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Geometry

Geometry Practice Test: Practice Test 6

Practice Test 6 for Geometry: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A circle is the set of all points a fixed distance from its center. What is the equation of the circle with center (2,−5)(2, -5)(2,−5) and radius 333?​

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Question 1

A circle is the set of all points a fixed distance from its center. What is the equation of the circle with center (2,−5)(2, -5)(2,−5) and radius 333?​

  1. (x−2)2+(y+5)2=9(x-2)^2+(y+5)^2=9(x−2)2+(y+5)2=9 (correct answer)
  2. (x+2)2+(y−5)2=9(x+2)^2+(y-5)^2=9(x+2)2+(y−5)2=9
  3. (x−2)2+(y+5)2=3(x-2)^2+(y+5)^2=3(x−2)2+(y+5)2=3
  4. x2+y2=9x^2+y^2=9x2+y2=9

Explanation: This question directly applies the definition of a circle to write its equation. A circle is the set of all points that are equidistant from a fixed center point, creating a perfect round shape. Using the distance formula, every point (x, y) on a circle with center (2, -5) and radius 3 satisfies √[(x - 2)² + (y - (-5))²] = 3. Squaring both sides to eliminate the square root gives us (x - 2)² + (y + 5)² = 9, which is the standard form of the circle equation. Notice how the center coordinates appear in the parentheses with opposite signs: center (2, -5) gives us (x - 2) and (y + 5). A common mistake is writing (y - 5) instead of (y + 5) when the y-coordinate is negative—always be careful with signs. To master circle equations, remember that they encode the simple geometric idea that every point maintains the same distance from the center.

Question 2

A company models shipping boxes as rectangular prisms. A box with dimensions 12 inches by 9 inches by 6 inches costs $2.40 to ship. If shipping cost is proportional to volume, what would it cost to ship a box with dimensions 15 inches by 12 inches by 8 inches?

  1. $4.00
  2. $4.44 (correct answer)
  3. $3.60
  4. $5.00

Explanation: Both boxes are modeled as rectangular prisms. Original volume: V1=12×9×6=648V_1 = 12 \times 9 \times 6 = 648V1​=12×9×6=648 cubic inches. New volume: V2=15×12×8=1440V_2 = 15 \times 12 \times 8 = 1440V2​=15×12×8=1440 cubic inches. Since cost is proportional to volume: Cost2Cost1=V2V1\frac{\text{Cost}_2}{\text{Cost}_1} = \frac{V_2}{V_1}Cost1​Cost2​​=V1​V2​​. Therefore: Cost2=2.40×1440648=2.40×209≈$4.44\text{Cost}_2 = 2.40 \times \frac{1440}{648} = 2.40 \times \frac{20}{9} \approx \$4.44Cost2​=2.40×6481440​=2.40×920​≈$4.44. Choice A uses an incorrect ratio. Choice C assumes linear scaling of one dimension. Choice D rounds incorrectly.

Question 3

Two circles, ⊙R\odot R⊙R and ⊙U\odot U⊙U, are drawn with different radii and with centers marked at RRR and UUU. Which transformation maps ⊙R\odot R⊙R to a circle similar to ⊙U\odot U⊙U while keeping the idea of similarity (same shape) explicit?

  1. Dilate ⊙R\odot R⊙R about its center RRR by an appropriate scale factor, then translate so RRR moves to UUU. (correct answer)
  2. Translate ⊙R\odot R⊙R so RRR moves to UUU, and the radius will automatically change to match ⊙U\odot U⊙U.
  3. Reflect ⊙R\odot R⊙R across a line through UUU to increase its radius to match ⊙U\odot U⊙U.
  4. Use the circumference formula to compute both circumferences and conclude the circles are similar.

Explanation: The skill here is understanding circle similarity in geometry. Circles are similar via dilation, which scales the radius while preserving shape. The centers R and U guide the sequence of transformations needed. Applying a dilation about R to match the radius, followed by a translation to move R to U, maps ⊙R to ⊙U. This justifies similarity as the transformations ensure same shape and adjusted size. A common distractor is option B, which wrongly assumes translation alone changes the radius. To transfer this strategy, think in terms of transformations like dilations and translations, not formulas.

Question 4

In a proof that all circles are similar, a student writes: "Since all circles have the same shape, they are similar by definition." What is the primary flaw in this reasoning?

  1. The statement is circular reasoning that doesn't establish the existence of a similarity transformation between specific circles (correct answer)
  2. The statement is incorrect because circles can have different eccentricities depending on their radii and positions
  3. The statement fails to account for the fact that similarity requires preservation of both angle and ratio measures
  4. The statement is invalid because it doesn't specify which circle is being used as the reference standard

Explanation: A valid proof of similarity must demonstrate the existence of a specific similarity transformation (combination of rigid motions and dilation) that maps one circle to another. Simply stating that circles have the same shape is circular reasoning that doesn't provide the required constructive proof. Choice B is incorrect since circles don't have varying eccentricity. Choice C misses the main issue of circular reasoning. Choice D incorrectly suggests a reference standard is needed.

Question 5

Which transformation is represented by the matrix A=(10 0−1)A=\begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix}A=(1​0 0​−1​)?

  1. Reflection across the yyy-axis
  2. Reflection across the xxx-axis (correct answer)
  3. Rotation 90∘90^\circ90∘ counterclockwise about the origin
  4. Uniform scaling by factor −1-1−1

Explanation: This question tests your ability to use 2×2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×2 matrix can represent a linear transformation of the plane: to transform a point (x, y), write it as column vector [x; y] and multiply by transformation matrix T = [a b; c d] using matrix multiplication: T[x; y] = [a b; c d][x; y] = [ax+by; cx+dy] = [x'; y'] where (x', y') is the transformed point. Common transformation matrices include: ROTATION by angle θ counterclockwise = (cos⁡(θ)−sin⁡(θ)sin⁡(θ)cos⁡(θ))\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}(cos(θ)sin(θ)​−sin(θ)cos(θ)​) (example: 90° rotation uses θ=90° giving (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}(01​−10​) since cos(90°)=0 and sin(90°)=1), REFLECTION across x-axis = (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}(10​0−1​) (keeps x same, negates y), REFLECTION across y-axis = (−1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}(−10​01​) (negates x, keeps y same), SCALING by factor k = (k00k)\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}(k0​0k​) (multiplies both coordinates by k, enlarges by factor k). The matrix A=[1 0; 0 -1] transforms (x,y) to (x, -y), which is exactly a reflection across the x-axis, as y is negated while x remains unchanged. Choice B correctly identifies this as reflection across the x-axis by recognizing the diagonal form with positive x and negative y scaling. A distractor like choice A might confuse it with y-axis reflection, which would have [-1 0; 0 1] instead. Remember the strategy: identify transformations by pattern—reflections have ±1±1±1 on diagonal with one negative; test with a point like (1,1) to see the effect, and you'll master this quickly!

Question 6

Consider the transformation f(x,y)=(−y,x)f(x,y) = (-y, x)f(x,y)=(−y,x). Which statement best describes the properties of this transformation function?

  1. It preserves both distance and angle measures, representing a 90°90°90° counterclockwise rotation about the origin. (correct answer)
  2. It preserves distance but not angle measures, representing a reflection across the line y=−xy = -xy=−x.
  3. It preserves angle measures but not distances, representing a uniform scaling by factor 2\sqrt{2}2​.
  4. It preserves neither distance nor angle measures, representing a horizontal stretch combined with vertical compression.

Explanation: The transformation f(x,y)=(−y,x)f(x,y) = (-y, x)f(x,y)=(−y,x) is a 90°90°90° counterclockwise rotation about the origin. Rotations are rigid transformations that preserve both distances and angle measures. We can verify: (1,0)→(0,1)(1,0) \to (0,1)(1,0)→(0,1) and (0,1)→(−1,0)(0,1) \to (-1,0)(0,1)→(−1,0), which represents a 90°90°90° counterclockwise rotation.

Question 7

Quadrilateral ABCD has vertices A(0,1)A(0,1)A(0,1), B(4,4)B(4,4)B(4,4), C(7,0)C(7,0)C(7,0), and D(3,−3)D(3,-3)D(3,−3) connected in that order. What is the perimeter of the figure?

  1. 5+5+5+5=205+5+5+5=205+5+5+5=20
  2. (4−0)2+(4−1)2+(7−4)2+(0−4)2+(3−7)2+(−3−0)2+(0−3)2+(1−(−3))2\sqrt{(4-0)^2+(4-1)^2}+\sqrt{(7-4)^2+(0-4)^2}+\sqrt{(3-7)^2+(-3-0)^2}+\sqrt{(0-3)^2+(1-(-3))^2}(4−0)2+(4−1)2​+(7−4)2+(0−4)2​+(3−7)2+(−3−0)2​+(0−3)2+(1−(−3))2​ (correct answer)
  3. (4−0)2+(4−1)2+(7−4)2+(0−4)2+(3−7)2+(−3−0)2\sqrt{(4-0)^2+(4-1)^2}+\sqrt{(7-4)^2+(0-4)^2}+\sqrt{(3-7)^2+(-3-0)^2}(4−0)2+(4−1)2​+(7−4)2+(0−4)2​+(3−7)2+(−3−0)2​
  4. 12∣0(4−0)+4(0+3)+7(−3−1)+3(1−4)∣\frac{1}{2}\left|0(4-0)+4(0+3)+7(-3-1)+3(1-4)\right|21​∣0(4−0)+4(0+3)+7(−3−1)+3(1−4)∣

Explanation: The skill is finding the perimeter of a quadrilateral using coordinates. The vertices are A(0,1)A(0,1)A(0,1), B(4,4)B(4,4)B(4,4), C(7,0)C(7,0)C(7,0), and D(3,−3)D(3,-3)D(3,−3). Side lengths are computed using the distance formula between consecutive points. The perimeter is the sum of these four side lengths. Each simplifies to 5, summing to 20, justifying the expression in the correct choice. A distractor misconception is summing only three sides, as in choice C. To transfer this strategy, always compute individual side lengths before summing for the perimeter.

Question 8

Consider a sphere of radius RRR and a cylinder of radius RRR and height 2R2R2R that contains two congruent cones as follows: each cone has its base as the circle in the cylinder’s mid-plane and its apex at one end of the cylinder (one apex at the top center, one at the bottom center). Slices are taken by planes parallel to the bases at the same height in all solids.

Which statement justifies the sphere’s volume formula using Cavalieri’s principle (without relying only on memorized formulas)?

  1. The sphere’s volume equals the cylinder’s volume because both have radius RRR.
  2. The sphere’s volume equals the cylinder’s volume minus two cones’ volumes because their cross-sections match in area at every height. (correct answer)
  3. The sphere’s volume equals the cylinder’s volume minus two cones’ volumes because the solids have the same surface area.
  4. The sphere’s volume equals 43πR3\frac{4}{3}\pi R^334​πR3 because that is the known formula for spheres.

Explanation: The skill is deriving the volume of a sphere using Cavalieri’s principle. We compare the sphere of radius R to the solid formed by a cylinder of radius R and height 2R minus two congruent cones, each with apex at the center and base at the end of the cylinder. At every height h from the center, the cross-sectional area of the sphere is equal to that of the cylinder minus the relevant cone's cross-sectional area in that half. Since the cross-sectional areas are equal at every corresponding height, Cavalieri’s principle states that the volumes are equal. Therefore, the sphere's volume equals the volume of the cylinder minus the volumes of the two cones. A common misconception is that solids with the same surface area have the same volume, but Cavalieri’s principle relies on matching cross-sectional areas, not surface areas. To apply this strategy to other solids, always compare cross-sectional areas at the same corresponding heights.

Question 9

A linear transformation TTT is represented by the matrix A=(−100−1).A=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}.A=(−10​0−1​). On the coordinate plane, the triangle with vertices (0,0)(0,0)(0,0), (2,0)(2,0)(2,0), and (0,1)(0,1)(0,1) is shown. Which statement describes the geometric effect of applying TTT to the triangle?

  1. It rotates the triangle 180∘180^\circ180∘ about the origin and keeps the area the same. (correct answer)
  2. It collapses the triangle to a line because the determinant is negative.
  3. It leaves the triangle unchanged because the determinant equals 111.
  4. It scales the triangle’s area by a factor of −2-2−2 because both entries are −1-1−1.

Explanation: Geometric matrix interpretation reveals rotations like 180 degrees that invert positions but preserve areas. Identity fixes, zero collapses. Positive determinant of 1 means area preserved with orientation. A rotates the triangle 180 degrees, mapping to opposite quadrant with same area. Justified by A scaling by -1 in both directions, equivalent to 180 rotation, |det|=1. Distractor: negative det collapses, but det=1 is positive, no collapse. Interpret determinant as area change, here 1 keeping it same.

Question 10

In right triangle △JKL\triangle JKL△JKL, the right angle is at KKK. The hypotenuse is JL=15JL=15JL=15, and ∠J=50∘\angle J = 50^\circ∠J=50∘. What is the length of leg JKJKJK?

  1. 15cos⁡(50∘)15\cos(50^\circ)15cos(50∘) (correct answer)
  2. 15sin⁡(50∘)\dfrac{15}{\sin(50^\circ)}sin(50∘)15​
  3. 15sin⁡(50∘)15\sin(50^\circ)15sin(50∘)
  4. 15tan⁡(50∘)15\tan(50^\circ)15tan(50∘)

Explanation: This problem requires finding a leg adjacent to a given angle in a right triangle. We have a right angle at K, hypotenuse JL = 15, and angle J = 50°. Since we know an angle and the hypotenuse, trigonometry is the appropriate method. To find leg JK (adjacent to angle J), we use cosine: cos(50°) = adjacent/hypotenuse = JK/15. Solving for JK: JK = 15·cos(50°). This correctly uses cosine to relate the adjacent side to the hypotenuse. A common mistake would be using sine (15·sin(50°)), which would give the opposite leg KL instead. Always identify which leg you're finding relative to the given angle before choosing the trig function.

Question 11

Two circles are drawn on a coordinate plane: Circle ⊙P\odot P⊙P has center P(−2,1)P(-2,1)P(−2,1) and Circle ⊙Q\odot Q⊙Q has center Q(4,−1)Q(4,-1)Q(4,−1). Their radii are different, and the centers are marked. Which transformation maps ⊙P\odot P⊙P to a circle similar to ⊙Q\odot Q⊙Q using similarity transformations (not formulas)?

  1. Translate ⊙P\odot P⊙P so that PPP moves to QQQ, then dilate about QQQ to match the radius. (correct answer)
  2. Reflect ⊙P\odot P⊙P across the xxx-axis to make it the same size as ⊙Q\odot Q⊙Q.
  3. Rotate ⊙P\odot P⊙P about the origin until it lands on ⊙Q\odot Q⊙Q.
  4. Use the area formula to show the ratio of areas equals the square of the ratio of radii.

Explanation: The skill here is understanding circle similarity in geometry. Circles are similar via dilation, which scales the radius while maintaining the round shape. The centers P and Q, located at different coordinates, must be aligned before scaling. Applying a translation to move P to Q, followed by a dilation about Q to adjust the radius, maps ⊙P to ⊙Q. This justifies similarity as the composition of translation and dilation is a similarity transformation that matches both position and size. A common distractor is option D, which relies on area formulas instead of transformations, missing the geometric mapping aspect. To transfer this strategy, think in terms of transformations like dilations and translations, not formulas.

Question 12

A non-right triangle △DEF\triangle DEF△DEF is shown. Sides DE=aDE=aDE=a and DF=bDF=bDF=b form the included angle at DDD, labeled CCC. A dashed altitude from EEE meets side DFDFDF at a right angle. Which expression represents the area of the triangle?

  1. A=12absin⁡(C)A=\tfrac12 ab\sin(C)A=21​absin(C) (correct answer)
  2. A=12absin⁡(∠E)A=\tfrac12 ab\sin(\angle E)A=21​absin(∠E)
  3. A=12abcos⁡(C)A=\tfrac12 ab\cos(C)A=21​abcos(C)
  4. A=absin⁡(C)A=ab\sin(C)A=absin(C)

Explanation: This question tests the derivation of triangle area using two sides and their included angle. The area of a triangle equals half the product of base times height. In triangle DEF with sides DE = a and DF = b forming included angle C at vertex D, we drop an altitude from E to side DF. The height of this altitude equals the length of side DE times the sine of angle C, giving height = a·sin(C). Using DF as the base (length b), the area becomes A = ½·b·(a·sin(C)) = ½ab·sin(C). The correct formula includes the ½ factor and uses sine of the included angle C at vertex D, not angle E as suggested in choice B. A common error is using cosine instead of sine, which would give the adjacent side length rather than the perpendicular height. When finding area with two sides and their included angle, always use A = ½ab·sin(C).

Question 13

A parabola is defined as the set of points equidistant from the focus F(3,−2)F(3,-2)F(3,−2) and the directrix y=2y=2y=2. The parabola opens downward.

Which equation represents the parabola?

  1. (x−3)2=−8(y+2)(x-3)^2=-8(y+2)(x−3)2=−8(y+2)
  2. (x−3)2=−8y(x-3)^2=-8y(x−3)2=−8y (correct answer)
  3. (x+3)2=−8(y+2)(x+3)^2=-8(y+2)(x+3)2=−8(y+2)
  4. (x−3)2=8(y+2)(x-3)^2=8(y+2)(x−3)2=8(y+2)

Explanation: The skill is deriving the equation of a parabola from its focus and directrix. A parabola is geometrically defined as the set of points equidistant from the focus at (3,-2) and the directrix y = 2. For any point (x,y) on the parabola, the distance to the focus equals the distance to the directrix, expressed as √((x-3)² + (y+2)²) = |y - 2|. Setting up this equality and squaring both sides eliminates the square root, leading to (x-3)² + (y+2)² = (y-2)², which simplifies through expansion and cancellation to (x-3)² = -8y. This final form is justified as it places the vertex at (3,0), midway between the focus and directrix, with the coefficient -8 corresponding to 4p where p=-2 for downward opening. A common distractor misconception is including an unnecessary y-shift like (y+2), altering the vertex position. To transfer this strategy, always start from the distance definition and carefully simplify when deriving parabola equations.

Question 14

In the right triangle shown, ∠C\angle C∠C is a right angle and ∠A=θ\angle A=\theta∠A=θ. Which reasoning proves sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1sin2θ+cos2θ=1?​

  1. Since sin⁡θ=BC/AB\sin\theta=BC/ABsinθ=BC/AB and cos⁡θ=AC/AB\cos\theta=AC/ABcosθ=AC/AB, squaring and adding gives (BC2+AC2)/AB2=AB2/AB2=1(BC^2+AC^2)/AB^2=AB^2/AB^2=1(BC2+AC2)/AB2=AB2/AB2=1. (correct answer)
  2. Since sin⁡θ=BC/AC\sin\theta=BC/ACsinθ=BC/AC and cos⁡θ=AC/BC\cos\theta=AC/BCcosθ=AC/BC, squaring and adding gives 111 by cancellation.
  3. Since BC2+AB2=AC2BC^2+AB^2=AC^2BC2+AB2=AC2, dividing by AB2AB^2AB2 gives sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1sin2θ+cos2θ=1.
  4. Because θ\thetaθ is acute, sin⁡θ=cos⁡θ\sin\theta=\cos\thetasinθ=cosθ, so sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1sin2θ+cos2θ=1.

Explanation: This problem asks us to prove the Pythagorean identity for a right triangle with right angle at C and angle θ at A. With this setup, sin θ = opposite/hypotenuse = BC/AB and cos θ = adjacent/hypotenuse = AC/AB. Squaring these ratios gives us sin²θ = BC²/AB² and cos²θ = AC²/AB². The Pythagorean Theorem for this triangle states that BC² + AC² = AB² (legs squared sum to hypotenuse squared). When we add our squared trigonometric ratios: sin²θ + cos²θ = BC²/AB² + AC²/AB² = (BC² + AC²)/AB² = AB²/AB² = 1. Choice B incorrectly defines sin θ = BC/AC and cos θ = AC/BC, which are not the standard definitions and would not lead to the identity. Remember to always identify the hypotenuse correctly—it's the side opposite the right angle.

Question 15

Two circles have equations (x−2)2+(y+1)2=9(x-2)^2 + (y+1)^2 = 9(x−2)2+(y+1)2=9 and (x+3)2+(y−4)2=25(x+3)^2 + (y-4)^2 = 25(x+3)2+(y−4)2=25. To prove these circles are similar, which transformation sequence correctly maps the first circle to the second?

  1. Translation by (−5,5)(-5, 5)(−5,5) then dilation by factor 53\frac{5}{3}35​ centered at (−3,4)(-3, 4)(−3,4) (correct answer)
  2. Dilation by factor 53\frac{5}{3}35​ centered at (2,−1)(2, -1)(2,−1) then translation by (−5,5)(-5, 5)(−5,5)
  3. Translation by (5,−5)(5, -5)(5,−5) then dilation by factor 35\frac{3}{5}53​ centered at (−3,4)(-3, 4)(−3,4)
  4. Rotation of 90°90°90° about origin then dilation by factor 53\frac{5}{3}35​ centered at (−3,4)(-3, 4)(−3,4)

Explanation: The first circle has center (2,−1)(2, -1)(2,−1) and radius 3; the second has center (−3,4)(-3, 4)(−3,4) and radius 5. Translation vector from (2,−1)(2, -1)(2,−1) to (−3,4)(-3, 4)(−3,4) is (−5,5)(-5, 5)(−5,5). After translation, dilation by 53\frac{5}{3}35​ (centered at the new position) scales radius from 3 to 5. Choice B dilates first, changing the required translation. Choice C uses wrong translation direction and scale factor. Choice D adds unnecessary rotation.

Question 16

In the right triangle shown, ∠A\angle A∠A is a right angle. The hypotenuse is BC‾\overline{BC}BC. If AB=8AB=8AB=8 and BC=17BC=17BC=17, what is the length of ACACAC?

  1. 353\sqrt{353}353​
  2. 999
  3. 252525
  4. 151515 (correct answer)

Explanation: This problem asks us to find a leg of a right triangle using the Pythagorean theorem. We're given leg AB = 8 and hypotenuse BC = 17, with angle A being the right angle. Since we need to find a leg when given the other leg and hypotenuse, we use: a² = c² - b². Setting up: AC² = BC² - AB² = 17² - 8² = 289 - 64 = 225, so AC = √225 = 15. The answer is 15, not √225, because we must simplify the square root. A student might add the squares instead (getting √353) by incorrectly treating this as finding a hypotenuse. Always check: is the unknown side the hypotenuse (longest side) or a leg before deciding whether to add or subtract.

Question 17

A cylindrical can has diameter 10 cm10\text{ cm}10 cm and height 12 cm12\text{ cm}12 cm. What is the volume of the can in cubic centimeters?

  1. 600π cm3600\pi\text{ cm}^3600π cm3
  2. 120π cm3120\pi\text{ cm}^3120π cm3
  3. 300π cm3300\pi\text{ cm}^3300π cm3 (correct answer)
  4. 240π cm3240\pi\text{ cm}^3240π cm3

Explanation: This problem involves finding the volume of a cylindrical can given its diameter. The solid is a right circular cylinder with diameter 10 cm and height 12 cm. Since the volume formula V = πr²h requires radius, we must first convert: radius = diameter/2 = 10/2 = 5 cm. Applying the formula: V = π(5)²(12) = π(25)(12) = 300π cubic centimeters. This represents the can's total capacity. A common mistake is using diameter directly in the formula, giving π(10)²(12) = 1200π, which is four times too large. Always convert diameter to radius before applying cylinder volume formulas.

Question 18

On the coordinate plane, segment AB‾\overline{AB}AB has endpoints A(0,2)A(0,2)A(0,2) and B(8,−6)B(8,-6)B(8,−6). Point PPP divides the directed segment from AAA to BBB internally in the ratio AP:PB=5:3AP:PB=5:3AP:PB=5:3. Which coordinates represent the partition point?

  1. (5,−3)(5,-3)(5,−3) (correct answer)
  2. (4,−2)(4,-2)(4,−2)
  3. (3,−1)(3,-1)(3,−1)
  4. (10,−8)(10,-8)(10,−8)

Explanation: The skill is partitioning a line segment in a given ratio. The endpoints are A(0,2) and B(8,-6), with the ratio AP:PB = 5:3. This means point P is a weighted average where A has weight 3 and B has weight 5, since the weights are the opposite parts of the ratio. Applying the ratio, the coordinates are x = (30 + 58)/8 = 40/8 = 5 and y = (32 + 5(-6))/8 = -24/8 = -3, so P is at (5,-3). This result is justified because it positions P five-eighths of the way from A to B, consistent with the ratio 5:3. A common distractor misconception is misapplying the weights, leading to (4,-2) by incorrectly averaging without proper ratio consideration. To transfer this strategy, think in terms of weights, not distances.

Question 19

Circles ⊙R\odot R⊙R and ⊙S\odot S⊙S are shown with centers marked. The radii are labeled RA‾=6\overline{RA}=6RA=6 and SB‾=9\overline{SB}=9SB=9. Which transformation maps one circle to the other to show they are similar?

  1. Translate ⊙R\odot R⊙R to move RRR to SSS, then dilate by scale factor 32\tfrac{3}{2}23​. (correct answer)
  2. Reflect ⊙R\odot R⊙R across line RS‾\overline{RS}RS to increase its radius from 666 to 999.
  3. They are congruent because both have radii drawn from their centers.
  4. Compute circumferences to confirm similarity, then no transformation is needed.

Explanation: The skill here is understanding circle similarity. Circles are similar because one can be mapped to another via a dilation that scales all distances by a constant factor, preserving shape. In this problem, the circles have centers R and S with radii 6 and 9, respectively. Applying a translation to move R to S, followed by a dilation centered at S with scale factor 3/2, maps ⊙R to ⊙S. This justifies similarity because the sequence of isometry and dilation forms a similarity transformation that overlays the circles. A distractor like choice B suggests reflection changes radius, but reflections preserve distances and do not alter sizes. To approach similar problems, think in terms of transformations like dilations rather than relying on formulas.

Question 20

A transformation TTT maps point (2,3)(2, 3)(2,3) to (5,7)(5, 7)(5,7) and point (4,1)(4, 1)(4,1) to (7,5)(7, 5)(7,5). If TTT is a translation, what is the image of point (0,−2)(0, -2)(0,−2) under this transformation?

  1. (3,2)(3, 2)(3,2) (correct answer)
  2. (3,4)(3, 4)(3,4)
  3. (2,3)(2, 3)(2,3)
  4. (−3,−6)(-3, -6)(−3,−6)

Explanation: For a translation, the transformation function is T(x,y)=(x+a,y+b)T(x,y) = (x+a, y+b)T(x,y)=(x+a,y+b) for some constants aaa and bbb. From (2,3)→(5,7)(2,3) \to (5,7)(2,3)→(5,7), we get a=3a = 3a=3 and b=4b = 4b=4. We can verify with the second point: (4,1)→(4+3,1+4)=(7,5)(4,1) \to (4+3, 1+4) = (7,5)(4,1)→(4+3,1+4)=(7,5). Therefore, T(0,−2)=(0+3,−2+4)=(3,2)T(0,-2) = (0+3, -2+4) = (3,2)T(0,−2)=(0+3,−2+4)=(3,2).

Question 21

A right circular cylinder stands upright. A plane slices the cylinder perpendicular to the circular bases and passes through the cylinder’s central axis (so the plane contains the axis). Which description correctly identifies the resulting shape?

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

Explanation: This problem involves finding the cross-section of a 3D solid. The original solid is a right circular cylinder standing upright with circular bases. The slicing plane is perpendicular to the circular bases and passes through the central axis. As the plane cuts through, it follows the height and spans the diameter of the bases. The resulting cross-section is a rectangle with sides equal to the height and diameter. A common misconception is thinking it forms a circle, but that's for a parallel slice. To visualize, imagine unfolding the cylinder along the axis step by step to see the rectangular outline.

Question 22

Triangle ABCABCABC has vertices A(1,1)A(1,1)A(1,1), B(3,1)B(3,1)B(3,1), and C(2,4)C(2,4)C(2,4). It is transformed by the matrix K=(2002).K=\begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix}.K=(20​02​). What are the coordinates of C′C'C′ (the image of CCC)?​

  1. (4,4)(4,4)(4,4)
  2. (4,8)(4,8)(4,8) (correct answer)
  3. (2,8)(2,8)(2,8)
  4. (1,2)(1,2)(1,2)

Explanation: This question tests your ability to use 2×2 matrices to represent and perform plane transformations (rotations, reflections, scaling) by multiplying transformation matrices with point coordinate vectors. A 2×2 matrix can represent a linear transformation of the plane: to transform a point (x, y), write it as column vector [x; y] and multiply by transformation matrix T = [a b; c d] using matrix multiplication: T[x; y] = [a b; c d][x; y] = [ax+by; cx+dy] = [x'; y'] where (x', y') is the transformed point. The matrix K = [2 0; 0 2] represents uniform scaling by factor 2. Applying this to vertex C(2, 4): K[2; 4] = [2 0; 0 2][2; 4] = [2·2 + 0·4; 0·2 + 2·4] = [4 + 0; 0 + 8] = [4; 8], giving C'(4, 8). Choice B correctly shows this result where both coordinates are doubled: x: 2 → 4 and y: 4 → 8. Choice A incorrectly has y' = 4 (not doubled), choice C incorrectly has x' = 2 (not doubled), and choice D shows the original point divided by 2 instead of multiplied. Matrix multiplication for transformations: SCALING matrices have equal diagonal entries [k 0; 0 k]—both coordinates multiplied by same k. The scaling transformation K = [2 0; 0 2] doubles all distances from the origin, enlarging the triangle by factor 2!

Question 23

A triangle △PQR\triangle PQR△PQR is shown, and a second triangle △STU\triangle STU△STU appears to be a rotated and dilated image of it. Angle markings show ∠P\angle P∠P and ∠S\angle S∠S each have a single arc, and ∠Q\angle Q∠Q and ∠T\angle T∠T each have a double arc. No side lengths are labeled, and the diagram is not drawn to scale.

Which reasoning uses similarity transformations correctly?

  1. A dilation and a rotation can map △PQR\triangle PQR△PQR to △STU\triangle STU△STU, so the triangles are similar by AA. (correct answer)
  2. A reflection maps △PQR\triangle PQR△PQR to △STU\triangle STU△STU, so the triangles are congruent.
  3. Since two angles match, all corresponding sides are equal.
  4. Because the triangles look the same size, they must be similar.

Explanation: The AA criterion is a key method for proving triangle similarity. If two angles of one triangle are congruent to two angles of another triangle, the triangles are similar. In this diagram, angle P is marked congruent to angle S with a single arc, and angle Q is marked congruent to angle T with a double arc. Therefore, by the AA similarity criterion, triangle PQR is similar to triangle STU. Similar triangles have corresponding sides that are proportional, meaning their ratios are equal, but the sides themselves are not necessarily equal in length. A common misconception is to think that a reflection alone proves congruence, but without confirming equal sizes, similarity is the appropriate conclusion. To apply this in other problems, always check for matching angles first before examining side lengths or proportions.

Question 24

A right triangle has legs of length aaa and bbb, and the angle opposite leg aaa measures θ\thetaθ. If the area of this triangle is 24 square units and tan⁡θ=34\tan \theta = \frac{3}{4}tanθ=43​, what is the length of the hypotenuse?

  1. 101010 units (correct answer)
  2. 888 units
  3. 121212 units
  4. 656\sqrt{5}65​ units

Explanation: From tan θ = 3/4, we have a/b = 3/4, so a = 3k and b = 4k for some positive k. The area is (1/2)ab = (1/2)(3k)(4k) = 6k² = 24, so k² = 4 and k = 2. Therefore a = 6 and b = 8. The hypotenuse is √(6² + 8²) = √(36 + 64) = √100 = 10. Choice B gives the length of leg b. Choice C assumes k = 2 incorrectly in the area formula. Choice D results from incorrectly using the Pythagorean theorem.

Question 25

A hyperbola has foci at F1(0,−6)F_1(0,-6)F1​(0,−6) and F2(0,6)F_2(0,6)F2​(0,6). For any point P(x,y)P(x,y)P(x,y) on the hyperbola, ∣PF2−PF1∣=8|PF_2-PF_1|=8∣PF2​−PF1​∣=8. Which equation represents the hyperbola? (No asymptotes are given.)

  1. y216−x220=1\dfrac{y^2}{16}-\dfrac{x^2}{20}=116y2​−20x2​=1 (correct answer)
  2. x216−y220=1\dfrac{x^2}{16}-\dfrac{y^2}{20}=116x2​−20y2​=1
  3. y220−x216=1\dfrac{y^2}{20}-\dfrac{x^2}{16}=120y2​−16x2​=1
  4. (y−6)216−x220=1\dfrac{(y-6)^2}{16}-\dfrac{x^2}{20}=116(y−6)2​−20x2​=1

Explanation: This problem asks for a vertical hyperbola equation from the focus definition. A hyperbola consists of points P(x,y) where the absolute difference of distances to two foci is constant. The foci F₁(0,-6) and F₂(0,6) are vertical with center at (0,0), and c = 6. The condition |PF₂ - PF₁| = 8 means 2a = 8, so a = 4. For hyperbolas, c² = a² + b², giving 36 = 16 + b², so b² = 20. Since the foci are vertical, the transverse axis is vertical, and the equation is y²/a² - x²/b² = 1, yielding y²/16 - x²/20 = 1. A common mistake is using the horizontal form x²/a² - y²/b² = 1 when foci are vertical. To avoid errors, identify the direction of the transverse axis from the foci positions before writing the equation.