Right Triangles - Math
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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?
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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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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?
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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?
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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?
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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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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?
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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?
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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?
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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?
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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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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?
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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?
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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?
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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?
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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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Given a right triangle where the two legs have lengths of 3 and 4 respectively, what is the length of the hypotenuse?
Given a right triangle where the two legs have lengths of 3 and 4 respectively, what is the length of the hypotenuse?
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The hypotenuse can be found using Pythagorean Theorem, which is a2 + b2 = c2, so we plug in a = 3 and b = 4 to get c.
c2 =25, so c = 5
The hypotenuse can be found using Pythagorean Theorem, which is a2 + b2 = c2, so we plug in a = 3 and b = 4 to get c.
c2 =25, so c = 5
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What is the hypotenuse of a right triangle with sides 5 and 8?
What is the hypotenuse of a right triangle with sides 5 and 8?
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Because this is a right triangle, we can use the Pythagorean Theorem which says _a_2 + _b_2 = _c_2, or the squares of the two sides of a right triangle must equal the square of the hypotenuse. Here we have a = 5 and b = 8.
_a_2 + _b_2 = _c_2
52 + 82 = _c_2
25 + 64 = _c_2
89 = _c_2
c = √89
Because this is a right triangle, we can use the Pythagorean Theorem which says _a_2 + _b_2 = _c_2, or the squares of the two sides of a right triangle must equal the square of the hypotenuse. Here we have a = 5 and b = 8.
_a_2 + _b_2 = _c_2
52 + 82 = _c_2
25 + 64 = _c_2
89 = _c_2
c = √89
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The lengths of the sides of a triangle are consecutive odd numbers and the triangle's perimeter is 57 centimeters. What is the length, in centimeters, of its longest side?
The lengths of the sides of a triangle are consecutive odd numbers and the triangle's perimeter is 57 centimeters. What is the length, in centimeters, of its longest side?
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First, define the sides of the triangle. Because the side lengths are consecutive odd numbers, if we define the shortest side will be as 
, the next side will be defined as 
, and the longest side will be defined as 
. We can then find the perimeter of a triangle using the following formula:
Substitute in the known values and variables.
Subtract 6 from both sides of the equation.
Divide both sides of the equation by 3.
Solve.
This is not the answer; we need to find the length of the longest side, or 
.
Substitute in the calculated value for 
and solve.
The longest side of the triangle is 21 centimeters long.
First, define the sides of the triangle. Because the side lengths are consecutive odd numbers, if we define the shortest side will be as 
Substitute in the known values and variables.
Subtract 6 from both sides of the equation.
Divide both sides of the equation by 3.
Solve.
This is not the answer; we need to find the length of the longest side, or 
Substitute in the calculated value for 
The longest side of the triangle is 21 centimeters long.
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Angela drives 30 miles north and then 40 miles east. How far is she from where she began?
Angela drives 30 miles north and then 40 miles east. How far is she from where she began?
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By drawing Angela’s route, we can connect her end point and her start point with a straight line and will then have a right triangle. The Pythagorean theorem can be used to solve for how far she is from the starting point: a2+b2=c2, 302+402=c2, c=50. It can also be noted that Angela’s route represents a multiple of the 3-4-5 Pythagorean triple.
By drawing Angela’s route, we can connect her end point and her start point with a straight line and will then have a right triangle. The Pythagorean theorem can be used to solve for how far she is from the starting point: a2+b2=c2, 302+402=c2, c=50. It can also be noted that Angela’s route represents a multiple of the 3-4-5 Pythagorean triple.
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Dan drives 5 miles north and then 8 miles west to get to school. If he walks, he can take a direct path from his house to the school, cutting down the distance. How long is the path from Dan's house to his school?
Dan drives 5 miles north and then 8 miles west to get to school. If he walks, he can take a direct path from his house to the school, cutting down the distance. How long is the path from Dan's house to his school?
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We are really looking for the hypotenuse of a triangle that has legs of 5 miles and 8 miles.
Apply the Pythagorean Theorem:
a2 + b2 = c2
25 + 64 = c2
89 = c2
c = 9.43 miles
We are really looking for the hypotenuse of a triangle that has legs of 5 miles and 8 miles.
Apply the Pythagorean Theorem:
a2 + b2 = c2
25 + 64 = c2
89 = c2
c = 9.43 miles
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