We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Intermediate Geometry : Intermediate Geometry

Study concepts, example questions & explanations for Intermediate Geometry

Create An Account

All Intermediate Geometry Resources

8 Diagnostic Tests 250 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #5 : How To Find The Area Of An Equilateral Triangle

Asd

Suppose we triple the sides of this equilateral triangle to . What is the area of the new triangle in terms of ?

Possible Answers:

$\frac{\sqrt{3}}{2}a^2$

a^2

$\frac{3}{4}a^2$

$\frac{9\sqrt{3}}{4}a^2$

$\frac{\sqrt{3}}{4}a^2$

Correct answer:

$\frac{9\sqrt{3}}{4}a^2$

Explanation:

The formula for the area of an equilateral triangle is $\frac{\sqrt{3}}{4}x^2$ if is the length of one of the triangle's sides.

In this problem, the length of one of the triangle's sides is being tripled, so we can substitute into the equation for and solve for the triangle's new area in terms of :

$Area=\frac{\sqrt{3}}{4}x^2=\frac{\sqrt{3}}{4}(3a)^2=\frac{\sqrt{3}}{4}9a^2=\frac{9\sqrt{3}}{4}a^2$

Report an Error

Example Question #2 : How To Find The Area Of An Equilateral Triangle

Asd

What is half the area of the above triangle if $a = 4\:cm$ ?

Possible Answers:

$2\sqrt{3}\:cm^2$

$4\sqrt{3}\:cm^2$

$4\:cm^2$

$\sqrt{3}\:cm^2$

$3\:cm^2$

Correct answer:

$2\sqrt{3}\:cm^2$

Explanation:

The formula for the area of an equilateral triangle is $\frac{\sqrt{3}}{4}a^2$ . For this problem's triangle, $a=4\:cm$ , so we can substitute into the equation for and solve for the area of the triangle:

$Area=\frac{\sqrt{3}}{4}a^2=\frac{\sqrt{3}}{4}(4)^2=\frac{\sqrt{3}}{4}16=4\sqrt{3}\:cm^2$

At this point, we need to divide by , since the problem asks for half of the triangle's area:

$\frac{4\sqrt{3}}{2}=2\sqrt{3}\:cm^2$

Report an Error

Example Question #671 : Plane Geometry

Asd

If a=10 , what is the area of the triangle?

Possible Answers:

$\sqrt{3}\:cm^2$

$2\sqrt{3}\:cm^2$

$10\:cm^2$

$25\sqrt{3}\:cm^2$

$25\:cm^2$

Correct answer:

$25\sqrt{3}\:cm^2$

Explanation:

We know this triangle is equilateral since each of its sides has the same length, . The formula of the area of an equilateral triangle is $\frac{\sqrt{3}}{4}x^2$ if is the length of one of the triangle's sides.

Since our side length is $10\:cm$ , we can substitute that value into the equation for and solve for the area of the triangle:

$Area=\frac{\sqrt{3}}{4}x^2=\frac{\sqrt{3}}{4}(10)^2=\frac{\sqrt{3}}{4}100=25\sqrt{3}\:cm^2$

Report an Error

Example Question #672 : Plane Geometry

Asd

If $a=5\:cm$ , what is the area of this equilateral triangle?

Possible Answers:

$\sqrt{3}\:cm^2$

$5\sqrt{3}\:cm^2$

$15\sqrt{3}\:cm^2$

$\frac{25\sqrt{3}}{4}\:cm^2$

$20\sqrt{3}\:cm^2$

Correct answer:

$\frac{25\sqrt{3}}{4}\:cm^2$

Explanation:

Given that the sides of our equilateral triangle are each $5\:cm$ long, we can just plug the value in to the formula for the area of an equilateral triangle and solve for the area of the triangle:

$Area=\frac{\sqrt{3}}{4}a^2$ if is a side of the triangle.

$Area=\frac{\sqrt{3}}{4}a^2=\frac{\sqrt{3}}{4}5^2=\frac{\sqrt{3}}{4}25=\frac{25\sqrt{3}}{4}\:cm^2$

Report an Error

Example Question #231 : Triangles

An equilateral triangle has sides of length 6cm. If the height of the triangle is 4.5cm what is the area of the triangle?

Possible Answers:

$18 \; cm^{2}$

$27 \; cm^{2}$

$13.5 \; cm^{2}$

$9 \; cm^{2}$

$6.75 \; cm^{2}$

Correct answer:

$13.5 \; cm^{2}$

Explanation:

To find the area of any triangle we can use the formula 1/2 (base x height) , that is the base times the height divided by two. It is important to remember any of the sides of an equaliateral triangle can be used as the base when the hieght is given. The area can be found by (6 x 4.5) divided by 2; which gives 13.5 square centimeters.

Report an Error

Example Question #231 : Triangles

Find the area of an equilateral triangle with a perimeter of 24cm. Leave answer in simplest radical form.

Possible Answers:

$32\ cm^{2}$

$32\sqrt3\ cm^{2}$

$64\ cm^{2}$

$16\sqrt3\ cm^{2}$

Correct answer:

$16\sqrt3\ cm^{2}$

Explanation:

To find the area of an equilateral triangle, one must find the base and the height.

$Area=\frac{1}{2}bh$

All the sides of an equilateral triangle are congruent, so if the perimeter of the equilaterail triangle is 24, then each side must equal one third of that total which is 8cm.

This will produce a triangle that includes the following information below:

1 ans2

Dropping an altitude down the center of the equilateral triangle will result in two 30-60-90 triangles with a hypotenuse of 8.

1 ans 3

In every 30-60-90 triangle the following formulas apply:

hypotenuse=2(shortleg)

$longleg=shortleg\sqrt3$

When we plug in the given information on the triangle we get:

8=2(shortleg)

Dividing both sides by 2 gives the below result.

shortleg=4

We can now plug this into the long leg formula to get the height of the triangle:

$longleg=4\sqrt3$

1 anspng

Now that we have all of the information needed to find the area we plug these values into the area formula.

$Area=\frac{1}{2}bh$

$Area=\frac{1}{2}(8)(4\sqrt3)$

$Area=16\sqrt3$

Report an Error

Example Question #11 : How To Find The Area Of An Equilateral Triangle

A circle with a radius of is inscribed in an equilateral triangle with side lengths of as shown in the figure below.

Find the area of the shaded region.

Possible Answers:

1.99

3.02

3.68

2.01

Correct answer:

3.02

Explanation:

In order to find the area of the shaded region, we must first find the area of the circle and the area of the equilateral triangle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Plug in the given radius to find the area of the circle.

$\text{Area of Circle}=\pi\times 2^2=4\pi$

Next, recall how to find the area of an equilateral triangle:

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(\text{side})^2$

Plug in the length of the side of the triangle to find the area.

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(6)^2=9\sqrt3$

In order to find the area of the shaded region, we will need to subtract the area of the circle from the area of the triangle.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Circle}$

$\text{Area of Shaded Region}=9\sqrt3-4\pi=3.02$

Make sure to round to places after the decimal.

Report an Error

Example Question #12 : How To Find The Area Of An Equilateral Triangle

A circle with a radius of is inscribed in an equilateral triangle with side lengths of as shown in the figure below.

Find the area of the shaded region.

Possible Answers:

5.87

6.80

7.25

6.33

Correct answer:

6.80

Explanation:

In order to find the area of the shaded region, we must first find the area of the circle and the area of the equilateral triangle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Plug in the given radius to find the area of the circle.

$\text{Area of Circle}=\pi\times 3^2=9\pi$

Next, recall how to find the area of an equilateral triangle:

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(\text{side})^2$

Plug in the length of the side of the triangle to find the area.

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(9)^2=\frac{81\sqrt3}{4}$

In order to find the area of the shaded region, we will need to subtract the area of the circle from the area of the triangle.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Circle}$

$\text{Area of Shaded Region}=\frac{81\sqrt3}{4}-9\pi=6.80$

Make sure to round to places after the decimal.

Report an Error

Example Question #13 : How To Find The Area Of An Equilateral Triangle

A circle with a radius of is inscribed in an equilateral triangle with side lengths of as shown in the figure below.

Find the area of the shaded region.

Possible Answers:

13.08

10.27

12.09

16.33

Correct answer:

12.09

Explanation:

In order to find the area of the shaded region, we must first find the area of the circle and the area of the equilateral triangle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Plug in the given radius to find the area of the circle.

$\text{Area of Circle}=\pi\times 4^2=16\pi$

Next, recall how to find the area of an equilateral triangle:

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(\text{side})^2$

Plug in the length of the side of the triangle to find the area.

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(12)^2=36\sqrt3$

In order to find the area of the shaded region, we will need to subtract the area of the circle from the area of the triangle.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Circle}$

$\text{Area of Shaded Region}=36\sqrt3-16\pi=12.09$

Make sure to round to places after the decimal.

Report an Error

Example Question #14 : How To Find The Area Of An Equilateral Triangle

A circle with a radius of is inscribed in an equilateral triangle with side lengths of as shown in the figure below.

Find the area of the shaded region.

Possible Answers:

100.36

99.07

97.86

94.67

Correct answer:

94.67

Explanation:

In order to find the area of the shaded region, we must first find the area of the circle and the area of the equilateral triangle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Plug in the given radius to find the area of the circle.

$\text{Area of Circle}=\pi\times 5^2=25\pi$

Next, recall how to find the area of an equilateral triangle:

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(\text{side})^2$

Plug in the length of the side of the triangle to find the area.

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(20)^2=100\sqrt3$

In order to find the area of the shaded region, we will need to subtract the area of the circle from the area of the triangle.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Circle}$

$\text{Area of Shaded Region}=100\sqrt3-25\pi=94.67$

Make sure to round to places after the decimal.

Report an Error

View Tutors

Josephine
Certified Tutor

University of Maryland-College Park, Bachelor of Science, Environmental Science.

View Tutors

Benjamin
Certified Tutor

Portland State University, Bachelor of Science, Computer Science.

View Tutors

Abby
Certified Tutor

Ripon College, Bachelor in Arts, Psychology.

All Intermediate Geometry Resources

8 Diagnostic Tests 250 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Statistics Tutors in San Diego, Spanish Tutors in San Francisco-Bay Area, ISEE Tutors in Chicago, GRE Tutors in Dallas Fort Worth, Computer Science Tutors in Dallas Fort Worth, English Tutors in Houston, GMAT Tutors in Los Angeles, GMAT Tutors in San Francisco-Bay Area, Biology Tutors in Phoenix, Math Tutors in Phoenix

Popular Courses & Classes

SAT Courses & Classes in San Diego, ISEE Courses & Classes in San Diego, LSAT Courses & Classes in Miami, Spanish Courses & Classes in Philadelphia, SSAT Courses & Classes in Houston, SSAT Courses & Classes in Atlanta, LSAT Courses & Classes in Denver, MCAT Courses & Classes in Seattle, ACT Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in San Diego

Popular Test Prep

ISEE Test Prep in San Francisco-Bay Area, ISEE Test Prep in Dallas Fort Worth, GMAT Test Prep in Philadelphia, SAT Test Prep in Miami, SSAT Test Prep in Houston, LSAT Test Prep in Miami, MCAT Test Prep in Atlanta, SSAT Test Prep in New York City, ISEE Test Prep in Miami, MCAT Test Prep in Chicago

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

Intermediate Geometry : Intermediate Geometry

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #5 : How To Find The Area Of An Equilateral Triangle

Example Question #2 : How To Find The Area Of An Equilateral Triangle

Example Question #671 : Plane Geometry

Example Question #672 : Plane Geometry

Example Question #231 : Triangles

Example Question #231 : Triangles

Example Question #11 : How To Find The Area Of An Equilateral Triangle

Example Question #12 : How To Find The Area Of An Equilateral Triangle

Example Question #13 : How To Find The Area Of An Equilateral Triangle

Example Question #14 : How To Find The Area Of An Equilateral Triangle

All Intermediate Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: