All PSAT Math Resources
Example Questions
Example Question #31 : Triangles
If and , what is the length of ?
AB is the leg adjacent to Angle A and BC is the leg opposite Angle A.
Since we have a triangle, the opposites sides of those angles will be in the ratio .
Here, we know the side opposite the sixty degree angle. Thus, we can set that value equal to .
which also means
Example Question #2 : How To Find The Length Of The Side Of A Right Triangle
Solve for x.
6
7
2
12
6
Use the Pythagorean Theorem. Let a = 8 and c = 10 (because it is the hypotenuse)
Example Question #11 : Right Triangles
Note: Figure NOT drawn to scale.
Refer to the above diagram. Evaluate .
The altitude perpendicular to the hypotenuse of a right triangle divides that triangle into two smaller triangles similar to each other and the large triangle. Therefore, the sides are in proportion. The hypotenuse of the triangle is equal to
Therefore, we can set up, and solve for in, a proportion statement involving the shorter side and hypotenuse of the large triangle and the larger of the two smaller triangles:
Example Question #32 : Triangles
What is the hypotenuse of a right triangle with sides 5 and 8?
8√13
12
5√4
√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 #21 : Right Triangles
If and , how long is side ?
Not enough information to solve
This problem is solved using the Pythagorean theorem . In this formula and are the legs of the right triangle while is the hypotenuse.
Using the labels of our triangle we have:
Example Question #1 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem
If one of the short sides of a 45-45-90 triangle equals 5, how long is the hypotenuse?
√15
5
π
√10
5√2
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
√(34)
3π/4
4π/5
√(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 #1 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem
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?
5
6
7
9
25
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 #35 : 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 + (x – 2)2 = x2
(x – 2)2 + x2 = (x + 2)2
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 #33 : Triangles
What is the hypotenuse of a right triangle with sides 5 and 8?
5√4
8√13
√89
15
12
√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
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