,

$MX = \frac{1}{2} \cdot MF = \frac{1}{2} \cdot 30 = 15$ .

For the same reason,

YN = 1 5

Adding the segment lengths:

MN = MX + XY + NY = 15 + 60 + 15 = 90 .

Report an Error

Example Question #4 : Finding Sides

Regular Pentagon ABCDE has perimeter 60.

To the nearest tenth, give the length of diagonal $\overline{BD}$ .

Possible Answers:

11.6

15.7

19.4

15 .0

14.1

Correct answer:

19.4

Explanation:

The perimeter of the regular pentagon is 60, so each side measures one fifth of this, or 12. Also, each interior angle of a regular pentagon measures $108^{\circ }$ .

The pentagon, along with diagonal $\overline{BD}$ , is shown below:

Pentagon 2

A triangle $\bigtriangleup BCD$ is formed with AB=BC = 16 , and included angle measure $\angle B= 108^{\circ }$ . The length of the remaining side can be calculated using the Law of Cosines:

$c^{2} = a^{2} + b^{2} -2 ab \cos \gamma$

where and are the lengths of two sides, $\gamma$ the measure of their included angle, and the length of the side opposite that angle.

Setting $a = BC= 12 , b= CD = 12, \gamma = m \angle C = 108^{\circ }$ , and c= BD , substitute and evaluate :

$(BD)^{2} = 12 ^{2} + 12 ^{2} -2 (12) (12) \cos 108^{\circ }$

$\approx 144+144 -288 (-0.3090)$

$\approx 288 + 89.00$

$(BD)^{2} \approx 377.00$

Taking the square root of both sides:

$BD\approx\sqrt{ 377.00 } \approx 19.4$ ,

the correct choice.

Report an Error

View SAT Subject Test in Mathematics Level 2 Tutors

Reuben
Certified Tutor

University of Alberta, Bachelor of Science, Mathematics.

View SAT Subject Test in Mathematics Level 2 Tutors

Irina
Certified Tutor

New York University, Bachelors, Biochemistry.

View SAT Subject Test in Mathematics Level 2 Tutors

Nafis
Certified Tutor

University of Leipzig, Germany., Bachelor of Science, Physics. University of Pittsburgh-Pittsburgh Campus, Doctor of Philosop...

All SAT II Math II Resources

2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

MCAT Tutors in Denver, Physics Tutors in Dallas Fort Worth, Math Tutors in Denver, French Tutors in Dallas Fort Worth, French Tutors in Denver, ISEE Tutors in Atlanta, GMAT Tutors in Seattle, Physics Tutors in Chicago, ISEE Tutors in San Diego, Biology Tutors in Washington DC

Popular Courses & Classes

MCAT Courses & Classes in Phoenix, SAT Courses & Classes in New York City, SAT Courses & Classes in Chicago, ACT Courses & Classes in Denver, GMAT Courses & Classes in Denver, SAT Courses & Classes in Houston, ACT Courses & Classes in Seattle, GRE Courses & Classes in Washington DC, GRE Courses & Classes in New York City, ACT Courses & Classes in Los Angeles

Popular Test Prep

ISEE Test Prep in Washington DC, SAT Test Prep in Dallas Fort Worth, SAT Test Prep in Denver, ACT Test Prep in Denver, LSAT Test Prep in San Diego, MCAT Test Prep in Chicago, ACT Test Prep in Seattle, LSAT Test Prep in New York City, GMAT Test Prep in Seattle, ACT Test Prep in Los Angeles

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

SAT II Math II : Finding Sides

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #41 : Geometry

Example Question #2 : Finding Sides

Example Question #1 : Finding Sides

Example Question #1 : Finding Sides

Example Question #3 : Finding Sides

Example Question #6 : Finding Sides

Example Question #2 : Finding Sides

Example Question #352 : Sat Subject Test In Math Ii

Example Question #51 : Geometry

Example Question #4 : Finding Sides

All SAT II Math II Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: