Geometry

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For the Classic ACT exam:
The ACT Mathematics test is a timed exam...60 questions in 60 minutes
This implies that you have to solve each question in one minute.
Each of the first 20 questions (less challenging) will typically take less than a minute a solve.
Each of the next 20 questions (medium challenging) may take about a minute to solve.
Each of the last 20 questions (more challenging) may take more than a minute to solve.
The goal is to maximize your time.
You use the time saved on the questions you solve in less than a minute to solve questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.
Also: please note that unless specified otherwise, geometric figures are drawn to scale. So, you can figure out the correct answer by eliminating the incorrect options.
Other suggestions are listed in the solutions/explanations as applicable.

These are the solutions to the ACT past questions on the topics in Geometry.
When applicable, the TI-84 Plus CE calculator (also applicable to TI-84 Plus calculator) solutions are provided for some questions.
The link to the video solutions will be provided for you. Please subscribe to the YouTube channel to be notified of upcoming livestreams. You are welcome to ask questions during the video livestreams.
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Please NOTE: For applicable questions (questions that require the angle in degrees), if you intend to use the TI-84 Family:
Please make sure you set the MODE to DEGREE.
It is RADIAN by default. So, it needs to be set to DEGREE.

Calculator Mode

Formulas

(1.) $\overleftrightarrow{AB}$
Line AB

(2.) $\overline{AB}$
Line Segment AB

(3.) $\overrightarrow{AB}$
Ray AB

(4.) Given: n collinear points:
the number of different rays that can be named is: 2(n − 1) rays.

(5.) Given: n collinear points:
the number of different lines that can be named if order is important is found from the Permutation formula.
Two points are used to name a line.
Permutation: order is important: perm 2 from n
$P(n, 2) = \dfrac{n!}{(n - 2)!}$

(6.) Given: n collinear points:
the number of different lines that can be named if order is not important is found from the Combination formula.
Two points are used to name a line.
Combination: order is not important: comb 2 from n
$C(n, 2) = \dfrac{n!}{(n - 2)! * 2!}$

(7.) Given: n coplanar points:
the number of different ways to name the plane if order is important is found from the Permutation formula.
Three points are used to name a plane.
Permutation: order is important: perm 3 from n
$P(n, 3) = \dfrac{n!}{(n - 3)!}$

(8.) Given: n coplanar points:
the number of different ways to name the plane if order is not important is found from the Combination formula.
Three points are used to name a plane.
Combination: order is not important: comb 3 from n
$C(n, 3) = \dfrac{n!}{(n - 3)! * 3!}$

(9.) Regular Polygons
Where: the number of sides is greater than 4
Let:
number of sides = n
side length = s
radius (the distance from the center to a vertex) = r
apothem (shortest distance from the center to one of the sides) = a
each interior angle = θ
perimeter = P
area = A

$ \Sigma \theta = 180^\circ(n - 2) \\[3ex] \theta = \dfrac{180^\circ(n - 2)}{n} \\[5ex] a = r\cos\left(\dfrac{180^\circ}{n}\right) \\[6ex] a = \dfrac{s}{2\tan \left(\dfrac{180^\circ}{n}\right)} \\[8ex] s = 2r\sin\left(\dfrac{180^\circ}{n}\right) \\[6ex] s = 2a\tan\left(\dfrac{180^\circ}{n}\right) \\[7ex] r = \dfrac{a}{\cos\left(\dfrac{180^\circ}{n}\right)} \\[8ex] r = \dfrac{s}{2\sin\left(\dfrac{180^\circ}{n}\right)} \\[8ex] P = ns \\[3ex] P = 2rn\sin\left(\dfrac{180^\circ}{n}\right) \\[6ex] P = 2an\tan\left(\dfrac{180^\circ}{n}\right) \\[6ex] A = \dfrac{\pi r^2}{2}\sin\left(\dfrac{360^\circ}{n}\right) \\[7ex] A = \dfrac{aP}{2} \\[5ex] A = \dfrac{ans}{2} \\[5ex] A = \dfrac{\pi a^2}{\cos^2\left(\dfrac{180^\circ}{n}\right)} \sin\left(\dfrac{360^\circ}{n}\right) \\[7ex] $ (10.) Section Formula
Given two points say A(x₁, y₁) and B(x₂, y₂): if a point say C(x, y) divides the line segment |AB| in the ratio: m:n, then the coordinates of C is given by:

$ C = \dfrac{mB + nA}{m + n} \\[5ex] C(x, y) = \left(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}\right) $

Symbols and Meanings for Circles
(h, k) = coordinates of the center of a circle
r = radius of a circle
d = diameter of a circle
C = circumference of a circle (also known as the perimeter)
A = area of a circle
π = pi = $\dfrac{22}{7}$
θ = central angle
° = DEG = degrees
RAD = radians
L = length of arc
A_sec = area of sector
P_sec = perimeter of sector
A_seg = area of segment
P_seg = perimeter of segment
(x₁, y₁) = first endpoint of the diameter of a circle
(x₂, y₂) = second endpoint of the diameter of a circle

Radius, Diameter, Circumference, Area

$ \underline{Circle} \\[3ex] d = 2r \\[3ex] r = \dfrac{d}{2} \\[5ex] C = \pi d \\[3ex] d = \dfrac{C}{\pi} \\[5ex] C = 2\pi r \\[3ex] r = \dfrac{C}{2\pi} \\[5ex] A = \pi r^2 \\[3ex] r = \sqrt{\dfrac{A}{\pi}} \\[5ex] A = \dfrac{\pi d^2}{4} \\[5ex] d = \sqrt{\dfrac{4A}{\pi}} \\[5ex] A = \dfrac{C^2}{4\pi} \\[5ex] C = 2\sqrt{A\pi} \\[5ex] \underline{Semicircle} \\[3ex] d = 2r \\[3ex] r = \dfrac{d}{2} \\[5ex] C = \pi r \\[3ex] C = \dfrac{\pi d}{2} \\[5ex] r = \dfrac{C}{\pi} \\[5ex] d = \dfrac{2C}{\pi} \\[5ex] A = \dfrac{\pi r^2}{2} \\[5ex] r = \sqrt{\dfrac{2A}{\pi}} \\[5ex] A = \dfrac{\pi d^2}{8} \\[5ex] d = \sqrt{\dfrac{8A}{\pi}} \\[7ex] \underline{\theta\;\;in\;\;DEG} \\[3ex] L = \dfrac{2\pi r\theta}{360} \\[5ex] \theta = \dfrac{180L}{\pi r} \\[5ex] r = \dfrac{180L}{\pi \theta} \\[5ex] A_{sec} = \dfrac{\pi r^2\theta}{360} \\[5ex] P_{sec} = \dfrac{r(\pi\theta + 360)}{180} \\[5ex] \theta = \dfrac{360A_{sec}}{\pi r^2} \\[5ex] r = \dfrac{360A_{sec}}{\pi\theta} \\[5ex] A_{sec} = \dfrac{Lr}{2} \\[5ex] A_{sec} = \dfrac{Lr}{2} \\[5ex] r = \dfrac{2A_{sec}}{L} \\[5ex] L = \dfrac{2A_{sec}}{r} \\[5ex] \underline{\theta\;\;in\;\;RAD} \\[3ex] L = r\theta \\[5ex] \theta = \dfrac{L}{r} \\[5ex] r = \dfrac{L}{\theta} \\[5ex] A_{sec} = \dfrac{r^2\theta}{2} \\[5ex] \theta = \dfrac{2A_{sec}}{r^2} \\[5ex] r = \sqrt{\dfrac{2A_{sec}}{\theta}} \\[5ex] $ (1.) Standard Form of the Equation of a Circle
$(x - h)^2 + (y - k)^2 = r^2$
where:
$x, y$ are the variables
$(h, k)$ are the coordinates of the center of the circle
$r$ is the radius of the circle

(2.) General Form of the Equation of a Circle
$x^2 + y^2 + 2gx + 2fy + c = 0$
where:
$x, y$ are the variables
$c$ is the coefficient of $x$
$d$ is the coefficient of $y$
$c, d, e$ are values/constants

(3.) Given: The Center Coordinates of a Circle and an Endpoint on the Circumference of the Circle
The coordinates of the center of the circle = $(h, k)$
The endpoint on the circumference of the circle = $(x_1, y_1)$
The radius of the circle can be found by the Distance Formula
The radius of the circle = $r$
r = $\sqrt{(x_1 - h)^2 + (y_1 - k)^2}$
The diameter of the circle = $d$
The diameter of the circle is twice the radius.
$d = 2 * r$
The second endpoint of the diameter of the circle can also be found
The second endpoint of the diameter of the circle = $(x_2, y_2)$

$ x_2 = x_1 + r \\[3ex] y_2 = y_1 + r \\[3ex] (x_2, y_2) = (x_1 + r, y_1 + r) \\[3ex] $ (4.) Given: The Endpoints of the Diameter of the Circle
$(x_1, y_1)$ = first endpoint of the diameter of a circle
$(x_2, y_2)$ = second endpoint of the diameter of a circle
The center of the circle is found using the Midpoint Formula
$(h, k)$ are the coordinates of the center of the circle

$ h = \dfrac{x_1 + x_2}{2} \\[5ex] k = \dfrac{y_1 + y_2}{2} \\[5ex] $ Euler's Theorem: The number of faces (F ), vertices (V ), and edges (E ) of a polyhedron are related by the formula: F + V = E + 2
⇒ F + V − E = 2

In naming a line, two points are used.
In naming a plane, three points are used.

We can measure angles in:
Degrees (DEG, $^\circ$)
Radians (RAD)
Gradians (GRAD)
Degrees, Minutes, and Seconds ($^\circ \:'\:''$)

$DRG$ means $Degree-Radian-Gradian$ in some calculators
$180^\circ = \pi \:\:RAD = 200 \:\:GRAD$

To convert from:
radians to degrees, multiply by $\dfrac{180}{\pi}$

degrees to radians, multiply by $\dfrac{\pi}{180}$

$DMS$ means $Degree-Minute-Second$ in some calculators

$ 1^\circ = 60' \\[3ex] 1^\circ = 3600'' \\[3ex] 1^\circ = 60' = 3600'' \\[3ex] 1' = 60'' \\[3ex] $ To convert from:
degrees to minutes, multiply by $60$
minutes to degrees, divide by $60$
degrees to seconds, multiply by $3600$
seconds to degrees, divide by $3600$
minutes to seconds, multiply by $60$
seconds to minutes, divide by $60$

Show students these angular measures in their scientific calculators.
Show them how to convert from one angular measure to another.

Ellipses

(h, k) = coordinates of the center of an ellipse
Ellipse: a = horizontal distance from the center to the boundary
Ellipse: b = vertical distance from the center to the boundary
Horizontal Ellipse: a = half the length of the major axis OR the length of the semi-major axis
Horizontal Ellipse: b = half the length of the minor axis OR the length of the semi-minor axis
Vertical Ellipse: a = half the length of the minor axis OR the length of the semi-minor axis
Vertical Ellipse: b = half the length of the major axis OR the length of the semi-major axis
c = linear eccentricity = distance from the center of the ellipse to the foci
$f_1,f_2$ = foci points
$f$ = distance between the foci
e = eccentricity
A = area
P = perimeter
L = length of the latus rectum
$L_1,\;L_2,\;L_3,\;L_4$ = endpoints of the latus rectum
D = distance between the two directrixes
C = fractional factorials binomial coefficient
λ = square of the ratio of the difference of the semi-axis to the sum of the semi-axis
n = number of terms

(1.) Standard Form of the Equation of an Ellipse

$ \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1 \\[5ex] $ If $a \gt b$, the ellipse is a horizontal ellipse.
If $a \lt b$, the ellipse is a vertical ellipse.

(2.) Eccentricity

$ e = \dfrac{\sqrt{semiMajor\;\;axis\;\;length^2 - semiMinor\;\;axis\;\;length^2}}{semiMajor\;\;axis\;\;length} \\[3ex] $ In other words:
Eccentricity: Horizontal Ellipse:
Major axis is the horizontal axis
Semi-major axis length = a

$ e = \dfrac{\sqrt{a^2 - b^2}}{a} \\[3ex] $ Eccentricity: Vertical Ellipse:
Major axis is the vertical axis
Semi-major axis length = b

$ e = \dfrac{\sqrt{b^2 - a^2}}{b} \\[5ex] $

Theorems

(1.) The sum of the interior angles of a triangle is 180°

(2.) The sum of angles on a straight line is 180°

(3.) The exterior angle of a triangle is the sum of the two interior opposite angles.

(4.)

Statements

(1.) Regular Polygon Symmetry: A regular polygon has equal side lengths and congruent interior angle vertices.
This symmetry leads to this statement:
Any line drawn from the center of the center (radius) to a vertex of a regular polygon bisects the interior angle of the polygon.

(1.) Numbered below are the 14 angles formed by parallel lines l and m and transversals p and q.
Lines l, p, and q intersect at a single point.
Which of the following congruence statements must be true?

Number 1

$ A.\;\; \angle 1 \cong \angle 5 \\[3ex] B.\;\; \angle 2 \cong \angle 1 \\[3ex] C.\;\; \angle 3 \cong \angle 10 \\[3ex] D.\;\; \angle 4 \cong \angle 2 \\[3ex] E.\;\; \angle 5 \cong \angle 8 \\[3ex] $

$ \angle 5 \cong \angle 8 ... alternate\;\;exterior\;\angle s\;\;are\;\;congruent \\[3ex] $

(2.) A section of highway is represented by a line segment with endpoints (30, 50) and (90, 100) in the standard (x, y) coordinate plane.
Exactly halfway along this highway section is a road sign.
What are the coordinates of the road sign?

$ F.\;\; (30, 25) \\[3ex] G.\;\; (60, 50) \\[3ex] H.\;\; (60, 75) \\[3ex] J.\;\; (70, 65) \\[3ex] K.\;\; (75, 60) \\[3ex] $

The coordinates of the road sign is found by the Midpoint Formula.

$ Endpoint\;1 = (30, 50) \\[3ex] x_1 = 30 \\[3ex] y_1 = 50 \\[3ex] Endpoint\;2 = (90, 100) \\[3ex] x_2 = 90 \\[3ex] y_2 = 100 \\[3ex] Midpoint = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right) \\[5ex] = \left(\dfrac{30 + 90}{2}, \dfrac{50 + 100}{2}\right) \\[5ex] = \left(\dfrac{120}{2}, \dfrac{150}{2}\right) \\[5ex] = (60, 75) $

(3.) A circle with the equation $x^2 + y^2 = 16$ is graphed in the standard (x, y) coordinate plane.
At what points does the circle intersect the x-axis?

$ F.\;\; (-2, 0) \;\;and\;\; (2, 0) \\[3ex] G.\;\; (-4, 0) \;\;and\;\; (4, 0) \\[3ex] H.\;\; (-8, 0) \;\;and\;\; (8, 0) \\[3ex] J.\;\; (-16, 0) \;\;and\;\; (16, 0) \\[3ex] K.\;\; (-32, 0) \;\;and\;\; (32, 0) \\[3ex] $

Based on the Standard Form of the Equation of a Circle, compare: $(x - h)^2 + (y - k)^2 = r^2$ with $x^2 + y^2 = 16$
This implies that:

$ x^2 + y^2 = 16 \\[4ex] (x - 0)^ + (y - 0)^2 = 16 \\[4ex] \implies \\[3ex] center = (h, k) = (0, 0) \\[3ex] r^2 = 16 \\[4ex] r = \pm \sqrt{16} \\[3ex] r = \pm 4 \\[3ex] $ The radius cannot be a negative number.
However, we shall consider the two values in finding the intersection on the x-axis.

Number 3

This implies that:
Intersecion on the x-axis = (-4, 0) and (4, 0)

(4.) For $\overleftrightarrow{RT}$ shown below, point S is on $\overline{RT}$, the length of $\overline{RS}$ is 6 cm, and the length of $\overline{ST}$ is 20 cm.
What is the distance, in centimeters, between T and the midpoint of $\overline{RS}$?

Number 4

$ A.\;\; 13 \\[3ex] B.\;\; 16 \\[3ex] C.\;\; 20 \\[3ex] D.\;\; 23 \\[3ex] E.\;\; 26 \\[3ex] $

$ \text{Midpoint of}\; \overline{RS} = \dfrac{1}{2} \cdot 6 = 3 \\[5ex] \text{Distance between T and the midpoint of}\; \overline{RS} \\[3ex] = \text{Distance between the midpoint of}\; \overline{RS} + \text{length of}\; \overline{ST} \\[3ex] = 3 + 20 \\[3ex] = 23 $

(5.) The ellipse in the standard (x, y) coordinate plane below is centered at the origin.
The endpoints of the major and minor axes of the ellipse are labeled.
Which of the following equations determines this ellipse?

Number 5

$ A.\;\; (x - 8)^2 + (y - 5)^2 = 1 \\[4ex] B.\;\; (x + 8)^2 + (y + 5)^2 = 1 \\[4ex] C.\;\; \dfrac{x^2}{8} + \dfrac{y^2}{5} = 1 \\[6ex] D.\;\; \dfrac{x^2}{16} + \dfrac{y^2}{10} = 1 \\[6ex] E.\;\; \dfrac{x^2}{64} + \dfrac{y^2}{25} = 1 \\[6ex] $

This is a Horizontal Ellipse

The easiest approach to solve these kinds of questions on ellipses because ACT is a timed test.

$ a = 8 \\[3ex] b = 5 \\[3ex] (h, k) = (0, 0) \\[3ex] \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1 \\[6ex] \dfrac{(x - 0)^2}{8^2} + \dfrac{(y - 0)^2}{5^2} = 1 \\[6ex] \dfrac{x^2}{64} + \dfrac{y^2}{25} = 1 \\[5ex] $ Student: I would like to learn to solve it without considering the 1-minute per question time of the ACT.
May you please explain the solution?
Show all work.
Teacher: Sure, let's do it.

$ \underline{Center} \\[3ex] Center = (h, k) = origin = (0, 0) \\[3ex] $ We can calculate the length of both axis using at least two approaches: the major axis and the minor axis
Use any approach you prefer.

$ \underline{First\;\;Approach:\;\;Distance\;\;Formula} \\[3ex] distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[5ex] \text{Major Axis = x-axis} \\[3ex] Endpoints = (-8, 0)\;\;and\;\;(8, 0) \\[3ex] Point\;1 = (-8, 0) \\[3ex] x_1 = -8 \\[4ex] y_1 = 0 \\[4ex] Point\;2 = (8, 0) \\[3ex] x_2 = 8 \\[4ex] y_2 = 0 \\[4ex] distance = \sqrt{(8 - -8)^2 + (0 - 0)^2} \\[4ex] = \sqrt{(8 + 8)^2 + (0)^2} \\[4ex] = \sqrt{16^2 + 0^2} \\[4ex] = \sqrt{256 + 0} \\[4ex] = \sqrt{256} \\[3ex] = 16 \\[3ex] a = \dfrac{16}{2} \\[5ex] a = 8 \\[5ex] \text{Minor Axis = y-axis} \\[3ex] Endpoints = (0, 5)\;\;and\;\;(0, -5) \\[3ex] Point\;1 = (0, 5) \\[3ex] x_1 = 0 \\[4ex] y_1 = 5 \\[4ex] Point\;2 = (0, -5) \\[3ex] x_2 = 0 \\[4ex] y_2 = -5 \\[4ex] distance = \sqrt{(0 - 0)^2 + (-5 - 5)^2} \\[4ex] = \sqrt{0^2 + (-10)^2} \\[4ex] = \sqrt{0 + 100} \\[4ex] = \sqrt{100} \\[3ex] = 10 \\[3ex] b = \dfrac{10}{2} \\[5ex] b = 5 \\[5ex] \underline{Second\;\;Approach:\;\;Coordinates} \\[3ex] \text{Major Axis = x-axis} \\[3ex] Half-length = a \\[3ex] Endpoints = (-8, 0)\;\;and\;\;(8, 0) \\[3ex] x-coordinates\;\;of\;\;Endpoints = -8\;\;and\;\;8 \\[3ex] x-coordinate\;\;of\;\;Center = 0 \\[3ex] \implies \\[3ex] -8 + a = 0 \;\;\;OR\;\;\; 8 - a = 0 \\[3ex] a = 0 + 8 \;\;\;OR\;\;\; 8 - 0 = a \\[3ex] a = 8 \\[5ex] \text{Minor Axis = y-axis} \\[3ex] Half-length = b \\[3ex] Endpoints = (0, 5)\;\;and\;\;(0, -5) \\[3ex] y-coordinates\;\;of\;\;Endpoints = 5\;\;and\;\;-5 \\[3ex] y-coordinate\;\;of\;\;Center = 0 \\[3ex] \implies \\[3ex] 5 - b = 0 \;\;\;OR\;\;\; -5 + b = 0 \\[3ex] 5 + 0 = b \;\;\;OR\;\;\; b = 0 + 5 \\[3ex] b = 5 \\[5ex] \underline{Standard\;\;Form} \\[3ex] \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1 \\[5ex] \dfrac{(x - 0)^2}{8^2} + \dfrac{(y - 0)^2}{5^2} = 1 \\[5ex] \dfrac{x^2}{64} + \dfrac{y^2}{25} = 1 \\[5ex] LCD = 64 \cdot 25 = 1600 \\[3ex] 1600 \cdot \dfrac{x^2}{64} + 1600 \cdot \dfrac{y^2}{25} = 1600(1) \\[5ex] 25x^2 + 64y^2 = 1600 $

(6.) The graph in the standard (x, y) coordinate plane below has 5 lines of symmetry that all intersect at the origin.
What is the smallest positive clockwise rotation about the origin that can be applied to this graph with the result being this graph?

Number 6

$ F.\;\; 72^\circ \\[4ex] G.\;\; 90^\circ \\[4ex] H.\;\; 120^\circ \\[4ex] J.\;\; 144^\circ \\[4ex] K.\;\; 216^\circ \\[4ex] $

5 lines of symmetry intersecting at the origin implies that the graph is symmetrical at regular intervals of each rotation.
A full rotation = 360°
This is equivalent to 5 rotations corresponding to the 5 lines of symmetry

$ 5\;rotations \rightarrow 360^\circ \\[4ex] 1\;rotation \rightarrow \dfrac{360}{5} = 72^\circ \\[3ex] $ The smallest positive clockwise rotation about the origin = 72°

(7.) The points K, L, M, and N are on a number line in that order such that KM = 15 units, LN = 26 units, and KN = 32 units.
The coordinate of L is 0, and the coordinate of K is negative.
What is the coordinate of M?

$ F.\;\; 6 \\[3ex] G.\;\; 9 \\[3ex] H.\;\; 11 \\[3ex] J.\;\; 15 \\[3ex] K.\;\; 17 \\[3ex] $

Let the coordinate of K = −n
Let the coordinate of M = p

Let us represent the information diagrammatically:
Number 7

Let us use a table to solve the question for easier comprehension.

$\|KL\|$	$\|KL\|$	$\|KM\|$	$\|KM\|$
$ 0 - (-n) \\[3ex] 0 + n \\[3ex] n $	$ 32 - 26 \\[3ex] 6 $	$ p - (-n) \\[3ex] p + n \\[3ex] $	15
$n = 6$		$ p + n = 15 \\[3ex] p + 6 = 15 \\[3ex] p = 15 - 6 \\[3ex] p = 9 $

The coordinate of M is 9.

(8.) Which of the following points is NOT on the graph of the function $y = |2x| - 4$ in the standard (x, y) coordinate plane?

$ F.\;\; (-2, 0) \\[3ex] G.\;\; (-1, 2) \\[3ex] H.\;\; (1, -2) \\[3ex] J.\;\; (2, 0) \\[3ex] K.\;\; (4, 4) \\[3ex] $

Let us test each coordinate until we get an output that does not correspond to the input

$y = |2x| - 4$
$(-2, 0)$	$ x = -2 \\[3ex] y = 0 \\[3ex] \|2(-2)\| - 4 \\[3ex] \|-4\| - 4 \\[3ex] 4 - 4 \\[3ex] 0 ...\checkmark $
$(-1, 2)$	$ x = -1 \\[3ex] y = 2 \\[3ex] \|2(-1)\| - 4 \\[3ex] \|-2\| - 4 \\[3ex] 2 - 4 \\[3ex] -2...STOP \\[3ex] -2 \ne 2 ...\text{Option G. is the answer} $

Student: Is it not recommended to test each option just to be sure?
Teacher: Because of the time limit for each question on the ACT, we need not test each option.
Student: Can we just complete it just to be sure that all the options checks well?
Teacher: Sure, let's do it.

$y = |2x| - 4$
$(1, -2)$	$ x = 1 \\[3ex] y = -2 \\[3ex] \|2(1)\| - 4 \\[3ex] \|2\| - 4 \\[3ex] 2 - 4 \\[3ex] -2 ...\checkmark $
$(2, 0)$	$ x = 2 \\[3ex] y = 0 \\[3ex] \|2(2)\| - 4 \\[3ex] \|4\| - 4 \\[3ex] 4 - 4 \\[3ex] 0 ...\checkmark $
$(4, 4)$	$ x = 4 \\[3ex] y = 4 \\[3ex] \|2(4)\| - 4 \\[3ex] \|8\| - 4 \\[3ex] 8 - 4 \\[3ex] 4 ...\checkmark $

(9.) A circle is tangent to the x-axis and to the y-axis.
The coordinates of its center are both positive.
The area of the circle is 64π.
What are the coordinates of the point of tangency on the y-axis?

$ F.\;\; (0, 4) \\[3ex] G.\;\; (0, 8) \\[3ex] H.\;\; (0, 16) \\[3ex] J.\;\; (0, 32) \\[3ex] K.\;\; (0, 64) \\[3ex] $

The coordinates of its center are both positive: this implies that the center is in the First Quadrant
It is better to represent this information diagramaatically:

Number 9

To get the coordinates of the point of tangency on the y-axis, we need to find the radius
The radius is the y-coordinate of the point of tangency on the y-axis

$ Area = \pi \cdot radius^2 \\[4ex] 64\pi = \pi \cdot radius^2 \\[4ex] radius^2 = 64 \\[4ex] radius = +\sqrt{64}...\text{The length of the radius must be positive} \\[3ex] radius = 8 \\[3ex] \therefore \text{The coordinates of the point} = (0, 8) ...\text{First Quadrant} $

(10.) Point (5, −1), which is graphed in the standard (x, y) coordinate plane below, will be reflected across the x-axis.
What will be the coordinates of the image of P?

Number 10

$ F.\;\; (-5, -1) \\[3ex] G.\;\; (-5, 1) \\[3ex] H.\;\; (-1, 5) \\[3ex] J.\;\; (5, -1) \\[3ex] K.\;\; (5, 1) \\[3ex] $

Reflected across the x-axis is Vertical Reflection
This implies that only the y-coordinate changes (by multiplying by −1).
The x-coordinate does not change.

$ (5, -1) \hspace{1em} \underrightarrow{VERE} \hspace{1em} (5, -1 \cdot 1) \\[3ex] \hspace{7em} \rightarrow (5, 1) $

(11.) The following shapes are a circle and some regular polygons that have their centers marked.
All units are given in feet.
Which of these shapes has the greatest area?

Number 11

F. is a circle with radius, 1 unit
G. is a square with apothem, 1 unit
H. is a square with radius, 1 unit
J. is an octagon with radius, 1 unit
K. is an octagon with apothem, 1 unit

As the number of sides of the polygon increases, the shape becomes more circular, making the area approach that of a circle.
This typically implies that the area of the circle is greater than the area of the octagon.
So, options J. and K. are out.

For the same radius of 1 unit, a circle has a greater area than a square because a circle encloses the maximum area possible for a given perimeter due to its continuous curvature, while a square has corners that do not use the radius as efficiently.
So, option H, is out.

We are now left with the two options of F. and G.
For the Square: The apothem is a shorter distance within the square (from center to midpoint of a side) which allows for a longer side length.
For the Circle: The radius of the circle defines the circle's circumference, leading to a smaller enclosed area when compared to the larger side length of the square.
Hence, a square with an apothem of 1 unit has a greater area than a circle with a radius of 1 unit.
Option F. is out.

This implies that the correct option is G.

Student: Mr. C, I want to see these areas worked out so I know the correct option.
Teacher: No problem. We shall work it out.
However, please note that each question on the ACT is limited to 1 minute.
Working it out will take more than a minute.
So, keep that in mind when taking your ACT.

$ \underline{Option\;F:\;\;Circle} \\[3ex] r = 1\;unit \\[3ex] A = \pi r^2 \\[3ex] A = \pi \cdot (1)^2 \\[4ex] A = 3.141592654\;square\;\;units \\[5ex] \underline{Option\;G:\;\;Square} \\[3ex] a = 1\;unit \\[3ex] L = 2a = 2(1) = 2 \\[3ex] A = L^2 = 2^2 \\[3ex] A = 4\;square\;\;units \\[5ex] \underline{Option\;H:\;\;Square} \\[3ex] r = 1\;unit \\[3ex] \implies\;\;4\;\;Right\;\;\triangle s \\[3ex] Area\;\;of\;\;1\;\;Right\;\;\triangle \\[3ex] b = r = 1\;cm \\[3ex] h = 1\;cm \\[3ex] A = \dfrac{1}{2} \cdot b \cdot h \\[5ex] A = \dfrac{1}{2} \cdot 1 \cdot 1 \\[5ex] A = \dfrac{1}{2}\;units \\[5ex] Area\;\;of\;\;4\;\;Right\;\;\triangle s \\[3ex] A = \dfrac{1}{2} \cdot 4 \\[5ex] A = 2\;square\;\;units \\[5ex] \underline{Option\;J:\;\;Octagon} \\[3ex] r = 1\;unit \\[3ex] n = 8 \\[3ex] A = \dfrac{\pi r^2}{2}\sin\left(\dfrac{360^\circ}{n}\right) \\[7ex] A = \dfrac{\pi \cdot (1)^2}{2}\sin\left(\dfrac{360^\circ}{8}\right) \\[7ex] A = \dfrac{\pi}{2}\sin 45^\circ \\[5ex] A = 1.110720735\;square\;\;units \\[5ex] \underline{Option\;K:\;\;Octagon} \\[3ex] a = 1\;unit \\[3ex] A = \dfrac{\pi a^2}{\cos^2\left(\dfrac{180^\circ}{n}\right)} \sin\left(\dfrac{360^\circ}{n}\right) \\[7ex] A = \dfrac{\pi \cdot (1)^2}{\cos^2\left(\dfrac{180^\circ}{8}\right)} \sin\left(\dfrac{360^\circ}{8}\right) \\[7ex] A = \dfrac{\pi\sin(45^\circ)}{\cos^2(22.5^\circ)} \\[7ex] A = 2.602580569\;square\;\;units \\[3ex] $ The results further confirm that Option G. is the correct answer.

(12.) As shown in the figure below, points A, B, and D lie on a line.
The measure of angle ABC ($m\angle ABC$) is x°, and ($m\angle CBD$) is (5x + 4)°.

Number 12

$ A.\;\; (5x + 4) = x \\[3ex] B.\;\; x - (5x + 4) = 90 \\[3ex] C.\;\; x + (5x + 4) = 90 \\[3ex] D.\;\; x + (5x + 4) = 180 \\[3ex] E.\;\; x + (5x + 4) = 360 \\[3ex] $

$ x + (5x + 4) = 180 ...\text{sum of angles on a straight line} $

(13.) In the figure below, $\overleftrightarrow{DE} \;||\; \overleftrightarrow{GF}$, point A lies on $\overline{DE}$, points C and B lie on $\overline{GF}$, $m\angle GCA = 120^\circ$, and $m\angle GBA = 50^\circ$.
What is $m\angle CAB$?
(Note: The figure is NOT drawn to scale.
The degree measure of $\angle STU$ is denoted $m\angle STU$.)

Number 13

$ F.\;\; 10^\circ \\[4ex] G.\;\; 20^\circ \\[4ex] H.\;\; 50^\circ \\[4ex] J.\;\; 60^\circ \\[4ex] K.\;\; 70^\circ \\[4ex] $

Let us update the information on the diagram

Number 13

$ m\angle GCA = m\angle CAB + m\angle GBA ...\text{exterior angle of a triangle = sum of two interior opposite angles} \\[3ex] 120^\circ = m\angle CAB + 50^\circ \\[4ex] m\angle CAB = 120 - 50 \\[3ex] m\angle CAB = 70^\circ $

(14.) In the figure below, C is the intersection of $\overline{AD}$ and $\overline{BE}$
What is the measure of $\angle BAC$?

Number 14

$ F.\;\; 85^\circ \\[3ex] G.\;\; 95^\circ \\[3ex] H.\;\; 115^\circ \\[3ex] J.\;\; 120^\circ \\[3ex] K.\;\; 140^\circ \\[3ex] $

$ \angle BCA = 55^\circ ...\text{vertical angles are congruent} \\[3ex] \angle BAC + \angle ABC + \angle BCA = 180^\circ ...\text{sum of angles in triangle BAC} \\[3ex] \angle BAC + 30^\circ + 55^\circ = 180^\circ \\[3ex] \angle BAC = 180 - 30 - 55 \\[3ex] \angle BAC = 95^\circ $

(15.) Points A(8,5) and C(−4,11) lie in the standard (x, y) coordinate plane.
Point B lies on $\overline{AC}$ such that AB:BC = 1:2
What are the coordinates of B?

$ F.\;\; (-8, 13) \\[3ex] G.\;\; (0, 9) \\[3ex] H.\;\; (2, 8) \\[3ex] J.\;\; (3, 6) \\[3ex] K.\;\; (4, 7) \\[3ex] $

We can solve this question using at least two approaches.
Use any approach you prefer.

$ \underline{\text{1st Approach: Section Formula}} \\[3ex] A(8, 5) \hspace{5em} C(-4, 11) \\[3ex] x_1 = 8 \hspace{5em} x_2 = -4 \\[4ex] y_1 = 5 \hspace{5em} y_2 = 11 \\[4ex] \text{Point B divides |AC| in the ratio of 1:2} \\[5ex] \text{Coordinates of B} = \left[\dfrac{1(-4) + 2(8)}{1 + 2}, \dfrac{1(11) + 2(5)}{1 + 2}\right] \\[5ex] = \left(\dfrac{-4 + 16}{3}, \dfrac{11 + 10}{3}\right) \\[5ex] = \left(\dfrac{12}{3}, \dfrac{21}{3}\right) \\[5ex] = (4, 7) \\[5ex] \underline{\text{2nd Approach: Vector Algebra}} \\[3ex] \overrightarrow{A} = (8, 5) \\[4ex] \overrightarrow{C} = (-4, 11) \\[4ex] \overrightarrow{AC} = \overrightarrow{C} - \overrightarrow{A} \\[4ex] = (-4, 11) - (8, 5) \\[3ex] = (-4 - 8, 11 - 5) \\[3ex] = (-12, 6) \\[5ex] \overrightarrow{AB} : \overrightarrow{BC} = 1 : 2 \\[4ex] \text{Sum of ratios} = 1 + 2 = 3 \\[3ex] \overrightarrow{AB} = \dfrac{1}{3} \cdot \overrightarrow{AC} \\[6ex] = \dfrac{1}{3} \cdot (-12, 6) \\[5ex] = \left(-\dfrac{12}{3}, \dfrac{6}{3}\right) \\[5ex] = (-4, 2) \\[5ex] \text{Coordinates of B} = \overrightarrow{A} + \overrightarrow{AB} \\[5ex] = (8, 5) + (-4, 2) \\[3ex] = (8 + -4, 5 + 2) \\[3ex] = (4, 7) $

(16.) For $\overleftrightarrow{RT}$ shown below, point S is on $\overline{RT}$, the length of $\overline{RS}$ is 10 cm, and the length of $\overline{ST}$ is 22 cm.
What is the distance, in centimeters, between T and the midpoint of $\overline{RS}$?

Number 16

$ A.\;\; 16 \\[3ex] B.\;\; 21 \\[3ex] C.\;\; 22 \\[3ex] D.\;\; 27 \\[3ex] E.\;\; 32 \\[3ex] $

$ \text{Midpoint of}\; \overline{RS} = \dfrac{1}{2} \cdot 10 = 5 \\[5ex] \text{Distance between T and the midpoint of}\; \overline{RS} \\[3ex] = \text{Distance between the midpoint of}\; \overline{RS} + \text{length of}\; \overline{ST} \\[3ex] = 5 + 22 \\[3ex] = 27 $

(17.) A certain rectangle in the standard (x, y) coordinate plane has a length of 7 coordinate units and a width of 5 coordinate units.
The point (0,0) is in the interior of this rectangle.
Two vertices of this rectangle are at (3, 2) and (3, −3).
What is the x-coordinate of the other 2 vertices?

$ A.\;\; -4 \\[3ex] B.\;\; -3 \\[3ex] C.\;\; -2 \\[3ex] D.\;\; 8 \\[3ex] E.\;\; 10 \\[3ex] $

Let us represent this information using a diagram.
Indicate the points given already.
Then, count 7 units to the left

Number 17

Notice that (0, 0) must be inside the rectangle, hence we have to count to the left from x = 3
But, what if you do not want to draw on a graph paper (or sketch a graph paper)?

$ Length = 7\;units \\[3ex] |x - 3| = 7 \\[3ex] x - 3 = 7 \\[3ex] x = 10 \\[5ex] OR \\[3ex] -(x - 3) = 7 \\[3ex] -1 \cdot -(x - 3) = -1 \cdot 7 \\[3ex] x - 3 = -7 x = -7 + 3 \\[3ex] x = -4 \\[3ex] $ But (0,0) should be included in the interior of the rectangle
Discard the x-coordinate of 10 because it does not include (0,0)
This implies that the x-coordinate of the other two vertices = −4

(18.) If an obtuse angle is bisected, the resulting angles must:

A. both be acute.
B. be complementary.
C. both be obtuse.
D. both be right.
E. be supplementary.

Teacher: Dear Student
Student: Yes Mr. C
Teacher: What is the greatest value of an obtuse angle that is a whole number?
Student: 179°
Teacher: Bisecting ...note the term: "bi"...this means "two"
Bisecting an angle means dividing the angle in two equal measures
Bisecting angle 179° gives...
Student: 89.5°...this is an acute angle
Teacher: Thank you.
What is the smallest whole number obtuse angle?
Student: 91°
Bisecting it gives 45.5°...this is also an acute angle

Therefore, if an obtuse angle is bisected, the resulting angles must: both be acute.

(19.) One of the following equations is an equation of the circle graphed in the standard (x, y) coordinate plane below.
Which one is it?

Number 19

$ F.\;\; x^2 + y^2 = 1 \\[4ex] G.\;\; x^2 + y^2 - 2x = 1 \\[4ex] H.\;\; x^2 + y^2 - 2y = 1 \\[4ex] J.\;\; x^2 + y^2 + 2x = 1 \\[4ex] K.\;\; x^2 + y^2 + 2y = 1 \\[4ex] $

We are not given the center of the circle.
Neither are we given the coordinates.
We can assume the coordinates of the points on the circumference visually
However, let us analyze the graphs of each option.

$F.\;\; x^2 + y^2 = 1$
Number 19-F

$G.\;\; x^2 + y^2 - 2x = 1$
Number 19-G

$H.\;\; x^2 + y^2 - 2y = 1$
Number 19-H

$J.\;\; x^2 + y^2 + 2x = 1$
Number 19-J

$K.\;\; x^2 + y^2 + 2y = 1$
Number 19-K

As you can see, Option J. is the correct option.
For this kind of question, I recommend that you select the answer based on your knowledge of graphs.
Carefully observe the coefficients of x or y in these graphs:
−2x in Option G.
−2y in Option H.
2x in Option J.
2y in Option K.

Top

(21.) In the standard (x, y) coordinate plane below, point C has coordinates (7, 5), and the midpoint of $\overline{AC}$ is point B, which has coordinates $\left(-\dfrac{3}{2}, 1\right)$.
What are the coordinates of point A?

Number 21

$ A.\;\; (-10, -3) \\[3ex] B.\;\; (4, 7) \\[3ex] C.\;\; \left(-\dfrac{17}{2}, 4\right) \\[5ex] D.\;\; \left(\dfrac{11}{2}, 6\right) \\[5ex] E.\;\; \left(\dfrac{11}{4}, 3\right) \\[5ex] $

$ Point\;A(x_1, y_1) = (?, ?) \hspace{3em} Point\;C(x_2, y_2) = (7, 5) \\[5ex] \text{Midpoint of } \overline{AC}\;(x, y) = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right) = \left(-\dfrac{3}{2}, 1\right) \\[5ex] \implies \\[3ex] \dfrac{y_1 + 5}{2} = 1 \\[5ex] y_1 + 5 = 2(1) \\[3ex] y_1 + 5 = 2 \\[3ex] y_1 = 2 - 5 \\[3ex] y_1 = -3 \\[3ex] \text{Correct answer is Option A} \\[3ex] $ Student: But, we are yet to find x₁
Also, could you explain why you found y₁ first rather than x₁?
Teacher: Remember we have to solve this question in about a minute.
So, we use the process of elimination to get our answer.
Option A. is the only option that has −3 as the y-coordinate
I found y₁ first because dealing with a positive integer: 1 is faster than dealing with a negative fraction: $-\dfrac{3}{2}$
Student: But can we complete it if you do not mind?
Teacher: Sure, let's do it.

$ \dfrac{x_1 + 7}{2} = -\dfrac{3}{2} \\[5ex] x_1 + 7 = -3 \\[3ex] x_1 = -3 - 7 \\[3ex] x_1 = -10 \\[3ex] \therefore \text{the coordinates of } A = (-10, -3) $

(23.) Regular octagon ABCDEFGH is inscribed in a circle, as shown below.
The sector of the circle bounded by radii $\overline{AJ}$ and $\overline{DJ}$ and by $\overset{\huge\frown}{AD}$ is shaded.
The area of the shaded sector is what fraction of the area of the circle?

Number 23

$ A.\;\; \dfrac{1}{8} \\[5ex] B.\;\; \dfrac{1}{4} \\[5ex] C.\;\; \dfrac{3}{10} \\[5ex] D.\;\; \dfrac{3}{8} \\[5ex] E.\;\; \dfrac{1}{2} \\[5ex] $

Regular Polygon Symmetry: A regular polygon has equal side lengths and congruent interior angle vertices.
This symmetry leads to this statement:
Any line drawn from the center of the center (radius) to a vertex of a regular polygon bisects the interior angle of the polygon.
So, isosceles triangles are formed as seen below.

Construction: Draw radii from J to vertices B and C
This creates isosceles triangles: AJB, BJC, CJD

Number 23

$ \underline{\text{Regular Octogan: ABCDEFGH}} \\[3ex] \text{number of sides, }n = 8 \\[3ex] \text{Sum of interior angles}, Sum \\[3ex] = 180(n - 2) \\[3ex] = 180(8 - 2) \\[3ex] = 180(6) \\[3ex] = 1080^\circ \\[5ex] \text{Each interior angle} = \dfrac{Sum}{n} \\[5ex] = \dfrac{1080}{8} \\[5ex] = 135^\circ \\[5ex] \implies \\[3ex] 2\alpha = 135^\circ \\[3ex] \underline{\triangle BJC} \\[3ex] \beta + \alpha + \alpha = 180^\circ ...\text{sum of angles in a triangle} \\[3ex] \beta + 2\alpha = 180 \\[3ex] \beta + 135 = 180 \\[3ex] \beta = 180 - 135 \\[3ex] \beta = 45^\circ \\[5ex] \underline{\text{Sahded Sector JAD}} \\[3ex] \text{central angle} = \beta + \beta + \beta \\[3ex] = 3\beta \\[3ex] = 3(45) \\[3ex] = 135^\circ \\[3ex] radius = r \\[3ex] Area, A_S = \dfrac{\pi r^2\theta}{360} \\[5ex] = \dfrac{3\pi r^2}{8} \\[5ex] \text{Circle} \\[3ex] radius = r ...\text{same radius as the radius of the sector} \\[3ex] Area, A_C = \pi r^2 \\[5ex] \text{Area of the shaded sector is what fraction of the Area of the circle?} \\[3ex] A_S = what * A_C \\[3ex] \dfrac{3\pi r^2}{8} = what * \pi r^2 \\[5ex] \text{Divide both sides by } \pi r^2 \\[3ex] \dfrac{3}{8} = what \\[5ex] what = \dfrac{3}{8} \\[5ex] $ The area of the shaded sector is $\dfrac{3}{8}$ of the area of the circle.

(25.) In the figure below, $\overline{AB}$ is congruent to $\overline{BC}$, and $\overline{AE}$ intersects $\overline{BF}$ at C.
What is the measure of $\angle B$?

Number 25

$ A.\;\; 26^\circ \\[3ex] B.\;\; 38^\circ \\[3ex] C.\;\; 52^\circ \\[3ex] D.\;\; 128^\circ \\[3ex] E.\;\; 154^\circ \\[3ex] $

$ \angle BCA = \angle ECF = 26^\circ ...\text{vertical angles are congruent} \\[3ex] \angle BAC = \angle BCA = 26^\circ ...\overline{AB} = \overline{BC} ...\text{base angles of isosceles triangle} \\[3ex] \angle ABC + \angle BAC + \angle BCA = 180^\circ ... \text{sum of angles of a triangle} \\[3ex] \angle ABC + 26 + 26 = 180 \\[3ex] \angle ABC = 180 - 26 - 26 \\[3ex] \angle ABC = 128 \\[3ex] \angle B = \angle ABC = 128^\circ ...diagram $