All PSAT Math Resources
Example Questions
Example Question #71 : Triangles
The lengths of the sides of a right triangle are consecutive integers, and the length of the shortest side is . Which of the following expressions could be used to solve for ?
Since the lengths of the sides are consecutive integers and the shortest side is , the three sides are , , and .
We then use the Pythagorean Theorem:
Example Question #72 : Triangles
Square is on the coordinate plane, and each side of the square is parallel to either the -axis or -axis. Point has coordinates and point has the coordinates .
Quantity A:
Quantity B: The distance between points and
Quantity A is greater.
Quantity B is greater.
The relationship cannot be determined from the information provided.
The two quantities are equal.
The two quantities are equal.
To find the distance between points and , split the square into two 45-45-90 triangles and find the hypotenuse. The side ratio of the 45-45-90 triangle is , so if the sides have a length of 5, the hypotenuse must be .
Example Question #513 : Geometry
Justin travels to the east and to the north. How far away from his starting point is he now?
This is solving for the hypotenuse of a triangle. Using the Pythagorean Theorem, which says that
Example Question #22 : Apply The Pythagorean Theorem To Find The Distance Between Two Points In A Coordinate System: Ccss.Math.Content.8.G.B.8
Susie walks north from her house to a park that is 30 meters away. Once she arrives at the park, she turns and walks west for 80 meters to a bench to feed some pigeons. She then walks north for another 30 meters to a concession stand. If Susie returns home in a straight line from the concession stand, how far will she walk from the concession stand to her house, in meters?
25
50
200
100
70
100
Susie walks 30 meters north, then 80 meters west, then 30 meters north again. Thus, she walks 60 meters north and 80 meters west. These two directions are 90 degrees away from one another.
At this point, construct a right triangle with one leg that measures 60 meters and a second leg that is 80 meters.
You can save time by using the 3:4:5 common triangle. 60 and 80 are and , respectively, making the hypotenuse equal to .
We can solve for the length of the missing hypotenuse by applying the Pythagorean theorem:
Substitute the following known values into the formula and solve for the missing hypotenuse: side .
Susie will walk 100 meters to reach her house.
Example Question #53 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem
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?
21
25
23
19
17
21
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.
Example Question #514 : Geometry
Each of the following answer choices lists the side lengths of a different triangle. Which of these triangles does not have a right angle?
cannot be the side lengths of a right triangle. does not equal . Also, special right triangle and rules can eliminate all the other choices.
Example Question #71 : Right Triangles
Note: Figure NOT drawn to scale.
Refer to the above diagram. Give the ratio of the perimeter of to that of .
The altitude of a right triangle from the vertex of its right triangle to its hypotenuse divides it into two similar triangles.
, as the length of the altitude corresponding to the hypotenuse, is the geometric mean of the lengths of the parts of the hypotenuse it forms; that is, it is the square root of the product of the two:
.
The ratio of the smaller sides of these similar triangles is
The ratio of the smaller side of to that of is
or : 1,
so this is also the ratio of the perimeter of to that of .
Example Question #12 : Triangles
The perimeter of a right triangle is 40 units. If the lengths of the sides are , , and units, then what is the area of the triangle?
Because the perimeter is equal to the sum of the lengths of the three sides of a triangle, we can add the three expressions for the lengths and set them equal to 40.
Perimeter:
Simplify the x terms.
Simplify the constants.
Subtract 8 from both sides.
Divide by 4
One side is 8.
The second side is
.
The third side is
.
Thus, the sides of the triangle are 8, 15, and 17.
The question asks us for the area of the triangle, which is given by the formula (1/2)bh. We are told it is a right triangle, so we can use one of the legs as the base, and the other leg as the height, since the legs will intersect at right angles. The legs of the right triangle must be the smallest sides (the longest must be the hypotenuse), which in this case are 8 and 15. So, let's assume that 8 is the base and 15 is the height.
The area of a triangle is (1/2)bh. We can substitute 8 and 15 for b and h.
.
The answer is 60 units squared.
Example Question #72 : Right Triangles
Figure not drawn to scale.
In the figure above, rays PA and PB are tangent to circle O at points A and B, respectively. If the diameter of circle O is 16 units and the length of line segment PO is 17 units, what is the area, in square units, of the quadrilateral PAOB?
240
60
120
136
68
120
Because PA and PB are tangent to circle O, angles PAO and PBO must be right angles; therefore, triangles PAO and PBO are both right triangles.
Since AO and OB are both radii of circle O, they are congruent. Furthermore, because PA and PB are external tangents originating from the same point, they must also be congruent.
So, in triangles PAO and PBO, we have two sides that are congruent, and we have a congruent angle (all right angles are congruent) between them. Therefore, by the Side-Angle-Side (SAS) Theorem of congruency, triangles PAO and PBO are congruent.
Notice that quadrilateral PAOB can be broken up into triangles PAO and PBO. Since those triangles are congruent, each must comprise one half of the area of quadrilateral PAOB. As a result, if we find the area of one of the triangles, we can double it in order to find the area of the quadrilateral.
Let's determine the area of triangle PAO. We have already established that it is a right triangle. We are told that PO, which is the hypotenuse of the triangle, is equal to 17. We are also told that the diameter of circle O is 16, which means that every radius of the circle is 8, because a radius is half the size of a diameter. Since segment AO is a radius, its length must be 8.
So, triangle PAO is a right triangle with a hypotenuse of 17 and a leg of 8. We can use the Pythagorean Theorem in order to find the other leg. According to the Pythagorean Theorem, if a and b are the lengths of the legs of a right triangle, and c is the length of the hypotenuse, then:
a2 + b2 = c2
Let us let b represent the length of PA.
82 + b2 = 172
64 + b2 = 289
Subtract 64 from both sides.
b2 = 225
Take the square root of both sides.
b = 15
This means that the length of PA is 15.
Now let's apply the formula for the area of a right triangle. Because the legs of a right triangle are perpendicular, one can be considered the base, and the other can be considered the height of the triangle.
area of triangle PAO = (1/2)bh
= (1/2)(8)(15) = 60
Ultimately, we must find the area of quadrilateral PAOB; however, we previously determined that triangles PAO and PBO each comprise half of the quadrilateral. Thus, if we double the area of PAO, we would get the area of quadrilateral PAOB.
Area of PAOB = 2(area of PAO)
= 2(60) = 120 square units
The answer is 120.
Example Question #2 : Triangles
If the hypotenuse of a triangle is 5 meters, which of the following is the closest value to the area of the triangle?
12
45
54
5
26
12
The answer is 12. In this circumstance, the area of the triangle cannot be smaller than its hypotenuse length, and cannot be bigger than its hypotenuse squared (that would be the area of a square).