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.

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.

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.

The skill is finding the perimeter of a quadrilateral using coordinates. The vertices are K(-1,1), L(2,2), M(4,-1), and N(1,-2). Side lengths are computed using the distance formula, which is the square root of the sum of the squares of the differences in x and y coordinates. The perimeter is the sum of all four side lengths. Calculating each distance gives KL = √10, LM = √13, MN = √10, and NK = √13, justifying the final value as √10 + √13 + √10 + √13. A common distractor misconception is adding without square roots, leading to 10 + 13 + 10 + 13. To transfer this strategy, compute side lengths before summing for perimeter or using them for area in other figures.

This problem asks for the area of triangle ABC with vertices A(1,1), B(7,1), and C(4,5). The vertices form a triangle with A(1,1), B(7,1), and C(4,5). Since A and B have the same y-coordinate (y=1), AB is horizontal with length |7-1| = 6. The height from C to line AB is the vertical distance |5-1| = 4. For a triangle with a horizontal base, Area = ½ × base × height = ½ × 6 × 4 = 12. A common error is forgetting the ½ factor, which would give 24 instead of 12. The transfer strategy is to look for horizontal or vertical sides, which simplify area calculations.

This problem asks which calculation finds the area of triangle DEF with vertices D(-2,-1), E(4,2), and F(1,6). The vertices are D(-2,-1), E(4,2), and F(1,6). To find the area of a triangle using coordinates, we use the shoelace formula: Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|. Substituting D, E, F coordinates: Area = ½|(-2)(2-6) + 4(6-(-1)) + 1((-1)-2)|. This matches answer choice B exactly. Answer A is missing the crucial ½ factor, while C calculates perimeter (sum of side lengths), and D incorrectly treats it as a rectangle. The transfer strategy is to recognize that triangle area from coordinates always requires the shoelace formula with the ½ factor.

This problem asks for the perimeter expression of quadrilateral JKLM. The vertices are J(0,0), K(4,1), L(3,5), and M(-1,4). We need to find the distance between consecutive vertices: JK = √[(4-0)² + (1-0)²] = √[16 + 1] = √17; KL = √[(3-4)² + (5-1)²] = √[1 + 16] = √17; LM = √[(-1-3)² + (4-5)²] = √[16 + 1] = √17; MJ = √[(0-(-1))² + (0-4)²] = √[1 + 16] = √17. The perimeter is JK + KL + LM + MJ = √17 + √17 + √17 + √17. A common mistake is assuming all sides equal 4 (confusing with a square of side 4). The key insight is that all four sides happen to have the same length √17, making this a rhombus.

This problem asks for the area of triangle RST with vertices R(-3,1), S(1,5), and T(5,0). The vertices are R(-3,1), S(1,5), and T(5,0). Using the shoelace formula for triangle area: Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|. Substituting: Area = ½|(-3)(5-0) + 1(0-1) + 5(1-5)| = ½|(-3)(5) + 1(-1) + 5(-4)| = ½|-15 - 1 - 20| = ½|-36| = ½(36) = 18. This matches answer A. Answer B forgets the ½ factor (giving 36), while C and D incorrectly assume a right triangle with base 8 and various heights. The transfer strategy is to always use the shoelace formula for general triangles in coordinate geometry.