SAT practice test

SAT Geometry and Trigonometry Practice Test

Area, volume, triangles, circles, right-triangle trig. 15% of Math. Here are 10 SAT geometry and trigonometry questions, each with the answer and the reason.

Updated 2026-11-14 by the Tigertest team. Original questions written to the digital SAT specifications. Not affiliated with College Board. Score estimates are unofficial.

1. 1.

   A cylindrical water tank has a diameter of $8$ feet and a height of $5$ feet. What is the volume of the tank, in cubic feet?

   • A.

     $40\pi$

   • B.

     $80\pi$

   • C.

     $160\pi$

   • D.

     $320\pi$

   Answer: B.

   The volume of a cylinder is $V = \pi r^2 h$, and the radius is half the diameter, so $r = 4$. Then $V = \pi (4^2)(5) = 80\pi$ cubic feet. The choice $320\pi$ comes from using the diameter $8$ as the radius: $\pi (8^2)(5)$. The choice $40\pi$ is $2\pi r h$, the lateral surface area, and $160\pi$ comes from mixing the diameter and the radius, $\pi (8)(4)(5)$.

2. 2.

   A circle in the $xy$-plane has the equation $x^2 + y^2 - 6x + 4y - 12 = 0$. What is the radius of the circle?

   • A.

     $2\sqrt{3}$

   • B.

     $5$

   • C.

     $12$

   • D.

     $25$

   Answer: B.

   Complete the square for each variable: $(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$, which gives $(x - 3)^2 + (y + 2)^2 = 25$. In the standard form $(x - h)^2 + (y - k)^2 = r^2$, so $r^2 = 25$ and $r = 5$. The choice $25$ is $r^2$, not $r$. The choice $2\sqrt{3}$, which equals $\sqrt{12}$, uses the constant $12$ without adding the $9$ and $4$ needed to complete the squares.

3. 3.

   In a right triangle, the measures of the two acute angles are $x^\circ$ and $(3x + 10)^\circ$. If $\sin(x^\circ) = \cos((3x + 10)^\circ)$, what is the value of $x$?

   • A.

     $10$

   • B.

     $20$

   • C.

     $22.5$

   • D.

     $40$

   Answer: B.

   The sine of an angle equals the cosine of its complement, so $\sin(x^\circ) = \cos((3x + 10)^\circ)$ when $x + (3x + 10) = 90$. Then $4x + 10 = 90$, so $4x = 80$ and $x = 20$. Check: the angles are $20^\circ$ and $70^\circ$, which sum to $90^\circ$ as the acute angles of a right triangle must. The choice $40$ comes from setting the difference of the angles, $(3x + 10) - x = 2x + 10$, equal to $90$ instead of their sum. The choice $22.5$ comes from $4x = 90$, dropping the $10$.

4. 4.

   A circle has radius $12$. An arc of the circle has a central angle measuring $150^\circ$. What is the length of the arc?

   • A.

     $5\pi$

   • B.

     $10\pi$

   • C.

     $20\pi$

   • D.

     $60\pi$

   Answer: B.

   Arc length is the fraction of the circle's circumference cut off by the central angle. The circumference is $2\pi(12) = 24\pi$, and $\frac{150}{360} = \frac{5}{12}$ of it is $\frac{5}{12}(24\pi) = 10\pi$. The choice $60\pi$ is the area of the sector, $\frac{5}{12}(144\pi)$, which answers a different question. The choice $5\pi$ uses $\pi r$ instead of $2\pi r$ for the circumference, and $20\pi$ doubles the correct length.

5. 5.

   A circle has a radius of $6$ centimeters. What is the area, in square centimeters, of a sector of the circle with a central angle of $120^\circ$?

   • A.

     $4\pi$

   • B.

     $12\pi$

   • C.

     $24\pi$

   • D.

     $36\pi$

   Answer: B.

   A sector's area is the fraction of the circle set by its central angle times the circle's area. The circle's area is $\pi(6)^2 = 36\pi$, and $\frac{120}{360} = \frac{1}{3}$, so the sector's area is $\frac{1}{3}(36\pi) = 12\pi$. The choice $4\pi$ is the arc length, $\frac{1}{3}(2\pi \cdot 6)$, which uses the circumference instead of the area. The choice $36\pi$ is the whole circle.

6. 6.

   In right triangle $JKL$, angle $K$ is the right angle and $\sin J = \frac{9}{41}$. Which of the following is also equal to $\frac{9}{41}$?

   • A.

     $\cos J$

   • B.

     $\sin L$

   • C.

     $\cos L$

   • D.

     $\tan L$

   Answer: C.

   Angles $J$ and $L$ are the two acute angles of a right triangle, so they are complementary, and the sine of an angle equals the cosine of its complement: $\sin J = \cos L$. In side terms, the side opposite $J$ is $\overline{KL}$, which is also the side adjacent to $L$, so $\sin J = \frac{KL}{JL} = \cos L = \frac{9}{41}$. The choice $\sin L$ is the strongest trap: it equals $\cos J = \frac{40}{41}$ (because $9^2 + 40^2 = 41^2$), not $\frac{9}{41}$. Also, $\tan L = \frac{40}{9}$.

7. 7.

   Points $A$, $B$ and $C$ lie on a circle with center $O$, and point $C$ is not on minor arc $AB$. The measure of central angle $AOB$ is $112^\circ$. What is the measure of inscribed angle $ACB$?

   • A.

     $56^\circ$

   • B.

     $68^\circ$

   • C.

     $112^\circ$

   • D.

     $124^\circ$

   Answer: A.

   An inscribed angle measures half of its intercepted arc. Angle $ACB$ intercepts minor arc $AB$, whose measure equals the central angle, $112^\circ$, so angle $ACB$ measures $\frac{112}{2} = 56^\circ$. The choice $112^\circ$ treats the inscribed angle as equal to the central angle; only an angle with its vertex at the center has the same measure as its arc. The choice $124^\circ$ is the inscribed angle from a point on the minor arc, half of the $248^\circ$ major arc, and $68^\circ$ is $180 - 112$.

8. 8.

   Triangle $ABC$ is similar to triangle $DEF$, where $A$, $B$ and $C$ correspond to $D$, $E$ and $F$, respectively. If $AB = 9$, $DE = 15$, and the area of triangle $ABC$ is $54$ square units, what is the area, in square units, of triangle $DEF$?

   • A.

     $19.44$

   • B.

     $90$

   • C.

     $150$

   • D.

     $250$

   Answer: C.

   The scale factor from $ABC$ to $DEF$ is $\frac{DE}{AB} = \frac{15}{9} = \frac{5}{3}$. Lengths scale by the factor $k$, but areas scale by $k^2$, because area multiplies two lengths: $54 \times \left(\frac{5}{3}\right)^2 = 54 \times \frac{25}{9} = 150$. The choice $90$ multiplies by $\frac{5}{3}$ only once, scaling the area like a length. The choice $250$ cubes the factor, which is how volumes scale, and $19.44$ squares the factor upside down, $\left(\frac{3}{5}\right)^2$.

9. 9.

   In the $xy$-plane, the graph of $2x^2 + 2y^2 + 12x - 8y - 46 = 0$ is a circle. What are the center and the radius of the circle?

   • A.

     Center $(-3, 2)$, radius $6$

   • B.

     Center $(3, -2)$, radius $6$

   • C.

     Center $(-6, 4)$, radius $7\sqrt{2}$

   • D.

     Center $(-3, 2)$, radius $\sqrt{23}$

   Answer: A.

   Divide every term by $2$ so the squared terms have coefficient $1$: $x^2 + y^2 + 6x - 4y = 23$. Complete each square by adding $\left(\frac{6}{2}\right)^2 = 9$ and $\left(\frac{-4}{2}\right)^2 = 4$ to both sides: $(x + 3)^2 + (y - 2)^2 = 36$. The center is $(-3, 2)$ and the radius is $\sqrt{36} = 6$. The choice with center $(-6, 4)$ completes the square without first dividing by $2$; the choice with radius $\sqrt{23}$ forgets to add $9$ and $4$ to the right side; and $(3, -2)$ flips the signs of the center.

10. 10.

    In the xy-plane, the graph of $x^2 + y^2 - 6x + 4y = 12$ is a circle. What is the radius of the circle?

    • A.

      $\sqrt{12}$

    • B.

      $5$

    • C.

      $12$

    • D.

      $25$

    Answer: B.

    Complete the square in both variables: $(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$, which gives $(x - 3)^2 + (y + 2)^2 = 25$. The right side is $r^2$, so $r = 5$. Choice D is $r^2$ rather than $r$, and choice A takes the square root of the original constant $12$ without first completing the square.