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These are the solutions to Mathematics questions on Geometry.
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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] \text{Sum of interior angles} = 180(n - 2) \\[4ex] \text{Each interior angle} = \dfrac{180(n - 2)}{n} \\[6ex] \text{Sum of exterior angles} = 360^\circ \\[4ex] \text{Each interior angle} = \dfrac{360}{n} \\[6ex] $ (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) $



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



(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.

$ \underline{\text{Assume a Horizontal Ellipse}} \\[3ex] \text{Length of Major Axis} = 2a \\[3ex] \text{Length of Minor Axis} = 2b \\[3ex] \text{Distance between two Foci} = 2\sqrt{a^2 - b^2} \\[5ex] $ (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] $

(1.) Standard Form of a Horizontal Hyperbola

$ \dfrac{(x - h)^2}{a^2} - \dfrac{(y - k)^2}{b^2} = 1 \\[5ex] $ where:
(h, k) is the coordinates of the center of the hyperbola
the transverse axis is along the x-axis (horizontal transverse axis)

(2.) Standard Form of a Vertical Hyperbola

$ \dfrac{(y - h)^2}{b^2} - \dfrac{(x - k)^2}{a^2} = 1 \\[5ex] $ where:
(h, k) is the coordinates of the center of the hyperbola
the transverse axis is along the y-axis (vertical transverse axis)

(3.) Asymptotes of a Hyperbola

$ \pm \dfrac{b}{a}x $

Theorems

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

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

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

(4.) The sum of the angles at a point is 360°

(5.) Interior Angle Sum Theorem: The sum of the interior angles of a polygon (regular and irregular polygons) is $180(n - 2)$ where n is the number of sides.
This is because any polygon of n sides can be split into n − 2 triangles.

(6.) Exterior Angle Sum Theorem: The sum of the exterior angles of a polygon (regular and irregular polygons) is 360°

(7.) Angle Bisector Theorem: Any point on the bisector of an angle is equidistant from the two sides of the angle.

(8.) For any point on an ellipse, the sum of the distances from the foci to that point is equal to the length of the major axis.

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.) What is the value of x in the diagram?

Number 1

$ A.\;\; 39^\circ \\[3ex] B.\;\; 32^\circ \\[3ex] C.\;\; 30^\circ \\[3ex] D.\;\; 28^\circ \\[3ex] $

Let us indicate points on the diagram

Number 1

$ \underline{\triangle ABC} \\[3ex] \angle BAC + x + 74 = 180^\circ ...\text{sum of angles in a triangle} \\[3ex] \angle BAC = 180 - x - 74 \\[3ex] \angle BAC = 106 - x \\[5ex] \text{Parallel Lines AB and DE} \\[3ex] \angle EDF = \angle BAC = 106 - x ...\text{corresponding angles are congruent} \\[5ex] \underline{\triangle DEF} \\[3ex] \angle EFG = \angle EDF + 2x ...\text{exterior angle of a triangle is the sum of the two interior angles} \\[3ex] 134 = (106 - x) + 2x \\[3ex] 106 - x + 2x = 134 \\[3ex] x = 134 - 106 \\[3ex] x = 28^\circ $
(2.) Hallar el área de la región sombreada si AOB y COD son sectores circulares, donde $\theta = \dfrac{2\pi}{9}$ y BC = 3 m.

Number 2

$ A.\;\; \dfrac{\pi}{2}\;m^2 \\[5ex] B.\;\; \pi\;m^2 \\[3ex] C.\;\; \dfrac{\pi}{3}\;m^2 \\[5ex] D.\;\; \dfrac{2\pi}{3}\;m^2 \\[5ex] E.\;\; \dfrac{\pi}{4}\;m^2 \\[5ex] $

Number 2

$ \underline{\text{Area of a Sector where Central Angle is in RAD}} \\[3ex] A = \dfrac{r^2\theta}{2} \\[5ex] where \\[3ex] A = \text{area of the sector} \\[3ex] r = \text{radius of the sector} \\[3ex] \theta = \text{central angle} = \dfrac{\pi}{9} \\[5ex] OA = OB = r_1 = \text{radius of Sector AOB} \\[3ex] OC = OD = r_2 = \text{radius of Sector COD} \\[5ex] \underline{\triangle OCB} \\[3ex] r_2^2 = r_1^2 + 3^2 ...\text{Pythagorean Theorem} \\[3ex] r_2^2 = r_1^2 + 9 \\[5ex] \underline{\text{Area of Sector AOB}}, A_1 \\[3ex] A_1 = \dfrac{r_1^2\theta}{2} \\[5ex] \underline{\text{Area of Sector COD}}, A_2 \\[3ex] A_2 = \dfrac{r_2^2\theta}{2} \\[5ex] \text{Substitute for }r_2^2 \\[3ex] A_2 = \dfrac{\theta(r_1^2 + 9)}{2} \\[5ex] \underline{\text{Area of the shaded region}} \\[3ex] = A_2 - A_1 \\[3ex] = \dfrac{\theta(r_1^2 + 9)}{2} - \dfrac{r_1^2\theta}{2} \\[5ex] = \dfrac{\theta r_1^2 + 9\theta - \theta r_1^2}{2} \\[5ex] = \dfrac{9\theta}{2} \\[5ex] = \dfrac{9}{2} \cdot \theta \\[5ex] = \dfrac{9}{2} \cdot \dfrac{\pi}{9} \\[5ex] = \pi\;m^2 $
(3.) Determine all the angles in the figure.
State your reasons accordingly.
To check your knowledge of the special properties of angles between parallel lines:
Use each reason at least once.
Do not use the same reason more than twice.


Number 3


$ a = 121^\circ ...\text{alternate exterior angles are congruent} \\[5ex] c = 121^\circ ...\text{vertical angles are congruent} \\[5ex] f = 121^\circ ...\text{corresponding angles are congruent} \\[5ex] d + 121^\circ = 180^\circ ...\text{angles on a straight line are supplementary} \\[3ex] d = 180 - 121 \\[3ex] d = 59^\circ \\[5ex] b + c = 180^\circ ...\text{consecutive interior angles are supplementary} \\[3ex] b = 180 - c \\[3ex] b = 180 - 121 \\[3ex] b = 59^\circ \\[5ex] g = b = 59^\circ ...\text{alternate interior angles are congruent} \\[5ex] e = d = 59^\circ ...\text{alternate exterior angles are congruent} $
(4.) In the diagram below, ML || PQ and NP || QR.
If $\angle LMN = 40^\circ$ and $\angle MNP = 55^\circ$, find $\angle PQR$

Number 4


Construction: Extend line MN to Point Q
Extend line PN to a point on line ML

Number 4

$ \angle MQP = \angle LMN = 40^\circ...\text{alternate interior angles are congruent} \\[5ex] \angle MNQ + \angle MNP = 180^\circ ...\text{sum of angles on a straight line} \\[3ex] \angle MNQ + 55 = 180 \\[3ex] \angle MNQ = 180 - 55 \\[3ex] \angle MNQ = 125^\circ \\[5ex] \angle NQP = \angle MQP = 40^\circ ...\text{Diagram} \\[3ex] \angle PNQ = \angle MNQ = 125^\circ...\text{Diagram} \\[5ex] \angle NPQ + \angle PNQ + \angle MQP = 180^\circ ...\text{sum of angles in }\triangle PNQ \\[3ex] \angle NPQ + 125 + 40 = 180 \\[3ex] \angle NPQ = 180 - 125 - 40 \\[3ex] \angle NPQ = 15^\circ \\[5ex] \angle PQR = \angle NPQ = 15^\circ ...\text{alternate interior angles are congruent} $
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