Call Now to Set Up Tutoring:

(888) 888-0446

Intermediate Geometry : Triangles

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 #141 : Triangles

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Possible Answers:

$\frac{227}{2}$ $\displaystyle \frac{227}{2}$

$\frac{235}{2}$ $\displaystyle \frac{235}{2}$

$\frac{215}{2}$ $\displaystyle \frac{215}{2}$

$\frac{225}{2}$ $\displaystyle \frac{225}{2}$

Correct answer:

$\frac{225}{2}$ $\displaystyle \frac{225}{2}$

Explanation:

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

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$ $\displaystyle \text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=10 \times 15$ $\displaystyle \text{Area of Rectangle}=10 \times 15$

$\text{Area of Rectangle}= 150$ $\displaystyle \text{Area of Rectangle}= 150$

Next, recall how to find the area of a triangle:

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$ $\displaystyle \text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{10}{2}$ $\displaystyle \text{height}=\frac{10}{2}$

$\text{height}=5$ $\displaystyle \text{height}=5$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 5\times15$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}\times 5\times15$

$\text{Area of Triangle}=\frac{75}{2}$ $\displaystyle \text{Area of Triangle}=\frac{75}{2}$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

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

$\text{Area of Shaded Region}=150-\frac{75}{2}$ $\displaystyle \text{Area of Shaded Region}=150-\frac{75}{2}$

Solve.

$\text{Area of Shaded Region}=\frac{225}{2}$ $\displaystyle \text{Area of Shaded Region}=\frac{225}{2}$

Report an Error

Example Question #142 : Triangles

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Possible Answers:

$\displaystyle 70$

$\displaystyle 74$

$\displaystyle 78$

$\displaystyle 64$

Correct answer:

$\displaystyle 78$

Explanation:

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

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$ $\displaystyle \text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=8 \times 13$ $\displaystyle \text{Area of Rectangle}=8 \times 13$

$\text{Area of Rectangle}= 104$ $\displaystyle \text{Area of Rectangle}= 104$

Next, recall how to find the area of a triangle:

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$ $\displaystyle \text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{8}{2}$ $\displaystyle \text{height}=\frac{8}{2}$

$\text{height}=4$ $\displaystyle \text{height}=4$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 13\times 4$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}\times 13\times 4$

$\text{Area of Triangle}=26$ $\displaystyle \text{Area of Triangle}=26$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

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

$\text{Area of Shaded Region}=104-26$ $\displaystyle \text{Area of Shaded Region}=104-26$

Solve.

$\text{Area of Shaded Region}=78$ $\displaystyle \text{Area of Shaded Region}=78$

Report an Error

Example Question #23 : How To Find The Area Of An Acute / Obtuse Triangle

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Possible Answers:

300 $\displaystyle 300$

350 $\displaystyle 350$

200 $\displaystyle 200$

250 $\displaystyle 250$

Correct answer:

300 $\displaystyle 300$

Explanation:

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

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$ $\displaystyle \text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=16 \times 25$ $\displaystyle \text{Area of Rectangle}=16 \times 25$

$\text{Area of Rectangle}= 400$ $\displaystyle \text{Area of Rectangle}= 400$

Next, recall how to find the area of a triangle:

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$ $\displaystyle \text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{16}{2}$ $\displaystyle \text{height}=\frac{16}{2}$

$\text{height}=8$ $\displaystyle \text{height}=8$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 25\times 8$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}\times 25\times 8$

$\text{Area of Triangle}=100$ $\displaystyle \text{Area of Triangle}=100$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

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

$\text{Area of Shaded Region}=400-100$ $\displaystyle \text{Area of Shaded Region}=400-100$

Solve.

$\text{Area of Shaded Region}=300$ $\displaystyle \text{Area of Shaded Region}=300$

Report an Error

Example Question #24 : How To Find The Area Of An Acute / Obtuse Triangle

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Possible Answers:

$\displaystyle 90$

100 $\displaystyle 100$

$\displaystyle 80$

120 $\displaystyle 120$

Correct answer:

120 $\displaystyle 120$

Explanation:

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

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$ $\displaystyle \text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=8 \times 20$ $\displaystyle \text{Area of Rectangle}=8 \times 20$

$\text{Area of Rectangle}= 160$ $\displaystyle \text{Area of Rectangle}= 160$

Next, recall how to find the area of a triangle:

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$ $\displaystyle \text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{8}{2}$ $\displaystyle \text{height}=\frac{8}{2}$

$\text{height}=4$ $\displaystyle \text{height}=4$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 20\times 4$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}\times 20\times 4$

$\text{Area of Triangle}=40$ $\displaystyle \text{Area of Triangle}=40$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

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

$\text{Area of Shaded Region}=160-40$ $\displaystyle \text{Area of Shaded Region}=160-40$

Solve.

$\text{Area of Shaded Region}=120$ $\displaystyle \text{Area of Shaded Region}=120$

Report an Error

Example Question #25 : How To Find The Area Of An Acute / Obtuse Triangle

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Possible Answers:

384 $\displaystyle 384$

346 $\displaystyle 346$

378 $\displaystyle 378$

296 $\displaystyle 296$

Correct answer:

378 $\displaystyle 378$

Explanation:

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

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$ $\displaystyle \text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=18 \times 28$ $\displaystyle \text{Area of Rectangle}=18 \times 28$

$\text{Area of Rectangle}= 504$ $\displaystyle \text{Area of Rectangle}= 504$

Next, recall how to find the area of a triangle:

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$ $\displaystyle \text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{18}{2}$ $\displaystyle \text{height}=\frac{18}{2}$

$\text{height}=9$ $\displaystyle \text{height}=9$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 28\times 9$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}\times 28\times 9$

$\text{Area of Triangle} =126$ $\displaystyle \text{Area of Triangle} =126$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

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

$\text{Area of Shaded Region}=504-126$ $\displaystyle \text{Area of Shaded Region}=504-126$

Solve.

$\text{Area of Shaded Region}=378$ $\displaystyle \text{Area of Shaded Region}=378$

Report an Error

Example Question #143 : Triangles

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Possible Answers:

$\frac{246}{4}$ $\displaystyle \frac{246}{4}$

$\frac{297}{4}$ $\displaystyle \frac{297}{4}$

$\frac{301}{4}$ $\displaystyle \frac{301}{4}$

$\frac{287}{4}$ $\displaystyle \frac{287}{4}$

Correct answer:

$\frac{297}{4}$ $\displaystyle \frac{297}{4}$

Explanation:

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

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$ $\displaystyle \text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=9\times 11$ $\displaystyle \text{Area of Rectangle}=9\times 11$

$\text{Area of Rectangle}=99$ $\displaystyle \text{Area of Rectangle}=99$

Next, recall how to find the area of a triangle:

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$ $\displaystyle \text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{9}{2}$ $\displaystyle \text{height}=\frac{9}{2}$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 11\times \frac{9}{2}$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}\times 11\times \frac{9}{2}$

$\text{Area of Triangle}=\frac{99}{4}$ $\displaystyle \text{Area of Triangle}=\frac{99}{4}$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

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

$\text{Area of Shaded Region}=99-\frac{99}{4}$ $\displaystyle \text{Area of Shaded Region}=99-\frac{99}{4}$

Solve.

$\text{Area of Shaded Region}=\frac{297}{4}$ $\displaystyle \text{Area of Shaded Region}=\frac{297}{4}$

Report an Error

Example Question #141 : Triangles

A triangle is placed in a parallelogram so they share a base.

If the height of the triangle is half that of the parallelogram, find the area of the shaded region.

Possible Answers:

$\displaystyle 84$

$\displaystyle 72$

$\displaystyle 48$

$\displaystyle 66$

Correct answer:

$\displaystyle 72$

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$ $\displaystyle \text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=12\times8=96$ $\displaystyle \text{Area of Parallelogram}=12\times8=96$

Next, recall how to find the area of a triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$ $\displaystyle \text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{8}{2}=4$ $\displaystyle \text{Height of Triangle}=\frac{8}{2}=4$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(12 \times 4)=24$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}(12 \times 4)=24$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=96-24=72$ $\displaystyle \text{Area of Shaded Region}=96-24=72$

Report an Error

Example Question #581 : Intermediate Geometry

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is one-fifth of that of the parallelogram, find the area of the shaded region.

Possible Answers:

134 $\displaystyle 134$

126 $\displaystyle 126$

120 $\displaystyle 120$

118 $\displaystyle 118$

Correct answer:

126 $\displaystyle 126$

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$ $\displaystyle \text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=14\times 10=140$ $\displaystyle \text{Area of Parallelogram}=14\times 10=140$

Next, recall how to find the area of a triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{5}$ $\displaystyle \text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{5}$

$\text{Height of Triangle}=\frac{10}{5}=2$ $\displaystyle \text{Height of Triangle}=\frac{10}{5}=2$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(14 \times 2)=14$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}(14 \times 2)=14$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=140-14=126$ $\displaystyle \text{Area of Shaded Region}=140-14=126$

Report an Error

Example Question #145 : Triangles

A triangle is placed so that it shares a base with a parallelogram.

If the height of the triangle is one-third that of the parallelogram, find the area of the shaded region.

Possible Answers:

2.5 $\displaystyle 2.5$

10.5 $\displaystyle 10.5$

8.5 $\displaystyle 8.5$

12.5 $\displaystyle 12.5$

Correct answer:

12.5 $\displaystyle 12.5$

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$ $\displaystyle \text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=3\times 5=15$ $\displaystyle \text{Area of Parallelogram}=3\times 5=15$

Next, recall how to find the area of a triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{3}$ $\displaystyle \text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{3}$

$\text{Height of Triangle}=\frac{3}{3}=1$ $\displaystyle \text{Height of Triangle}=\frac{3}{3}=1$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(5\times 1)=\frac{5}{2}=2.5$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}(5\times 1)=\frac{5}{2}=2.5$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=15-2.5=12.5$ $\displaystyle \text{Area of Shaded Region}=15-2.5=12.5$

Report an Error

Example Question #591 : Plane Geometry

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is half that of the parallelogram, find the area of the shaded region.

Possible Answers:

$\displaystyle 88$

$\displaystyle 92$

$\displaystyle 84$

$\displaystyle 90$

Correct answer:

$\displaystyle 84$

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$ $\displaystyle \text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=14\times8=112$ $\displaystyle \text{Area of Parallelogram}=14\times8=112$

Next, recall how to find the area of a triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$ $\displaystyle \text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{8}{2}=4$ $\displaystyle \text{Height of Triangle}=\frac{8}{2}=4$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(14 \times 4)=28$ $\displaystyle \text{Area of Triangle}=\frac{1}{2}(14 \times 4)=28$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=112-28=84$ $\displaystyle \text{Area of Shaded Region}=112-28=84$

Report an Error

View Tutors

Laurentine
Certified Tutor

Rutgers University-Newark, Bachelor of Science, Biomedical Engineering.

View Tutors

Munnesa
Certified Tutor

SUNY College at Buffalo, Bachelor of Science, Chemistry. SUNY College at Buffalo, Master of Science, Biological and Physical ...

View Tutors

Vincent
Certified Tutor

Institute of Management and Technology, Bachelor of Engineering, Electrical Engineering. Enugu State University of Science an...

All Intermediate Geometry Resources

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

Popular Subjects

SSAT Tutors in Phoenix, ISEE Tutors in Phoenix, Computer Science Tutors in San Francisco-Bay Area, French Tutors in New York City, Reading Tutors in San Diego, Math Tutors in New York City, Biology Tutors in Philadelphia, Reading Tutors in New York City, SAT Tutors in Philadelphia, Spanish Tutors in Miami

Popular Courses & Classes

SAT Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in Chicago, GRE Courses & Classes in Atlanta, Spanish Courses & Classes in Houston, LSAT Courses & Classes in Los Angeles, LSAT Courses & Classes in Philadelphia, ISEE Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in Atlanta, MCAT Courses & Classes in Boston, Spanish Courses & Classes in Boston

Popular Test Prep

MCAT Test Prep in Los Angeles, GRE Test Prep in Miami, LSAT Test Prep in San Diego, GMAT Test Prep in Dallas Fort Worth, GRE Test Prep in Houston, ISEE Test Prep in San Diego, MCAT Test Prep in New York City, MCAT Test Prep in Chicago, ACT Test Prep in Houston, GRE 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:

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 : Triangles

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #141 : Triangles

Example Question #142 : Triangles

Example Question #23 : How To Find The Area Of An Acute / Obtuse Triangle

Example Question #24 : How To Find The Area Of An Acute / Obtuse Triangle

Example Question #25 : How To Find The Area Of An Acute / Obtuse Triangle

Example Question #143 : Triangles

Example Question #141 : Triangles

Example Question #581 : Intermediate Geometry

Example Question #145 : Triangles

Example Question #591 : Plane Geometry

All Intermediate Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: