Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

38% (02:36) correct
61% (01:10) wrong based on 192 sessions

The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true?

I. c>a+b II. c/a/b=10/6/2 III. c^2>a^2+b^2

A. I and III only B. II and III only C. I only D. II only E. III only

OA does not explain how you could get an area < 1, with side > 15?

This problem was replaced by the following question:

The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true?

I. c>a+b II. c/a/b=10/6/2 III. c^2>a^2+b^2

A. I and III only B. II and III only C. I only D. II only E. III only

According to the relationship of the sides of a triangle: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Thus I and II can never be true: one side (c) can not be larger than the sum of the other two sides (a and b). Note that II is basically the same as I: if c=10, a=6 and b=2 then c>a+b, which can never be true. Thus even not considering the angles, we can see that only answer choice E (III only) is left (all other options are out because each of them has either I or II in them).

Answer: E.

Now, if interested why III is true: as the angles in a triangle are x, 3x, and 5x degrees then x+3x+5x=180 --> x=20, 3x=60, and 5x=100. Next, if angle opposite c were 90 degrees, then according to Pythagoras theorem c^2=a^+b^2, but since the angel opposite c is more than 90 degrees (100) then c is larger, hence c^2>a^+b^2.
_________________

The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true?

I. c>a+b II. c/a/b=10/6/2 III. c^2>a^2+b^2

A. I and III only B. II and III only C. I only D. II only E. III only

OA does not explain how you could get an area < 1, with side > 15?

This problem was replaced by the following question:

The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true?

I. c>a+b II. c/a/b=10/6/2 III. c^2>a^2+b^2

A. I and III only B. II and III only C. I only D. II only E. III only

According to the relationship of the sides of a triangle: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Thus I and II can never be true: one side (c) can not be larger than the sum of the other two sides (a and b). Note that II is basically the same as I: if c=10, a=6 and b=2 then c>a+b, which can never be true. Thus even not considering the angles, we can see that only answer choice E (III only) is left (all other options are out because each of them has either I or II in them).

Answer: E.

Now, if interested why III is true: as the angles in a triangle are x, 3x, and 5x degrees then x+3x+5x=180 --> x=20, 3x=60, and 5x=100. Next, if angle opposite c were 90 degrees, then according to Pythagoras theorem c^2=a^+b^2, but since the angel opposite c is more than 90 degrees (100) then c is larger, hence c^2>a^+b^2.

Is it true for all cases the side opposite to the angle >90 in a triangle will follow the same rule c^2>a^2 + b^2 ?

OA does not explain how you could get an area < 1, with side > 15?

This problem was replaced by the following question:

The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true?

I. c>a+b II. c/a/b=10/6/2 III. c^2>a^2+b^2

A. I and III only B. II and III only C. I only D. II only E. III only

According to the relationship of the sides of a triangle: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Thus I and II can never be true: one side (c) can not be larger than the sum of the other two sides (a and b). Note that II is basically the same as I: if c=10, a=6 and b=2 then c>a+b, which can never be true. Thus even not considering the angles, we can see that only answer choice E (III only) is left (all other options are out because each of them has either I or II in them).

Answer: E.

Now, if interested why III is true: as the angles in a triangle are x, 3x, and 5x degrees then x+3x+5x=180 --> x=20, 3x=60, and 5x=100. Next, if angle opposite c were 90 degrees, then according to Pythagoras theorem c^2=a^+b^2, but since the angel opposite c is more than 90 degrees (100) then c is larger, hence c^2>a^+b^2.

Is it true for all cases the side opposite to the angle >90 in a triangle will follow the same rule c^2>a^2 + b^2 ?