All SAT Math Resources
Example Questions
Example Question #21 : Triangles
If one of the short sides of a 45-45-90 triangle equals 5, how long is the hypotenuse?
5√2
√15
5
π
√10
5√2
Using the Pythagorean theorem, x2 + y2 = h2. And since it is a 45-45-90 triangle the two short sides are equal. Therefore 52 + 52 = h2 . Multiplied out 25 + 25 = h2.
Therefore h2 = 50, so h = √50 = √2 * √25 or 5√2.
Example Question #1 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem
The height of a right circular cylinder is 10 inches and the diameter of its base is 6 inches. What is the distance from a point on the edge of the base to the center of the entire cylinder?
None of the other answers
4π/5
3π/4
√(34)
√(43)/2
√(34)
The best thing to do here is to draw diagram and draw the appropiate triangle for what is being asked. It does not matter where you place your point on the base because any point will produce the same result. We know that the center of the base of the cylinder is 3 inches away from the base (6/2). We also know that the center of the cylinder is 5 inches from the base of the cylinder (10/2). So we have a right triangle with a height of 5 inches and a base of 3 inches. So using the Pythagorean Theorem 32 + 52 = c2. 34 = c2, c = √(34).
Example Question #1221 : Basic Geometry
A right triangle with sides A, B, C and respective angles a, b, c has the following measurements.
Side A = 3in. Side B = 4in. What is the length of side C?
9
25
7
6
5
5
The correct answer is 5. The pythagorean theorem states that a2 + b2 = c2. So in this case 32 + 42 = C2. So C2 = 25 and C = 5. This is also an example of the common 3-4-5 triangle.
Example Question #24 : Triangles
The lengths of the three sides of a right triangle form a set of consecutive even integers when arranged from least to greatest. If the second largest side has a length of x, then which of the following equations could be used to solve for x?
(x – 2)2 + x2 = (x + 2)2
(x + 2)2 + (x – 2)2 = x2
x 2 + (x + 2)2 = (x + 4)2
(x – 1)2 + x2 = (x + 1)2
(x – 2) + x = (x + 2)
(x – 2)2 + x2 = (x + 2)2
We are told that the lengths form a series of consecutive even integers. Because even integers are two units apart, the side lengths must differ by two. In other words, the largest side length is two greater than the second largest, and the second largest length is two greater than the smallest length.
The second largest length is equal to x. The second largest length must thus be two less than the largest length. We could represent the largest length as x + 2.
Similarly, the second largest length is two larger than the smallest length, which we could thus represent as x – 2.
To summarize, the lengths of the triangle (in terms of x) are x – 2, x, and x + 2.
In order to solve for x, we can make use of the fact that the triangle is a right triangle. If we apply the Pythagorean Theorem, we can set up an equation that could be used to solve for x. The Pythagorean Theorem states that if a and b are the lengths of the legs of the triangle, and c is the length of the hypotenuse, then the following is true:
a2 + b2 = c2
In this particular case, the two legs of our triangle are x – 2 and x, since the legs are the two smallest sides; therefore, we can say that a = x – 2, and b = x. Lastly, we can say c = x + 2, since x + 2 is the length of the hypotenuse. Subsituting these values for a, b, and c into the Pythagorean Theorem yields the following:
(x – 2)2 + x2 = (x + 2)2
The answer is (x – 2)2 + x2 = (x + 2)2.
Example Question #51 : Sat Mathematics
What is the hypotenuse of a right triangle with sides 5 and 8?
5√4
12
8√13
√89
15
√89
Because this is a right triangle, we can use the Pythagorean Theorem which says a2 + b2 = c2, 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.
a2 + b2 = c2
52 + 82 = c2
25 + 64 = c2
89 = c2
c = √89
Example Question #26 : Triangles
A right triangle has side lengths of 21 and 72. What is the length of the hypotenuse?
80
70
75
84
67
75
By the Pythagorean Theorem, 212 + 722 = hyp2. Then hyp2 = 5625, and the hypotenuse = 75. If you didn't know how to figure out that 752 = 5625, that's okay. Look at the answer choices. We could easily have squared them and chosen the answer choice that, when squared, equals 5625.
Example Question #1222 : Basic Geometry
Sam and John both start at the same point. Sam walks 30 feet north while John walks 40 feet west. How far apart are they at their new locations?
Sam and John have walked at right angles to each other, so the distance between them is the hypotenuse of a triangle. The distance can be found using the Pythagorean Theorem.
Example Question #1223 : Basic Geometry
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?
2 + √2 miles
2 – √2 miles
Cannot be determined
√2 miles
1 mile
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.
Example Question #1224 : Basic Geometry
Which of the following sets of sides cannnot belong to a right triangle?
2, 2, 2√2
3, 4 ,5
6, 7, 8
2, 2√3, 4
5, 12, 13
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.
Example Question #91 : Plane Geometry
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?
5 miles
4√2 miles
2√13 miles
7 miles
10 miles
2√13 miles
This can be solved with the Pythagorean Theorem.
62 + 42 = c2
52 = c2
c = √52 = 2√13