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Geometry

Geometry Practice Test: Practice Test 8

Practice Test 8 for Geometry: real questions and explanations from the Varsity Tutors practice-test pool.

0 / 25 answered

Question 1 of 25

Use the figure. A plane cuts through the triangular prism parallel to the triangular base but not passing through any vertices. If the cut is made at $\frac{2}{5}$ of the height from the base, how does the resulting cross-section compare to the original triangular base?

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Question 1

Identical triangle with same area, angles, and side lengths as original base (correct answer)
Similar triangle with area $\frac{2}{5}$ of original and proportional side lengths
Similar triangle with side lengths $\frac{2}{5}$ of original and area $\frac{4}{25}$ of original
Congruent triangle rotated by specific angle but maintaining all original measurements

Explanation: In a prism, any cross-section parallel to the base is congruent to the base regardless of the height at which the cut is made. The cross-section maintains the same shape, size, area, and all measurements as the original base. Choice B incorrectly applies scaling rules from pyramids. Choice C also incorrectly applies pyramid scaling. Choice D incorrectly suggests rotation affects the measurements.

Question 2

What is the equation of the circle with center $(-2,-1)$ and radius $3$ ? (Justify using that all points on the circle are the same distance from the center.)

$(x+2)^2+(y+1)^2=3$
$(x-2)^2+(y-1)^2=9$
$(x+2)^2+(y+1)^2=9$ (correct answer)
$x^2+y^2=9$

Explanation: To write a circle's equation given its center and radius, we apply the geometric definition directly. A circle is the set of all points that are the same distance from a fixed center point. If the center is (-2, -1) and radius is 3, then any point (x, y) on the circle satisfies the distance formula: √[(x-(-2))² + (y-(-1))²] = 3. Squaring both sides gives (x+2)² + (y+1)² = 9. The standard form (x-h)² + (y-k)² = r² makes the center and radius immediately visible. A common error involves the signs: since we have center (-2, -1), the equation uses (x-(-2)) = (x+2) and (y-(-1)) = (y+1). When given center and radius, think of the distance relationship first, then translate to algebra.

Question 3

Which solid is formed by rotating a right triangle about one of its legs (that leg is the axis of rotation)?

A cone (correct answer)
A cylinder
A sphere
A prism

Explanation: This problem examines solids of revolution created by rotating 2D shapes. The original shape is a right triangle with one 90-degree angle and two perpendicular legs. The axis of rotation is one of the legs (perpendicular sides), around which the triangle spins. As the right triangle rotates about one leg, the hypotenuse traces out a slanted circular surface while the other leg sweeps from the axis outward to create the base. The resulting solid is a cone with the rotating leg as its height and the other leg determining the base radius. Students might incorrectly choose cylinder, which comes from rotating rectangles, not triangles. To understand this, imagine spinning a right triangular flag on its pole—the flag creates a cone shape.

Question 4

In right triangle $\triangle GHI$ shown, $\angle H$ is a right angle. The leg $GH=4$ and hypotenuse $GI=9$ . What is the measure of $\angle G$ (nearest degree)?

$\sin^{-1}\!\left(\dfrac{4}{9}\right)$
$\cos^{-1}\!\left(\dfrac{4}{9}\right)$ (correct answer)
$\tan^{-1}\!\left(\dfrac{4}{9}\right)$
$\sin\!\left(\dfrac{4}{9}\right)$

Explanation: This problem requires finding an angle when given two sides of a right triangle. We have leg GH = 4 (adjacent to angle G) and hypotenuse GI = 9. Since we have the adjacent side and hypotenuse, we use cosine: cos(G) = adjacent/hypotenuse = 4/9. To find angle G, we use the inverse cosine: G = cos⁻¹(4/9). This gives the angle measure in degrees. A common error is using regular cosine cos(4/9) instead of inverse cosine, or choosing the wrong trig function. When finding angles from sides, use inverse trig functions with the correct ratio.

Question 5

A point $P(5,2)$ is transformed by $T=\begin{pmatrix}-1&0\ 0&1\end{pmatrix}.$ What is the image point $P'$ ?

$(5,-2)$
$(-5,2)$ (correct answer)
$(-5,-2)$
$(2,5)$

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 $\begin{pmatrix} x \\ y \end{pmatrix}$ and multiply by transformation matrix $T = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ using matrix multiplication: $T \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix} = \begin{pmatrix} x' \\ y' \end{pmatrix}$ where $(x', y')$ is the transformed point. Common transformation matrices include: ROTATION by angle $θ$ counterclockwise = $\begin{pmatrix} \cos(θ) & -\sin(θ) \\ \sin(θ) & \cos(θ) \end{pmatrix}$ (example: 90° rotation uses $θ=90°$ giving $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ since $\cos(90°)=0$ and $\sin(90°)=1$ ), REFLECTION across x-axis = $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ (keeps x same, negates y), REFLECTION across y-axis = $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ (negates x, keeps y same), SCALING by factor k = $\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$ (multiplies both coordinates by k, enlarges by factor k). COMPOSITIONS of transformations: multiply matrices in reverse order (rightmost applied first)—to rotate then scale, compute (scaling matrix)·(rotation matrix). The resulting product matrix represents the combined transformation in one step! Applying $T = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ to $(5, 2)$ gives $x' = -1 \cdot 5 + 0 \cdot 2 = -5$ , $y' = 0 \cdot 5 + 1 \cdot 2 = 2$ , resulting in $(-5, 2)$ , a reflection across the y-axis. Choice B correctly performs the multiplication, negating only the x-coordinate as per the matrix. Distractors like choice A might forget the negation or confuse with x-axis reflection, but test with a point like $(1,0)$ to see it goes to $(-1,0)$ , confirming y-axis mirror. Matrix multiplication for transformations: Given transformation matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and point $(x, y)$ : (1) Write point as column vector $\begin{pmatrix} x \\ y \end{pmatrix}$ . (2) Multiply: first row of matrix times vector gives x'-coordinate = $a \cdot x + b \cdot y$ . Second row times vector gives y'-coordinate = $c \cdot x + d \cdot y$ . (3) Result is transformed point $(x', y') = (ax + by, cx + dy)$ . Example: $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ applied to $(5, 3)$ : $x' = 0 \cdot 5 + (-1) \cdot 3 = -3$ , $y' = 1 \cdot 5 + 0 \cdot 3 = 5$ , so image is $(-3, 5)$ . That's a 90° counterclockwise rotation! Identifying transformations from matrices: ROTATION matrices have form $\begin{pmatrix} \cos(θ) & -\sin(θ) \\ \sin(θ) & \cos(θ) \end{pmatrix}$ —look for this pattern with cos and -sin in first row, sin and cos in second row. Common: $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ is 90° rotation, $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ is 180° rotation. REFLECTION matrices have form $\begin{pmatrix} ±1 & 0 \\ 0 & ±1 \end{pmatrix}$ with exactly one negative— $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ reflects across x-axis (y negated), $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ reflects across y-axis (x negated). SCALING matrices have equal diagonal entries $\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$ —both coordinates multiplied by same k, or different entries $\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$ for non-uniform scaling. Pattern recognition allows identification without calculation! Checking: after transforming point, verify result makes sense geometrically (rotation should preserve distance from origin, reflection should mirror across axis, scaling should change distances proportionally). You're excelling at reflections—try visualizing the plane to reinforce!

Question 6

A cube is shown. A slicing plane cuts the cube so that it is perpendicular to the top face and parallel to one pair of opposite side faces.

Which shape results from the cross-section shown?

Rectangle (correct answer)
Triangle
Circle
Pentagon

Explanation: This problem involves finding cross-sections of three-dimensional solids, a key skill in geometry that also relates to solids of revolution. The original solid is a cube, with six square faces and equal edges. The slicing plane is perpendicular to the top face and parallel to one pair of opposite side faces. Cutting this way, the plane aligns with the cube's height and one direction, intersecting four faces. This results in a rectangle, matching the cube's face dimensions in height and width. A distractor misconception is thinking of a triangle, perhaps if angled, but the perpendicular and parallel orientation ensures a rectangle. To apply, visualize pushing the plane through the cube step by step, noting how it carves out the rectangular slice.

Question 7

A company is cutting a circular logo (radius 5 cm) from a square sticker sheet. The logo must fit entirely inside the square, and the square’s side length must be an integer number of centimeters to match a cutting template. Which square side length best meets the geometric requirements?

9 cm
10 cm (correct answer)
11 cm
12 cm

Explanation: This problem involves using geometry to determine the minimum square size to contain a circular logo. The constraint is that a circle of radius 5 cm must fit entirely inside a square with integer side length. For a circle to fit inside a square, the square's side must be at least as long as the circle's diameter. The diameter is 2(5) = 10 cm, so the square needs side length at least 10 cm. Since we need an integer value and 10 cm exactly fits the requirement, option B (10 cm) is correct. Option A (9 cm) is too small to contain the 10 cm diameter circle. Options C and D are larger than necessary but would work; however, the problem asks for the size that "best meets" the requirements, implying the minimum sufficient size. The strategy is to identify the minimum geometric requirement (diameter ≤ side) and apply it directly.

Question 8

A pilot flies from airport A to airport B (150 miles away) and then to airport C. The angle at airport B is $125°$ , and the distance from B to C is 200 miles. To find the direct distance from A to C using the Law of Cosines, what is the correct setup?

$AC^2 = 150^2 + 200^2 - 2(150)(200)\cos(125°)$ (correct answer)
$AC^2 = 150^2 + 200^2 - 2(150)(200)\cos(55°)$
$AC = \frac{150 \sin(125°)}{\sin C}$ where angle C is unknown
$AC^2 = 150^2 + 200^2 + 2(150)(200)\cos(125°)$

Explanation: The Law of Cosines states $c^2 = a^2 + b^2 - 2ab\cos C$ , where $C$ is the included angle between sides $a$ and $b$ . Here, the angle at B ( $125°$ ) is between the sides AB (150 miles) and BC (200 miles), so $AC^2 = 150^2 + 200^2 - 2(150)(200)\cos(125°)$ . Choice B incorrectly uses the supplement of $125°$ . Choice C attempts to use Law of Sines without sufficient information. Choice D has the wrong sign in the formula.

Question 9

Circle $\odot M$ has center $M$ and Circle $\odot N$ has center $N$ . The circles have different radii, and the centers are marked. Which statement explains why the circles are similar without using circumference or area formulas?

They are similar because both are circles, and a dilation can scale $\odot M$ to match the radius of $\odot N$ . (correct answer)
They are similar because their radii are different, so they cannot be congruent.
They are similar because $\pi$ is the same for all circles, so the circles must be similar.
They are similar because the dilation must be centered at $N$ to map $\odot M$ to $\odot N$ exactly.

Explanation: The skill here is understanding circle similarity in geometry. Circles are similar because a dilation can scale the radius of one to match the other, preserving the shape. The centers M and N are referenced to consider how transformations affect positioning. Applying a dilation, possibly combined with a translation if needed, maps ⊙M to a circle matching ⊙N in size. This justifies similarity as all circles share the same shape, scalable via such transformations. A common distractor is option D, which incorrectly specifies the dilation center at N for an exact mapping without considering general cases. To transfer this strategy, think in terms of transformations like dilations and translations, not formulas.

Question 10

Triangle LMN has vertices $L(-2,0)$ , $M(2,0)$ , and $N(1,3)$ on the coordinate plane. Which value correctly represents the area?

$\frac{1}{2}\cdot 4\cdot 3=6$ (correct answer)
$4+\sqrt{(1-2)^2+(3-0)^2}+\sqrt{(1+2)^2+(3-0)^2}$
$\frac{1}{2}\cdot 3\cdot 3=4.5$
$\frac{1}{2}\cdot 2\cdot 3=3$

Explanation: The skill is finding the area of a triangle using coordinates. The vertices are L(-2,0), M(2,0), and N(1,3). Side lengths are computed using the distance formula, but base and height are evident. The area logic is half base times height. With base 4 and height 3, this gives 6, justifying the value in the correct choice. A distractor misconception is using incorrect height, as in choice D with 2. To transfer this strategy, identify base and height or use shoelace after noting coordinates.

Question 11

In the diagram, $\triangle PQR$ is a right triangle with $\angle Q$ marked as $90^\circ$ . The acute angles are labeled $\angle P=\theta$ and $\angle R=\varphi$ (so they are complementary). Which identity follows from the diagram (using right-triangle definitions and the fact that the two acute angles are complementary)?

$\cos(\theta)=\sin(\varphi)$ (correct answer)
$\cos(\theta)=\cos(\varphi)$
$\cos(\theta)=\tan(\varphi)$
$\cos(\theta)=\sin(90^\circ)$

Explanation: The skill here is understanding the sine-cosine relationship for complementary angles in a right triangle. In right triangle PQR with right angle at Q, angles θ at P and φ at R are complementary because their sum is 90 degrees, as the third angle is 90 degrees. For angle θ at P, the opposite side is QR, the adjacent side is PQ, and the hypotenuse is PR; for angle φ at R, the opposite side is PQ, the adjacent side is QR, and the hypotenuse is PR. Cosine of θ is adjacent over hypotenuse (PQ/PR), while sine of φ is opposite over hypotenuse (PQ/PR), showing they are equal. Therefore, cos(θ) = sin(φ), which follows from the diagram as the identity in choice A. A common distractor misconception is thinking cos(θ) = cos(φ), but since θ and φ are different, their cosines are generally not equal. To transfer this strategy, redraw the triangle and label the opposite and adjacent sides relative to each angle to see how they swap roles.

Question 12

Two right triangles $\triangle ABC$ and $\triangle A'B'C'$ are shown. Each has a right angle at $C$ and $C'$ respectively, and $\angle A \cong \angle A'$ . Which relationship depends only on the angle and not on the size of the triangle?

$\dfrac{BC}{AB}=\sin(\angle A)$ (correct answer)
$AB-BC$ is the same in both triangles
$AB$ is the same in both triangles
$BC+AC$ is the same in both triangles

Explanation: This problem explores which relationships remain constant in similar right triangles. When triangles ABC and A'B'C' both have right angles at C and C' respectively, and angle A is congruent to angle A', the triangles are similar by AA similarity. In any right triangle with angle A, the sine of A equals the ratio of the opposite side to the hypotenuse, which is BC/AB. This ratio depends only on angle A, not on the triangle's size, because similar triangles have proportional sides. Options B, C, and D involve specific lengths or sums that change with triangle size and are not ratios. The key insight is that trigonometric ratios are defined through similarity to be angle-dependent but size-independent. Students often mistakenly think that individual side lengths or their differences remain constant, but only ratios of sides are preserved under similarity.

Question 13

A small museum is installing a rectangular display case inside a rectangular alcove that is $9\text{ ft}$ wide and $6\text{ ft}$ deep. Building code requires a clear walkway of at least $1\text{ ft}$ along each wall of the alcove (left, right, top, and bottom). Which design satisfies all constraints?

Case $7\text{ ft} \times 4\text{ ft}$ , centered in the alcove. (correct answer)
Case $8\text{ ft} \times 4\text{ ft}$ , centered in the alcove.
Case $7\text{ ft} \times 5\text{ ft}$ , centered in the alcove.
Case $8\text{ ft} \times 5\text{ ft}$ , centered in the alcove.

Explanation: This problem requires using geometry to design a display case that fits within spatial constraints. The alcove is 9 ft wide by 6 ft deep, and we need 1 ft clearance along each wall. To find the maximum case dimensions, subtract 2 ft from each alcove dimension (1 ft clearance on each side): width = 9 - 2 = 7 ft, depth = 6 - 2 = 4 ft. Option A (7 ft × 4 ft) exactly meets these maximum dimensions while satisfying all clearance requirements. Option B violates the width constraint (8 ft is too wide), while options C and D violate the depth constraint (5 ft is too deep). A common mistake is subtracting only 1 ft total instead of 1 ft from each side. When solving design problems with clearance requirements, always account for clearance on all sides by subtracting twice the clearance distance from each dimension.

Question 14

Rhombus $DEFG$ has vertices $D(1, 2)$ , $E(4, 6)$ , $F(8, 3)$ , and $G(5, -1)$ . To verify this quadrilateral is indeed a rhombus using coordinate geometry, which two properties must be proven?

All four sides are congruent, and opposite angles are supplementary to adjacent angles
All four sides are congruent, and the diagonals bisect each other at right angles (correct answer)
Opposite sides are parallel and congruent, and all four angles measure $90°$
Opposite sides are parallel and congruent, and the diagonals are congruent in length

Explanation: A rhombus is defined as a quadrilateral with all four sides congruent. Additionally, the diagonals of a rhombus bisect each other at right angles. Let's verify: |DE| = √[(4-1)² + (6-2)²] = √[9+16] = 5. |EF| = √[(8-4)² + (3-6)²] = √[16+9] = 5. |FG| = √[(5-8)² + (-1-3)²] = √[9+16] = 5. |GD| = √[(1-5)² + (2-(-1))²] = √[16+9] = 5. All sides are congruent. Diagonals DF and EG intersect at ((1+8)/2, (2+3)/2) = (4.5, 2.5) and ((4+5)/2, (6+(-1))/2) = (4.5, 2.5), confirming they bisect each other. Slope of DF = (3-2)/(8-1) = 1/7. Slope of EG = (-1-6)/(5-4) = -7. Since (1/7)(-7) = -1, diagonals are perpendicular. Choice A is wrong because supplementary angles aren't the defining property. Choice C describes a rectangle. Choice D describes a rectangle where diagonals are congruent.

Question 15

Two solids are compared: (1) a sphere of radius $r$ , and (2) a cylinder of radius $r$ and height $2r$ with two congruent cones removed (each cone has base radius $r$ and height $r$ , with tips meeting at the center). Slices are taken by horizontal planes at height $h$ from the center. Which claim is NOT supported by the slicing argument used with Cavalieri’s principle?

If the cross-sectional areas match for every height $h$ , then the volumes match.
If the cross-sectional areas match at a single height $h$ , then the volumes match. (correct answer)
At each height $h$ , comparing the sphere’s disk to the cylinder’s disk minus two cone disks can justify equal volumes.
The argument depends on taking slices at corresponding heights measured from the same reference level (the center).

Explanation: This question asks which claim is NOT supported by the Cavalieri's principle argument for sphere volume. The comparison involves a sphere and a cylinder minus two cones, with slices at height h from the center. Option A correctly states that matching cross-sectional areas at every height implies equal volumes (Cavalieri's principle). Option C correctly describes the comparison of disk areas. Option D correctly emphasizes measuring from the same reference level. However, option B incorrectly claims that matching areas at just a single height suffices for equal volumes - this is false, as Cavalieri's principle requires matching at ALL heights. The misconception in B is thinking one cross-section determines the entire volume relationship. The correct application requires verifying equal cross-sectional areas throughout the entire range of heights.

Question 16

Triangle $PQR$ is shown in the plane. Point $M$ lies on segment $PQ$ and point $N$ lies on segment $PR$ . The diagram marks $PM \cong MQ$ (matching tick marks on the two parts of $PQ$ ) and $PN \cong NR$ (matching tick marks on the two parts of $PR$ ). Segment $MN$ is drawn. No angle measures, no parallel markings, and no lengths are given, and the diagram is not drawn to scale.

Which statement must be true?

$MN \parallel QR$ (correct answer)
$MN \perp QR$
$M$ is the midpoint of $QR$
$MN \cong QR$

Explanation: The skill involves theorems about triangles, focusing on properties of segments connecting midpoints. The midsegment theorem states that a segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length. The diagram identifies points M and N as midpoints of PQ and PR, respectively, with matching tick marks confirming PM congruent to MQ and PN congruent to NR. Applying the theorem, segment MN connects these midpoints, so it must be parallel to the third side QR. This conclusion is justified as the midsegment theorem directly applies to midpoints on two sides, ensuring parallelism. A distractor misconception might involve assuming perpendicularity without any right-angle indicators. To approach similar diagrams, match midpoint markings to theorems like the midsegment theorem for parallelism or length relationships.

Question 17

A city park is modeled as a rectangle $1.5$ miles long and $0.8$ miles wide. The park has a population density of $2{,}400$ people per square mile during a festival. Which calculation gives the total number of people in the park?

$2{,}400 \times (1.5+0.8)$
$2{,}400 \div (1.5\times 0.8)$
$2{,}400 \times (1.5\times 0.8)$ (correct answer)
$(2{,}400\times 1.5)+0.8$

Explanation: This problem requires applying density in geometric modeling to find the total number of people in a rectangular park. Density is a measure of how much of something exists per unit of space—in this case, 2,400 people per square mile. The relevant geometric measure is the park's area, which equals length times width: 1.5 × 0.8 square miles. To find the total number of people, we multiply the density by the area: 2,400 × (1.5 × 0.8). This gives us the total because density times area equals the total amount. A common misconception is adding the dimensions (1.5 + 0.8) instead of multiplying them, which would give perimeter-based thinking rather than area. To avoid errors, always check that your units work out: people/square mile × square miles = people.

Question 18

Triangle $UVW$ is mapped onto triangle $U'V'W'$ on the coordinate plane by two rigid transformations. Order matters. Which sequence correctly produces the image?

Rotate $180^\circ$ about the origin, then translate up 1 unit. (correct answer)
Translate up 1 unit, then rotate $180^\circ$ about the origin.
Rotate $90^\circ$ clockwise about the origin, then translate up 1 unit.
Rotate $180^\circ$ about the origin only.

Explanation: This is a sequencing rigid transformations problem where we need to map triangle UVW onto triangle U'V'W'. The transformations needed are a rotation and a translation. The order matters significantly because rotations about the origin change both orientation and position. The correct sequence is to rotate 180° about the origin first, then translate up 1 unit. This works because the 180° rotation flips the triangle to the opposite side of the origin with reversed orientation, then the translation moves it up slightly to the final position. If we translated up first then rotated (choice B), the elevated triangle would rotate to the wrong final position because the rotation would move it in a semicircle from its elevated starting point. The principle is to complete rotations about the origin before translations for predictable results.

Question 19

Triangle ABC has angles measuring 45°, 60°, and 75°. Triangle DEF undergoes a sequence of similarity transformations (rotation, reflection, and dilation with scale factor 3) to produce triangle GHI. If triangle GHI has angles measuring 45°, 60°, and 75°, which statement about the relationship between triangles ABC and GHI is most accurate?

The triangles are similar by AA criterion, and the transformations preserve angle measures while scaling side lengths proportionally (correct answer)
The triangles are congruent because similarity transformations always preserve both angle measures and side lengths exactly
The triangles are not similar because the dilation changes the side lengths, violating the definition of similarity
The triangles are similar only if the rotation and reflection map corresponding vertices to the same relative positions

Explanation: Since similarity transformations preserve angle measures, triangles ABC and GHI have all corresponding angles equal (45°, 60°, 75°). By the AA criterion, having two pairs of equal corresponding angles guarantees similarity (the third pair is automatically equal). The dilation scales all sides proportionally while preserving angles, which is exactly what defines similarity. Choice B is wrong because similarity transformations don't preserve side lengths exactly (only proportionally). Choice C incorrectly suggests that changing side lengths violates similarity. Choice D incorrectly focuses on vertex positioning rather than angle equality.

Question 20

9 cm
10 cm (correct answer)
11 cm
12 cm

Question 21

Quadrilateral $RSTU$ has vertices $R(-3,-2)$ , $S(1,0)$ , $T(3,-4)$ , and $U(-1,-6)$ . A student claims $RSTU$ is a parallelogram. Which argument correctly uses coordinate geometry?

Show $RS=TU$ and $ST=UR$ , so opposite sides are parallel.
Show $m_{RS}=m_{TU}$ and $m_{ST}=m_{UR}$ , so opposite sides are parallel. (correct answer)
Show $m_{RS}\cdot m_{ST}=-1$ , so opposite sides are parallel.
Show $RT=SU$ , so the diagonals are parallel and it is a parallelogram.

Explanation: Coordinate proofs establish parallelogram properties using coordinate-based calculations. The student claims RSTU is a parallelogram. To verify, we translate the property into algebraic conditions where opposite sides have equal slopes. Applying slope reasoning, m_RS = 1/2 = m_TU and m_ST = -2 = m_UR. This justifies opposite sides are parallel, confirming the parallelogram. A distractor misconception is using perpendicular slopes to imply parallelism. The transfer strategy is converting geometric parallelism into equations by setting opposite side slopes equal.

Question 22

A party cone is filled with candy. The cone has radius $6\text{ cm}$ and height $10\text{ cm}$ . Which calculation correctly applies the volume formula?

$V=\pi(6)^2(10)$
$V=\tfrac{1}{3}\pi(6)^2(10)$ (correct answer)
$V=\tfrac{4}{3}\pi(6)^3$
$V=2\pi(6)(10)+2\pi(6)^2$

Explanation: This problem involves finding the volume of a party cone filled with candy. The solid is a cone with radius 6 cm and height 10 cm. The volume formula for a cone is V = (1/3)πr²h, where r is the radius and h is the height. The correct calculation is V = (1/3)π(6)²(10), which matches option B. This formula gives one-third the volume of a cylinder with the same base and height. Option A incorrectly uses the cylinder formula without the 1/3 factor, while option D uses the surface area formula. To solve volume problems correctly, first identify whether the solid is a cone, cylinder, or sphere before selecting the appropriate formula.

Question 23

A metal rod is modeled as a cylinder with radius $2\text{ cm}$ and height $15\text{ cm}$ . The metal has density $7.8\ \text{g/cm}^3$ . Which calculation gives the mass of the rod?

$7.8 \times (2\times 15)$
$7.8 \times \pi(2^2)(15)$ (correct answer)
$7.8 \div \pi(2^2)(15)$
$7.8 + \pi(2^2)(15)$

Explanation: This problem requires applying density in geometric modeling to find the mass of a cylindrical rod. Density measures mass per unit volume—in this case, 7.8 grams per cubic centimeter. The relevant geometric measure is the cylinder's volume, given by the formula π × radius² × height: π(2²)(15) cubic centimeters. To find the rod's mass, multiply the density by the volume: 7.8 × π(2²)(15). This calculation works because density times volume equals total mass. A common mistake is using the wrong volume formula, such as 2 × 15 (treating it as a rectangle), which ignores the circular cross-section. Always verify units match: g/cm³ × cm³ = g, confirming you're calculating mass correctly.

Question 24

A class is discussing why the volume of a sphere of radius $r$ can be justified without calculus by comparing it to a cylinder of radius $r$ and height $2r$ with two congruent cones removed (each cone has base radius $r$ and height $r$ ). Which statement justifies the sphere’s volume formula?

If the cross-sectional areas are equal for every height, then the volumes are equal by Cavalieri’s principle. (correct answer)
If the cross-sectional circumferences are equal for every height, then the volumes are equal by Cavalieri’s principle.
If the solids share the same top and bottom points, then the volumes are equal by Cavalieri’s principle.
If the formula for the sphere is memorized, then the comparison is automatically justified.

Explanation: This final question reinforces the core principle behind the sphere volume derivation. Cavalieri's principle states that if two solids have equal cross-sectional areas when sliced by parallel planes at every height, then they have equal volumes. For the sphere and cylinder-minus-cones comparison, at height h the areas are both π(r²-h²), confirming equal volumes. Option A correctly states this principle. Option B incorrectly focuses on circumferences rather than areas, C wrongly emphasizes endpoint alignment, and D avoids the geometric reasoning entirely. The key misconception to address is confusing other measurements (like perimeter) with area - Cavalieri's principle specifically requires equal cross-sectional areas at all heights for the volume conclusion.

Question 25

In the right triangle shown, the acute angle at $A$ is $\theta$ . The hypotenuse is labeled $1$ , the leg adjacent to $\theta$ is labeled $\cos\theta$ , and the leg opposite $\theta$ is labeled $\sin\theta$ . Which reasoning proves $\sin^2\theta+\cos^2\theta=1$ ?

Since $\sin\theta+\cos\theta=1$ for a unit hypotenuse, squaring gives $\sin^2\theta+\cos^2\theta=1$ .
Using $\sin\theta=\frac{\text{adjacent}}{1}$ and $\cos\theta=\frac{\text{opposite}}{1}$ , then $\sin^2\theta+\cos^2\theta=1$ .
By the Pythagorean Theorem, $(\cos\theta)^2+(\sin\theta)^2=1^2$ , so $\sin^2\theta+\cos^2\theta=1$ . (correct answer)
Because the triangle looks like a $45^\circ$ - $45^\circ$ - $90^\circ$ triangle, $\sin^2\theta+\cos^2\theta=1$ .

Explanation: This question asks you to prove the Pythagorean identity sin²θ + cos²θ = 1 using a right triangle with hypotenuse 1. In a right triangle, sine is defined as opposite/hypotenuse and cosine as adjacent/hypotenuse. When the hypotenuse equals 1, we get sin θ = opposite/1 = opposite and cos θ = adjacent/1 = adjacent. By the Pythagorean Theorem, (adjacent)² + (opposite)² = (hypotenuse)² = 1². Therefore, (cos θ)² + (sin θ)² = 1, which gives us sin²θ + cos²θ = 1. Choice A incorrectly assumes sin θ + cos θ = 1 without justification, while Choice B reverses the definitions of sine and cosine. To verify this identity for any angle, always return to the right triangle definition and apply the Pythagorean Theorem.