The correct answer is B. Using the shoelace formula for quadrilateral PQTS with vertices P(1,2), Q(7,2), T(4,9), S(1,6): Area = ½|x₁(y₂-y₄) + x₂(y₃-y₁) + x₃(y₄-y₂) + x₄(y₁-y₃)| = ½|1(2-6) + 7(9-2) + 4(6-2) + 1(2-9)| = ½|(-4) + 49 + 16 + (-7)| = ½|36| = 18. Choice A results from incorrectly calculating the rectangle area as 3×5. Choice C comes from adding the rectangle area (24) and subtracting 3 incorrectly. Choice D is simply the area of rectangle PQRS, ignoring point T.

The correct answer is A. Reflections preserve distances, so the perimeter remains unchanged. Using the distance formula: AB = √[(8-2)² + (7-3)²] = √52, BC = √[(5-8)² + (11-7)²] = √25 = 5, and AC = √[(5-2)² + (11-3)²] = √73. After reflection across the y-axis, A'(-2,3), B'(-8,7), C'(-5,11) form a congruent triangle with identical side lengths, so the difference is 0. Choice B incorrectly uses the difference in x-coordinates of one vertex. Choice C doubles this error. Choice D represents three times the coordinate difference, possibly from adding coordinate changes.

The skill is finding the area of a rectangle using coordinates. The vertices are R(-2,-3), S(4,-3), T(4,1), and U(-2,1). Side lengths are computed by differences in x and y for length and width. The area logic is length times width. This gives 6 times 4 equals 24, justifying the calculation in the correct choice. A distractor misconception is confusing area with perimeter, as in choice B. To transfer this strategy, compute length and width before multiplying for area.

The correct answer is C. The original triangle OPQ has area ½(6)(8) = 24. After 90° counterclockwise rotation about origin: O'(0,0), P'(0,6), Q'(-8,0). The rotated triangle O'P'Q' also has area 24. However, the triangles overlap only at the origin, so the total area of their union is 24 + 24 = 48. Choice A gives the area of just one triangle. Choice B incorrectly assumes some overlap beyond the origin point. Choice D might result from adding the areas and including additional area incorrectly.

The correct answer is A. First find the area of rectangle WXYZ: length = 8-2 = 6, width = 7-1 = 6, so area = 36 square units. Next find the area of triangle WPZ with vertices W(2,1), P(5,4), Z(8,1). Using the triangle area formula: Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)| = ½|2(4-1) + 5(1-1) + 8(1-4)| = ½|6 + 0 - 24| = ½|18| = 9 square units. The area inside the rectangle but outside the triangle is 36 - 9 = 27 square units. Choice B results from incorrectly calculating the triangle area as 6. Choice C results from calculating the triangle area as 3. Choice D gives the full rectangle area without subtracting the triangle.

The skill is finding the perimeter of a triangle using coordinates. The vertices are A(-1,2), B(3,1), and C(1,5). Side lengths are computed using the distance formula, which is the square root of the sum of the squares of the differences in x and y coordinates. The perimeter is the sum of all three side lengths. Calculating each distance gives AB = √17, BC = √20, and CA = √13, justifying the final value as √17 + √20 + √13. A common distractor misconception is adding the squared distances without taking square roots, resulting in 17 + 20 + 13. To transfer this strategy, compute side lengths before summing for perimeter or using them for area in other figures.

When you encounter transformation problems in geometry, the key insight is understanding how different transformations affect area and other measurements.

Translation is a transformation that slides a figure from one position to another without changing its size, shape, or orientation. When triangle ABC is translated 3 units left and 2 units down, every point moves the same distance in the same direction. The new vertices A'B'C' are simply the original vertices with each x-coordinate decreased by 3 and each y-coordinate decreased by 2.

Since translation preserves all distances and angles, the translated triangle A'B'C' is congruent to the original triangle ABC. Congruent figures have identical areas, so the ratio of their areas is $$1:1$$.

Looking at the incorrect choices: Answer A) $$9:4$$ might tempt you if you incorrectly squared the translation distances (3² and 2²), but translation distances don't affect area ratios. Answer B) $$3:2$$ could mislead you into thinking the ratio relates directly to the translation amounts (3 units and 2 units), but again, translation preserves area. Answer C) $$5:1$$ has no logical connection to this problem and represents a common distractor with no mathematical basis.

Remember this crucial distinction: translations, rotations, and reflections (rigid transformations) preserve area, while dilations (scaling transformations) change area by the square of the scale factor. When you see translation problems, immediately think "same area, same shape" – the ratio will always be $$1:1$$.

This problem asks us to find the area of a triangle using coordinates. The vertices are D(1,2), E(7,2), and F(7,8). Since D and E have the same y-coordinate (2), DE is horizontal with length |7-1| = 6. Since E and F have the same x-coordinate (7), EF is vertical with length |8-2| = 6. This forms a right triangle with legs of length 6 each, so the area is (1/2) × 6 × 6 = 18. A common error is computing the full rectangle area (36) without dividing by 2. For right triangles with coordinate vertices, use the two perpendicular sides as base and height.

The skill is finding the perimeter of a quadrilateral using coordinates. The vertices are W(1,-2), X(5,0), Y(3,4), and Z(-1,2). Side lengths are computed using the distance formula between consecutive points. The perimeter is the sum of all four side lengths. Each side calculates to √20, but the expression summing the four distance formulas justifies the perimeter. A distractor misconception is omitting one side, as in choice B with only three terms. To transfer this strategy, always compute individual side lengths before summing for the perimeter.

The skill is finding the area of a triangle using coordinates. The vertices are P(-1,2), Q(4,1), and R(2,5). Side lengths are computed using the distance formula, but for area, the shoelace formula is efficient. The area logic involves listing coordinates and computing half the absolute value of the summed products as shown. This yields 9, justifying the value in the correct choice. A distractor misconception is confusing area with perimeter, as in choice C which sums distances. To transfer this strategy, compute side lengths if needed, but use shoelace directly for area.