Geometry (Subtest II)
Euclidean Reasoning
Geometry on Subtest II rewards precise theorem use. Triangle congruence criteria are SSS, SAS, ASA, AAS, and HL (right triangles only); SSA is not valid. Similar triangles have proportional sides and equal angles, with AA as the usual shortcut. The Pythagorean theorem, its converse, and special right triangles (45-45-90 and 30-60-90) support most numerical work.
Circles
Inscribed angles subtending the same arc are equal, and an inscribed angle is half its intercepted central angle. A tangent is perpendicular to the radius at the point of tangency. The power-of-a-point relations cover two chords (AP·PB = CP·PD), a secant and tangent (tangent² = external · whole secant), and two secants.
Coordinates, Transformations, and Trigonometry
Distance, midpoint, slope, and conic equations are standard: (x−h)² + (y−k)² = r², plus parabolas, ellipses, and hyperbolas in standard form. Rigid motions (translation, rotation, reflection) preserve congruence; adding a dilation produces similarity. In trigonometry, sin²θ + cos²θ = 1, and the law of cosines c² = a² + b² − 2ab cos C extends Pythagoras to any triangle; the law of sines is a/sin A = b/sin B = c/sin C.
Worked Example
A triangle has sides 5, 7, and 8. Find the angle opposite the side of length 8 and the area. By the law of cosines: 8² = 5² + 7² − 2(5)(7)cos C, so 64 = 74 − 70cos C, giving cos C = 10/70 = 1/7. Thus C ≈ 81.79°. For the area, first find sin C = √(1 − 1/49) = √48/7 ≈ 0.9897, then Area = ½(5)(7)sin C ≈ 17.5(0.9897) ≈ 17.32 square units. Heron's formula with s = 10 gives √(10·5·3·2) = √300 ≈ 17.32, confirming the result.
Traps That Cost Marks
- Using the ambiguous SSA case without checking for two possible triangles.
- Doubling instead of halving when relating inscribed and central angles.
- Mixing degrees and radians in trigonometric calculations.
- Assuming area scales linearly under dilation; it scales by the square of the factor.
- Confusing congruent with similar, or applying HL to non-right triangles.