In triangle A B C, the measure of angle A is 52^{\circ} and A C=35. In triangle P Q R, the measure of angle P is 52^{\circ} and P R=105. Which additional piece of information is sufficient to prove that triangle A B C is similar to triangle P Q R?

C. The measures of angle B and angle R are 34^{\circ} and 94^{\circ}, respectively.

D. The measures of angle B and angle Q are 52^{\circ} and 34^{\circ}, respectively.

In triangle FGH, the measure of angle F is 63^\circ. Which additional piece of information is sufficient to prove that triangle FGH is a right triangle?

C. The measure of angle H is 27^\circ.

A. The length of \overline{FH} is 63.

B. The length of \overline{GH} is 126.

D. The sum of the measures of angles F, G, and H is 180^\circ.

A circle in the x y-plane has its center at (14,19) and has a radius of 7 k. Which equation represents this circle?

B. (x-14)^{2}+(y-19)^{2}=49 k^{2}

D. (x-14)^{2}+(y-19)^{2}=7 k^{2}

In triangles LMN and RST, angles L and R each have measure 60^\circ, LN = 10, and RT = 30. Which additional piece of information is sufficient to prove that triangle LMN is similar to triangle RST?

D. The measures of angles M and T are 70^\circ and 50^\circ, respectively.

C. The measures of angles M and S are 70^\circ and 60^\circ, respectively.

SAT Geometry and Trigonometry Practice Questions

Geometry and Trigonometry contributes about 5 to 7 questions to the digital SAT: area and volume, angle relationships, triangle similarity and congruence, circle equations, and right-triangle trigonometry (SOH-CAH-TOA territory; the test does not go into radians-based trig identities).

A reference sheet with common formulas is available on every real test, so memorization is not the bottleneck. Setup is: drawing the described figure yourself, labeling what you know, and picking the relationship that connects it to what is asked.

Question 1Lines, angles, and triangles · medium
In triangle $A B C$ , the measure of angle $A$ is $52^{\circ}$ and $A C=35$ . In triangle $P Q R$ , the measure of angle $P$ is $52^{\circ}$ and $P R=105$ . Which additional piece of information is sufficient to prove that triangle $A B C$ is similar to triangle $P Q R$ ?
A $A B=30$ and $P Q=30$ .
B $A B=30$ and $Q R=90$ .
CThe measures of angle $B$ and angle $R$ are $34^{\circ}$ and $94^{\circ}$ , respectively.
DThe measures of angle $B$ and angle $Q$ are $52^{\circ}$ and $34^{\circ}$ , respectively.
Show answer & explanation
Correct answer: C
The measures of angle $B$ and angle $R$ are $34^{\circ}$ and $94^{\circ}$ , respectively.
Option A is incorrect
$AB = 30$ and $PQ = 30$ makes the ratio $AB/PQ = 1$ while $AC/PR = \tfrac{35}{105} = \tfrac{1}{3}$ . The side ratios around the equal angles disagree, so this information cannot prove similarity — it actually rules it out.
Option B is incorrect
$QR$ is not the side that corresponds to $AB$ (that would be $PQ$ ), and $QR$ does not touch the known angle $P$ . Sides that aren't a matched corresponding pair around the equal angle can't feed the SAS similarity test.
Option C is correct

Similarity by AA needs two matching angle pairs. One pair is given: angle $A$ and angle $P$ are both $52°$ .

Choice C supplies angle $B = 34°$ and angle $R = 94°$ . In triangle $ABC$ : the third angle is $180° - 52° - 34° = 94°$ , so its angles are $52°, 34°, 94°$ .

In triangle $PQR$ : $P = 52°$ and $R = 94°$ force $Q = 180° - 52° - 94° = 34°$ . Its angles are also $52°, 34°, 94°$ , with $A \leftrightarrow P$ , $B \leftrightarrow Q$ , $C \leftrightarrow R$ .

Two (in fact three) pairs of equal angles — the triangles are similar by AA.

Answer: C.
Option D is incorrect
With angle $B = 52°$ , triangle $ABC$ 's angles are $52°, 52°, 76°$ ; with angle $Q = 34°$ , triangle $PQR$ 's are $52°, 34°, 94°$ . Different angle sets — these triangles are provably not similar.
Question 2Area and Volume · medium
A geologist estimates that the volume of a boulder is greater than 14.6 cubic feet but less than 18.6 cubic feet. The geologist also estimates that the boulder weighs 160 pounds per cubic foot of volume. Which inequality represents this situation, where $x$ represents the estimated total weight, in pounds, of the boulder?
A $160-18.6<x<160-14.6$
B $160+14.6<x<160+18.6$
C $160(14.6)<x<160(18.6)$
D $\frac{160}{18.6}<x<\frac{160}{14.6}$
Show answer & explanation
Correct answer: C
$160(14.6)<x<160(18.6)$
Option A is incorrect
Subtracting volume from a rate mixes units (pounds per cubic foot minus cubic feet means nothing), and it produces a boulder weighing under 146 pounds total — less than one cubic foot's worth.
Option B is incorrect
Adding a volume to a rate is dimensionally meaningless. Weight at a fixed density scales by multiplication, not addition.
Option C is correct

Weight = (pounds per cubic foot) × (cubic feet): $x = 160 \cdot V$ , where $V$ is the volume.

The volume is bounded: $14.6 < V < 18.6$ .

Multiplying every part of the inequality by the positive number 160 preserves the directions: $160(14.6) < x < 160(18.6)$ .

Numerically that is $2336 < x < 2976$ pounds — a sensible weight for a boulder, which is a quick sanity check the other choices fail.

Answer: C.
Option D is incorrect
Dividing flips the relationship: more volume would then mean less weight. It also reverses the bounds, putting the smaller volume on the heavier side.
Question 3Right triangles and trigonometry · easy
In triangle $FGH$ , the measure of angle $F$ is $63^\circ$ . Which additional piece of information is sufficient to prove that triangle $FGH$ is a right triangle?
AThe length of $\overline{FH}$ is 63.
BThe length of $\overline{GH}$ is 126.
CThe measure of angle $H$ is $27^\circ$ .
DThe sum of the measures of angles $F$ , $G$ , and $H$ is $180^\circ$ .
Show answer & explanation
Correct answer: C
The measure of angle $H$ is $27^\circ$ .
Option A is incorrect
A single side length says nothing about any angle's measure — triangles of every shape have a side of length 63.
Option B is incorrect
Same problem: one side length, even a conveniently doubled-looking one, does not determine any angle.
Option C is correct

To prove the triangle is right, we need one angle to be exactly $90°$ . We know angle $F = 63°$ , so we need information that forces another angle's measure.

Choice C gives angle $H = 27°$ . Then angle $G = 180° - 63° - 27° = 90°$ .

That makes $G$ a right angle, so triangle $FGH$ is a right triangle.

Answer: C.
Option D is incorrect
The three angles of every triangle sum to $180°$ — this is always true and therefore adds zero new information about this particular triangle.
Question 4Circles · medium
A circle in the $x y$ -plane has its center at $(14,19)$ and has a radius of $7 k$ . Which equation represents this circle?
A $(x-14)^{2}+(y-19)^{2}=49 k$
B $(x-14)^{2}+(y-19)^{2}=49 k^{2}$
C $(x-14)^{2}+(y-19)^{2}=7 k$
D $(x-14)^{2}+(y-19)^{2}=7 k^{2}$
Show answer & explanation
Correct answer: B
$(x-14)^{2}+(y-19)^{2}=49 k^{2}$
Option A is incorrect
This squares the 7 but not the $k$ : $(7k)^2$ means $7k \cdot 7k = 49k^2$ , not $49k$ .
Option B is correct

The standard circle equation is $(x - h)^2 + (y - k)^2 = r^2$ — note the right side is the radius squared.

The center $(14, 19)$ gives the left side $(x-14)^2 + (y-19)^2$ .

The radius is $7k$ , so the right side is $(7k)^2$ . Squaring a product squares both factors: $(7k)^2 = 49k^2$ .

Answer: B.
Option C is incorrect
This puts the radius itself on the right side, but the circle equation requires the radius squared.
Option D is incorrect
This squares the $k$ but not the 7 — the whole product $7k$ gets squared, both factors included.
Question 5Lines, angles, and triangles · hard
In triangles $LMN$ and $RST$ , angles $L$ and $R$ each have measure $60^\circ$ , $LN = 10$ , and $RT = 30$ . Which additional piece of information is sufficient to prove that triangle $LMN$ is similar to triangle $RST$ ?
A $MN = 7$ and $ST = 7$
B $MN = 7$ and $ST = 21$
CThe measures of angles $M$ and $S$ are $70^\circ$ and $60^\circ$ , respectively.
DThe measures of angles $M$ and $T$ are $70^\circ$ and $50^\circ$ , respectively.
Show answer & explanation
Correct answer: D
The measures of angles $M$ and $T$ are $70^\circ$ and $50^\circ$ , respectively.
Option A is incorrect
$MN/ST = 7/7 = 1$ conflicts with the known ratio $LN/RT = \tfrac{1}{3}$ — and these sides sit opposite the known angles anyway, which is not a valid similarity configuration.
Option B is incorrect
The ratio matches ( $7/21 = \tfrac{1}{3}$ ), but $MN$ and $ST$ are opposite the $60°$ angles, not adjacent to them. Two sides and a non-included angle (SSA) is famously ambiguous and proves nothing about similarity.
Option C is incorrect
With $M = 70°$ and $S = 60°$ : triangle $LMN$ has angles $60°, 70°, 50°$ , but triangle $RST$ gets $R = 60°, S = 60°, T = 60°$ — equilateral. The angle sets differ, so the triangles are not similar.
Option D is correct

One equal angle pair is given ( $L = R = 60°$ ) plus one side ratio ( $LN/RT = 10/30 = \tfrac{1}{3}$ ). AA similarity needs a second angle pair, matched to the right vertices.

Choice D gives angle $M = 70°$ and angle $T = 50°$ . Triangle $LMN$ : $60° + 70°$ leaves $N = 50°$ , so its angles are $60°, 70°, 50°$ .

Triangle $RST$ : $R = 60°$ and $T = 50°$ leave $S = 70°$ . Same three angles, with $L \leftrightarrow R$ , $M \leftrightarrow S$ , $N \leftrightarrow T$ — similar by AA.

Answer: D.
Question 6Area and Volume · medium
A sphere has a radius of $\tfrac{8}{3}$ feet. What is the volume, in cubic feet, of the sphere?
A $\tfrac{3\pi}{8}$
B $\tfrac{32\pi}{9}$
C $\tfrac{17\pi}{3}$
D $\tfrac{2048\pi}{81}$
Show answer & explanation
Correct answer: D
$\tfrac{2048\pi}{81}$
Option A is incorrect
$\tfrac{3\pi}{8}$ inverts the radius fraction instead of cubing it — $\tfrac{8}{3}$ flipped upside down with a $\pi$ attached is not a volume.
Option B is incorrect
$\tfrac{32\pi}{9}$ is $\tfrac{4}{3}\pi \cdot \tfrac{8}{3}$ — the radius used once, not cubed. Volume scales with the cube of the radius.
Option C is incorrect
$\tfrac{17\pi}{3}$ cannot come out of $\tfrac{4}{3}\pi r^3$ at all — it results from combining the numbers additively ( $\tfrac{8}{3} + 3 = \tfrac{17}{3}$ ) instead of cubing and multiplying.
Option D is correct

The volume of a sphere is $V = \tfrac{4}{3}\pi r^3$ (this formula is on the SAT reference sheet).

Cube the radius first: $\left(\tfrac{8}{3}\right)^3 = \tfrac{512}{27}$ .

Multiply: $V = \tfrac{4}{3} \cdot \tfrac{512}{27} \cdot \pi = \tfrac{2048}{81}\pi$ cubic feet.

Answer: D.
Question 7Right triangles and trigonometry · medium
In right triangle $RST$ , the sum of the measures of angle $R$ and angle $S$ is $90^\circ$ . The value of $\sin(R)$ is $\frac{2\sqrt{10}}{7}$ . What is the value of $\cos(S)$ ?
A $\frac{3\sqrt{10}}{20}$
B $\frac{2\sqrt{10}}{7}$
C $\frac{7\sqrt{10}}{20}$
D $\frac{2\sqrt{10}}{5}$
Show answer & explanation
Correct answer: B
$\frac{2\sqrt{10}}{7}$
Option A is incorrect
Since $R$ and $S$ are complementary, $\cos(S)$ must equal $\sin(R) = \tfrac{2\sqrt{10}}{7}$ exactly. This value differs, so it cannot be $\cos(S)$ — it comes from reconstructing the triangle's sides and mis-assigning which side is which.
Option B is correct

Angles $R$ and $S$ sum to $90°$ , so they are complementary.

The co-function identity says the sine of an angle equals the cosine of its complement: $\cos(S) = \cos(90° - R) = \sin(R)$ .

Therefore $\cos(S) = \sin(R) = \tfrac{2\sqrt{10}}{7}$ — no triangle-side computation needed at all.

Answer: B.
Option C is incorrect
Also not equal to $\sin(R)$ . This comes from building the side lengths and then reading off the ratio for the wrong angle — the co-function identity makes all of that unnecessary.
Option D is incorrect
Also not equal to $\sin(R)$ . Rescaling the ratio (changing the hypotenuse from 7 to 5) changes the value; the identity $\cos(S) = \sin(R)$ requires them to match exactly.
Question 8Circles · medium
In the $xy$ -plane, an equation of circle A is $(x - 2)^2 + (y - 3)^2 = 9$ . Circle B has the same center as circle A but has a radius that is twice the radius of circle A. Which equation represents circle B?
A $(x - 2)^2 + (y - 3)^2 = 18$
B $(x - 2)^2 + (y - 3)^2 = 36$
C $(x - 2)^2 + (y - 3)^2 = 54$
D $(x - 2)^2 + (y - 3)^2 = 81$
Show answer & explanation
Correct answer: B
$(x - 2)^2 + (y - 3)^2 = 36$
Option A is incorrect
18 is the 9 doubled — but 9 is the radius squared, not the radius. Doubling the radius from 3 to 6 multiplies the squared radius by four, from 9 to 36.
Option B is correct

Circle A's equation ends in $= 9$ , and $9 = r^2$ , so circle A's radius is 3.

Circle B's radius is twice that: 6.

Same center, so only the right side changes: $r^2 = 6^2 = 36$ , giving $(x-2)^2 + (y-3)^2 = 36$ .

The trap is doubling the 9: doubling the radius multiplies $r^2$ by 4, not 2.

Answer: B.
Option C is incorrect
54 comes from multiplying 9 by the new radius 6 — mixing $r^2$ and $r$ together. The right side must be the new radius squared: $6^2 = 36$ .
Option D is incorrect
81 is $9^2$ — squaring the right side treats 9 as if it were the radius. Circle A's radius is 3, and circle B's is 6, giving 36.
Question 9Area and Volume · hard
One gallon of paint will cover 310 square feet of a surface. A room has a total wall area of $w$ square feet. Which equation represents the total amount of paint $P$ , in gallons, needed to paint the walls of the room twice?
A $P = \frac{w}{155}$
B $P = 620w$
C $P = \frac{w}{310}$
D $P = 310w$
Show answer & explanation
Correct answer: A
$P = \frac{w}{155}$
Option A is correct

Painting the walls once takes $\tfrac{w}{310}$ gallons: total area divided by coverage per gallon.

Painting twice doubles the paint: $P = \tfrac{2w}{310}$ .

Simplify: $\tfrac{2w}{310} = \tfrac{w}{155}$ .

Answer: A.
Option B is incorrect
This multiplies instead of divides — it would demand 620 gallons for every square foot of wall. Coverage divides area by the per-gallon rate.
Option C is incorrect
$\tfrac{w}{310}$ is the paint for painting the walls once. The room is painted twice, which doubles it to $\tfrac{2w}{310} = \tfrac{w}{155}$ .
Option D is incorrect
Multiplying $310w$ treats the coverage rate as gallons per square foot instead of square feet per gallon — the units are upside down.
Question 10Right triangles and trigonometry · medium
The measure of angle $B$ is $\frac{3 \pi}{4}$ radians. What is the value of $\sin (B)$ ?
A $-\frac{\sqrt{3}}{2}$
B $-\frac{\sqrt{2}}{2}$
C $\frac{1}{2}$
D $\frac{\sqrt{2}}{2}$
Show answer & explanation
Correct answer: D
$\frac{\sqrt{2}}{2}$
Option A is incorrect
$-\tfrac{\sqrt{3}}{2}$ uses the wrong reference angle ( $\tfrac{\pi}{3}$ instead of $\tfrac{\pi}{4}$ ) and the wrong sign — sine is positive for second-quadrant angles like $\tfrac{3\pi}{4}$ .
Option B is incorrect
Right magnitude, wrong sign. $\tfrac{3\pi}{4}$ is in the second quadrant, where sine is positive; $-\tfrac{\sqrt{2}}{2}$ would be $\sin$ of $\tfrac{5\pi}{4}$ or $\tfrac{7\pi}{4}$ .
Option C is incorrect
$\tfrac{1}{2}$ is the sine of $\tfrac{\pi}{6}$ — a reference-angle mix-up. The reference angle of $\tfrac{3\pi}{4}$ is $\tfrac{\pi}{4}$ , whose sine is $\tfrac{\sqrt{2}}{2}$ .
Option D is correct

$\tfrac{3\pi}{4}$ radians is $135°$ — a second-quadrant angle.

Its reference angle is $\pi - \tfrac{3\pi}{4} = \tfrac{\pi}{4}$ (that is, $45°$ ), and $\sin\tfrac{\pi}{4} = \tfrac{\sqrt{2}}{2}$ .

Sine is positive in the second quadrant, so $\sin\tfrac{3\pi}{4} = +\tfrac{\sqrt{2}}{2}$ .

Answer: D.

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