Triangles - Math
Card 0 of 880
If 2 sides of the triangle are have lengths equal to 8 and 14, what is one possible length of the third side?
If 2 sides of the triangle are have lengths equal to 8 and 14, what is one possible length of the third side?
The sum of the lengths of 2 sides of a triangle must be greater than—but not equal to—the length of the third side. Further, the third side must be longer than the difference between the greater and the lesser of the other two sides; therefore, 20 is the only possible answer.
The sum of the lengths of 2 sides of a triangle must be greater than—but not equal to—the length of the third side. Further, the third side must be longer than the difference between the greater and the lesser of the other two sides; therefore, 20 is the only possible answer.
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In 
the length of AB is 15 and the length of side AC is 5. What is the least possible integer length of side BC?
In
Rule - the length of one side of a triangle must be greater than the differnce and less than the sum of the lengths of the other two sides.
Given lengths of two of the sides of the 
are 15 and 5. The length of the third side must be greater than 15-5 or 10 and less than 15+5 or 20.
The question asks what is the least possible integer length of BC, which would be 11
Rule - the length of one side of a triangle must be greater than the differnce and less than the sum of the lengths of the other two sides.
Given lengths of two of the sides of the
The question asks what is the least possible integer length of BC, which would be 11
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Two similiar triangles exist where the ratio of perimeters is 4:5 for the smaller to the larger triangle. If the larger triangle has sides of 6, 7, and 12 inches, what is the perimeter, in inches, of the smaller triangle?
Two similiar triangles exist where the ratio of perimeters is 4:5 for the smaller to the larger triangle. If the larger triangle has sides of 6, 7, and 12 inches, what is the perimeter, in inches, of the smaller triangle?
The larger triangle has a perimeter of 25 inches. Therefore, using a 4:5 ratio, the smaller triangle's perimeter will be 20 inches.
The larger triangle has a perimeter of 25 inches. Therefore, using a 4:5 ratio, the smaller triangle's perimeter will be 20 inches.
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To get from his house to the hardware store, Bob must drive 3 miles to the east and then 4 miles to the north. If Bob was able to drive along a straight line directly connecting his house to the store, how far would he have to travel then?
To get from his house to the hardware store, Bob must drive 3 miles to the east and then 4 miles to the north. If Bob was able to drive along a straight line directly connecting his house to the store, how far would he have to travel then?
Since east and north directions are perpendicular, the possible routes Bob can take can be represented by a right triangle with sides a and b of length 3 miles and 5 miles, respectively. The hypotenuse c represents the straight line connecting his house to the store, and its length can be found using the Pythagorean theorem: _c_2 = 32+ 42 = 25. Since the square root of 25 is 5, the length of the hypotenuse is 5 miles.
Since east and north directions are perpendicular, the possible routes Bob can take can be represented by a right triangle with sides a and b of length 3 miles and 5 miles, respectively. The hypotenuse c represents the straight line connecting his house to the store, and its length can be found using the Pythagorean theorem: _c_2 = 32+ 42 = 25. Since the square root of 25 is 5, the length of the hypotenuse is 5 miles.
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A right triangle has legs of 15m and 20m. What is the length of the hypotenuse?
A right triangle has legs of 15m and 20m. What is the length of the hypotenuse?
The Pythagorean theorem is a2 + b2 = c2, where a and b are legs of the right triangle, and c is the hypotenuse.
(15)2 + (20)2 = c2 so c2 = 625. Take the square root to get c = 25m
The Pythagorean theorem is a2 + b2 = c2, where a and b are legs of the right triangle, and c is the hypotenuse.
(15)2 + (20)2 = c2 so c2 = 625. Take the square root to get c = 25m
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Length AB = 4
Length BC = 3
If a similar triangle has a hypotenuse length of 25, what are the lengths of its two legs?
Length AB = 4
Length BC = 3
If a similar triangle has a hypotenuse length of 25, what are the lengths of its two legs?
Similar triangles are in proportion.
Use Pythagorean Theorem to solve for AC:
Pythagorean Theorem: _AB_2 + _BC_2 = _AC_2
42 + 32 = _AC_2
16 + 9 = _AC_2
25 = _AC_2
AC = 5
If the similar triangle's hypotenuse is 25, then the proportion of the sides is AC/25 or 5/25 or 1/5.
Two legs then are 5 times longer than AB or BC:
5 * (AB) = 5 * (4) = 20
5 * (BC) = 5 * (3) = 15
Similar triangles are in proportion.
Use Pythagorean Theorem to solve for AC:
Pythagorean Theorem: _AB_2 + _BC_2 = _AC_2
42 + 32 = _AC_2
16 + 9 = _AC_2
25 = _AC_2
AC = 5
If the similar triangle's hypotenuse is 25, then the proportion of the sides is AC/25 or 5/25 or 1/5.
Two legs then are 5 times longer than AB or BC:
5 * (AB) = 5 * (4) = 20
5 * (BC) = 5 * (3) = 15
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If a = 7 and b = 4, which of the following could be the perimeter of the triangle?
I. 11
II. 15
III. 25
If a = 7 and b = 4, which of the following could be the perimeter of the triangle?
I. 11
II. 15
III. 25
Consider the perimeter of a triangle:
P = a + b + c
Since we know a and b, we can find c.
In I:
11 = 7 + 4 + c
11 = 11 + c
c = 0
Note that if c = 0, the shape is no longer a trial. Thus, we can eliminate I.
In II:
15 = 7 + 4 + c
15 = 11 + c
c = 4.
This is plausible given that the other sides are 7 and 4.
In III:
25 = 7 + 4 + c
25 = 11 + c
c = 14.
It is not possible for one side of a triangle to be greater than the sum of both of the other sides, so eliminate III.
Thus we are left with only II.
Consider the perimeter of a triangle:
P = a + b + c
Since we know a and b, we can find c.
In I:
11 = 7 + 4 + c
11 = 11 + c
c = 0
Note that if c = 0, the shape is no longer a trial. Thus, we can eliminate I.
In II:
15 = 7 + 4 + c
15 = 11 + c
c = 4.
This is plausible given that the other sides are 7 and 4.
In III:
25 = 7 + 4 + c
25 = 11 + c
c = 14.
It is not possible for one side of a triangle to be greater than the sum of both of the other sides, so eliminate III.
Thus we are left with only II.
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Two sides of an isosceles triangle are 20 and 30. What is the difference of the largest and the smallest possible perimeters?
Two sides of an isosceles triangle are 20 and 30. What is the difference of the largest and the smallest possible perimeters?
The trick here is that we don't know which is the repeated side. Our possible triangles are therefore 20 + 20 + 30 = 70 or 30 + 30 + 20 = 80. The difference is therefore 80 – 70 or 10.
The trick here is that we don't know which is the repeated side. Our possible triangles are therefore 20 + 20 + 30 = 70 or 30 + 30 + 20 = 80. The difference is therefore 80 – 70 or 10.
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A right triangle has a total perimeter of 12, and the length of its hypotenuse is 5. What is the area of this triangle?
A right triangle has a total perimeter of 12, and the length of its hypotenuse is 5. What is the area of this triangle?
The area of a triangle is denoted by the equation 1/2 b x h.
b stands for the length of the base, and h stands for the height.
Here we are told that the perimeter (total length of all three sides) is 12, and the hypotenuse (the side that is neither the height nor the base) is 5 units long.
So, 12-5 = 7 for the total perimeter of the base and height.
7 does not divide cleanly by two, but it does break down into 3 and 4,
and 1/2 (3x4) yields 6.
Another way to solve this would be if you recall your rules for right triangles, one of the very basic ones is the 3,4,5 triangle, which is exactly what we have here
The area of a triangle is denoted by the equation 1/2 b x h.
b stands for the length of the base, and h stands for the height.
Here we are told that the perimeter (total length of all three sides) is 12, and the hypotenuse (the side that is neither the height nor the base) is 5 units long.
So, 12-5 = 7 for the total perimeter of the base and height.
7 does not divide cleanly by two, but it does break down into 3 and 4,
and 1/2 (3x4) yields 6.
Another way to solve this would be if you recall your rules for right triangles, one of the very basic ones is the 3,4,5 triangle, which is exactly what we have here
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In order to get to work, Jeff leaves home and drives 4 miles due north, then 3 miles due east, followed by 6 miles due north and, finally, 7 miles due east. What is the straight line distance from Jeff’s work to his home?
In order to get to work, Jeff leaves home and drives 4 miles due north, then 3 miles due east, followed by 6 miles due north and, finally, 7 miles due east. What is the straight line distance from Jeff’s work to his home?
Jeff drives a total of 10 miles north and 10 miles east. Using the Pythagorean theorem (a2+b2=c2), the direct route from Jeff’s home to his work can be calculated. 102+102=c2. 200=c2. √200=c. √100√2=c. 10√2=c
Jeff drives a total of 10 miles north and 10 miles east. Using the Pythagorean theorem (a2+b2=c2), the direct route from Jeff’s home to his work can be calculated. 102+102=c2. 200=c2. √200=c. √100√2=c. 10√2=c
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Jim leaves his home and walks 10 minutes due west and 5 minutes due south. If Jim could walk a straight line from his current position back to his house, how far, in minutes, is Jim from home?
Jim leaves his home and walks 10 minutes due west and 5 minutes due south. If Jim could walk a straight line from his current position back to his house, how far, in minutes, is Jim from home?
By using Pythagorean Theorem, we can solve for the distance “as the crow flies” from Jim to his home:
102 + 52 = _x_2
100 + 25 = _x_2
√125 = x, but we still need to factor the square root
√125 = √25*5, and since the √25 = 5, we can move that outside of the radical, so
5√5= x
By using Pythagorean Theorem, we can solve for the distance “as the crow flies” from Jim to his home:
102 + 52 = _x_2
100 + 25 = _x_2
√125 = x, but we still need to factor the square root
√125 = √25*5, and since the √25 = 5, we can move that outside of the radical, so
5√5= x
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If one of the short sides of a 45-45-90 triangle equals 5, how long is the hypotenuse?
If one of the short sides of a 45-45-90 triangle equals 5, how long is the hypotenuse?
Using the Pythagorean theorem, _x_2 + _y_2 = _h_2. And since it is a 45-45-90 triangle the two short sides are equal. Therefore 52 + 52 = _h_2 . Multiplied out 25 + 25 = _h_2.
Therefore _h_2 = 50, so h = √50 = √2 * √25 or 5√2.
Using the Pythagorean theorem, _x_2 + _y_2 = _h_2. And since it is a 45-45-90 triangle the two short sides are equal. Therefore 52 + 52 = _h_2 . Multiplied out 25 + 25 = _h_2.
Therefore _h_2 = 50, so h = √50 = √2 * √25 or 5√2.
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Daria and Ashley start at the same spot and walk their two dogs to the park, taking different routes. Daria walks 1 mile north and then 1 mile east. Ashley walks her dog on a path going northeast that leads directly to the park. How much further does Daria walk than Ashley?
Daria and Ashley start at the same spot and walk their two dogs to the park, taking different routes. Daria walks 1 mile north and then 1 mile east. Ashley walks her dog on a path going northeast that leads directly to the park. How much further does Daria walk than Ashley?
First let's calculate how far Daria walks. This is simply 1 mile north + 1 mile east = 2 miles. Now let's calculate how far Ashley walks. We can think of this problem using a right triangle. The two legs of the triangle are the 1 mile north and 1 mile east, and Ashley's distance is the diagonal. Using the Pythagorean Theorem we calculate the diagonal as √(12 + 12) = √2. So Daria walked 2 miles, and Ashley walked √2 miles. Therefore the difference is simply 2 – √2 miles.
First let's calculate how far Daria walks. This is simply 1 mile north + 1 mile east = 2 miles. Now let's calculate how far Ashley walks. We can think of this problem using a right triangle. The two legs of the triangle are the 1 mile north and 1 mile east, and Ashley's distance is the diagonal. Using the Pythagorean Theorem we calculate the diagonal as √(12 + 12) = √2. So Daria walked 2 miles, and Ashley walked √2 miles. Therefore the difference is simply 2 – √2 miles.
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Which of the following sets of sides cannnot belong to a right triangle?
Which of the following sets of sides cannnot belong to a right triangle?
To answer this question without plugging all five answer choices in to the Pythagorean Theorem (which takes too long on the GRE), we can use special triangle formulas. Remember that 45-45-90 triangles have lengths of x, x, x√2. Similarly, 30-60-90 triangles have lengths x, x√3, 2x. We should also recall that 3,4,5 and 5,12,13 are special right triangles. Therefore the set of sides that doesn't fit any of these rules is 6, 7, 8.
To answer this question without plugging all five answer choices in to the Pythagorean Theorem (which takes too long on the GRE), we can use special triangle formulas. Remember that 45-45-90 triangles have lengths of x, x, x√2. Similarly, 30-60-90 triangles have lengths x, x√3, 2x. We should also recall that 3,4,5 and 5,12,13 are special right triangles. Therefore the set of sides that doesn't fit any of these rules is 6, 7, 8.
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Max starts at Point A and travels 6 miles north to Point B and then 4 miles east to Point C. What is the shortest distance from Point A to Point C?
Max starts at Point A and travels 6 miles north to Point B and then 4 miles east to Point C. What is the shortest distance from Point A to Point C?
This can be solved with the Pythagorean Theorem.
62 + 42 = _c_2
52 = _c_2
c = √52 = 2√13
This can be solved with the Pythagorean Theorem.
62 + 42 = _c_2
52 = _c_2
c = √52 = 2√13
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Paul leaves his home and jogs 3 miles due north and 4 miles due west. If Paul could walk a straight line from his current position back to his house, how far, in miles, is Paul from home?
Paul leaves his home and jogs 3 miles due north and 4 miles due west. If Paul could walk a straight line from his current position back to his house, how far, in miles, is Paul from home?
By using the Pythagorean Theorem, we can solve for the distance “as the crow flies” from Paul to his home:
32 + 42 = _x_2
9 + 16 = _x_2
25 = _x_2
5 = x
By using the Pythagorean Theorem, we can solve for the distance “as the crow flies” from Paul to his home:
32 + 42 = _x_2
9 + 16 = _x_2
25 = _x_2
5 = x
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Which set of side lengths CANNOT correspond to a right triangle?
Which set of side lengths CANNOT correspond to a right triangle?
Because we are told this is a right triangle, we can use the Pythagorean Theorem, _a_2 + _b_2 = _c_2. You may also remember some of these as special right triangles that are good to memorize, such as 3, 4, 5.
Here, 6, 8, 11 will not be the sides to a right triangle because 62 + 82 = 102.
Because we are told this is a right triangle, we can use the Pythagorean Theorem, _a_2 + _b_2 = _c_2. You may also remember some of these as special right triangles that are good to memorize, such as 3, 4, 5.
Here, 6, 8, 11 will not be the sides to a right triangle because 62 + 82 = 102.
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Points A and B lie on a circle centered at Z, where central angle <AZB measures 140°. What is the measure of angle <ZAB?
Points A and B lie on a circle centered at Z, where central angle <AZB measures 140°. What is the measure of angle <ZAB?
Because line segments ZA and ZB are radii of the circle, they must have the same length. That makes triangle ABZ an isosceles triangle, with <ZAB and <ZBA having the same measure. Because the three angles of a triangle must sum to 180°, you can express this in the equation:
140 + 2x = 180 --> 2x = 40 --> x = 20
Because line segments ZA and ZB are radii of the circle, they must have the same length. That makes triangle ABZ an isosceles triangle, with <ZAB and <ZBA having the same measure. Because the three angles of a triangle must sum to 180°, you can express this in the equation:
140 + 2x = 180 --> 2x = 40 --> x = 20
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In triangle ABC, Angle A = x degrees, Angle B = 2x degrees, and Angle C = 3x+30 degrees. How many degrees is Angle B?
In triangle ABC, Angle A = x degrees, Angle B = 2x degrees, and Angle C = 3x+30 degrees. How many degrees is Angle B?
Because the interior angles of a triangle add up to 180°, we can create an equation using the variables given in the problem: x+2x+(3x+30)=180. This simplifies to 6X+30=180. When we subtract 30 from both sides, we get 6x=150. Then, when we divide both sides by 6, we get x=25. Because Angle B=2x degrees, we multiply 25 times 2. Thus, Angle B is equal to 50°. If you got an answer of 25, you may have forgotten to multiply by 2. If you got 105, you may have found Angle C instead of Angle B.
Because the interior angles of a triangle add up to 180°, we can create an equation using the variables given in the problem: x+2x+(3x+30)=180. This simplifies to 6X+30=180. When we subtract 30 from both sides, we get 6x=150. Then, when we divide both sides by 6, we get x=25. Because Angle B=2x degrees, we multiply 25 times 2. Thus, Angle B is equal to 50°. If you got an answer of 25, you may have forgotten to multiply by 2. If you got 105, you may have found Angle C instead of Angle B.
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Triangle FGH has equal lengths for FG and GH; what is the measure of ∠F, if ∠G measures 40 degrees?
Triangle FGH has equal lengths for FG and GH; what is the measure of ∠F, if ∠G measures 40 degrees?
It's good to draw a diagram for this; we know that it's an isosceles triangle; remember that the angles of a triangle total 180 degrees.
Angle G for this triangle is the one angle that doesn't correspond to an equal side of the isosceles triangle (opposite side to the angle), so that means ∠F = ∠H, and that ∠F + ∠H + 40 = 180,
By substitution we find that ∠F * 2 = 140 and angle F = 70 degrees.
It's good to draw a diagram for this; we know that it's an isosceles triangle; remember that the angles of a triangle total 180 degrees.
Angle G for this triangle is the one angle that doesn't correspond to an equal side of the isosceles triangle (opposite side to the angle), so that means ∠F = ∠H, and that ∠F + ∠H + 40 = 180,
By substitution we find that ∠F * 2 = 140 and angle F = 70 degrees.
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