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39% (02:32) correct
61% (01:11) wrong based on 203 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 ?