Solved Examples: Basic Geometry



Samuel Dominic Chukwuemeka (SamDom For Peace) Formulas:
(1.) Geometry Formulas
(2.) Mensuration Formulas

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the 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 any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

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(1.) A student made several mathematical statements/sentences or asked questions regarding basic geometry.
As a teacher, how would you respond?

(a.) Number 1a
The line above is $\overleftrightarrow{ABC}$

(b.) A line segment has a finite number of points because it has two endpoints.

(c.) $\overline{AB}$ and $\overline{CD}$ are parallel because they do not intersect. Number 1c

(d.) An angle is the intersection of two line segments (or the intersection of the endpoints of two rays).
Can we also define an angle as the union of two​ half-lines and a common point?

(e.) To make the measure of an angle​ greater, just extend the rays.

(f.) $\overline{AB}$ ≠ $\overline{BA}$ because:
$\overline{AB}$ starts at A and ends at​ B, and $\overline{BA}$ starts at B and ends at A.

(g.) There can only be 360 different rays emanating from a point since there are only 360 ° in a circle.

(h.) It is actually impossible to measure an​ angle, since each angle is the union of two rays that extend infinitely and therefore continue forever.

(i.) If any two planes that do not intersect are​ parallel, then any two lines that do not intersect should also be parallel.

(j.) If two​ lines, a and​ b, are perpendicular to the same​ line, lines a and b must be parallel.

(k.) A ray has two endpoints.

(l.) $\overrightarrow{MN}$ = $\overrightarrow{NM}$

(m.) For any points M and​ N, $\overleftrightarrow{MN}$ = $\overleftrightarrow{NM}$

(n.) A line segment contains an infinite number of points.

(o.) Skew lines are coplanar.

(p.) If two distinct planes​ intersect, their intersection is a line segment.

(q.) These two statements mean the same thing. I can write either way.

$ 1st\;\;Statement:\;\; AB \cong CD \\[3ex] 2nd\;\;Statement:\;\; \overline{AB} \cong \overline{CD} \\[3ex] $ (r.)

(s.)

(t.)

(u.)

(v.)

(w.)

(x.)

(y.)

(z.)


(a.) The line should be named as: $\overleftrightarrow{AC}$
Two points determine a​ line, so only two points are used to name the line.

(b.) A line segment has two endpoints. However, it also includes all the points between the endpoints.
The number of points in a line segment is comparable to the number of real numbers between 0 and 1.
Just as the number of real numbers between 0 and 1 is infinite, the number of points between the endpoints of a line segment is infinite.

(c.) The diagram shows that the line segment with endpoints A and B does not intersect the line segment with endpoints C and D.
Lines AB and CD extend beyond those​ points, and lines AB and CD appear to intersect at a point above the figure.
Parallel lines are coplanar lines (lines that lie on the same plane) that do not intersect. Hence, lines AB and CD are not parallel.

(d.) An angle is formed by two sides that share only an endpoint.
Because​ half-lines are rays without​ endpoints, they cannot have a common endpoint.

(e.) Extending the rays does not change the angle measure.

(f.) Please note that:
(1st.) Lines and line segments do not have a​ direction, and any two points uniquely determine both lines and line segments.
(2nd.) Lines​ do not have endpoints. Line segments do.

(g.) As a degree can be subdivided​ infinitely, there are an infinite number of points that lie on a circle that can be used for the second point of a ray.

(h.) Angle measurements do not relate to length.
Angles are measured by the amount of​ "opening" between their sides (the amount of turn between the sides) around their vertex.

(i.) Two distinct lines are parallel if they do not intersect and are in a single plane.
Lines that do not intersect and are not in a single plane are called skew lines.

(j.) Lines a and b could be​ parallel, but they do not have to​ be, because they could form different planes with the perpendicular​ line, and parallel lines must be coplanar.

(k.) A ray has only one endpoint.

(l.) $\overrightarrow{MN}$ ≠ $\overrightarrow{NM}$
This is because $\overrightarrow{MN}$ has an endpoint at point M and extends in the direction of point N while $\overrightarrow{NM}$ has an endpoint at point N and extends in the direction of point M.

(m.) The statement is correct.
Both lines $\overleftrightarrow{MN}$ and $\overleftrightarrow{NM}$ extend in the directions of point M and point N.

(n.) The statement is correct.

(o.) Skew lines are not coplanar.
They cannot be contained in a single plane.

(p.) If two distinct planes​ intersect, their intersection is a line.
Their intersection is not a line segment.

(q.) That is incorrect.
A line is different from a line segment in that a line has no endpoints (extends infinitely in both direcitions) while a line segment has two endpoints.

$ 1st\;\;Statement:\;\; AB \cong CD...Lines \\[3ex] 2nd\;\;Statement:\;\; \overline{AB} \cong \overline{CD}...Line\;\;Segments \\[3ex] $

(r.)

(s.)

(t.)

(u.)

(v.)

(w.)

(x.)

(y.)

(z.)
(2.) Determine if these statements are true or false.
If​ any is false, explain why.

(a.) A line segment contains an infinite number of points.

(b.) Three noncollinear points are always coplanar.

(c.) The union of two rays is always a line.

(d.) Two intersecting lines are coplanar.

(e.) Two planes can intersect in exactly one point.


(a.) The statement is true.

(b.) The statement is true.

(c.) The statement is false because the union of two rays can be a single ray.

(d.) The statement is true.

(e.) The statement is false because they can intersect in a​ line, the empty​ set, or a plane.
(3.)

(4.) If the measure of ∠A is 100° more than three times the measure of ∠B, and the sum of the measures of the two angles is 180° determine the measure of each angle.


$ Let: \\[3ex] m\angle A = x \\[3ex] m\angle B = y \\[3ex] x = 100 + 3y ...eqn.(1) \\[3ex] x + y = 180^\circ...eqn.(2) \\[3ex] Substitute\;\;eqn.(1) \;\;into\;\;eqn.(2) \\[3ex] OR\;\;Substitute\;\;(100 + 3y)\;\;for\;\;x\;\;in\;\;eqn.(2) \\[3ex] 100 + 3y + y = 180 \\[3ex] 4y = 180 - 100 \\[3ex] 4y = 80 \\[3ex] y = \dfrac{80}{4} \\[5ex] y = 20^\circ \\[3ex] Substitute\;\;20\;\;for\;\;y\;\;in\;\;eqn.(1) \\[3ex] x = 100 + 3(20) \\[3ex] x = 160^\circ \\[3ex] $ The measure of angle A is 160°
The measure of angle B is 20°
(5.) In the figure below:

Number 5

If $m\angle AOB = \dfrac{1}{9} m\angle COD$, determine
(a.) $m\angle AOB$
(b.) $m\angle COD$


$ \underline{Diagram} \\[3ex] \angle AOB = x \\[3ex] \angle BOC = 90^\circ \\[3ex] \angle COD = y \\[3ex] \underline{Given} \\[3ex] \angle AOB = \dfrac{1}{9} \angle COD \\[5ex] x = \dfrac{1}{9} * y \\[5ex] 9x = y \\[3ex] y = 9x \\[5ex] \angle AOB + \angle BOC + \angle COD = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] x + 90 + y = 180 \\[3ex] Substitute\;\;9x\;\;for\;\;y \\[3ex] x + 90 + 9x = 180 \\[3ex] 10x + 90 = 180 \\[3ex] 10x = 180 - 90 \\[3ex] 10x = 90 \\[3ex] x = \dfrac{90}{10} \\[5ex] x = 9 \\[3ex] y = 9x = 9(9) \\[3ex] y = 81 \\[3ex] (a.)\;\; m\angle AOB = x = 9^\circ \\[3ex] (b.)\;\; m\angle COD = y = 81^\circ $
(6.) Use the figure to answer these questions.

Number 6

(a.) Determine the measure of angle BOC
(b.) Explain why the exact values of x and y cannot be determined.


$ (a.) \\[3ex] x + x + x + y + y + y = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 3x + 3y = 180 \\[3ex] 3(x + y) = 180 \\[3ex] x + y = \dfrac{180}{3} \\[5ex] x + y = 60^\circ \\[3ex] m\angle BOC = x + y ...diagram \\[3ex] m\angle BOC = 60^\circ \\[3ex] $ (b.) The exact values of x and y cannot be​ determined, because the figure only gives enough information to find the sum of x and y precisely.
To find x and​ y, more information about the relationship between x and​ y, or a value such as $m\angle EOD$ would be required.
(7.) WASSCE The curved surface areas of two cones are equal.
The base radius of one is $5\:cm$ and its slant height is $12\:cm$
Calculate the height of the second cone if its base radius is $6\:cm$


$ \underline{First\:\:Cone} \\[3ex] r = 5\:cm \\[3ex] l = 12\:cm \\[3ex] LSA = \pi rl = \pi (5)(12) = 60\pi cm^2 \\[3ex] \underline{Second\:\:Cone} \\[3ex] r = 6\:cm \\[3ex] LSA = 60\pi cm^2...from\:\:the\:\:question:\:\:curved\:\:surface\:\:areas\:\:are\:\:equal \\[3ex] h = ? \\[3ex] h = \dfrac{\sqrt{LSA^2 - \pi^2 r^4}}{\pi r} \\[5ex] = \dfrac{\sqrt{(60\pi)^2 - \pi^2 * 6^4}}{\pi * 6} \\[5ex] = \dfrac{\sqrt{3600\pi^2 - 1296\pi^2}}{6\pi} \\[5ex] = \dfrac{\sqrt{2304\pi^2}}{6\pi} \\[5ex] = \dfrac{48\pi}{6\pi} \\[5ex] = 8 \\[3ex] h = 8\:cm $
(8.) Given three collinear points​ A, B,​ C, with B between A and​ C, four different rays can be named using these​ points: $\overrightarrow{AB}$, $\overrightarrow{BA}$, $\overrightarrow{BC}$, and $\overrightarrow{CB}$.
How many different rays can be named given five collinear points?


Given: n collinear points:
the number of different rays that can be named is: 2(n − 1) rays. Therefore, given five collinear points, the number of different rays that can be named is:
2(5 − 1) = 2(4) = 8 rays
(9.) WASSCE A container, in the form of a cone resting on its vertex, is full when $4.158$ litres of water is poured into it.
(a) If the radius of its base is $21\:cm$,
(i) represent the information in a diagram;
(ii) calculate the height of the container.

(b) A certain amount of water is drawn out of the container such that the surface diameter of the water drops to $28\:cm$.
Calculate the volume of the water drawn out. $\left(Take\:\:\pi = \dfrac{22}{7}\right)$


(a)
(i)
The diagram can be represented as:
Number 9

$ (ii) \\[3ex] r = 21\:cm \\[3ex] V = 4.158\:litres = 4.158\:dm^3 \\[3ex] Convert\:\:to\:\:cm^3 \\[3ex] 4.158\:dm^3 \\[3ex] = 4.158\:dm * dm * dm * \dfrac{0.1\:m * 0.1\:m * 0.1\:m}{1\:dm * 1\:dm * 1\:dm} * \dfrac{1\:cm * 1\:cm * 1\:cm}{0.01\:m * 0.01\:m * 0.01\:m} \\[5ex] = \dfrac{4.158 * 0.1 * 0.1 * 0.1 * cm^3}{0.01 * 0.01 * 0.01} \\[5ex] = \dfrac{0.001458\:cm^3}{0.000001} \\[5ex] = 4158\:cm^3 \\[3ex] h = \dfrac{3V}{\pi r^2} \\[5ex] = 3V \div \pi r^2 \\[3ex] = 3(4158) \div \left(\dfrac{22}{7} * 21 * 21\right) \\[5ex] = 3(4158) \div (22 * 3 * 21) \\[3ex] = \dfrac{3 * 4158}{22 * 3 * 21} \\[5ex] = \dfrac{4158}{462} \\[5ex] = 9\:cm \\[3ex] (b) \\[3ex] Volume\:\:of\:\:cone = 4158\:cm^3 \\[3ex] Water\:\:was\:\:drawn\:\:out \\[3ex] \underline{Remaining\:\:volume\:\:of\:\:water} \\[3ex] d_{remain} = 28\:cm \\[3ex] r_{remain} = \dfrac{d_{remain}}{2} \\[5ex] r_{remain} = \dfrac{28}{2} = 14\:cm \\[5ex] \dfrac{h_{remain}}{h} = \dfrac{r_{remain}}{r}...Similar\:\:\triangle s \\[5ex] \dfrac{h_{remain}}{9} = \dfrac{14}{21} \\[5ex] h_{remain} = \dfrac{9 * 14}{21} = 6\:cm \\[5ex] V_{remain} = \dfrac{\pi * r_{remain}^2 * h_{remain}}{3} \\[5ex] V_{remain} = \dfrac{1}{3} * \dfrac{22}{7} * 14 * 14 * 6 \\[5ex] V_{remain}= 22 * 2 * 14 * 2 \\[3ex] V_{remain} = 1232\:cm^3 \\[3ex] \underline{Volume\:\:of\:\:water\:\:drawn\:\:out} \\[3ex] V_{drawn} + V_{remain} = V \\[3ex] V_{drawn} = V - V_{remain} \\[3ex] V_{drawn} = 4158 - 1232 \\[3ex] V_{drawn} = 2926\:cm^3 $
(10.) In the diagram: Number 10

Name the rays that contain these line segments.
(a.) $\overline{NQ}$
(b.) $\overline{ST}$


(a.) The rays that contain the line segment $\overline{NQ}$ are:

$ \overrightarrow{NQ} \\[3ex] \overrightarrow{NP} \\[3ex] \overrightarrow{RQ} \\[3ex] \overrightarrow{RP} \\[3ex] $ (b.) The rays that contain the line segment $\overline{ST}$ are:

$ \overrightarrow{ST} \\[3ex] \overrightarrow{SU} \\[3ex] \overrightarrow{RT} \\[3ex] \overrightarrow{RU} $
(11.) Points​ A, B,​ C, and D are collinear.
In how many ways can the line be named using only these​ points?
Assume that different order means different​ name.


We can solve the question using at least two approaches.
Use any approach you prefer.
Assume that different order means different​ name: means that order is important.

1st Approach: Algebraically: By Formula
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
$P(n, 2) = \dfrac{n!}{(n - 2)!}$
n = 4

$ P(4, 2) = \dfrac{4!}{(4 - 2)!} \\[5ex] = \dfrac{4 * 3 * 2!}{2!} \\[5ex] = 12\;\;ways \\[3ex] $ 1st Approach: Graphically: By Diagram
Number 11

The number of different ways to name the lines using only the points are:
$\overline{AB}$     $\overline{AC}$     $\overline{AD}$
$\overline{BA}$     $\overline{CA}$     $\overline{DA}$

$\overline{BC}$     $\overline{BD}$
$\overline{CB}$     $\overline{DB}$

$\overline{CD}$
$\overline{DC}$

12 ways.
(12.) ACT How long, in centimeters, is $1$ side of a square whose perimeter is equal to the circumference of a circle with a radius of $2$ centimeters?

$ F.\;\; \pi \\[3ex] G.\;\; \dfrac{\pi}{2} \\[5ex] H.\;\; 4\pi \\[3ex] J.\;\; 16\pi \\[3ex] K.\;\; 4 \\[3ex] $

$ \underline{Square} \\[3ex] Let\;\;side = x \\[3ex] Perimeter = 4 * side \\[3ex] P = 4x \\[3ex] \underline{Circle} \\[3ex] radius = 2\;cm \\[3ex] Circumference = 2 * \pi * radius \\[3ex] C = 2 * \pi * 2 \\[3ex] C = 4\pi \\[3ex] \underline{Question} \\[3ex] P = C \\[3ex] 4x = 4\pi \\[3ex] x = \dfrac{4\pi}{4} \\[5ex] x = \pi \\[3ex] $ The length of a side of the square is $\pi$ centimeters
(13.) In the figure below:

Number 13

If $m\angle AOB$ is 15° less than 9$m\angle BOC$, determine
(a.) $m\angle AOB$
(b.) $m\angle BOC$


$ \underline{Diagram} \\[3ex] \angle COB = \angle BOC = x \\[3ex] \angle BOA = \angle AOB = y \\[3ex] \angle COA = \angle AOC = 90^\circ \\[3ex] \underline{Given} \\[3ex] \angle AOB = 9\angle BOC - 15 \\[5ex] y = 9x - 15 \\[5ex] \angle AOB + \angle BOC = \angle AOC \\[3ex] y + x = 90 \\[3ex] Substitute\;\;(9x - 15)\;\;for\;\;y \\[3ex] \implies \\[3ex] 9x - 15 + x = 90 \\[3ex] 10x = 90 + 15 \\[3ex] 10x = 105 \\[3ex] x = \dfrac{105}{10} \\[5ex] x = 10.5 \\[3ex] y = 9x - 15 = 9(10.5) - 15 = 79.5 \\[3ex] (a.)\;\; m\angle BOC = x = 10.5^\circ \\[3ex] (b.)\;\; m\angle AOB = y = 79.5^\circ $
(14.)


(15.)

(16.) ACT A rectangular stage is $90$ feet long and $30$ feet wide.
What is the area, in square yards, of this stage?

$ A.\;\; 30\sqrt{3} \\[3ex] B.\;\; 300 \\[3ex] C.\;\; 675 \\[3ex] D.\;\; 900 \\[3ex] E.\;\; 1,700 \\[3ex] $

$ Area\:\;in\:\:square\:\:feet = 90\:feet * 30\:feet = 2700\:square\:feet \\[3ex] Convert\:\: 2700\:feet^2 \:\:to\:\: yard^2 \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 2700\:ft^2\:\:to\:\:yd^2 \\[3ex] 2700\:ft * ft * \dfrac{.....yd}{.....ft} * \dfrac{.....yd}{.....ft} \\[5ex] 2700\:ft * ft * \dfrac{1\:yd}{3\:ft} * \dfrac{1\:yd}{3\:ft} \\[5ex] = \dfrac{2700 * 1 * 1}{3 * 3} \\[5ex] = \dfrac{2700}{9} \\[5ex] = 300\:yd^2 \\[3ex] $ Second Method: Proportional Reasoning Method

Let $c$ = converted unit from $ft^2$ to $yd^2$

$ 3\:ft = 1\:yd \\[3ex] Square\:\:both\:\:sides \\[3ex] (3\:ft)^2 = (1\:yd)^2 \\[3ex] 3^2 * ft^2 = 1^2 * yd^2 \\[3ex] 9\:ft^2 = 1\:yd^2 \\[3ex] $
$ft^2$ $yd^2$
$9$ $1$
$2700$ $c$

$ \dfrac{c}{1} = \dfrac{2700}{9} \\[5ex] c = 300\:yd^2 \\[3ex] \therefore 2700\:ft^2 = 300\:yd^2 \\[3ex] $ The area of the rectangular stage is $300$ square yards
(17.)

(18.) In the diagram: Number 18

Name the rays that contain these line segments.
(a.) $\overline{NQ}$
(b.) $\overline{TU}$


(a.) The rays that contain the line segment $\overline{NQ}$ are:

$ \overrightarrow{NQ} \\[3ex] \overrightarrow{NP} \\[3ex] \overrightarrow{RQ} \\[3ex] \overrightarrow{RP} \\[3ex] $ (b.) The rays that contain the line segment $\overline{TU}$ are:

$ \overrightarrow{TU} \\[3ex] \overrightarrow{SU} \\[3ex] \overrightarrow{RU} $
(19.) An angle of depression on an object is the angle between the horizontal and the line from the​ observer's eye​ (line of​ sight) to the object.
Draw an illustration of a cliff on a​ shore, the water​ below, and an angle of depression.

Number 19


An an illustration that correctly demonstrates an example of an angle of depression is:

Number 19
(20.)





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(21.)

(22.)


(23.) The drawing shown here shows a method of using a floor for a protractor and a door to make an angle.

Number 23

List one pro and one con to using this aid.


A pro to using this aid is: It is a convenient visual aid to show how angles of different measures appear in a​ real-world context.

A con to using this aid is: The reading is only accurate on the side of the door with the​ hinges; the thickness of the door creates an additional source of error compared to the insignificant error made with a pencil line and a more traditional protractor.
(24.)

(25.)

(26.)

(27.)

(28.) A line n is perpendicular to plane ​α, and plane β contains n​ (n is in plane β​).

Number 28

Must planes α and β be​ perpendicular? Explain why or why not.


Yes, because n is perpendicular to any line in α through point P.​
Therefore, it is perpendicular to the line in α through point P that is perpendicular to $\overline{AB}$
(29.)

(30.)

(31.) Points​ A, B,​ C, D, and E are coplanar and no three are collinear.
In how many ways can the plane be named using only these​ points?


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
$P(n, 3) = \dfrac{n!}{(n - 3)!}$
n = 5

$ P(5, 3) = \dfrac{5!}{(5 - 3)!} \\[5ex] = \dfrac{5 * 4 * 3 * 2!}{2!} \\[5ex] = 60\;\;ways $
(32.) Let S be the set of​ lines, skew to a given​ line, and P be the set of lines parallel to that line.
Draw a Venn diagram for sets S and P.


Sets S and P are disjoint sets. They do not have any common point.
Hence, the Venn diagram for the disjoint sets is:

Number 32
(33.)

(34.)

(35.)

(36.)

(37.)

(38.)

(39.) The figure below is a rectangular box in which EFGH and ABCD are rectangles.
$\overline{BF}$ is perpendicular to planes EFGH and ABCD.

Number 39

(a.) Name one pair of skew lines.
(b.) What is the intersection of planes EFG and DCG​?
(c.) Plane DCG is perpendicular to what line?
(d.) What is the intersection of $\overleftrightarrow{FG}$ and plane ABC​?
(e.) Explain why planes EAD and DCG are perpendicular.


(a.) These are pairs of skew lines:
(i.) $\overleftrightarrow{CD}$ and $\overleftrightarrow{EH}$
(ii.) $\overleftrightarrow{AD}$ and $\overleftrightarrow{GH}$
(iii.) $\overleftrightarrow{AD}$ and $\overleftrightarrow{EF}$
(iv.) $\overleftrightarrow{BD}$ and $\overleftrightarrow{GH}$
(v.) $\overleftrightarrow{BD}$ and $\overleftrightarrow{FG}$
(vi.) $\overleftrightarrow{AB}$ and $\overleftrightarrow{EH}$
(vii.) $\overleftrightarrow{AB}$ and $\overleftrightarrow{FH}$


(b.) The intersection of planes EFG and DCG is: $\overleftrightarrow{GH}$

(c.) Plane DCG is perpendicular to $\overleftrightarrow{FG}$

(d.) The intersection of $\overleftrightarrow{FG}$ and plane ABC is an empty set.

(e.) The intersection of the planes is a single line and the planes form a right dihedral angle.
(40.) Use the pyramid to answer the following questions.

Number 40

(a.) What is the intersection of $\overline{AB}$ and $\overline{AD}$​?
(b.) What is the intersection of $\overline{AB}$ and $\overline{CE}$​?
(c.) What is the intersection of $\overline{CD}$ and $\overline{BC}$​?
(d.) What is the intersection of $\overline{CD}$ and $\overline{AE}$​?
(e.) What is the intersection of planes ​BCE, ADE​, and​ ABC?
(f.) What is the intersection of $\overleftrightarrow{BE}$ and $\overleftrightarrow{BC}$​?
(g.) What is the intersection of $\overleftrightarrow{AE}$ and $\overleftrightarrow{BE}$​?
(h.) What is the intersection of $\overleftrightarrow{AB}$ and $\overleftrightarrow{BC}$​?
(i.) What is the intersection of $\overleftrightarrow{CD}$ and $\overleftrightarrow{DE}$​?
(j.) Name a pair of skew lines.
(k.) Name a pair of parallel lines.
(l.) Find a plane not determined by one of the triangular faces or by the base.


(a.) The intersection of $\overline{AB}$ and $\overline{AD}$ is Point A

(b.) The intersection of $\overline{AB}$ and $\overline{CE}$ is the empty set, φ

(c.) The intersection of $\overline{CD}$ and $\overline{BC}$ is Point C

(d.) The intersection of $\overline{CD}$ and $\overline{AE}$ is the empty set, φ

(e.) The intersection of planes ​BCE, ADE​, and​ ABC is the empty set, φ

(f.) The intersection of $\overleftrightarrow{BE}$ and $\overleftrightarrow{BC}$ is Point B

(g.) The intersection of $\overleftrightarrow{AE}$ and $\overleftrightarrow{BE}$ is Point E

(h.) The intersection of $\overleftrightarrow{AB}$ and $\overleftrightarrow{BC}$ is Point B

(i.) The intersection of $\overleftrightarrow{CD}$ and $\overleftrightarrow{DE}$ is Point D

(j.) These are pairs of skew lines:
(i.) $\overleftrightarrow{BC}$ and $\overleftrightarrow{DE}$
(ii.) $\overleftrightarrow{AB}$ and $\overleftrightarrow{DE}$
(iii.) $\overleftrightarrow{CD}$ and $\overleftrightarrow{AE}$
(iv.) $\overleftrightarrow{CD}$ and $\overleftrightarrow{BE}$
(v.) $\overleftrightarrow{AD}$ and $\overleftrightarrow{CE}$

(k.) These are pairs of parallel lines:
(i.) $\overleftrightarrow{AD}$ and $\overleftrightarrow{BC}$
(ii.) $\overleftrightarrow{AB}$ and $\overleftrightarrow{DC}$

(l.) A plane not determined by one of the triangular faces or by the base is plane
(i.) AEC
(ii.) BED




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(41.) Paul was studying right angles and wondered if during his math class the minute and hour hands of the clock formed a right angle.
(a.) If his class meets from 7​:00 P.M. to 7:50 P.M.​, is a right angle​ formed?
(b.) If​ so, figure out to the nearest second when the hands form the right angle.
​(Round to the nearest integer as​ needed.)


m = number of minutes
h = number of hours
From 7:00 P.M. to 7:50 P.M.
h = 7
To find the minutes at which a right angle is forned, we use the formula for the: Angle Between Hour hand and Minute hand
$ 1st\;\;\angle = |30h - 5.5m| \\[3ex] $ The angle should be a right angle: an angle of 90° ⇒

$ 30h - 5.5m = 90 \\[3ex] 30(7) - 5.5m = 90 \\[3ex] 210 - 5.5m = 90 \\[3ex] 210 - 90 = 5.5m \\[3ex] 5.5m = 120 \\[3ex] m = \dfrac{120}{5.5} \\[5ex] m = 21.81818182\;minutes \\[3ex] $ (a.) Yes, a right angle is formed at 21.81818182 minutes into the class.

To find the number of seconds, we multiply the decimal part of the number of minutes by 60
This is because there are 60 seconds in 1 minute

$ m = 21.81818182\;minutes \\[3ex] m = 21\;minutes + 0.81818182\;minutes \\[3ex] Integer\;\;part = 21 \\[3ex] Decimal\;\;part = 0.81818182 \\[3ex] Number\;\;of\;\;seconds \\[3ex] = 0.81818182(60) \\[3ex] = 49.09090909 \\[3ex] \approx 49\;seconds \\[3ex] $ (b.) A right angle is formed at: 7:21:49 P.M.
(42.)

(43.)

(44.)