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SAT Math — Geometry & Trigonometry Practice Questions

Practice Math — Geometry & Trigonometry questions for the Scholastic Assessment Test. Every question includes a full explanation of why the correct answer is right and why the tempting distractors are wrong.

25 questions available · hard difficulty · SAT · Free, no registration required

Sample Math — Geometry & Trigonometry Questions with Answers

10 example questions with full explanations. Use the interactive practice above to work through the complete set.

Question 1medium

A right triangle has legs of length 5 and 12. The statement 'the length of the hypotenuse is 13 and the perimeter of the triangle is 30' is true.

  • A.True
  • B.False

Correct answer: A

Using the Pythagorean theorem: c² = 5² + 12² = 25 + 144 = 169, so c = 13. The perimeter is then 5 + 12 + 13 = 30. Both parts of the statement are correct, so the answer is True. This is a classic 5-12-13 Pythagorean triple, and combining the hypotenuse calculation with the perimeter sum tests two related skills in one step.

Question 2medium

In a circle with center O, a sector has a central angle of 120° and the circle has a radius of 6. The statement 'the arc length of the sector is 4π' is true.

  • A.True
  • B.False

Correct answer: A

Arc length = (central angle / 360°) × 2πr = (120/360) × 2π(6) = (1/3) × 12π = 4π. The statement is correct, so the answer is True. A common error is to use the sector area formula (πr² × θ/360) instead of the arc length formula, which would give 12π — that misconception makes this a meaningful medium-difficulty check.

Question 3medium

Triangle ABC is similar to triangle DEF with a scale factor of 3:5 (ABC to DEF). If the area of triangle ABC is 27 square units, the statement 'the area of triangle DEF is 75 square units' is true.

  • A.True
  • B.False

Correct answer: A

For similar figures with linear scale factor 3:5, the ratio of areas is (3:5)² = 9:25. Setting up the proportion: 27/Area(DEF) = 9/25, so Area(DEF) = 27 × 25/9 = 75 square units. The statement is True. A common mistake is to multiply by the linear scale factor (5/3) rather than its square (25/9), which incorrectly gives 45 — that error would lead a student to choose False.

Question 4medium

Line ℓ passes through the points (2, 5) and (6, 11). The statement 'a line perpendicular to ℓ has a slope of −2/3' is true.

  • A.True
  • B.False

Correct answer: A

The slope of line ℓ is (11 − 5)/(6 − 2) = 6/4 = 3/2. A perpendicular line has a slope equal to the negative reciprocal of 3/2, which is −2/3. The statement is True. Common errors include only taking the reciprocal (giving 2/3) or only negating (giving −3/2), both of which would cause a student to incorrectly choose False.

Question 5medium

In right triangle PQR, angle Q is the right angle, angle P measures 35°, and the hypotenuse PR has length 10. The statement 'the side QR opposite to angle P has length 10 sin(35°)' is true.

  • A.True
  • B.False

Correct answer: A

In right triangle PQR with the right angle at Q, side QR is opposite angle P and the hypotenuse is PR = 10. By SOH-CAH-TOA, sin(P) = opposite/hypotenuse = QR/PR, so QR = PR × sin(35°) = 10 sin(35°). The statement is True. A common error is to confuse which side is opposite versus adjacent, leading students to use cos(35°) instead, or to incorrectly identify PR as a leg rather than the hypotenuse.

Question 6medium

A cylinder has a radius of 4 and a height of 9. A cone has the same radius and the same height. The statement 'the volume of the cylinder is exactly three times the volume of the cone' is true.

  • A.True
  • B.False

Correct answer: A

The volume of the cylinder is πr²h = π(4²)(9) = 144π. The volume of the cone is (1/3)πr²h = (1/3)(144π) = 48π. Since 144π = 3 × 48π, the cylinder's volume is exactly three times the cone's volume. The statement is True. This relationship holds for any cylinder and cone sharing the same radius and height — the factor of 1/3 in the cone formula is precisely what creates this ratio. Students who misremember the cone formula (omitting the 1/3) would incorrectly choose False.

Question 7hard

In triangle PQR, angle Q = 90°, PQ = 7, and angle P = 60°. The triangle is inscribed in a rectangle such that vertices Q and R lie on the base of the rectangle, and vertex P lies on the top side directly above Q. What is the area of the rectangle?

  • A.7√3
  • B.49√3
  • C.98√3
  • D.14√3

Correct answer: B

In right triangle PQR with angle Q = 90° and angle P = 60°, angle R = 30°. Side PQ = 7 (leg adjacent to P), so tan(60°) = QR/PQ → QR = 7·tan(60°) = 7√3. The rectangle has P directly above Q, so the rectangle's height equals PQ = 7 and the rectangle's width equals QR = 7√3. Area = 7 × 7√3 = 49√3. Option A (7√3) is just the length of QR, not the area. Option D (14√3) results from using 2·PQ as one dimension. Option C (98√3) doubles the correct area, a common error when confusing the full rectangle with a triangle area.

Question 8hard

If sin(x°) = cos(3x − 10)°, and 0 < x < 90, what is the value of x?

  • A.20
  • B.30
  • C.25
  • D.22.5

Correct answer: C

Using the complementary angle identity, sin(x°) = cos(90° − x°). So cos(90 − x)° = cos(3x − 10)°. Setting the arguments equal: 90 − x = 3x − 10 → 100 = 4x → x = 25. Verify: sin(25°) = cos(65°) and cos(3·25−10)° = cos(65°) ✓. Option A (x=20) gives 3(20)−10 = 50, and 90−20 = 70 ≠ 50. Option B (x=30) gives 3(30)−10 = 80, and 90−30 = 60 ≠ 80. Option D (x=22.5) is a distractor from solving 90−x = 3x incorrectly without the −10 term: 90 = 4x → x = 22.5, ignoring the constant.

Question 9hard

In triangle ABC, angle C = 90°, BC = 5, and AC = 12. Point D is on hypotenuse AB such that CD is perpendicular to AB. What is the length of CD?

  • A.5√2/2
  • B.12/5
  • C.65/12
  • D.60/13

Correct answer: D

The hypotenuse AB = √(BC² + AC²) = √(25 + 144) = √169 = 13. The altitude from the right angle to the hypotenuse has length CD = (BC · AC)/AB = (5 · 12)/13 = 60/13. This is the geometric mean relationship in a right triangle: the altitude to the hypotenuse equals the product of the two legs divided by the hypotenuse. Distractor B(12/5) confuses one leg with the altitude formula. Distractor C(65/12) inverts the formula, using the hypotenuse squared over a leg. Distractor A suggests a 45-45-90 assumption that does not apply here.

Question 10hard

A solid is formed by attaching a cone of height 4 and base radius 3 on top of a cylinder of height 10 and radius 3. A second solid is a sphere. If the volume of the sphere equals the total volume of the composite solid, what is the radius of the sphere? (Use π where needed and simplify.)

  • A.∛(306/4)
  • B.∛(102/4)
  • C.3∛(102/4)
  • D.3

Correct answer: A

Volume of cylinder = π r² h = π(9)(10) = 90π. Volume of cone = (1/3)π r² h = (1/3)π(9)(4) = 12π. Total volume = 102π. Volume of sphere = (4/3)π R³ = 102π → R³ = 102π · 3/(4π) = 306/4 → R = ∛(306/4). Distractor B omits the factor of 3 when solving for R³. Distractor C incorrectly factors out 3 from the cube root. Distractor D results from guessing the radius equals the shared base radius without calculation.