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Geometry

Geometry Practice Test: Practice Test 11

Practice Test 11 for Geometry: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A Ferris wheel has a diameter of 50 m, and its center is 30 m above the ground. The wheel makes one full rotation every 8 minutes. A rider starts at the top of the wheel at time t=0t=0t=0 minutes. Which set of parameters correctly describes a cosine model for the rider’s height above ground (amplitude, midline, period)?

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Question 1

A Ferris wheel has a diameter of 50 m, and its center is 30 m above the ground. The wheel makes one full rotation every 8 minutes. A rider starts at the top of the wheel at time t=0t=0t=0 minutes. Which set of parameters correctly describes a cosine model for the rider’s height above ground (amplitude, midline, period)?

  1. Amplitude =50=50=50 m, midline =30=30=30 m, period =8=8=8 min
  2. Amplitude =25=25=25 m, midline =30=30=30 m, period =8=8=8 min (correct answer)
  3. Amplitude =25=25=25 m, midline =55=55=55 m, period =8=8=8 min
  4. Amplitude =25=25=25 m, midline =30=30=30 m, period =4=4=4 min

Explanation: This question tests your ability to model real-world periodic phenomena using trigonometric functions by identifying key parameters—amplitude (maximum variation from center), period (time for complete cycle), and midline (center value). Periodic phenomena that repeat in regular cycles can be modeled with sine or cosine functions of the form f(t) = A·sin(B(t-C)) + D or f(t) = A·cos(B(t-C)) + D, where A is AMPLITUDE (half the total variation, calculated as (max - min)/2—represents how far values deviate from center), D is MIDLINE or vertical shift (the center line, calculated as (max + min)/2—the average value around which oscillation occurs), the PERIOD is 2π/B (time or distance for one complete cycle—how often pattern repeats), and C is phase shift (horizontal shift, where cycle starts—often 0 for simplified models). Example: tides vary from 2 ft (low) to 10 ft (high) with 12-hour period between consecutive low tides: AMPLITUDE = (10-2)/2 = 4 ft (tide varies 4 ft above and below center), MIDLINE = (10+2)/2 = 6 ft (center line is 6 ft, tide oscillates around this), PERIOD = 12 hours (pattern repeats every 12 hours), so function could be h(t) = 4cos(2π/12·t) + 6 = 4cos(πt/6) + 6 where t is hours. For this Ferris wheel with diameter 50 m (so radius 25 m, height ranging from 30 - 25 = 5 m to 30 + 25 = 55 m) and one rotation every 8 minutes, the amplitude is (55 - 5)/2 = 25 m, midline is (55 + 5)/2 = 30 m (the center height), and period is 8 minutes (full cycle time); since the rider starts at the top (maximum height), a cosine model fits well without a phase shift. Choice B correctly identifies these parameters by properly calculating amplitude as half the range, midline as the average, and period from the rotation time. Choice A incorrectly uses the full diameter as amplitude instead of the radius (half), while choice C mistakes the midline for the maximum height, and choice D halves the period unnecessarily. Parameter extraction recipe: (1) Find MAXIMUM value from scenario (highest tide, warmest temperature, top of Ferris wheel, peak of wave). (2) Find MINIMUM value (lowest tide, coldest temperature, bottom of wheel, trough of wave). (3) Calculate AMPLITUDE = (max - min) ÷ 2 (half the total variation). Example: max 85°F, min 35°F → amplitude = (85-35)/2 = 25°F. (4) Calculate MIDLINE = (max + min) ÷ 2 (average of extremes). Example: (85+35)/2 = 60°F midline. (5) Identify PERIOD from how often pattern repeats (time between consecutive maximums or minimums, or stated cycle time). Example: temperature repeats every 12 months → period = 12 months. These three parameters (amplitude, midline, period) fully describe the periodic behavior! Quick checks: Does amplitude make sense? (Should be positive, half the total variation). Does midline split the difference? (Should be exactly between max and min). Does period match cycle description? (daily = 24 hours, yearly = 12 months or 365 days, stated rotation time). If values seem wrong, recheck calculations!

Question 2

In the shown right triangle △MNO\triangle MNO△MNO, ∠N\angle N∠N is marked as a right angle, with acute angles θ=∠M\theta=\angle Mθ=∠M and ϕ=∠O\phi=\angle Oϕ=∠O. Which statement correctly relates cos⁡(θ)\cos(\theta)cos(θ) and sin⁡(ϕ)\sin(\phi)sin(ϕ)?

  1. cos⁡(θ)=sin⁡(ϕ)\cos(\theta)=\sin(\phi)cos(θ)=sin(ϕ) (correct answer)
  2. cos⁡(θ)=cos⁡(ϕ)\cos(\theta)=\cos(\phi)cos(θ)=cos(ϕ)
  3. cos⁡(θ)=sin⁡(θ)\cos(\theta)=\sin(\theta)cos(θ)=sin(θ)
  4. cos⁡(θ)=0\cos(\theta)=0cos(θ)=0

Explanation: This problem focuses on the cosine-sine relationship for complementary angles. In triangle MNO with right angle at N, angles θ = ∠M and φ = ∠O are complementary because θ + φ = 90°. For angle θ, the opposite side is NO and the adjacent side is MN, giving cos(θ) = MN/MO. For angle φ, the opposite side is MN and the adjacent side is NO, giving sin(φ) = MN/MO. Since both equal MN/MO, we have cos(θ) = sin(φ). The distractor cos(θ) = cos(φ) wrongly assumes complementary angles have equal cosine values. Remember that in a right triangle, each acute angle's adjacent side is the other acute angle's opposite side.

Question 3

A maker is cutting a circular tabletop from a square sheet of wood that is 48 in48\text{ in}48 in on each side. The circle must fit entirely inside the square, and the maker also needs a 2 in2\text{ in}2 in safety margin between the circle and each edge of the sheet. Which option best meets the geometric requirements?

  1. Tabletop diameter 44 in44\text{ in}44 in. (correct answer)
  2. Tabletop diameter 46 in46\text{ in}46 in.
  3. Tabletop diameter 48 in48\text{ in}48 in.
  4. Tabletop diameter 45 in45\text{ in}45 in.

Explanation: This problem requires inscribing a circle within a square while maintaining safety margins. The square sheet is 48 in × 48 in with a 2 in margin required from each edge. The available space for the circle is a 44 in × 44 in square (after subtracting 2 in from each side). The maximum diameter for a circle inscribed in this reduced square is 44 in. Option A (diameter 44 in) exactly meets this constraint. Options B (46 in) and C (48 in) would violate the safety margin by extending too close to the sheet edges. Option D (45 in) also exceeds the 44 in limit. A common error is subtracting the margin only once instead of from both sides, leading to an overestimate of available space. When working with safety margins, always account for the margin on all sides by subtracting twice the margin distance from each dimension.

Question 4

Triangle PQRPQRPQR has a right angle at QQQ. If triangle PQRPQRPQR is similar to triangle STUSTUSTU with similarity ratio 3:23:23:2, and the area of triangle STUSTUSTU is 24 square units, what is the length of the hypotenuse of triangle PQRPQRPQR if the hypotenuse of triangle STUSTUSTU is 8 units?

  1. 101010 units
  2. 121212 units (correct answer)
  3. 161616 units
  4. 181818 units

Explanation: Since the triangles are similar with ratio 3:23:23:2, corresponding linear dimensions are in the ratio 3:23:23:2. The hypotenuse of △STU\triangle STU△STU is 8 units, so the hypotenuse of △PQR\triangle PQR△PQR is 8⋅32=128 \cdot \frac{3}{2} = 128⋅23​=12 units. Note that areas are in the ratio of the square of the similarity ratio, which would be 9:49:49:4, making the area of △PQR\triangle PQR△PQR equal to 24⋅94=5424 \cdot \frac{9}{4} = 5424⋅49​=54 square units, but this doesn't affect the hypotenuse calculation.

Question 5

In right triangle △XYZ\triangle XYZ△XYZ drawn in the plane, ∠Z\angle Z∠Z is marked as a right angle. The acute angles are labeled θ\thetaθ at XXX and ϕ\phiϕ at YYY (so they are complementary). Which expression represents sin⁡(θ)\sin(\theta)sin(θ) in terms of cos⁡(ϕ)\cos(\phi)cos(ϕ)?

  1. sin⁡(θ)=cos⁡(ϕ)\sin(\theta)=\cos(\phi)sin(θ)=cos(ϕ) (correct answer)
  2. sin⁡(θ)=cos⁡(θ)\sin(\theta)=\cos(\theta)sin(θ)=cos(θ)
  3. sin⁡(θ)=sin⁡(ϕ)\sin(\theta)=\sin(\phi)sin(θ)=sin(ϕ)
  4. sin⁡(θ)=cos⁡(90∘)\sin(\theta)=\cos(90^\circ)sin(θ)=cos(90∘)

Explanation: The skill here is understanding the relationship between sine and cosine for complementary angles in a right triangle. In a right triangle, the two acute angles θ and φ add up to 90 degrees because the third angle is 90 degrees, making them complementary. For angle θ at X, the opposite side is YZ and the adjacent side is XZ; for φ at Y, the opposite side is XZ and the adjacent side is YZ. Sine is defined as opposite over hypotenuse, and cosine as adjacent over hypotenuse, so these definitions connect the functions across the angles. Therefore, sin(θ) equals cos(φ) as both are the ratio of the same side over the hypotenuse. A common distractor misconception is assuming sin(θ) = sin(φ), but complementary angles have different measures unless they are both 45 degrees. To transfer this strategy, redraw the triangle and label the opposite and adjacent sides for each angle to see how their roles swap.

Question 6

Quadrilateral ABCDABCDABCD has vertices A(0,0)A(0,0)A(0,0), B(4,0)B(4,0)B(4,0), C(3,3)C(3,3)C(3,3), and D(−1,3)D(-1,3)D(−1,3) in that order. What is the perimeter of the figure?

  1. 141414
  2. 4+3+4+34+3+4+34+3+4+3
  3. 4+10+4+104+\sqrt{10}+4+\sqrt{10}4+10​+4+10​ (correct answer)
  4. 4+18+4+184+\sqrt{18}+4+\sqrt{18}4+18​+4+18​

Explanation: The skill is finding the perimeter of a quadrilateral using coordinates. The vertices are A(0,0), B(4,0), C(3,3), and D(-1,3). Side lengths are computed using the distance formula, which is the square root of the sum of the squares of the differences in x and y coordinates. The perimeter is the sum of all four side lengths. Calculating each distance gives AB = 4, BC = √10, CD = 4, and DA = √10, justifying the final value as 4 + √10 + 4 + √10. A common distractor misconception is using √18 instead of √10 by doubling a difference. To transfer this strategy, compute side lengths before summing for perimeter or using them for area in other figures.

Question 7

Two circles are shown with different centers. Circle ⊙G\odot G⊙G has radius 333 and circle ⊙H\odot H⊙H has radius 999. Which statement explains why the circles are similar without using circumference or area formulas?

  1. They are similar because a dilation with scale factor 333 maps ⊙G\odot G⊙G to a circle of radius 999. (correct answer)
  2. They are similar because the circles have different centers, so they cannot be related by transformations.
  3. They are similar because their circumferences differ by 6π6\pi6π.
  4. They are similar because they look like perfect circles in the diagram.

Explanation: The skill here is understanding circle similarity in geometry without relying on measurements. Circles are similar because a dilation can scale one to any desired size, preserving the shape. The centers of G and H are different, but similarity focuses on shape, not position. Applying a dilation with scale factor 333 maps circle G to a new circle with radius 999, matching circle H's size. This scaling shows the shapes are proportionally identical, justifying similarity. A common distractor is choice C, which mentions circumference difference, but that's not a transformation-based reason. To solve similar problems, think in terms of transformations to adjust scale, not formulas.

Question 8

On the coordinate plane, triangle ABCABCABC (solid) is mapped to triangle A′B′C′A'B'C'A′B′C′ (dashed). The labeled point-image pairs include A(−2,1)→A′(1,1)A(-2,1)\to A'(1,1)A(−2,1)→A′(1,1) and B(−1,4)→B′(2,4)B(-1,4)\to B'(2,4)B(−1,4)→B′(2,4). Which mapping rule matches the diagram?

  1. Reflect across the yyy-axis: (x,y)↦(−x,y)(x,y)\mapsto(-x,y)(x,y)↦(−x,y).
  2. Translate 3 units right: (x,y)↦(x+3,y)(x,y)\mapsto(x+3,y)(x,y)↦(x+3,y). (correct answer)
  3. Translate 3 units up: (x,y)↦(x,y+3)(x,y)\mapsto(x,y+3)(x,y)↦(x,y+3).
  4. Each input point maps to two outputs: (x,y)↦(x+3,y)(x,y)\mapsto(x+3,y)(x,y)↦(x+3,y) and (x−3,y)(x-3,y)(x−3,y).

Explanation: The skill is representing geometric transformations as functions. A transformation is defined as an input-output rule that assigns to each point (x, y) in the plane a unique output point. In the diagram, points move horizontally to the right by 3 units while keeping the y-coordinate the same, as seen from A(-2,1) to A'(1,1) and B(-1,4) to B'(2,4). This mapping preserves distances, angles, and orientations since the figure remains congruent to the original. The correct answer is B because adding 3 to the x-coordinate matches both point pairs exactly. A distractor like D suggests multiple outputs, but transformations as functions map each input to exactly one output. To analyze similar problems, track one point at a time to identify the pattern in coordinates.

Question 9

Which conclusion follows from the diagram?

Triangle OXYOXYOXY is a right triangle with right angle at XXX (right-angle box at XXX). Ray OZOZOZ lies inside ∠XOY\angle XOY∠XOY and splits it into two angles: ∠XOZ=θ\angle XOZ=\theta∠XOZ=θ and ∠ZOY=φ\angle ZOY=\varphi∠ZOY=φ. From point ZZZ on ray OZOZOZ, perpendiculars are dropped to OXOXOX and to XYXYXY meeting them at MMM and NNN (right-angle boxes at MMM and NNN). No other information is marked.

Which statement correctly describes a geometric strategy that can lead to a sum identity (rather than assuming it)?

  1. Use right triangles △OZM\triangle OZM△OZM and △ZXN\triangle ZXN△ZXN to relate projections along OXOXOX and XYXYXY, then combine to express a projection corresponding to cos⁡(θ+φ)\cos(\theta+\varphi)cos(θ+φ). (correct answer)
  2. Assume cos⁡(θ+φ)\cos(\theta+\varphi)cos(θ+φ) equals cos⁡θcos⁡φ−sin⁡θsin⁡φ\cos\theta\cos\varphi-\sin\theta\sin\varphicosθcosφ−sinθsinφ and label the segments to match.
  3. Because θ\thetaθ and φ\varphiφ share vertex OOO, conclude cos⁡(θ+φ)=cos⁡θ+cos⁡φ\cos(\theta+\varphi)=\cos\theta+\cos\varphicos(θ+φ)=cosθ+cosφ.
  4. Because the diagram is not to scale, no trigonometric identity can be derived from it.

Explanation: The skill focuses on proving cosine sum identities through projections in a right triangle. The geometric setup is right triangle OXY with right angle at X, ray OZ splitting angle at O into theta and phi, and perpendiculars from Z to OX and XY at M and N. This decomposes the angle at O as theta + phi. Side relationships are tracked via projections in right triangles OZM and ZXN along OX and XY. Combining these projections justifies cos(theta + phi) = cos theta cos phi - sin theta sin phi. A distractor misconception is adding cosines directly due to shared vertex, as in choice C, without projection adjustments. To transfer this strategy, visualize geometric perpendiculars and projections in triangles before algebraic work.

Question 10

In the diagram, △LMN\triangle LMN△LMN is shown. Segments LM‾\overline{LM}LM and LN‾\overline{LN}LN have matching tick marks indicating LM≅LNLM\cong LNLM≅LN. No other markings are given, and the diagram is not drawn to scale.

Which conclusion follows from the diagram?

  1. ∠LNM≅∠LMN\angle LNM\cong \angle LMN∠LNM≅∠LMN (correct answer)
  2. MN≅LMMN\cong LMMN≅LM
  3. MN‾\overline{MN}MN bisects ∠L\angle L∠L
  4. LM‾⊥LN‾\overline{LM}\perp \overline{LN}LM⊥LN

Explanation: Theorems about triangles include isosceles properties linking sides to angles. The isosceles theorem states congruent sides imply congruent opposite angles. Tick marks mark LM congruent to LN in the diagram. Applying it, base angles at M and N are congruent. Justification is the theorem's symmetry from vertex L. Distractor misconceptions assume side equality without basis, as in choice B. Transfer strategy: match side ticks to isosceles for angle congruences.

Question 11

A circle with center OOO is drawn. A tangent line ℓ\ellℓ touches the circle at S. Radius OSOSOS is drawn, and the right angle at SSS is marked. Which angle relationship is guaranteed?

  1. ∠OSℓ=90∘\angle OS\ell = 90^\circ∠OSℓ=90∘. (correct answer)
  2. ∠OSC=90∘\angle O S C = 90^\circ∠OSC=90∘ for any point CCC on the circle.
  3. ∠SOℓ=90∘\angle SO\ell = 90^\circ∠SOℓ=90∘.
  4. ∠OSS′=90∘\angle OSS' = 90^\circ∠OSS′=90∘ where S′S'S′ is any point on ℓ\ellℓ.

Explanation: The skill involves understanding properties of tangents to circles. A tangent to a circle is a line that touches the circle at exactly one point. The point where the tangent touches the circle is called the point of tangency, here point S. At the point of tangency, the radius to that point is perpendicular to the tangent line. Therefore, the angle between OS and ℓ is a right angle, guaranteeing the relationship. A common misconception is that the right angle applies to any point on the circle, but it is specific to the tangency point. To solve similar problems, identify the radius to the point of tangency and apply the perpendicular property.

Question 12

A transformation matrix T=(2103)T = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}T=(20​13​) is applied to a triangle with vertices at (0,0)(0,0)(0,0), (2,0)(2,0)(2,0), and (0,1)(0,1)(0,1). What is the ratio of the area of the transformed triangle to the original triangle?

  1. 3:13:13:1
  2. 5:15:15:1
  3. 6:16:16:1 (correct answer)
  4. 7:17:17:1

Explanation: The ratio of areas under a linear transformation equals the absolute value of the determinant of the transformation matrix. For matrix T, det(T) = (2)(3) - (1)(0) = 6. Therefore, the area ratio is 6:1. Choice A uses only one diagonal element, choice B adds the matrix elements, and choice D uses the sum of all elements.

Question 13

In the plane, triangles △LMN\triangle LMN△LMN and △QRS\triangle QRS△QRS are drawn. The diagram marks ∠L≅∠Q\angle L \cong \angle Q∠L≅∠Q with one arc and ∠M≅∠R\angle M \cong \angle R∠M≅∠R with two arcs. No side lengths are shown, and the diagram is not drawn to scale. Which statement proves the triangles are similar?

  1. The triangles are similar because they look like the same shape in the diagram.
  2. The triangles are congruent because two corresponding angles are marked congruent.
  3. The triangles are similar by AA because two pairs of corresponding angles are marked congruent. (correct answer)
  4. The triangles are similar because LM=QRLM=QRLM=QR and MN=RSMN=RSMN=RS.

Explanation: The skill being assessed is the AA criterion for triangle similarity. The AA criterion states that if two pairs of corresponding angles in two triangles are congruent, then the triangles are similar, which can be verified through similarity transformations preserving angles. In this problem, the marked angles are angle L congruent to angle Q with one arc and angle M congruent to angle R with two arcs. Applying the AA criterion, triangle LMN is similar to triangle QRS with correspondence L to Q, M to R, and N to S. Because the triangles are similar, their corresponding sides are proportional, meaning the ratios of the lengths of corresponding sides are equal, but the sides themselves are not necessarily equal in length. A common misconception is to rely on visual appearance for similarity without confirming angle congruences, but diagrams not to scale require explicit markings. When approaching similar problems, always check for matching angles first to establish similarity before comparing side lengths or ratios.

Question 14

A sinusoidal function has maximum value 181818 and minimum value −2-2−2, and it completes one full cycle every 555 seconds. Which statement is correct about its amplitude, midline, and period?

  1. Amplitude =10=10=10, midline =8=8=8, period =5=5=5 (correct answer)
  2. Amplitude =20=20=20, midline =8=8=8, period =5=5=5
  3. Amplitude =10=10=10, midline =18=18=18, period =5=5=5
  4. Amplitude =10=10=10, midline =8=8=8, period =10=10=10

Explanation: This question tests your ability to model real-world periodic phenomena using trigonometric functions by identifying key parameters—amplitude (maximum variation from center), period (time for complete cycle), and midline (center value). Periodic phenomena that repeat in regular cycles can be modeled with sine or cosine functions of the form f(t) = A·sin(B(t-C)) + D or f(t) = A·cos(B(t-C)) + D, where A is AMPLITUDE (half the total variation, calculated as (max - min)/2—represents how far values deviate from center), D is MIDLINE or vertical shift (the center line, calculated as (max + min)/2—the average value around which oscillation occurs), the PERIOD is 2π/B (time or distance for one complete cycle—how often pattern repeats), and C is phase shift (horizontal shift, where cycle starts—often 0 for simplified models). For this function: maximum = 18, minimum = -2, so AMPLITUDE = (18 - (-2))/2 = 20/2 = 10 (function varies 10 units above and below center), MIDLINE = (18 + (-2))/2 = 16/2 = 8 (center value is 8), and PERIOD = 5 seconds (given directly). Choice A correctly identifies amplitude = 10, midline = 8, and period = 5 by properly calculating parameters. Choice B incorrectly uses the full range (20) as amplitude, Choice C incorrectly uses the maximum value (18) as midline, and Choice D incorrectly doubles the period. Parameter extraction with negative minimum: (1) MAX = 18, MIN = -2, (2) AMPLITUDE = (18 - (-2)) ÷ 2 = 20 ÷ 2 = 10, (3) MIDLINE = (18 + (-2)) ÷ 2 = 16 ÷ 2 = 8, (4) PERIOD = 5 seconds as stated!

Question 15

A pendulum’s horizontal displacement from center is modeled by x(t)=12cos⁡(t)x(t)=12\cos(t)x(t)=12cos(t), where ttt is in radians. For what values of ttt in the interval 0≤t≤2π0\le t\le 2\pi0≤t≤2π is the displacement x(t)=−3x(t)= -3x(t)=−3?

(Do not use memorized general solution sets beyond this interval.)

  1. t=arccos⁡ ⁣(−14)t=\arccos\!\left(-\dfrac{1}{4}\right)t=arccos(−41​) only
  2. t=arccos⁡ ⁣(−14)t=\arccos\!\left(-\dfrac{1}{4}\right)t=arccos(−41​) and t=2π−arccos⁡ ⁣(−14)t=2\pi-\arccos\!\left(-\dfrac{1}{4}\right)t=2π−arccos(−41​) (correct answer)
  3. t=arcsin⁡ ⁣(−14)t=\arcsin\!\left(-\dfrac{1}{4}\right)t=arcsin(−41​) and t=π−arcsin⁡ ⁣(−14)t=\pi-\arcsin\!\left(-\dfrac{1}{4}\right)t=π−arcsin(−41​)
  4. t=π+arccos⁡ ⁣(−14)t=\pi+\arccos\!\left(-\dfrac{1}{4}\right)t=π+arccos(−41​) and t=2π−arccos⁡ ⁣(−14)t=2\pi-\arccos\!\left(-\dfrac{1}{4}\right)t=2π−arccos(−41​)

Explanation: This problem involves solving trigonometric equations in the context of a pendulum's horizontal displacement. The equation cos(t) = -1/4 comes from setting x(t) = 12 cos(t) equal to -3 and solving. To find t, we use the inverse cosine function, which gives one angle, but we must find both where cosine is negative in the cycle. Thus, t = arccos(-1/4) and t = 2π - arccos(-1/4) are the solutions. Both fit the interval 0 ≤ t ≤ 2π, occurring in the second and third quadrants. A distractor might use arcsin incorrectly, confusing the function modeled. Always check solutions against the context to ensure they align with the pendulum's motion.

Question 16

A spherical balloon has radius 7 in7\text{ in}7 in. What is the volume of the solid?

  1. 43π(7)3 in3\tfrac{4}{3}\pi(7)^3\text{ in}^334​π(7)3 in3 (correct answer)
  2. π(7)2 in3\pi(7)^2\text{ in}^3π(7)2 in3
  3. 13π(7)2 in3\tfrac{1}{3}\pi(7)^2\text{ in}^331​π(7)2 in3
  4. 43π(7)3 in2\tfrac{4}{3}\pi(7)^3\text{ in}^234​π(7)3 in2

Explanation: This problem asks for the volume of a spherical balloon. The solid is a sphere with radius 7 inches. The volume formula for a sphere is V = (4/3)πr³, where r is the radius. Applying the formula: V = (4/3)π(7)³ = (4/3)π(343) in³. The correct answer includes the proper cubic units (in³) for volume. Option B incorrectly uses πr², which is the area of a circle, not the volume of a sphere, while option D has the wrong units (in² instead of in³). When calculating sphere volume, remember to cube the radius and multiply by (4/3)π, not just π.

Question 17

A right circular cone is shown with its base on a horizontal plane. A slicing plane cuts the cone and passes through the apex, and the plane is also perpendicular to the base (so the plane contains the cone’s axis).

Which shape results from the cross-section shown?

  1. Circle
  2. Isosceles triangle (correct answer)
  3. Rectangle
  4. Ellipse

Explanation: This question tests understanding of cross-sections and solids of revolution in geometry. The original solid is a right circular cone with a base on a horizontal plane. The slicing plane passes through the apex and is perpendicular to the base, containing the axis. The plane cuts along the height and through the base diameter. This produces an isosceles triangular cross-section with the base as the diameter and sides as generators. Confusing it with a circle ignores the axial cut through the apex. Picture the vertical slice step by step from apex to base to see the triangle emerge.

Question 18

Which statement correctly identifies the center and radius of the circle given by the equation x2+y2−6x+4y−12=0x^2+y^2-6x+4y-12=0x2+y2−6x+4y−12=0? (Use completing the square to interpret the geometry.)

  1. Center (3,−2)(3,-2)(3,−2) and radius 555 (correct answer)
  2. Center (−3,2)(-3,2)(−3,2) and radius 555
  3. Center (3,−2)(3,-2)(3,−2) and radius 252525
  4. Center (0,0)(0,0)(0,0) and radius 555

Explanation: This problem requires deriving the center and radius from the general form of a circle equation. A circle is defined as all points equidistant from its center. To interpret x² + y² - 6x + 4y - 12 = 0, we complete the square for both variables to reach standard form (x-h)² + (y-k)² = r². For x terms: x² - 6x = (x-3)² - 9; for y terms: y² + 4y = (y+2)² - 4. Substituting gives (x-3)² + (y+2)² - 9 - 4 - 12 = 0, which simplifies to (x-3)² + (y+2)² = 25. This reveals center (3, -2) and radius √25 = 5. Students often mistake r² for r, thinking the radius is 25 instead of 5. Remember: completing the square transforms the distance relationship into readable geometric parameters.

Question 19

Consider slicing a sphere of radius rrr and a “comparison solid” formed by taking a cylinder of radius rrr and height 2r2r2r and removing two congruent cones (each with base radius rrr and height rrr) whose tips meet at the center. The slices are taken by planes perpendicular to the axis at the same signed height hhh from the center (with −r≤h≤r-r\le h\le r−r≤h≤r). Which statement justifies the sphere’s volume formula using Cavalieri’s principle?

  1. At every height hhh, the sphere and the comparison solid have equal cross-sectional areas, so they have equal volumes. (correct answer)
  2. At every height hhh, the sphere and the comparison solid have equal cross-sectional perimeters, so they have equal volumes.
  3. At height h=rh=rh=r, both solids have cross-sectional area 000, so their volumes must be equal.
  4. The cylinder and the two cones each have height 2r2r2r, so subtracting their volumes gives the sphere.

Explanation: This problem focuses on applying Cavalieri's principle to justify the sphere volume formula through cross-sectional comparison. The sphere and comparison solid (cylinder minus two cones) are sliced at the same signed height h from the center. At height h, the sphere's cross-section has area π(r²-h²), and the cylinder minus cones also has area πr² - 2πh² = π(r²-h²). Since these areas match at every height from -r to r, Cavalieri's principle guarantees equal volumes. Option A correctly identifies this relationship, while B incorrectly mentions perimeters instead of areas, C wrongly focuses on just one height, and D misunderstands the geometric setup. The critical concept is that equal cross-sectional areas at all heights imply equal volumes.

Question 20

A student is asked to prove that tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}tan(A+B)=1−tanAtanBtanA+tanB​. The student begins by writing tan⁡(A+B)=sin⁡(A+B)cos⁡(A+B)\tan(A + B) = \frac{\sin(A + B)}{\cos(A + B)}tan(A+B)=cos(A+B)sin(A+B)​. What should be the student's next step?

  1. Factor out common terms from the numerator and denominator before applying formulas
  2. Convert the tangent function directly to the quotient form without using sine and cosine
  3. Apply the Pythagorean identity to simplify the expression before substitution
  4. Substitute the angle addition formulas for both sine and cosine in the numerator and denominator (correct answer)

Explanation: When proving trigonometric identities involving tangent addition formulas, you're working with the fundamental relationship that tangent equals sine divided by cosine. The student correctly started with tan⁡(A+B)=sin⁡(A+B)cos⁡(A+B)\tan(A + B) = \frac{\sin(A + B)}{\cos(A + B)}tan(A+B)=cos(A+B)sin(A+B)​, which sets up the perfect foundation for using angle addition formulas. The next logical step is to substitute the angle addition formulas for both sine and cosine. You need to replace sin⁡(A+B)\sin(A + B)sin(A+B) with sin⁡Acos⁡B+cos⁡Asin⁡B\sin A \cos B + \cos A \sin BsinAcosB+cosAsinB and cos⁡(A+B)\cos(A + B)cos(A+B) with cos⁡Acos⁡B−sin⁡Asin⁡B\cos A \cos B - \sin A \sin BcosAcosB−sinAsinB. This gives you tan⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡Bcos⁡Acos⁡B−sin⁡Asin⁡B\tan(A + B) = \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B - \sin A \sin B}tan(A+B)=cosAcosB−sinAsinBsinAcosB+cosAsinB​. From here, you can divide both numerator and denominator by cos⁡Acos⁡B\cos A \cos BcosAcosB to eventually reach the desired form with individual tangent terms. Option A is incorrect because there are no common factors to extract before applying the formulas—you need the expanded forms first. Option B misses the point entirely since the student already chose the sine/cosine approach, which is perfectly valid and systematic. Option C is wrong because the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 θ + \cos^2 θ = 1sin2θ+cos2θ=1 doesn't directly help with angle addition—you need the specific addition formulas. Remember: when proving tangent addition formulas, always apply the sine and cosine addition formulas first, then manipulate algebraically to isolate the tangent terms. This systematic approach works reliably for all angle addition proofs.

Question 21

A square pyramid paperweight has a base that is a square with side length 8 cm8\text{ cm}8 cm and a height of 15 cm15\text{ cm}15 cm. Which expression represents the volume?

  1. V=82+15V=8^2+15V=82+15
  2. V=13(82)(15)V=\tfrac{1}{3}(8^2)(15)V=31​(82)(15) (correct answer)
  3. V=(82)(15)V=(8^2)(15)V=(82)(15)
  4. V=2(8)(15)+2(82)V=2(8)(15)+2(8^2)V=2(8)(15)+2(82)

Explanation: This problem involves finding the volume expression for a square pyramid paperweight. The solid is a pyramid with a square base of side length 8 cm and height 15 cm. The volume formula for a pyramid is V = (1/3) × base area × height, where the base area for a square is side². The correct expression is V = (1/3)(8²)(15) or V = (1/3)(64)(15). This gives one-third the volume of a rectangular prism with the same base and height. Option C incorrectly omits the 1/3 factor, while option D uses a surface area formula. When working with pyramids, always include the factor of 1/3 in the volume calculation.

Question 22

A right triangle △RST\triangle RST△RST has a right angle at SSS. The angle at RRR is labeled θ\thetaθ. The hypotenuse RT‾\overline{RT}RT is labeled hhh (no numerical value is given). The leg adjacent to θ\thetaθ is RS‾\overline{RS}RS and the opposite leg is ST‾\overline{ST}ST. Which argument uses the Pythagorean Theorem correctly to prove sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1sin2θ+cos2θ=1 for this triangle?

Image description (not drawn to scale): A right triangle with RRR at left, TTT at upper-right, and SSS at lower-right. A right-angle box marks ∠S=90∘\angle S=90^\circ∠S=90∘. Segment RTRTRT is the slanted side and is labeled hhh. Segment RSRSRS is the lower leg from RRR to SSS (adjacent to θ\thetaθ). Segment STSTST is the vertical leg from SSS to TTT (opposite θ\thetaθ). An angle arc at RRR labels ∠R=θ\angle R=\theta∠R=θ. No side lengths besides hhh are marked.

  1. Since sin⁡θ=RSRT\sin\theta=\dfrac{RS}{RT}sinθ=RTRS​ and cos⁡θ=STRT\cos\theta=\dfrac{ST}{RT}cosθ=RTST​, squaring gives sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1sin2θ+cos2θ=1.
  2. Using RS2+ST2=RT2RS^2+ST^2=RT^2RS2+ST2=RT2 and dividing by RT2RT^2RT2 gives (STRT)2+(RSRT)2=1\left(\dfrac{ST}{RT}\right)^2+\left(\dfrac{RS}{RT}\right)^2=1(RTST​)2+(RTRS​)2=1, so sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1sin2θ+cos2θ=1. (correct answer)
  3. Because RS+ST=RTRS+ST=RTRS+ST=RT in a right triangle, dividing by RTRTRT gives sin⁡θ+cos⁡θ=1\sin\theta+\cos\theta=1sinθ+cosθ=1 and then squaring gives the identity.
  4. Since θ\thetaθ is acute, RS=STRS=STRS=ST, so sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1sin2θ+cos2θ=1.

Explanation: The Pythagorean identity states that sin²θ + cos²θ = 1 for any angle θ. In a right triangle, sine of θ is the ratio of the opposite side to the hypotenuse, and cosine is the ratio of the adjacent side to the hypotenuse. Squaring these ratios yields sin²θ = (opposite / hypotenuse)² and cos²θ = (adjacent / hypotenuse)². Applying the Pythagorean theorem, opposite² + adjacent² = hypotenuse². Dividing both sides by hypotenuse² gives sin²θ + cos²θ = 1. A misconception in distractors is swapping the definitions of sine and cosine, but squaring still yields 1, though the proof requires correct ratios. To apply this elsewhere, return to the triangle's geometry and ensure definitions match opposite and adjacent sides.

Question 23

When constructing a regular octagon inscribed in a circle, a student first constructs perpendicular diameters to create four equally spaced points. What construction technique should be applied next to locate the remaining four vertices?

  1. Bisect each of the four arcs created by the perpendicular diameters using compass and straightedge (correct answer)
  2. Construct equilateral triangles using each diameter as a base to find the missing points
  3. Use the compass to mark points at distance equal to the diameter from each existing vertex
  4. Draw tangent lines to the circle at each existing point and find their intersection points

Explanation: A regular octagon requires 8 equally spaced points, each separated by 45°. After creating perpendicular diameters (4 points separated by 90°), the remaining points are found by bisecting each of the four equal arcs, creating points at the 45° intervals. Choice B is incorrect because equilateral triangles create 60° angles, not the needed 45°. Choice C is incorrect as using the diameter length would create points too far from the existing vertices. Choice D is incorrect because tangent line intersections do not yield points on the circle.

Question 24

A parabola is defined as the set of points equidistant from the focus F(0,1)F(0,1)F(0,1) and the directrix line y=−3y=-3y=−3. The parabola opens upward. Which expression represents all points (x,y)(x,y)(x,y) equidistant from the focus and directrix?

  1. x2+(y−1)2=∣y+3∣\sqrt{x^2+(y-1)^2}=|y+3|x2+(y−1)2​=∣y+3∣ (correct answer)
  2. x2+(y+3)2=∣y−1∣\sqrt{x^2+(y+3)^2}=|y-1|x2+(y+3)2​=∣y−1∣
  3. (x−1)2+y2=∣y+3∣\sqrt{(x-1)^2+y^2}=|y+3|(x−1)2+y2​=∣y+3∣
  4. x2+(y−1)2=∣x+3∣\sqrt{x^2+(y-1)^2}=|x+3|x2+(y−1)2​=∣x+3∣

Explanation: The skill here is deriving the equation of a parabola from its focus and directrix. A parabola is defined as the set of all points equidistant from a fixed point called the focus and a fixed line called the directrix. For any point (x,y) on this parabola, the distance to the focus F(0,1) equals the distance to the directrix y=-3. This equality is directly represented by the expression √(x² + (y-1)²) = |y+3|, without yet squaring. This form is justified as it captures the raw geometric definition before algebraic simplification to the parabola equation. A distractor like choice B swaps the focus and directrix distances, which would not satisfy the definition. To derive equations for other parabolas, always start by equating the distance to the focus and to the directrix, then square and simplify if needed.

Question 25

A rectangle of width x+yx+yx+y and height x+yx+yx+y (a square) is partitioned into four regions by one vertical and one horizontal segment. The left width is labeled xxx and the right width is labeled yyy; the top height is labeled xxx and the bottom height is labeled yyy. The two off-diagonal rectangles (top-right and bottom-left) are shaded, indicating their combined area.

Which expression represents the total shaded area for all real xxx and yyy?

  1. x2+y2x^2+y^2x2+y2
  2. 2xy2xy2xy (correct answer)
  3. (x+y)2(x+y)^2(x+y)2
  4. (x−y)2(x-y)^2(x−y)2

Explanation: This task involves applying polynomial identities through geometric shading. The model is a square of side x+y, divided into four rectangles by vertical and horizontal segments. The shaded off-diagonal regions match xy for top-right and yx for bottom-left, both xy. The shaded area sums these two regions, which are distinct without overlap. Thus, the total shaded equals xy + xy = 2xy, identifying the expression. Distractor A might add squares instead, ignoring the cross products. Use shading in area models to isolate and verify specific polynomial terms.