This problem involves the exterior angle theorem for triangles. The exterior angle theorem states that an exterior angle of a triangle equals the sum of the two non-adjacent interior angles. Here, the exterior angle at C is 132°, and this equals the sum of angles A and B. Since angle A = 47°, we can find angle B: 132° = 47° + angle B, so angle B = 132° - 47° = 85°. A common mistake is thinking the exterior angle equals 180° minus the adjacent interior angle. Remember that the exterior angle equals the sum of the two remote interior angles, which provides a quick way to solve such problems.

This problem involves an isosceles triangle where we need to find a base angle. In an isosceles triangle with PQ = PR, the base angles Q and R are congruent. The vertex angle at P is 36°, and since all angles sum to 180°, we have: 36° + angle Q + angle R = 180°. Since angle Q = angle R, we can write: 36° + 2(angle Q) = 180°, which gives us 2(angle Q) = 144°, so angle Q = 72°. A common error is confusing which angles are equal in an isosceles triangle - remember that the base angles (opposite the equal sides) are congruent. When working with isosceles triangles, always identify the equal sides first to determine which angles are congruent.

This problem asks for the length of a horizontal line segment in the coordinate plane. Points A(-2, 1) and B(4, 1) have the same y-coordinate, so AB is horizontal. For a horizontal segment, the length equals the absolute difference of x-coordinates: |4 - (-2)| = |4 + 2| = 6. We can verify using the distance formula: √[(4-(-2))² + (1-1)²] = √[6² + 0²] = √36 = 6. A common error is subtracting coordinates incorrectly, especially with negative values. When finding distances on the coordinate plane, horizontal segments have length |x₂ - x₁| and vertical segments have length |y₂ - y₁|.

This problem asks for an angle measure in a triangle with vertices given in the coordinate plane. Points A(0,0), B(6,0), and C(0,8) form a triangle where A is at the origin, B is on the positive x-axis, and C is on the positive y-axis. Since AB lies along the x-axis and AC lies along the y-axis, these two sides are perpendicular, making angle A = 90°. We can verify this is a right triangle by checking if AB² + AC² = BC²: AB = 6, AC = 8, and BC = √[(6-0)² + (0-8)²] = √[36 + 64] = √100 = 10, confirming 6² + 8² = 36 + 64 = 100 = 10². When vertices lie on coordinate axes, look for right angles at the origin.

This problem tests the triangle inequality theorem, which states that the sum of any two sides must be greater than the third side. With XY = 7 and YZ = 10, we need: (1) 7 + 10 > x, so x < 17; (2) 7 + x > 10, so x > 3; (3) 10 + x > 7, which is always true for positive x. Therefore, 3 < x < 17. Looking at the options: 2 is too small, 3 is not valid (must be greater than 3), 17 is not valid (must be less than 17), but 12 satisfies 3 < 12 < 17. A common error is using ≤ or ≥ instead of strict inequalities. Remember that the triangle inequality uses strict inequalities: the sum must be greater than, not equal to, the third side.

This problem involves similar triangles and finding a corresponding side length. When triangles are similar, corresponding sides are proportional. We have the ratio DE/AB = 18/12 = 3/2, which is the scale factor. Since B corresponds to E and C corresponds to F, side EF corresponds to side BC. Therefore, EF = BC × (scale factor) = 10 × (3/2) = 15. A common error is using the wrong correspondence or inverting the ratio. When working with similar triangles, always identify the correspondence first and use the same scale factor for all corresponding parts.

This problem involves a line parallel to one side of a triangle, creating proportional segments. When DE is parallel to BC, it creates similar triangles and proportional segments: AD/DB = AE/EC. We have AD = 4, DB = 6, and AE = 5. Setting up the proportion: 4/6 = 5/EC. Cross-multiplying: 4(EC) = 6(5) = 30, so EC = 30/4 = 7.5. A common error is using AD/AB instead of AD/DB in the proportion. Remember that when a line is parallel to one side of a triangle, it divides the other two sides proportionally.

This problem asks for the area of a triangle with vertices given in the coordinate plane. Points P(1,2), Q(5,2), and R(5,-1) form a right triangle since PQ is horizontal (both points have y = 2) and QR is vertical (both points have x = 5). The base PQ has length |5 - 1| = 4, and the height QR has length |2 - (-1)| = 3. The area is (1/2) × base × height = (1/2) × 4 × 3 = 6. A common error is forgetting the factor of 1/2 in the triangle area formula. When you can identify perpendicular sides in a coordinate plane triangle, use them as base and height for easy calculation.

This problem requires using the Pythagorean theorem to find the hypotenuse of a right triangle. With legs of length 9 and 12, the hypotenuse c satisfies: c² = 9² + 12² = 81 + 144 = 225. Taking the square root: c = √225 = 15. This is a multiple of the famous 3-4-5 right triangle (specifically, 3×3 = 9, 3×4 = 12, 3×5 = 15). A common error is adding the legs instead of squaring them first. Remember the Pythagorean theorem uses squares: a² + b² = c², not a + b = c.

This problem asks about an angle in an equilateral triangle. By definition, an equilateral triangle has all three sides equal and all three angles equal. Since the sum of angles in any triangle is 180°, and all three angles are equal in an equilateral triangle, each angle measures 180° ÷ 3 = 60°. Therefore, angle M = 60°. A common mistake is confusing equilateral with isosceles triangles, which only require two sides to be equal. Remember that "equilateral" means all sides are equal, which also makes all angles equal to 60°.