Geometry Practice Test: Practice Test 4

Question 1

1. $15$ square units (correct answer)
2. $20$ square units
3. $30$ square units
4. $45$ square units

Explanation: The triangle formed by connecting the midpoints of any triangle is similar to the original triangle with a similarity ratio of $1:2$. This follows from the theorem that a line connecting midpoints of two sides of a triangle is parallel to the third side and half its length. Since areas of similar figures are in the ratio of the square of the similarity ratio, the area ratio is $(1/2)^2 = 1/4$. Therefore, the area of the midpoint triangle is $60 \cdot \frac{1}{4} = 15$ square units.

Question 2

1. $\overline{A'B'}$ is $\tfrac{2}{3}$ as long as $\overline{AB}$.
2. $\overline{A'B'}$ is $\tfrac{3}{2}$ as long as $\overline{AB}$. (correct answer)
3. $\overline{A'B'}$ is the same length as $\overline{AB}$.
4. $\overline{A'B'}$ is longer than $\overline{AB}$ by $\tfrac{1}{2}$ unit.

Explanation: This question tests understanding of how dilations with scale factor k = 3/2 affect segment lengths. In a dilation, the scale factor multiplies all distances from the center, which means each segment length is multiplied by the scale factor. Since triangle ABC is dilated to form triangle A'B'C', the corresponding segments are AB and A'B'. Applying the scale factor k = 3/2, we get A'B' = (3/2) × AB, which means A'B' is 3/2 as long as AB. This matches answer choice B, confirming that the image segment is 1.5 times the original length. A common misconception (choice A) is to think the reciprocal 2/3 applies, but dilations multiply lengths by the scale factor, not its reciprocal. To solve dilation problems, always multiply the original length by the scale factor to find the image length.

Question 3

1. Student 1, because any consistent height measurement works even if the solids are centered differently.
2. Student 2, because the same height reference aligns corresponding cross-sections in both solids. (correct answer)
3. Both, because matching cross-sections at two heights is enough to guarantee equal volume.
4. Neither, because Cavalieri’s principle requires knowing the sphere’s volume formula first.

Explanation: The skill is deriving the volume of a sphere using an informal argument based on Cavalieri’s principle. The comparison involves a sphere of radius r and a cylinder of radius r and height 2r from which two congruent cones, each of radius r and height r, have been removed, with the cones' tips meeting at the cylinder's center. For any height h from the center, the cross-sectional area of the sphere equals the cross-sectional area of the cylinder minus the cross-sectional areas of the cones at that height, where typically only one cone contributes a non-zero area. By Cavalieri’s principle, solids with the same height and identical cross-sectional areas at every corresponding level have the same volume. This equality of cross-sections at every h establishes that the volume of the sphere equals the volume of the cylinder minus the volumes of the two cones. A distractor misconception is thinking that measuring heights from different references like the bottom still aligns cross-sections properly, but a common center is needed for correspondence. When applying this strategy to other volume derivations, always compare cross-sections at the same height from a consistent reference point, such as the center for symmetric solids.

Question 4

1. A reflection across the line $y = 2$ followed by a reflection across $x = 4$
2. A $180°$ rotation about point $H$ only
3. A $90°$ rotation about point $H$ since the rhombus has four-fold symmetry
4. A reflection across the line containing diagonal $DF$ or diagonal $EG$ (correct answer)

Explanation: When analyzing transformations that carry a figure onto itself, you're looking for the symmetries of that shape. A rhombus has specific symmetry properties that determine which transformations preserve it. First, let's find where the diagonals intersect. Diagonal $DF$ connects $(0,0)$ to $(8,4)$, and diagonal $EG$ connects $(3,4)$ to $(5,0)$. The intersection point $H$ is at $(4,2)$. In any rhombus, the diagonals bisect each other at right angles, creating natural lines of reflection symmetry. The correct answer is D because a rhombus has exactly two lines of reflection symmetry: the lines containing its diagonals. When you reflect the rhombus across diagonal $DF$ or diagonal $EG$, each vertex maps to another vertex of the rhombus, carrying the figure onto itself. Choice A combines two reflections that don't correspond to the rhombus's natural symmetries. The lines $y = 2$ and $x = 4$ pass through point $H$ but aren't the diagonal lines. Choice B is partially correct—a $180°$ rotation about $H$ does map the rhombus onto itself—but it's incomplete since reflection symmetries also work. Choice C incorrectly assumes the rhombus has four-fold rotational symmetry. Only squares have $90°$ rotational symmetry; general rhombuses only have $180°$ rotational symmetry. Strategy tip: Remember that rhombuses have exactly three types of symmetries: two diagonal reflections and one $180°$ rotation about the center. Don't confuse rhombus symmetries with square symmetries.

Question 5

1. $\frac{x^2}{25} + \frac{y^2}{16} = 1$ (correct answer)
2. $\frac{x^2}{16} + \frac{y^2}{25} = 1$
3. $\frac{x^2}{25} + \frac{y^2}{9} = 1$
4. $\frac{x^2}{9} + \frac{y^2}{25} = 1$

Explanation: For an ellipse, the sum of distances from any point to the two foci is constant. Using point (0,4): distance from (0,4) to (-3,0) is √(9+16) = 5, and distance from (0,4) to (3,0) is √(9+16) = 5. So 2a = 10, thus a = 5. Since c = 3 (distance from center to focus), we have b² = a² - c² = 25 - 9 = 16, so b = 4. The equation is x²/25 + y²/16 = 1. Choice B swaps a and b values. Choice C uses b² = 9 instead of 16. Choice D incorrectly uses a = 3 and b = 5.

Question 6

1. Amplitude $=70$ ft, midline $=40$ ft, period $=10$ min
2. Amplitude $=35$ ft, midline $=40$ ft, period $=10$ min (correct answer)
3. Amplitude $=35$ ft, midline $=75$ ft, period $=10$ min
4. Amplitude $=35$ ft, midline $=40$ ft, period $=5$ 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 the wheel point height from 5 ft to 75 ft with 10-minute rotation, amplitude is (75 - 5)/2 = 35 ft, midline is (75 + 5)/2 = 40 ft, and period is 10 minutes. Choice B correctly identifies these parameters by properly calculating amplitude as half the range, midline as the average, and period from rotation time. Choice A uses full range for amplitude, choice C sets midline near max, and choice D halves the period. 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 7

1. $(4\pi)(6^2)$
2. $\frac{4}{3}\pi(6^3)$ (correct answer)
3. $\pi(6^2)(6)$
4. $\frac{1}{3}\pi(6^2)(6)$

Explanation: This problem asks which calculation correctly applies the volume formula for a sphere. The solid is a sphere with radius 6 cm. The volume formula for a sphere is V = (4/3)πr³, where r is the radius. The correct calculation is V = (4/3)π(6³) = (4/3)π(216). This gives the volume in cubic centimeters. Option C represents the formula for a cylinder (πr²h), which is incorrect for a sphere. To solve volume problems accurately, first identify the three-dimensional shape before selecting the appropriate formula.

Question 8

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

Explanation: This question tests understanding of cross-sections and solids of revolution in geometry. The original solid is a right circular cylinder with circular bases. The slicing plane is horizontal and parallel to the circular bases. When the plane intersects the cylinder, it cuts through the curved surface at points equidistant from the axis. This results in a circular cross-section identical in shape to the bases but possibly smaller if not at the ends. A common misconception is thinking it forms an ellipse, which occurs only with tilted planes. To visualize, imagine slicing step by step parallel to the base to see the consistent circular shape.

Question 9

1. Endpoints $(-6,0)$ and $(6,0)$
2. Endpoints $(-3, \sqrt{27})$ and $(3, \sqrt{27})$
3. Endpoints $(-4, \sqrt{20})$ and $(4, \sqrt{20})$
4. Endpoints $(-5, \sqrt{11})$ and $(5, \sqrt{11})$ (correct answer)

Explanation: This problem involves using geometry to design a footbridge as a chord of a circular fountain. The constraints are: the fountain has radius 6 m centered at origin, the bridge must be a chord with length between 10 m and 11 m. For a horizontal chord at height h, the endpoints are at $(\pm \sqrt{36 - h^2}, h)$, giving chord length $2 \sqrt{36 - h^2}$. We need $10 \leq 2 \sqrt{36 - h^2} \leq 11$, which means $25 \leq 36 - h^2 \leq 30.25$, so $5.75 \leq h^2 \leq 11$. Option D has endpoints at $(\pm 5, \sqrt{11})$, giving chord length $2(5) = 10$ m, which satisfies the constraints. Option A gives length 12 m (too long), option B gives length 6 m (too short), and option C gives length 8 m (too short). The strategy is to verify both that points lie on the circle and that the chord length meets requirements.

Question 10

1. Knowing that $\angle J = 42°$ and $\angle K = 85°$, since similarity transformations preserve all angle measures completely (correct answer)
2. Knowing that $\angle J = 42°$ and $\angle K = 85°$, plus verification that the reflection line and dilation center are properly positioned
3. Knowing any two angle measures in triangle JKL, since the transformation process guarantees that corresponding angles will be equal
4. Knowing that $\angle J = 42°$ and $\angle K = 85°$, plus confirmation that the scale factor k produces proportional side lengths

Explanation: Similarity transformations always preserve angle measures, so if $\angle J' = 42°$ and $\angle K' = 85°$ in the image triangle, then $\angle J = 42°$ and $\angle K = 85°$ in the original triangle. This gives us two pairs of equal corresponding angles, satisfying the AA criterion. Choice B incorrectly suggests that geometric positioning affects angle preservation. Choice C is wrong because we need the specific angles to match, not just any two angles. Choice D adds unnecessary verification since proportional sides are guaranteed by the AA criterion establishing similarity.

Question 11

1. Match slice areas at the same height $h$; then the sphere’s volume equals the cylinder’s volume minus the two cones’ volumes. (correct answer)
2. Match slice circumferences at the same height $h$; then the sphere’s volume equals the cylinder’s volume minus the two cones’ volumes.
3. Match slice areas but allow different heights in the two solids; then the volumes are still equal.
4. Use the known sphere volume formula directly; no cross-sectional comparison is needed.

Explanation: This question tests proper application of Cavalieri's principle to derive the sphere's volume formula. The correct approach compares a sphere of radius r to a cylinder (radius r, height 2r) with two cones removed by matching cross-sectional areas at the same height h. At height h, the sphere's cross-section has area π(r²-h²), while the cylinder minus the two cone cross-sections yields πr² - πh² = π(r²-h²). Since these areas match at every height h in [-r,r], Cavalieri's principle confirms the volumes are equal. Option B incorrectly uses circumferences instead of areas, C allows different heights which violates the principle's requirement, and D avoids the comparison entirely. The misconception in B stems from confusing perimeter with area in volume calculations. The transfer strategy emphasizes matching areas, not perimeters, at corresponding heights.

Question 12

1. Use the product-to-sum formulas to simplify the right side first
2. Convert the right side to exponential form and compare with the left side
3. Expand both $\sin(x + y)$ and $\sin(x - y)$ using their respective formulas, then add the results (correct answer)
4. Apply the Pythagorean identity to both sides before expanding the angle formulas

Explanation: When you encounter trigonometric identities involving angle addition and subtraction, the most straightforward verification approach is to expand the complex expressions using fundamental formulas and see if they simplify to match. To verify this identity, you should expand the left side using the angle addition and subtraction formulas: $\sin(x + y) = \sin x \cos y + \cos x \sin y$ and $\sin(x - y) = \sin x \cos y - \cos x \sin y$. Adding these expressions gives you: $\sin(x + y) + \sin(x - y) = (\sin x \cos y + \cos x \sin y) + (\sin x \cos y - \cos x \sin y)$ The $\cos x \sin y$ terms cancel out, leaving $2\sin x \cos y$, which matches the right side exactly. Answer C is correct because it takes the most direct path to verification. Answer A is backwards—product-to-sum formulas would complicate rather than simplify $2\sin x \cos y$. Answer B introduces unnecessary complexity; while exponential forms work, they're far more elaborate than needed for this straightforward identity. Answer D makes no sense because the Pythagorean identity ($\sin^2 θ + \cos^2 θ = 1$) doesn't apply to sums of different angles. Remember: when verifying trigonometric identities, always choose the path that directly applies the most relevant formulas. For expressions involving $\sin(x ± y)$, immediately think of the angle addition and subtraction formulas—they're usually your fastest route to a solution.

Question 13

1. $\dfrac{(x-3)^2}{81}+\dfrac{(y-4)^2}{65}=1$ (correct answer)
2. $\dfrac{(x-3)^2}{65}+\dfrac{(y-4)^2}{81}=1$
3. $\dfrac{(x-3)^2}{81}-\dfrac{(y-4)^2}{65}=1$
4. $\dfrac{(x+1)^2}{81}+\dfrac{(y-4)^2}{65}=1$

Explanation: This problem involves deriving an ellipse equation with both horizontal and vertical shifts. An ellipse is defined by points P(x,y) where PF₁ + PF₂ equals a constant. The foci F₁(-1,4) and F₂(7,4) are horizontal with center at (3,4), and the distance between foci is 8, so c = 4. The condition PF₁ + PF₂ = 18 gives 2a = 18, so a = 9. Using a² = b² + c² for ellipses, we get 81 = b² + 16, so b² = 65. For a horizontal ellipse centered at (h,k), the equation is (x-h)²/a² + (y-k)²/b² = 1, yielding (x-3)²/81 + (y-4)²/65 = 1. A common mistake is incorrectly calculating the center - it must be the midpoint of the foci. Always verify your center coordinates before writing the shifted equation form.

Question 14

1. $\dfrac{(x+2)^2}{9}-\dfrac{(y+1)^2}{7}=1$ (correct answer)
2. $\dfrac{(x+2)^2}{9}+\dfrac{(y+1)^2}{7}=1$
3. $\dfrac{(y+1)^2}{9}-\dfrac{(x+2)^2}{7}=1$
4. $\dfrac{(x+2)^2}{16}-\dfrac{(y+1)^2}{7}=1$

Explanation: This problem asks for the equation of a hyperbola given its foci and distance difference. A hyperbola consists of all points P(x,y) where |PF₂ - PF₁| is constant. The foci F₁(-6,-1) and F₂(2,-1) share the same y-coordinate, making this a horizontal hyperbola centered at ((-6+2)/2, -1) = (-2,-1). Given |PF₂ - PF₁| = 6, we have 2a = 6, so a = 3 and a² = 9. The focal distance is 2c = 8, so c = 4, and for hyperbolas b² = c² - a² = 16 - 9 = 7, yielding (x+2)²/9 - (y+1)²/7 = 1. Students often mistakenly add the squared terms instead of subtracting. Remember that hyperbolas have a difference of squares in their standard form.

Question 15

1. A dilation centered at $O$ with scale factor $4$ maps the smaller circle to the larger circle. (correct answer)
2. They are congruent because they share the same center $O$.
3. They are similar because the larger circle has circumference $4$ times the smaller circle’s circumference.
4. They are similar because the larger circle surrounds the smaller circle in the picture.

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 share the same center O with radii 5 and 20. Applying a dilation centered at O with scale factor 4 maps the smaller circle directly to the larger one. This justifies similarity because the dilation, being a similarity transformation, maps one circle exactly onto the other while preserving shape. A distractor like choice B claims congruence due to the shared center, but different radii mean they are not congruent. To approach similar problems, think in terms of transformations like dilations rather than relying on formulas.

Question 16

1. Square the ratios and add: $\left(\dfrac{BC}{AB}\right)^2+\left(\dfrac{AC}{AB}\right)^2=\dfrac{BC^2+AC^2}{AB^2}=\dfrac{AB^2}{AB^2}=1$. (correct answer)
2. Square the ratios and add: $\left(\dfrac{BC}{AB}\right)^2+\left(\dfrac{AC}{AB}\right)^2=\dfrac{BC^2+AC^2}{AB}=1$.
3. Use $AB^2+AC^2=BC^2$ to get $\sin^2\theta+\cos^2\theta=1$.
4. Since the triangle is right, $\sin\theta=\cos\theta$, so $\sin^2\theta+\cos^2\theta=1$.

Explanation: The Pythagorean identity states that sin²θ + cos²θ = 1 for any angle θ. In a right triangle, sine of θ is opposite over hypotenuse, and cosine is adjacent over hypotenuse. Squaring the ratios yields sin²θ = (opposite / hypotenuse)² and cos²θ = (adjacent / hypotenuse)². Applying the Pythagorean theorem, opposite² + adjacent² = hypotenuse². Dividing by hypotenuse² gives sin²θ + cos²θ = 1. A common distractor is dividing by hypotenuse instead of its square, leading to incorrect equality. To reinforce, always consult the right triangle geometry for accurate ratio application.

Question 17

1. The vertex is at $(3,1)$. (correct answer)
2. The vertex is at $(1,3)$.
3. The vertex is at $(5,3)$.
4. The vertex is at $(3,5)$.

Explanation: The skill involves deriving the equation of a parabola given 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 on the parabola, the distance to the focus F(5,1) equals the distance to the directrix x=1. The vertex is the midpoint between the focus and directrix, calculated as x=(5+1)/2=3 and y=1. This identifies the vertex at (3,1), consistent with the parabola opening to the right. A distractor misconception is miscalculating the midpoint, such as averaging y-coordinates incorrectly leading to (5,3). To derive equations for other parabolas, always start from the distance definition and simplify step by step.

Question 18

1. Amplitude $=34$ ft, period $=14$ min (correct answer)
2. Amplitude $=68$ ft, period $=14$ min
3. Amplitude $=34$ ft, period $=7$ min
4. Amplitude $=40$ ft, period $=14$ 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). For this Ferris wheel: maximum height = 74 ft, minimum height = 6 ft, so AMPLITUDE = (74 - 6)/2 = 68/2 = 34 ft (passenger moves 34 ft above and below center), and PERIOD = 14 minutes (one complete rotation). Choice A correctly identifies amplitude = 34 ft and period = 14 min by properly calculating half the range for amplitude. Choice B incorrectly uses the full range (68 ft) as amplitude instead of half, Choice C incorrectly halves the period to 7 minutes, and Choice D miscalculates the amplitude. Parameter extraction: (1) Find MAX = 74 ft and MIN = 6 ft, (2) Calculate AMPLITUDE = (74 - 6) ÷ 2 = 34 ft, (3) PERIOD = 14 minutes directly from problem—this is the time for one complete rotation!

Question 19

1. 3 (correct answer)
2. 2
3. 5
4. 12

Explanation: When two chords intersect inside a circle, the products of their segments are equal: $AE \cdot EB = CE \cdot ED$. Substituting: $6 \cdot 4 = 8 \cdot ED$, so $24 = 8 \cdot ED$, giving $ED = 3$. Choice B (2) might result from using $AE - EB = ED$. Choice C (5) could come from incorrectly using $\frac{AE + EB}{2} - 1 = ED$. Choice D (12) might result from using $CE + EB = ED$.

Question 20

1. Use $YZ^2=XY^2+XZ^2-2(XY)(XZ)\cos 30^\circ$, which reduces to the Pythagorean Theorem when the included angle is $90^\circ$. (correct answer)
2. Use $YZ=XY+XZ-\cos 30^\circ$ because the cosine term is subtracted from the sum of the sides.
3. Use $YZ^2=XY^2+XZ^2$ because the included angle is given, so the triangle can be treated as right at $X$.
4. Use the Law of Sines: $\frac{YZ}{\sin 30^\circ}=\frac{11}{\sin Y}$ and choose $\angle Y$ by visual estimation.

Explanation: The skill involves proving and applying the Law of Cosines for sides. In triangle XYZ, sides XY and XZ enclose acute angle YXZ at 30 degrees. The derivation idea adds a cosine term to generalize Pythagorean for any angle. Apply the law using YZ² = 6² + 11² - 2(6)(11)cos 30°, reducing to Pythagorean at 90°. This is correct because it connects to the special case effectively. A distractor like choice C treats it as right-angled without basis. To transfer this strategy, always ask why the formula generalizes Pythagorean before how to calculate the side.

Question 21

1. $MN \parallel QR$ (correct answer)
2. $MN \perp QR$
3. $M$ is the midpoint of $QR$
4. $\angle PMN \cong \angle PNM$

Explanation: This question involves theorems about triangles, particularly the midsegment theorem. The midsegment theorem states that the segment joining the midpoints of two sides of a triangle is parallel to the third side. The diagram shows matching tick marks indicating $PM \cong MQ$ and $PN \cong NR$, meaning M and N are midpoints of $PQ$ and $PR$ respectively. Applying the theorem in triangle $PQR$, segment $MN$ connects these midpoints and thus must be parallel to $QR$. This is justified because the midsegment creates a smaller triangle similar to the original, enforcing parallelism. A distractor misconception is assuming perpendicularity or angle congruence without evidence from markings. To transfer this strategy, match midpoint markings to known triangle theorems like the midsegment theorem for parallelism.

Question 22

1. $\frac{(x-2)^2}{25} + \frac{(y-2)^2}{9} = 1$
2. $\frac{(x-2)^2}{16} + \frac{(y-2)^2}{25} = 1$
3. $\frac{(x-2)^2}{25} + \frac{(y-2)^2}{16} = 1$ (correct answer)
4. $\frac{(x-2)^2}{100} + \frac{(y-2)^2}{91} = 1$

Explanation: When you encounter an ellipse problem with given foci and a point on the ellipse, you need to use the fundamental definition: an ellipse is the set of all points where the sum of distances to two foci is constant. First, find the center by averaging the foci coordinates: $(\frac{-1+5}{2}, \frac{2+2}{2}) = (2, 2)$. Since both foci have the same y-coordinate, this is a horizontal ellipse. The distance between foci is $|5-(-1)| = 6$, so $2c = 6$ and $c = 3$. Now calculate the sum of distances from point $Q(2, 6)$ to both foci:

• Distance to $F_1(-1, 2)$: $\sqrt{(2-(-1))^2 + (6-2)^2} = \sqrt{9+16} = 5$
• Distance to $F_2(5, 2)$: $\sqrt{(2-5)^2 + (6-2)^2} = \sqrt{9+16} = 5$

Question 23

1. $8$
2. $\sqrt{194}$
3. $\sqrt{144}$ (correct answer)
4. $18$

Explanation: This problem involves finding a missing leg when given the hypotenuse and one leg of a right triangle. We have hypotenuse DF = 13 and leg DE = 5, and need to find leg EF. The Pythagorean theorem applies in the form a² + b² = c², where c is the hypotenuse. Rearranging to find the missing leg: EF² = DF² - DE² = 13² - 5² = 169 - 25 = 144. Therefore, EF = √144 = 12. A common mistake is adding instead of subtracting when finding a leg (13² + 5² = 194). Remember: when finding a leg, subtract the known leg squared from the hypotenuse squared.

Question 24

1. It collapses the plane to the origin because all off-diagonal entries are $0$.
2. It leaves every point fixed because it is the identity matrix. (correct answer)
3. It reverses orientation because the determinant is $-1$.
4. It changes area by a factor of $2$ because there are two ones on the diagonal.

Explanation: This question tests recognizing the identity matrix and its properties. The matrix $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ is the identity matrix, which leaves every point exactly where it is—it's the "do nothing" transformation. The determinant is $1 \times 1 - 0 \times 0 = 1$, confirming that areas are preserved without any scaling or flipping. Every vector $\begin{pmatrix}x\\y\end{pmatrix}$ maps to itself, so geometric figures maintain their shape, size, and position. The identity matrix plays the same role in matrix multiplication that the number 1 plays in regular multiplication—it's the neutral element. Students might think that having 1s on the diagonal means doubling (since there are two 1s), but the identity matrix is special. The key insight is: identity matrix = no change transformation, with determinant 1 confirming area preservation.

Question 25

1. $x^2 = 6y$
2. $x^2 = 12y$
3. $x^2 = 24y$ (correct answer)
4. $y^2 = 24x$

Explanation: With vertex at origin (0, 0) and opening upward, the standard form is $x^2 = 4py$ where p is the distance from vertex to focus. Given that the focus is 6 inches from the vertex, $p = 6$. Therefore, the equation is $x^2 = 4(6)y = 24y$. Choice A uses p directly instead of 4p. Choice B uses 4p/2 = 12 instead of 4p = 24. Choice D incorrectly assumes the parabola opens horizontally instead of vertically, and uses the wrong variable arrangement.