All SAT II Math II Resources
Example Questions
Example Question #11 : Area
Find the area of a triangle with a base length of and a height of .
Write the formula for the area of a triangle.
Substitute the dimensions.
The answer is:
Example Question #11 : Geometry
Find the area of a circle with a radius of .
Write the formula for the area of a circle.
Substitute the radius into the equation.
The area is:
Example Question #314 : Sat Subject Test In Math Ii
Determine the area of a rectangle if the length is and the height is .
The area of a rectangle is:
Substitute the length and height into the formula.
We will move the constant to the front and apply the FOIL method to simplify the binomials.
Distribute the fraction through all the terms of the trinomial.
The answer is:
Example Question #315 : Sat Subject Test In Math Ii
What's the area of triangle with a side of and a height of ?
Write the formula for the area of a triangle.
Substitute the dimensions.
Use the FOIL method to expand this.
Simplify the terms.
Combine like-terms.
The answer is:
Example Question #12 : Area
Determine the side of a square with an area of .
Write the formula for the area of a square.
Substitute the area into the equation.
Square root both sides.
The answer is:
Example Question #13 : Area
Find the area of a circle with a radius of .
The area of a circle is .
Substitute the radius and solve for the area.
The answer is:
Example Question #12 : Geometry
Determine the area of a triangle with a base of 6, and a height of .
Write the formula for the area of a triangle.
Substitute the base and height into the equation.
The answer is:
Example Question #13 : Geometry
Find the area of a circle with a diameter of .
Divide the diameter by two. This will be the radius.
Write the formula for the area of a circle.
The answer is:
Example Question #323 : Sat Subject Test In Math Ii
In the provided diagram, hexagon is regular; and are the midpoints of their respective sides. The perimeter of the hexagon is ; what is the area of Quadrilateral ?
Quadrilateral is a trapezoid, so we need to find the lengths of its bases and its height.
The perimeter of the hexagon is , so each side of the hexagon measures one sixth of this, or .
Construct the diameters of the hexagon, which meet at center ; construct the apothem from to , with point of intersection . The diagram is below:
The six triangles formed by the diameters are equilateral, so , and . Quadrilateral is a trapezoid with bases of length 10 and 20. Since has its endpoints at the midpoints of the legs of Trapezoid , it follows that is a midsegment, and has as its length .
The trapezoid has bases of length and ; we now need to find its height. This is the measure of , which is half the length of apothem . is the height of an equilateral triangle and, consequently, the long leg of a right triangle . By the 30-60-90 Theorem,
.
The area of a trapezoid of height and base lengths and is
;
Setting :
Example Question #14 : Area
To the nearest whole, give the area of a regular pentagon with a perimeter of fifty.
In a regular pentagon, called Pentagon , construct the five perpendicular segments from each vertex to its opposite side, as shown below:
The segments divide the pentagon into ten congruent triangles.
In particular, examine . , a radius of the pentagon, bisects , which, as the interior angle of a regular pentagon, has measure ; therefore, . is an apothem and therefore bisects ; since the pentagon has perimeter 50, has length one fifth of this, or 10, and .
Using trigonometry,
,
or, substituting,
Solving for :
The area of this triangle is half the product of the lengths of legs and :
Since the pentagon comprises ten triangles of this area, multiply:
To the nearest whole, this is 172.
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