SAT Geometry & Trigonometry Domain Test

Calculator is allowed on all questions. Figures are not necessarily drawn to scale.

Question 1

Points \(A\), \(V\), and \(B\) lie on a straight line, with ray \(VC\) drawn from \(V\). If the measure of angle \(AVC\) is \(65°\), what is the measure of angle \(CVB\)?

A B C V 65° ?




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Since \(A\), \(V\), \(B\) lie on a straight line, angles \(AVC\) and \(CVB\) are supplementary: \(180° - 65° = 115°\).

    1. 130°: Doubled the given angle instead of subtracting it from \(180°\).
    1. 155°: Arithmetic slip computing \(180 - 65\).
    1. 295°: Used \(360°\) instead of \(180°\), computing \(360-65\).

Answer: A

Question 2

Two lines intersect at point \(O\), forming two pairs of vertical angles. One angle measures \((3x + 10)°\) and its vertical angle measures \(70°\). What is the value of \(x\)?

O (3x+10)° 70°




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Vertical angles are congruent, so \(3x + 10 = 70\). Subtract 10: \(3x = 60\). Divide by 3: \(x = 20\).

    1. 15: Arithmetic slip dividing \(60\) by \(4\) instead of \(3\).
    1. 23.33: Forgot to subtract 10 first, dividing \(70\) by \(3\) directly.
    1. 26.67: Sign error, solving \(3x - 10 = 70\) instead of \(3x + 10 = 70\).

Answer: B

Question 3

What is the sum, in degrees, of the interior angles of a hexagon (a 6-sided polygon)?

Enter your answer:

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The sum of interior angles of an \(n\)-sided polygon is \((n-2) \times 180°\). For a hexagon, \(n=6\): \((6-2) \times 180° = 4 \times 180° = 720°\).

Answer: 720

Question 4

In a triangle, two of the angles measure \(50°\) and \(65°\). What is the measure of the third angle?

50° 65° ?




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The angles of a triangle sum to \(180°\): \(180 - 50 - 65 = 65°\).

    1. 50: Reported one of the given angles instead of solving for the third.
    1. 55: Arithmetic slip computing \(180-50-65\).
    1. 115: Added the two given angles instead of subtracting their sum from \(180°\).

Answer: C

Question 5

A triangle has a base of 10 and a height of 6. What is its area?

10 6




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Area \(= \dfrac{1}{2}(\text{base})(\text{height}) = \dfrac{1}{2}(10)(6) = 30\).

    1. 16: Added the base and height instead of multiplying, computing \(\frac{1}{2}(10) + \frac{1}{2}(6)\).
    1. 24: Arithmetic slip computing \(\frac{1}{2}(10)(6)\).
    1. 60: Forgot the \(\frac{1}{2}\) factor, computing base \(\times\) height only.

Answer: C

Question 6

A right triangle has legs of length 6 and 8. What is the length of its hypotenuse?

6 8

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By the Pythagorean theorem: \(6^2 + 8^2 = c^2\), so \(c^2 = 36+64=100\), \(c = 10\).

Answer: 10

Question 7

A circle has a radius of 5. What is its circumference?

O 5




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Circumference \(= 2\pi r = 2\pi(5) = 10\pi\).

    1. \(5\pi\): Forgot the factor of 2, computing \(\pi r\) instead of \(2\pi r\).
    1. \(20\pi\): Arithmetic slip, tripling the radius factor instead of doubling it.
    1. \(25\pi\): Used the area formula \(\pi r^2\) instead of the circumference formula.

Answer: B

Question 8

In right triangle \(ABC\), the right angle is at \(B\). If \(AB = 3\), \(BC = 4\), and \(AC = 5\), what is \(\sin(A)\)?

A B C 3 4 5




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Angle \(A\) has opposite side \(BC=4\) and hypotenuse \(AC=5\). \(\sin(A) = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{BC}{AC} = \dfrac{4}{5}\).

    1. \(\dfrac{3}{4}\): Computed the tangent ratio (opposite/adjacent) instead of sine.
    1. \(\dfrac{4}{3}\): Computed the reciprocal of tangent using the wrong leg placement.
    1. \(\dfrac{3}{5}\): Used the adjacent side (\(AB\)) instead of the opposite side (\(BC\)) in the sine ratio.

Answer: A

Question 9

In the figure, lines \(m\) and \(n\) are parallel and cut by transversal \(t\). If the measure of the angle shown at the top intersection is \(125°\), what is the measure of the marked angle at the bottom intersection?

m n t 125° ?




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The marked angles are co-interior (same-side interior) angles, which are supplementary: \(180° - 125° = 55°\).

    1. 35°: Mistakenly treated the angle pair as complementary, computing an unrelated subtraction from \(90°\).
    1. 65°: Arithmetic slip computing \(180-125\).
    1. 125°: Mistakenly treated co-interior angles as congruent, like corresponding angles.

Answer: B

Question 10

In a 30-60-90 triangle, the hypotenuse has length 14. What is the length of the shorter leg?

30° 60° 14




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In a 30-60-90 triangle, the shorter leg (opposite the \(30°\) angle) is half the hypotenuse: \(14/2 = 7\).

    1. 3.5: Divided the hypotenuse by \(4\) instead of \(2\), squaring the denominator by mistake.
    1. 4.67: Divided the hypotenuse by \(3\), confusing the ratio with a different special triangle.
    1. 9.9: Mistakenly treated the triangle as 45-45-90, computing \(14/\sqrt{2}\).

Answer: C

Question 11

Triangle \(ABC\) is similar to triangle \(DEF\), where angle \(A\) corresponds to angle \(D\). If \(AB = 6\), \(BC = 9\), \(AC = 12\), and \(DF = 20\), what is the length of \(EF\)?

A C B 6 9 12 D F E 20

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The scale factor is \(\dfrac{DF}{AC} = \dfrac{20}{12} = \dfrac{5}{3}\) (since \(AC\) corresponds to \(DF\)). Since \(BC\) corresponds to \(EF\): \(EF = BC \times \dfrac{5}{3} = 9 \times \dfrac{5}{3} = 15\).

Answer: 15

Question 12

A circle has a radius of 9. What is the length of an arc with a central angle of \(120°\)?

O 120° 9




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Arc length \(= \dfrac{\theta}{360°} \times 2\pi r = \dfrac{120}{360} \times 2\pi(9) = \dfrac{1}{3} \times 18\pi = 6\pi\).

    1. \(3\pi\): Forgot to double the radius, using the sector-area-style setup \(\dfrac{120}{360}\times \pi r\) instead of \(\dfrac{120}{360}\times 2\pi r\).
    1. \(4.5\pi\): Arithmetic slip in the fraction, using \(\dfrac{120}{480}\) instead of \(\dfrac{120}{360}\).
    1. \(12\pi\): Used the diameter (18) in place of the radius, without adjusting the formula.

Answer: C

Question 13

In a circle, a central angle measures \(80°\). What is the measure of the inscribed angle that intercepts the same arc?

O A B P 80°




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The Inscribed Angle Theorem states the inscribed angle is half the central angle intercepting the same arc: \(80°/2 = 40°\).

    1. 80°: Assumed the inscribed angle equals the central angle.
    1. 140°: Used a supplementary-style relationship (\(180°-40°\)) rather than the inscribed-angle theorem.
    1. 160°: Doubled the central angle instead of halving it.

Answer: A

Question 14

A rectangular prism has a length of 5, a width of 4, and a height of 3. What is its volume?

5 3 4




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Volume \(= l \times w \times h = 5 \times 4 \times 3 = 60\).

    1. 12: Added the three dimensions instead of multiplying them.
    1. 15: Multiplied only the length and height, forgetting the width.
    1. 20: Multiplied only the length and width, forgetting the height.

Answer: D

Question 15

If \(\cos(35°) = \sin(x°)\), where \(0 < x < 90\), what is the value of \(x\)?




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The complementary angle identity states \(\cos(\theta) = \sin(90° - \theta)\). So \(\cos(35°) = \sin(90°-35°) = \sin(55°)\), giving \(x=55\).

    1. 35: Mistakenly assumed sine and cosine of the same angle are always equal.
    1. 90: Reported the generic complement value (\(90°\)) without subtracting the given angle from it.
    1. 145: Mistakenly used a supplementary relationship (\(180°-35°\)) instead of a complementary one.

Answer: B

Question 16

A circle is defined by the equation \((x - 3)^2 + (y + 2)^2 = 16\). What is the radius of the circle?

Enter your answer:

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The standard form of a circle is \((x-h)^2+(y-k)^2=r^2\). Here \(r^2=16\), so \(r=\sqrt{16}=4\).

Answer: 4

Question 17

A square has a side length of 10. A circle with a radius of 5 is inscribed inside the square, touching all four sides. What is the area of the region inside the square but outside the circle, in terms of \(\pi\)?

10 5




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Area of the square: \(10^2 = 100\). Area of the circle: \(\pi(5)^2 = 25\pi\). The region inside the square but outside the circle: \(100 - 25\pi\).

    1. \(25\pi\): Forgot to subtract from the square’s area, reporting only the circle’s area.
    1. \(100 - 50\pi\): Arithmetic slip squaring the radius, using \(50\) instead of \(25\).
    1. \(100 + 25\pi\): Sign error, adding the circle’s area instead of subtracting it.

Answer: D

Question 18

A circle has center \(O\) and radius 6. Point \(P\) lies outside the circle, a distance of 10 from \(O\). Segment \(PT\) is tangent to the circle at point \(T\). What is the length of \(PT\)?

O T P 6 10

Enter your answer:

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A tangent line is perpendicular to the radius at the point of tangency, so triangle \(OTP\) is a right triangle with the right angle at \(T\). By the Pythagorean theorem: \(OT^2 + TP^2 = OP^2\), so \(6^2 + TP^2 = 10^2\), \(TP^2 = 100-36=64\), \(TP=8\).

Answer: 8

Question 19

A circle has a radius of 8. What is the length of an arc with a central angle of \(\dfrac{2\pi}{3}\) radians?




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For a central angle given in radians, arc length \(= r\theta = 8 \times \dfrac{2\pi}{3} = \dfrac{16\pi}{3}\).

    1. \(\dfrac{2\pi}{3}\): Forgot to multiply by the radius, reporting the angle itself.
    1. \(\dfrac{8\pi}{3}\): Used half the central angle by mistake, computing \(r \times \dfrac{\theta}{2}\) instead of \(r\theta\).
    1. \(\dfrac{32\pi}{3}\): Used the diameter (16) instead of the radius (8) in the arc length formula.

Answer: A

Question 20

What is the value of \(\sin(120°)\)?




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\(120°\) lies in the second quadrant, where sine is positive. Its reference angle is \(180°-120°=60°\), so \(\sin(120°) = \sin(60°) = \dfrac{\sqrt{3}}{2}\).

    1. \(-\dfrac{\sqrt{3}}{2}\): Used the correct reference angle but applied the wrong sign for the second quadrant, where sine is actually positive.
    1. \(\dfrac{1}{2}\): Used the wrong reference angle (\(30°\) instead of \(60°\)).
    1. \(-\dfrac{1}{2}\): Computed \(\cos(120°)\) instead of \(\sin(120°)\).

Answer: B