GMAT Math : Geometry

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

22 Diagnostic Tests 693 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #271 : Geometry

A right triangle has a hypotenuse of length 13 yards and a leg of length 5 yards. Give the perimeter of this triangle in inches.

Possible Answers:

$1,040\textrm{ in}$

$1,200\textrm{ in}$

$960\textrm{ in}$

$1,400\textrm{ in}$

$1,080\textrm{ in}$

Correct answer:

$1,080\textrm{ in}$

Explanation:

We can use the Pythagorean Theorem to calculate the length of the second leg in yards of the triangle by setting c = 13, a = 5 in this formula:

$b ^{2} = c ^{2 } - a ^{2}$

$b ^{2} = 13 ^{2 } - 5 ^{2} = 169 - 25 = 144$

$b = \sqrt{ 144} = 12$

The perimeter of the triangle is yards is

P = a + b + c = 5 + 12 + 13 = 30 yards. Multiply this by 36 to covert to inches:

$30 \times 36 = 1,080$ inches.

Report an Error

Example Question #3 : Calculating The Perimeter Of A Right Triangle

A right triangle has legs of length feet and and $6 \frac{2}{3}$ feet. Give the perimeter of this triangle in yards.

Possible Answers:

$7 \frac{1}{3} \textrm{ yd}$

$6 \textrm{ yd}$

$7 \textrm{ yd}$

$6 \frac{2}{3} \textrm{ yd}$

$5 \frac{2}{3} \textrm{ yd}$

Correct answer:

$6 \frac{2}{3} \textrm{ yd}$

Explanation:

We can use the Pythagorean Theorem to calculate the hypotenuse of the triangle by setting $a = 5, b = 6 \frac{2}{3}= \frac{20}{3}$ in this formula:

$c^{2} = a ^{2}+ b^{2}$

$c^{2} = 5 ^{2}+ \left ( \frac{20}{3} \right ) ^{2}$

$c^{2} = 25 + \frac{400}{9} = \frac{225}{9} + \frac{400}{9} = \frac{625}{9}$

$c = \sqrt{\frac{625}{9} } = \frac{25}{3} = 8 \frac{1}{3}$

The perimeter is $P = a + b + c = 5 + 6\frac{2}{3} + 8 \frac{1}{3} = 20$ feet.

Divide by 3 to convert to yards:

$20 \div 3 = 6 \frac{2}{3}$ yards

Report an Error

Example Question #2 : Calculating The Perimeter Of A Right Triangle

What is the perimeter of a 30-60-90 triangle?

1) One of the sides measures 10 inches.

2) One of the sides measures 20 inches.

Possible Answers:

Statement 2 ALONE is sufficient, but Statement 1 alone is not sufficient.

Statement 1 ALONE is sufficient, but Statement 2 alone is not sufficient.

BOTH statements TOGETHER are sufficient, but neither statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements 1 and 2 TOGETHER are not sufficient.

Correct answer:

BOTH statements TOGETHER are sufficient, but neither statement ALONE is sufficient.

Explanation:

The hypotenuse of a 30-60-90 triangle measures twice its shorter leg; its longer leg measures $\small \sqrt{3}$ times its shorter leg. Given one side length alone, there is no indication which of the three sides it measures; but given two, one of which is twice the other, as is the case here (10 and 20), 10 must be the shorter leg and 20 the hypotenuse. The longer leg is therefore $10\sqrt{3}$ , and the perimeter is

$P=10+20+10\sqrt{3}=30+10\sqrt{3}$ .

Therefore, both statements together are sufficent but neither alone is sufficient.

Report an Error

Example Question #24 : Triangles

A right triangle has a base of 4 and a height of 3. What is the perimeter of the triangle?

Possible Answers:

Correct answer:

Explanation:

We are given two sides of the right triangle, so in order to calculate the perimeter we must first find the length of the third side, the hypotenuse, using the Pythagorean theorem:

a^2+b^2=c^2

3^2+4^2=c^2

$c^2=25\rightarrow c=5$

Now that we know the length of the third side, we can add the lengths of the three sides to calculate the perimeter of the right triangle:

P=3+4+5=12

Report an Error

Example Question #25 : Right Triangles

Export-png__2_

$\Delta BCD$ is a right triangle and CD=4 and DB=3 . What is half the circle's circumference added to the triangle's perimeter?

Possible Answers:

$\frac{5}{2}\pi+12$

$\pi+12$

$3\pi+12$

$5\pi+12$

$\frac{\pi}{2}+6$

Correct answer:

$\frac{5}{2}\pi+12$

Explanation:

As you see that a right triangle as sides 3 and 4, you should always remember that, this right triangle is a Pythagorean Triple, in other words, its sides will be in the ratio 3n,4n,5n where is a constant, this will save you a lot of time. Here we can say that the hypotenuse will be 5. Therefore, the circumference will be $5\pi$ . To get the final answer, we should just divide the circumference by 2 and add the perimeter, being 3+4+5 or 12.

Report an Error

Example Question #272 : Geometry

In the above figure, triangle SPQ, SPR and PRQ are right triangles. Given the lengths of SP and PQ are 2cm. What is the area of triangle SPR?

Possible Answers:

Correct answer:

Explanation:

Since $\triangle SPQ,\triangle SPR,and\ \triangle PRQ$ are right triangles, we know that $\triangle SPQ$ is an isosceles right triangle. So we know that the lengths of $\overline{SP}$ and $\overline{PQ}$ are 2 cm, so we can get the length of $\overline{SQ}$ by using the Pythagorean Theorem:

$SQ =\sqrt{2^2+2^2}=\sqrt{8}$

is the midpoint of $\overline{SQ}$ , so the length of $\overline{SR}$ is $\sqrt{2}$ .

Now we can use the Pythagorean Theorem again to solve for $\overline{PR}$ : $\sqrt{ (2^2-(\sqrt2)^2)}=\sqrt2$ .

Finally, we have all the elements needed to solve for the area of $\triangle SPR$ :

$(1/2)*\sqrt2*\sqrt2=1$

Report an Error

Example Question #273 : Geometry

If a right triangle has a hypotenuse that is 10, and one side is 6, find the area of the triangle.

Possible Answers:

Correct answer:

Explanation:

Becaue this is a right triangle and since the hypotenuse is 10 and one side is 6, the other side will be 8. This can be found using the Pythagorean Theorem, or multiplying a 3-4-5 triangle by 2 to get a 6-8-10 triangle. Since the area of a right triangle is half of the product of the two sides, we have

$A=\frac{1}{2}L*W=\frac{1}{2}*6*8=24.$

Report an Error

Example Question #28 : Right Triangles

The hypotenuse of a $30 ^{\circ } - 60^{\circ } -90^{\circ }$ triangle is equal to the sidelength of a square. Give the ratio of the area of the square to that of the triangle.

Possible Answers:

$8\sqrt{3}:3$

16 :3

$16\sqrt{3}:3$

8 :3

4:1

Correct answer:

$8\sqrt{3}:3$

Explanation:

Let be the sidelength of the square. Then its area is $s ^{2}$ .

If the hypotenuse of a $30 ^{\circ } - 60^{\circ } -90^{\circ }$ triangle is , its shorter leg is half that, or $\frac{1}{2 }s$ ; its longer leg is $\sqrt{3}$ times the shorter leg, or $\frac{1}{2 }s \sqrt{3}$ . The area of the triangle is half the product of the legs, or

$\frac{1}{2} \cdot \frac{1}{2}s \cdot\frac{1}{2 }s \sqrt{3} =\frac{ \sqrt{3}}{8} s ^{2}$

The ratio of the area of the square to that of the triangle is

$s ^{2}: \frac{ \sqrt{3}}{8} s$ or $1: \frac{ \sqrt{3}}{8}$

or $8\sqrt{3}:3$

Report an Error

Example Question #274 : Geometry

A triangle on the coordinate plane has vertices (0,0), (c+1,0 ), (0,c-1) .

Which of the following expressions is equal to the area of the triangle?

Possible Answers:

$\frac{1}{2} c^{2} +\frac{1}{2}$

$c^{2} -1$

$c^{2}$

$\frac{1}{2} c^{2} -\frac{1}{2}$

$\frac{1}{2} c^{2}$

Correct answer:

$\frac{1}{2} c^{2} -\frac{1}{2}$

Explanation:

This is a right triangle with legs along the - and -axes, so the area of each can be calculated by taking one-half the product of the two legs.

The vertical leg has length c-1 ; the horizontal leg has length c+ 1 .

Now calculate the area:

$A = \frac{1}{2} bh = \frac{1}{2} (c-1)(c+1) = \frac{1}{2} (c^{2}-1) = \frac{1}{2} c^{2}- \frac{1}{2}$

Report an Error

Example Question #277 : Geometry

Calculate the area of the following right triangle, leave in terms of .

(Not drawn to scale.)

Possible Answers:

x^2+2x

x^2+3x+2

$\frac{x}{2}(x+2)$

$\frac{x}{2}(x+1)$

Correct answer:

$\frac{x}{2}(x+1)$

Explanation:

The equation for the area of a right triangle is:

$A=\frac{1}{2}b\times h$

In this case, our values are: b=x, h=(x+1)

Plugging this into the equation leaves us with:

$A=\frac{1}{2}(x)(x+1)$

which can be rewritten as $A=\frac{x}{2}(x+1)$

Report an Error

All GMAT Math Resources

22 Diagnostic Tests 693 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Reading Tutors in Atlanta, ACT Tutors in New York City, MCAT Tutors in New York City, GRE Tutors in Washington DC, SSAT Tutors in Boston, Statistics Tutors in San Francisco-Bay Area, GRE Tutors in Dallas Fort Worth, Physics Tutors in San Diego, Statistics Tutors in Washington DC, Algebra Tutors in Seattle

Popular Courses & Classes

SAT Courses & Classes in Denver, Spanish Courses & Classes in New York City, GRE Courses & Classes in Chicago, ISEE Courses & Classes in Chicago, SAT Courses & Classes in Houston, GRE Courses & Classes in Los Angeles, Spanish Courses & Classes in Los Angeles, GMAT Courses & Classes in Seattle, LSAT Courses & Classes in Philadelphia, ACT Courses & Classes in New York City

Popular Test Prep

GRE Test Prep in Dallas Fort Worth, GRE Test Prep in Phoenix, GRE Test Prep in Los Angeles, ACT Test Prep in New York City, SSAT Test Prep in Miami, MCAT Test Prep in San Francisco-Bay Area, MCAT Test Prep in Seattle, ACT Test Prep in Dallas Fort Worth, SAT Test Prep in Los Angeles, ISEE Test Prep in Seattle

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)

Email address:
Your name:
Feedback:

GMAT Math : Geometry

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #271 : Geometry

Example Question #3 : Calculating The Perimeter Of A Right Triangle

Example Question #2 : Calculating The Perimeter Of A Right Triangle

Example Question #24 : Triangles

Example Question #25 : Right Triangles

Example Question #272 : Geometry

Example Question #273 : Geometry

Example Question #28 : Right Triangles

Example Question #274 : Geometry

Example Question #277 : Geometry

All GMAT Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details