Additional Topics in Math SAMPLE

The figure above shows a metal hex nut with two regular hexagonal faces and a thickness of 1 cm. The length of each side of a hexagonal face is 2 cm. A hole with a diameter of 2 cm is drilled through the nut. The density of the metal is 7.9 grams per cubic cm. What is the mass of this nut, to the nearest gram? (Density is mass divided by volume.)

content: Additional Topics in Math key: 57

calculator usage: Calculator

The question above asks students to make connections between physical concepts such as mass and density and essential geometric ideas such as the Pythagorean Theorem and volume formulas. There are multiple approaches to solving this problem, but in any of them, the aim is to find the volume of the metal nut and then use the density of the metal to calculate the mass of the nut (57 grams). This is a multistep problem that requires students to devise a multistep strategy and carry out all the algebraic and numerical steps without error.

To solve this problem, students need to find the volume of the hex nut and then use the given fact that density is mass divided by volume. Finding the volume of the hex nut requires several steps. The first step is to calculate the area of one of the hexagonal faces (without the drilled hole). Each face is a regular hexagon, which can be divided into 6 equilateral triangles with side lengths of 2 cm. Using 30-60-90 triangle properties, the height of each equilateral triangle is 3 cm. In turn, the area of one equilateral triangle is 1₂bh= 1₂( )( )2 3 = 3 square cm, so the area of the

hexagonal face is 6 3 square cm. The volume of the hexagonal prism is the area of one face multiplied by the height (or thickness in this case),

(6 3 1)( ) =6 3 cubic cm. Then, to account for the drilled hole, students need to calculate the volume of a cylinder with diameter 2 cm (or radius 1 cm) and height 1 cm, V =

π

r h2 =

π

( )( )1 12 =

π

cubic cm, and subtract it from the volume of the hexagonal prism to yield 6 3 −

π

cubic cm. Finally, density is mass divided by volume, 7 9

6 3

. = .

−

mass

π

Multiplying

both sides of the equation by (6 3−

π

yields the mass of the hex nut as

7 9 6 3.

(

−

π)

grams. When the values for 3and

π

are substituted and the result is rounded to the nearest gram, the answer is approximately 57 grams. Note that it is critical for students to attend to the precision of their calculations when solving this problem, and not apply any intermediate rounding until the final answer is reached. Here, the use of a calculator provides the ability to attend to precision more effectively, and thus is highly encouraged.

SAMPLE 25

The semicircle above has a radius of r inches, and chord CD is parallel to the diameter AB. If the length of CD is 2₃ of the length of AB, what is the distance between the chord and the diameter in terms of r?

A) 1 3

π

r B) 2 3

π

r C) 2 2 r D) ₃5r

content: Additional Topics in Math key: D

calculator usage: No calculator

This problem requires students to make use of properties of circles and parallel lines in an abstract setting. Students will have to draw an additional line in order to find the relationship between the distance of the chord from the diameter and the radius of the semicircle. Again, this item provides an opportunity for using different approaches to find the distance required: one can use either the Pythagorean Theorem or the trigonometric ratios.

Choice D is correct. This represents the length of the distance between the chord and the diameter, using a radius of the circle to create a triangle, and then the Pythagorean Theorem to solve correctly:

r2 = x2 +

( )

2₃r 2, where r represents the radius of the circle and x represents the distance between the chord and the diameter. Choice A is not the correct answer. It does not represent the length of the distance between the chord and the diameter. The student who selects this answer may have tried to use the circumference formula to determine the distance rather than making use of the radius of the circle to create a triangle.

A B

Choice B is not the correct answer. It does not represent the length of the distance between the chord and the diameter. The student who selects this answer may have tried to use the circumference formula to determine the distance rather than making use of the radius of the circle to create a triangle.

Choice C is not the correct answer. It does not represent the length of the distance between the chord and the diameter. The student who selects this answer may have made a triangle within the circle, using a radius to connect the chord and the diameter, but then may have mistaken the triangle for a 45-45-90 triangle and tried to use this relationship to determine the distance.

SAMPLE 26

It is given that sinx a= , where x is the radian measure of an angle and

2 < <x .

π

_π

If sinw = −a, which of the following could be the value of w? A)

π

− x

B) x−

π

C)

π

+x

D) x−

π

content: Additional Topics in Math key: B

calculator usage: No calculator

In problems like this, students must reason how angles x and w are related based on their corresponding sine values and determine the radian measure of angle w, given the parameters of angle x.

Choice B is correct. If an angle with radian measure x such that ₂

π

< <x

π

is placed in standard position, its terminal side will fall in Quadrant II, and

sinx a= will be the y-coordinate of the point P where its terminal side intersects the unit circle. If sinw = −a, then when the angle with radian measure w is placed in standard position, its terminal side will intersect the unit circle at a point with y-coordinate equal to –a. There are two such points on the unit circle: the reflection of P across the x-axis, which

would correspond to an angle with radian measure –x (and also with radian measures ...−6

π

− −x, 4

π

− −x, 2

π

−x,2

π

−x,4

π

−x,6

π

−x...); and the reflection of P through the origin, which would correspond to an angle with radian measure x−

π

(and also with radian measures

...x −5

π

,x −3

π

, x+

π

,x+3

π

,x +5

π

...). Thus, of the choices given, only x −

π

could be the value of w.

Choice A is not the correct answer. In general, sin

( )

−x = −sinx and

sin

(

x +

π)

= −sin ,x so sin

(π

−x

)

= −sin

( )

−x = − −

(

sinx

)

=sin .x

Therefore, sin

(π

−x

)

=a, not −a.

Choice C is not the correct answer. In general, sin

(

π

+x

)

=sin ,x

so sin

(

π

+x

)

=a, not −a.

Choice D is not the correct answer. In general, sin

(

x −

π)

=sin ,x

Executive Summary

section i

Behind the Redesign

section ii

The Redesigned SAT: Evidentiary Foundation

section iii

Test Specifications: SAT Evidence-Based Reading

and Writing and SAT Essay

section iv

Test Specifications: SAT Math Test

In document These draft test specifications and sample items and other materials are just that drafts. As such, they will systematically evolve over time. (Page 194-199)