Geometry Domain Test

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

Question 1

Two lines intersect at a point. One of the four angles formed measures \(64°\). What is the measure of the angle that is vertical (opposite) to the \(64°\) angle?




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Vertical angles are formed opposite each other when two lines intersect. Vertical angles are always equal, so the angle opposite the \(64°\) angle also measures \(64°\).

    1. \(26°\): This is \(90° - 64°\), the complement of \(64°\) — not related to vertical angles.
    1. \(116°\): This is \(180° - 64°\), the supplement of \(64°\). It equals the adjacent angle, not the vertical angle.
    1. \(154°\): This is \(180° + 64° - 90°\) — no geometric basis; a common arithmetic error.

Answer: B

Question 2

The three interior angles of a triangle measure \(47°\), \(83°\), and \(x°\). What is the value of \(x\)?




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The interior angles of any triangle sum to \(180°\):

\[47 + 83 + x = 180\] \[130 + x = 180\] \[x = 50\]

    1. \(40\): Added incorrectly — computed \(47 + 83 = 120\) then subtracted from \(160\) instead of \(180\).
    1. \(83\): Set \(x\) equal to the other given angle — assumed the triangle is isosceles without basis.
    1. \(93\): Added \(47 + 83 = 130\) then added \(43\) instead of subtracting from \(180\).

Answer: G

Question 3

A right triangle has legs of length \(5\) and \(12\). What is the length of the hypotenuse?




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Apply the Pythagorean theorem:

\[c^2 = 5^2 + 12^2 = 25 + 144 = 169\] \[c = \sqrt{169} = 13\]

This is the classic \(5\)-\(12\)-\(13\) Pythagorean triple.

    1. \(7\): Subtracted the legs: \(12 - 5 = 7\).
    1. \(11\): Computed \(\sqrt{5^2 + 12^2}\) but estimated \(\sqrt{169}\) incorrectly.
    1. \(17\): Added the legs directly: \(5 + 12 = 17\).

Answer: C

Question 4

What is the area of a circle with radius \(9\)? (Express in terms of \(\pi\).)




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Area of a circle: \(A = \pi r^2\)

\[A = \pi (9)^2 = 81\pi\]

    1. \(9\pi\): Used \(A = \pi r\) instead of \(\pi r^2\) — forgot to square the radius.
    1. \(18\pi\): Used the circumference formula \(C = 2\pi r = 18\pi\) instead of area.
    1. \(162\pi\): Doubled the correct answer — may have computed \(2\pi r^2\).

Answer: H

Question 5

In the right triangle below, what is \(\sin(\theta)\)?

P Q R 8 15 θ




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First, find the hypotenuse using the Pythagorean theorem:

\[PR = \sqrt{PQ^2 + QR^2} = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\]

For angle \(\theta\) at vertex \(R\), the opposite side is \(PQ = 8\) and the hypotenuse is \(PR = 17\):

\[\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{8}{17}\]

    1. \(\dfrac{15}{17}\): This is \(\cos(\theta)\) — used the adjacent side instead of the opposite.
    1. \(\dfrac{8}{15}\): This is \(\tan(\theta)\) — divided opposite by adjacent instead of hypotenuse.
    1. \(\dfrac{17}{8}\): Inverted the sine ratio — placed hypotenuse in the numerator.

Answer: A

Question 6

What is the sum of the interior angles of a regular hexagon?




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The interior angle sum of a polygon with \(n\) sides is \((n - 2) \times 180°\).

For a hexagon, \(n = 6\):

\[(6 - 2) \times 180° = 4 \times 180° = 720°\]

    1. \(540°\): Interior angle sum of a pentagon (\(n = 5\)) — off by one side.
    1. \(900°\): Interior angle sum of a heptagon (\(n = 7\)) — off by one side in the other direction.
    1. \(1{,}080°\): Interior angle sum of an octagon (\(n = 8\)) — used \(n = 8\) instead of \(n = 6\).

Answer: G

Question 7

A triangle has a base of \(14\) units and a height of \(9\) units. What is the area of the triangle?




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Area of a triangle: \(A = \dfrac{1}{2} \times \text{base} \times \text{height}\)

\[A = \frac{1}{2} \times 14 \times 9 = \frac{126}{2} = 63\]

    1. \(23\): Added base and height: \(14 + 9 = 23\).
    1. \(126\): Computed \(14 \times 9 = 126\) without multiplying by \(\frac{1}{2}\).
    1. \(252\): Doubled the product: \(2 \times 14 \times 9 = 252\).

Answer: B

Question 8

In the figure below, lines \(\ell\) and \(m\) are parallel and are cut by transversal \(t\). If the marked angle at the upper intersection measures \(112°\), what is the measure of angle \(x\) at the lower intersection?

ℓ m t 112° x




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Angle \(x\) and the \(112°\) angle are co-interior angles (also called same-side interior or consecutive interior angles) — they are on the same side of the transversal, between the two parallel lines. Co-interior angles are supplementary:

\[x + 112° = 180°\] \[x = 68°\]

    1. \(112°\): Assumed \(x\) equals the given angle — this would be true for alternate interior or corresponding angles, not co-interior angles.
    1. \(124°\): No geometric basis; likely a computational error.
    1. \(136°\): Computed \(180° + 112° - 156°\) — an arithmetic error when applying the supplementary relationship.

Answer: F

Question 9

Triangle \(ABC\) is similar to triangle \(DEF\), with \(AB\) corresponding to \(DE\), \(BC\) corresponding to \(EF\), and \(AC\) corresponding to \(DF\). If \(AB = 8\), \(BC = 12\), \(AC = 20\), and \(DE = 6\), what is the length of \(EF\)?




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Since \(\triangle ABC \sim \triangle DEF\), corresponding sides are proportional:

\[\frac{DE}{AB} = \frac{EF}{BC}\]

\[\frac{6}{8} = \frac{EF}{12}\]

\[EF = 12 \times \frac{6}{8} = 12 \times \frac{3}{4} = 9\]

    1. \(6\): Set \(EF = DE\) — confused the scale factor with the side length.
    1. \(10\): Used \(AC\) as the reference side instead of \(AB\): \(20 \times \frac{6}{8}\)… actually computed \(\frac{6}{8} \times 20 \div 1.5\) — a multi-step arithmetic error.
    1. \(16\): Inverted the scale factor: \(12 \times \frac{8}{6} = 16\).

Answer: B

Question 10

A circle has a radius of \(10\). A central angle measures \(72°\). What is the length of the arc intercepted by this central angle? (Express in terms of \(\pi\).)




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Arc length formula:

\[\text{Arc length} = \frac{\theta}{360°} \times 2\pi r = \frac{72}{360} \times 2\pi(10) = \frac{1}{5} \times 20\pi = 4\pi\]

    1. \(2\pi\): Used \(\pi r\) instead of \(2\pi r\) in the circumference — missing the factor of 2.
    1. \(5\pi\): Used \(\dfrac{72}{360} \times \pi r^2\) — applied the sector area formula instead of arc length.
    1. \(8\pi\): Used \(\dfrac{72}{360} \times 4\pi r\) — introduced an extra factor of 2.

Answer: G

Question 11

What is the distance between points \((-2, 3)\) and \((4, -5)\) in the coordinate plane?




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Apply the distance formula:

\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]

\[d = \sqrt{(4-(-2))^2 + (-5-3)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\]

    1. \(\sqrt{28}\): Computed \(\sqrt{(4-(-2))^2 - (-5-3)^2} = \sqrt{36-64}\)… absolute value error — likely subtracted differences instead of adding squares.
    1. \(\sqrt{52}\): Used \((4+(-2))^2 + (-5+3)^2 = 2^2 + (-2)^2 + ...\) — added coordinates instead of taking differences.
    1. \(14\): Added the absolute coordinate differences without squaring: \(|6| + |-8| = 14\).

Answer: C

Question 12

An inscribed angle in a circle intercepts an arc of \(110°\). What is the measure of the inscribed angle?




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The Inscribed Angle Theorem: an inscribed angle equals half the intercepted arc.

\[\text{Inscribed angle} = \frac{1}{2} \times 110° = 55°\]

    1. \(27.5°\): Halved the answer again — applied the \(\frac{1}{2}\) relationship twice.
    1. \(110°\): Set the inscribed angle equal to the arc — confused it with a central angle (which does equal its arc).
    1. \(220°\): Doubled the arc instead of halving it.

Answer: G

Question 13

In a \(30°\)-\(60°\)-\(90°\) triangle, the hypotenuse has length \(24\). What is the length of the shorter leg?




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In a \(30°\)-\(60°\)-\(90°\) triangle, the side lengths follow the ratio \(1 : \sqrt{3} : 2\), where the hypotenuse is twice the shorter leg.

\[\text{shorter leg} = \frac{\text{hypotenuse}}{2} = \frac{24}{2} = 12\]

The longer leg (opposite \(60°\)) would be \(12\sqrt{3}\).

    1. \(8\sqrt{3}\): Applied the longer-leg ratio to the hypotenuse: \(\frac{24}{\sqrt{3}} = 8\sqrt{3}\) — divided by \(\sqrt{3}\) instead of \(2\).
    1. \(12\sqrt{3}\): This is the length of the longer leg (opposite \(60°\)), not the shorter leg.
    1. \(24\): Set the shorter leg equal to the hypotenuse — did not apply the \(1:2\) ratio.

Answer: B

Question 14

A right circular cylinder has a radius of \(5\) and a height of \(12\). What is the volume of the cylinder? (Express in terms of \(\pi\).)




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Volume of a cylinder: \(V = \pi r^2 h\)

\[V = \pi (5)^2 (12) = \pi (25)(12) = 300\pi\]

    1. \(60\pi\): Used \(V = \pi r \cdot h = \pi(5)(12)\) — forgot to square the radius.
    1. \(120\pi\): Used \(V = 2\pi r h = 2\pi(5)(12)\) — applied the lateral surface area formula instead of volume.
    1. \(600\pi\): Used \(V = 2\pi r^2 h = 2\pi(25)(12)\) — introduced an extra factor of \(2\).

Answer: H

Question 15

A tangent segment is drawn from external point \(P\) to a circle with center \(O\) and radius \(7\). If the distance from \(P\) to the center \(O\) is \(25\), what is the length of the tangent segment \(PT\)?




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A tangent line is perpendicular to the radius at the point of tangency \(T\), so \(\angle OTP = 90°\). Triangle \(OTP\) is a right triangle with hypotenuse \(OP = 25\) and leg \(OT = 7\).

By the Pythagorean theorem:

\[PT = \sqrt{OP^2 - OT^2} = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24\]

    1. \(18\): Computed \(25 - 7 = 18\) — subtracted directly instead of applying the Pythagorean theorem.
    1. \(20\): Recalled the \(7\)-\(24\)-\(25\) triple but confused which leg is unknown, or used \(\sqrt{25^2 - 7^2}\) with an arithmetic error.
    1. \(\sqrt{674}\): Added instead of subtracted: \(\sqrt{25^2 + 7^2} = \sqrt{625 + 49} = \sqrt{674}\) — treated \(OP\) as a leg rather than the hypotenuse.

Answer: C

Question 16

Which of the following is the equation of a circle with center \((3, -4)\) and radius \(5\)?




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The standard form of a circle with center \((h, k)\) and radius \(r\) is:

\[(x - h)^2 + (y - k)^2 = r^2\]

With center \((3, -4)\) and \(r = 5\):

\[(x - 3)^2 + (y - (-4))^2 = 5^2\] \[(x - 3)^2 + (y + 4)^2 = 25\]

    1. \((x+3)^2 + (y-4)^2 = 25\): Wrong signs on both terms — the signs inside the parentheses must be opposite to the center coordinates.
    1. \((x-3)^2 + (y+4)^2 = 5\): Correct signs but used \(r\) instead of \(r^2\) on the right side.
    1. \((x+3)^2 + (y-4)^2 = 5\): Wrong signs on both terms AND used \(r\) instead of \(r^2\).

Answer: H

Question 17

What are the coordinates of the point on the unit circle corresponding to an angle of \(\dfrac{4\pi}{3}\) radians?




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\(\dfrac{4\pi}{3}\) radians \(= 240°\), which lies in Quadrant III (between \(180°\) and \(270°\)).

The reference angle is \(240° - 180° = 60°\), which corresponds to \(\dfrac{\pi}{3}\).

The benchmark angle \(60°\) has unit circle coordinates \(\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)\).

In Quadrant III, both \(x\) and \(y\) are negative:

\[\left(-\frac{1}{2},\ -\frac{\sqrt{3}}{2}\right)\]

    1. \(\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)\): Correct reference-angle values but placed in Quadrant I — did not apply the Quadrant III sign change.
    1. \(\left(-\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)\): Correct \(x\)-sign for Q III but \(y\) is positive — this corresponds to \(120°\) in Quadrant II.
    1. \(\left(-\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right)\): Swapped \(x\) and \(y\) values and used mixed signs — corresponds to \(150°\).

Answer: D

Question 18

The equation of an ellipse is \(\dfrac{x^2}{49} + \dfrac{y^2}{16} = 1\). What is the distance between the two foci of this ellipse?




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For an ellipse \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) with \(a > b\), the foci lie along the major axis at distance \(c\) from the center, where \(c^2 = a^2 - b^2\).

Here \(a^2 = 49\) and \(b^2 = 16\):

\[c^2 = 49 - 16 = 33 \implies c = \sqrt{33}\]

The two foci are located at \((\pm\sqrt{33},\ 0)\), so the distance between them is:

\[2c = 2\sqrt{33}\]

    1. \(2\sqrt{65}\): Added instead of subtracted: \(c^2 = 49 + 16 = 65\) — used the formula for a hyperbola’s \(c\).
    1. \(6\): Computed \(\sqrt{49} - \sqrt{16} = 7 - 4 = 3\), then doubled to get \(6\) — subtracted the semi-axes instead of their squares.
    1. \(14\): Gave the length of the major axis (\(2a = 14\)) rather than the distance between foci.

Answer: F

Question 19

The figure below shows a rectangle with a semicircle attached to one of its shorter sides. The rectangle has length \(10\) and width \(6\), and the semicircle has a diameter equal to the width of the rectangle. What is the total area of the figure? (Express in terms of \(\pi\).)

10 6

Note: Figure not drawn to scale.




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Rectangle area:

\[A_{\text{rect}} = 10 \times 6 = 60\]

Semicircle area: The diameter equals the width \(= 6\), so the radius \(= 3\).

\[A_{\text{semi}} = \frac{1}{2}\pi r^2 = \frac{1}{2}\pi(3)^2 = \frac{9\pi}{2} = 4.5\pi\]

Total area:

\[A = 60 + 4.5\pi\]

    1. \(60 + 9\pi\): Used the full circle area \(\pi r^2 = 9\pi\) instead of the semicircle area \(\frac{1}{2}\pi r^2\).
    1. \(60 + 18\pi\): Used the full circle area with diameter as radius: \(\pi(6)^2 = 36\pi\), then halved incorrectly to get \(18\pi\).
    1. \(66 + 4.5\pi\): Added the diameter to the rectangle area (\(60 + 6 = 66\)) instead of computing the rectangle area as \(10 \times 6 = 60\).

Answer: A

Question 20

The equation of a hyperbola is \(\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1\). What are the coordinates of the vertices of this hyperbola?




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The standard form \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) is a hyperbola that opens left and right along the \(x\)-axis. The vertices are at \((\pm a,\ 0)\).

Here \(a^2 = 9\), so \(a = 3\). The vertices are at \((\pm 3,\ 0)\).

    1. \((0, \pm 4)\): Identified \(b = 4\) and placed vertices on the \(y\)-axis — this would be the vertices of the conjugate hyperbola \(\dfrac{y^2}{16} - \dfrac{x^2}{9} = 1\).
    1. \((\pm 5, 0)\): Computed \(c = \sqrt{a^2 + b^2} = \sqrt{9+16} = 5\) — found the foci, not the vertices.
    1. \((0, \pm 3)\): Correctly found \(a = 3\) but placed it on the wrong axis — the \(x^2\) term is positive, so the transverse axis is horizontal.

Answer: G