Math

SAT Suite of Assessments Skills Insight Tool

Below are the skills and knowledge that students in the content domain and performance score band selected above are typically able to demonstrate as well as examples of the kinds of questions that these students are likely able to answer correctly. To view skill/knowledge statements and example questions in other domains and/or performance score bands, update the selections above and click Go.

Skills

A student in this performance score band can typically demonstrate the following skills in this content domain:

Solve for missing values in objects modeled by various 2D and 3D geometric shapes by applying formulas for area, surface area, or volume
(SAT, PSAT/NMSQT, and PSAT 10 only) Solve complex problems by applying properties of similar and congruent triangles, theorems related to angles and triangles, or right triangle trigonometry, or (SAT only) use similarity to calculate values of trigonometric ratios
(SAT only) Solve problems using properties and theorems related to circles and parts of circles, such as radii, diameters, tangents, angles, arcs, arc length, and sectors

Example Questions

Example Question 1

Triangle $A B C$ is similar to triangle $D E F$ , where $A$ corresponds to $D$ and $C$ corresponds to $F$ . Angles $C$ and $F$ are right angles. If $tangent left parenthesis upper A right parenthesis equals StartRoot 3 EndRoot$ and $upper D upper F equals 125$ , what is the length of $\bar{D E}$ ?

$125 StartFraction StartRoot 3 EndRoot Over 3 EndFraction$
$125 StartFraction StartRoot 3 EndRoot Over 2 EndFraction$
$125 StartRoot 3 EndRoot$
$250$

Key: D

Key Explanation

Choice D is correct. Corresponding angles in similar triangles have equal measures. It's given that triangle $A B C$ is similar to triangle $D E F$ , where $A$ corresponds to $D$ , so the measure of angle $A$ is equal to the measure of angle $D$ . Therefore, if $tangent left parenthesis upper A right parenthesis equals StartRoot 3 EndRoot$ , then $tangent left parenthesis upper D right parenthesis equals StartRoot 3 EndRoot$ . It's given that angles $C$ and $F$ are right angles, so triangles $A B C$ and $D E F$ are right triangles. The adjacent side of an acute angle in a right triangle is the side closest to the angle that is not the hypotenuse. It follows that the adjacent side of angle $D$ is side $D F$ . The opposite side of an acute angle in a right triangle is the side across from the acute angle. It follows that the opposite side of angle $D$ is side $E F$ . The tangent of an acute angle in a right triangle is the ratio of the length of the opposite side to the length of the adjacent side. Therefore, $\tan (D) = \frac{E F}{D F}$ . If $upper D upper F equals 125$ , the length of side $E F$ can be found by substituting $StartRoot 3 EndRoot$ for $\tan (D)$ and $125$ for $D F$ in the equation $\tan (D) = \frac{E F}{D F}$ , which yields $\sqrt{3} = \frac{E F}{125}$ . Multiplying both sides of this equation by $125$ yields $125 StartRoot 3 EndRoot equals upper E upper F$ . Since the length of side $E F$ is $StartRoot 3 EndRoot$ times the length of side $D F$ , it follows that triangle $D E F$ is a special right triangle with angle measures $30 degree$ , $60 degree$ , and $90 degree$ . Therefore, the length of the hypotenuse, $\bar{D E}$ , is $2$ times the length of side $D F$ , or $upper D upper E equals 2 left parenthesis upper D upper F right parenthesis$ . Substituting $125$ for $D F$ in this equation yields $upper D upper E equals 2 left parenthesis 125 right parenthesis$ , or $upper D upper E equals 250$ . Thus, if $tangent left parenthesis upper A right parenthesis equals StartRoot 3 EndRoot$ and $upper D upper F equals 125$ , the length of $\bar{D E}$ is $250$ .

Distractor Explanations

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect. This is the length of $\bar{E F}$ , not $\bar{D E}$ .

Example Question 2

A circle in the xy-plane has its center at $(- 4, - 6)$ . Line $k$ is tangent to this circle at the point $(- 7, - 7)$ . What is the slope of line $k$ ?

$negative 3$
$- \frac{1}{3}$
$\frac{1}{3}$
$3$

Key: A

Key Explanation

Choice A is correct. A line that's tangent to a circle is perpendicular to the radius of the circle at the point of tangency. It's given that the circle has its center at $(- 4, - 6)$ and line $k$ is tangent to the circle at the point $(- 7, - 7)$ . The slope of a radius defined by the points $(q, r)$ and $(s, t)$ can be calculated as $\frac{t - r}{s - q}$ . The points $(- 7, - 7)$ and $(- 4, - 6)$ define the radius of the circle at the point of tangency. Therefore, the slope of this radius can be calculated as $\frac{(- 6) - (- 7)}{(- 4) - (- 7)}$ , or $\frac{1}{3}$ . If a line and a radius are perpendicular, the slope of the line must be the negative reciprocal of the slope of the radius. The negative reciprocal of $\frac{1}{3}$ is $negative 3$ . Thus, the slope of line $k$ is $negative 3$ .

Distractor Explanations

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect. This is the slope of the radius of the circle at the point of tangency, not the slope of line $k$ .

Choice D is incorrect and may result from conceptual or calculation errors.

Reading and Writing

Math

Math

Reading and Writing

Skills

Example Questions

Example Question 1

Key: D

Key Explanation

Distractor Explanations

Example Question 2

Key: A

Key Explanation

Distractor Explanations