All HiSET: Math Resources
Example Questions
Example Question #91 : Hi Set: High School Equivalency Test: Math
Find the area of a square with the following side length:
We can find the area of a circle using the following formula:
In this equation the variable, , represents the length of a single side.
Substitute and solve.
Example Question #92 : Hi Set: High School Equivalency Test: Math
The perimeter of a square is . In terms of , give the area of the square.
Since a square comprises four segments of the same length, the length of one side is equal to one fourth of the perimeter of the square, which is . The area of the square is equal to the square of this sidelength, or
.
Example Question #1 : Area
The volume of a sphere is equal to . Give the surface area of the sphere.
None of the other choices gives the correct response.
The volume of a sphere can be calculated using the formula
Solving for :
Set . Multiply both sides by :
Divide by :
Take the cube root of both sides:
Now substitute for in the surface area formula:
,
the correct response.
Example Question #91 : Hi Set: High School Equivalency Test: Math
Express the area of a square plot of land 60 feet in sidelength in square yards.
600 square yards
400 square yards
600 square yards
3,600 square yards
200 square yards
400 square yards
One yard is equal to three feet, so convert 60 feet to yards by dividing by conversion factor 3:
Square this sidelength to get the area of the plot:
,
the correct response.
Example Question #1 : Area
A square has perimeter . Give its area in terms of .
Divide the perimeter to get the length of one side of the square.
Divide each term by 4:
Square this sidelength to get the area of the square. The binomial can be squared by using the square of a binomial pattern:
Example Question #96 : Hi Set: High School Equivalency Test: Math
A cube has surface area 6. Give the surface area of the sphere that is inscribed inside it.
A cube with surface area 6 has six faces,each with area 1. As a result, each edge of the cube has length the square root of this, which is 1.
This is the diameter of the sphere inscribed in the cube, so the radius of the sphere is half this, or . Substitute this for in the formula for the surface area of a sphere:
,
the correct choice.
Example Question #92 : Hi Set: High School Equivalency Test: Math
Find the length of the hypotenuse of a right triangle whose legs are the following lengths:
The hypotenuse of a right triangle can be calculated using the Pythagorean Theorem. This theorem states that if we know the lengths of the two other legs of the triangle, then we can calculate the hypotenuse. It is written in the following way:
In this formula the legs are noted by the variables, and . The variable represents the hypotenuse.
Substitute and solve for the hypotenuse.
Simplify.
Take the square root of both sides of the equation.
Example Question #1 : Apply The Pythagorean Theorem
If the two legs of a right triangle are cm and cm, what is the length of the hypotenuse. Answer must be in SIMPLIFIED form (or lowest terms).
cm
cm
cm
cm
cm
Step 1: Recall the Pythagorean theorem statement and formula.
Statement: For any right triangle, the sums of the squares of the shorter sides is equal to the square of the longest side.
Formula: In a right triangle , If are the shorter sides and is the longest side.. then,
Step 2: Plug in the values given to us in the problem....
Evaluate:
Simplify:
Simplify:
Take the square root...
Step 3: Simplify the root...
The length of the hypotenuse in most simplified form is cm.
Example Question #3 : Apply The Pythagorean Theorem
Which of the following could be the lengths of the sides of a right triangle?
In each choice, the two shortest sides of the triangle are 9 and 12, so the third side can be found by applying the Pythagorean Theorem. Set in the Pythagorean equation and solve for :
Take the square root:
.
The correct choice is
.
Example Question #92 : Hi Set: High School Equivalency Test: Math
What is the result of reflecting the point over the y-axis in the coordinate plane?
Reflecting a point
over the y-axis geometrically is the same as negating the x-coordinate of the ordered pair to obtain
.
Thus, since our initial point was
and we want to reflect it over the y-axis, we obtain the reflection by negating the first term of the ordered pair to get
.