Drill 4 · Math · Geometry
ACT Math: Geometry (Drill 4) is a Math practice drill covering Geometry. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
This drill covers the 45-45-90 special right triangle, vertical and supplementary angle relationships, volume of a cone, finding the area of a triangle using coordinates, and basic right triangle trigonometry.
Question 1. An isosceles right triangle has legs of length 9. What is the length of the hypotenuse?
Explanation: An isosceles right triangle is a 45-45-90 triangle. The side ratios are 1 : 1 : √2. If each leg = 9, the hypotenuse = 9√2. You can also use the Pythagorean theorem: c = √92 + 92 = √81 + 81 = √162 = 9√2. Choice A results from simply doubling the leg: 9 × 2 = 18, confusing the 45-45-90 hypotenuse rule with the 30-60-90 rule. Choice B uses the √3 factor from the 30-60-90 triangle instead of √2. Choice D results from halving the leg instead of multiplying by √2.
Question 2. Two lines intersect at a point, forming four angles. One of the angles measures (3x + 10)° and the angle directly across from it (its vertical angle) measures (5x − 30)°. What is the measure of each of the other two angles?
Explanation: Vertical angles are equal: 3x + 10 = 5x − 30. Solve: 40 = 2x, so x = 20. Each vertical angle = 3(20) + 10 = 70°. The other two angles are supplementary to 70°: 180° − 70° = 110°. Choice B (70°) is the measure of the vertical angles themselves, not the other two angles the question asks about. Choice C results from solving the equation with an arithmetic error: treating 10 + 30 = 40 as 2x, then computing 3(20) = 60, then 60 − 10 = 50. Choice D is the value of x itself (20), not an angle measure.
Question 3. A cone has a radius of 6 and a height of 10. What is the volume of the cone, in terms of π?
Explanation: Volume of a cone = (1/3)πr2h = (1/3)π(6)2(10) = (1/3)π(36)(10) = (1/3)(360π) = 120π. Choice A results from forgetting the (1/3) factor: πr2h = π(36)(10) = 360π; this is the volume of a cylinder with the same dimensions. Choice C results from using (2/3) instead of (1/3): (2/3)(360π) = 240π. Choice D results from using the radius instead of r2: (1/3)π(6)(10) = 20π... or from computing (1/3)(180π) = 60π, using the diameter (12) as r in the formula.
Question 4. A triangle has vertices at P(0, 0), Q(8, 0), and R(3, 5). What is the area of the triangle?
Explanation: Since P and Q both lie on the x-axis, side PQ is the base: base = 8 − 0 = 8. The height is the perpendicular distance from R to the x-axis, which is simply the y-coordinate of R: height = 5. Area = (1/2)(base)(height) = (1/2)(8)(5) = 20. Choice A results from using the x-coordinate of R as the base: (1/2)(3)(5) = 7.5... or using the shorter distance QR projected: (1/2)(5+1)(5) = 15. Choice C results from computing (1/2)(8)(5) = 20 and then adding something, or using base = PQ length via distance formula: √64 + 25 × some factor ≈ 24. Choice D results from forgetting the (1/2): base × height = 8 × 5 = 40.
Question 5. In right triangle ABC, angle C = 90°, AB = 13, and BC = 5. What is the value of sin(A)?
Explanation: In right triangle ABC with C = 90°, AB is the hypotenuse = 13, and BC = 5. Find the missing leg AC using the Pythagorean theorem: AC2 + BC2 = AB2 → AC2 + 25 = 169 → AC2 = 144 → AC = 12. This is a 5-12-13 Pythagorean triple. For angle A: the side opposite A is BC = 5, and the hypotenuse is AB = 13. sin(A) = opposite / hypotenuse = 5/13. Choice B is cos(A) = adjacent / hypotenuse = AC / AB = 12/13. Choice C is tan(A) = opposite / adjacent = BC / AC = 5/12. Choice D is the reciprocal of sin(A) (i.e., csc A), which is greater than 1 and cannot be a valid sine value.