euclidean geometry assignment grade 12

Euclidean geometry Grade 12 Notes, Questions and Answers

Euclidean geometry Grade 12 Notes, Questions and Answers: Euclidean geometry is a mathematical system attributed to Alexandrian Greek mathematician Euclid, which he described in his textbook on geometry: the Elements. Euclid’s method consists in assuming a small set of intuitively appealing axioms and deducing many other propositions from these.

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Euclidean Geometry Grade 12 Questions and Answers (Downloadable pdf)

Below you will find a list of Euclidean Geometry Grade 12 Questions and Answers from previous years’ activities, tests, and notes:

Watch: Grade 12 Geometry Lesson 1| Euclidean Geometry: Introduction (Triangles and parallel lines)

Types of triangles 

Congruent triangles (emcj2).

Source:https://intl.siyavula.com/read/maths/grade-12/euclidean-geometry/08-euclidean-geometry-01

Hey, Grade 12 Learners . My Courses portal has everything you needed to ace your matric exams, tests, assessments, research tasks, and assignments. Feel free to explore all resources for grade 12 learners, such as Study Guides, Department of Basic Education Past Exam Papers with Memos , and Speech Topics.

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EUCLIDEAN GEOMETRY GRADE 12 NOTES - MATHEMATICS STUDY GUIDES

• Revise: Proportion and area of triangles
• Proportion theorems
• Similar polygons

12.1 Revise: Proportion and area of triangles

Ratio and proportion of areas and sides of triangles

• Triangles on the same base (or equal bases) and between parallel lines are equal in area. Area ∆ABC = ½(AB)h Area ∆ADB = ½(AB)h Area ∆ABC = Area ∆ADB

12.2 Proportion theorems

Solving problems using proportion

Determine the value of x if PQ ∣∣ BC. (4)

12.3 Similar polygons

• All the pairs of corresponding angles are equal (They are equiangular) and
• All the pairs of corresponding sides are in the same proportion. Both of these conditions must hold at the same time.

||| is the symbol we use to say one polygon ‘is similar to’ another polygon.

Triangles are special polygons:

• If two triangles are equiangular, then their sides will always be in the same proportion, so the triangles are similar.
• If the sides of two triangles are in the same proportion, then the triangles will be equiangular, so the triangles are similar.

equiangular ∆s ➝ similar ∆s corresponding sides ∆s in proportion ➝ ∆s are similar

• Prove ∆AEF ||| ∆AMD (5)
• Hence, find the numerical value of AF . (5)                                                        AD 
• Prove ∆ CDB ||| ∆ADE (4)
• Prove that ^D 4 = ^D 2 (3)
• Prove that ∆BHD ||| ∆FED (5)

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MATHS Gr. 12: Euclidean Geometry - Similarity Theorem

This is a grade 12 lesson on Similarity.

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Course Content

Introduction to euclidean geometry, similar triangles – video, similar triangles – pdf, midpoint theorem – video, midpoint theorem – pdf, pythagorean theorem – video, pythagorean theorem – pdf, euclidean geometry caps, quiz – euclidean geometry.

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• Chapter 8: Euclidean geometry
• 8.1 Revision
• 8.2 Ratio and proportion
• 8.3 Polygons
• 8.4 Triangles
• 8.5 Similarity
• 8.6 Pythagorean theorem
• 8.7 Summary
• End of chapter exercises
• Practice this chapter

Mathematics e-learning courses

Grade 12 – chapter 10 – euclidean geometry (proportionality and similarity).

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BRIGHT YOUNG BRAINS

Euclidean geometry.

Imagine the feeling you will get, knowing that approximately 30% of all your math paper 2 content is learnt and understood, when you complete this course. Any fears or hate that you have towards euclidean geometry, will be greatly reduced .

Course curriculum

Chapter 1: learning how to see the bigger picture.

Course Warm Up Exercise

Analysing a sketch in order to prove a cyclic quadrilateral

EXERCISE on proving a Similarity Theorem

How to properly analyse a sketch before looking at any questions

Identifying multiple angles that are equal to each other

Determining the size of an angle that has multiple ways of solving it

Proving that triangles are similar to each other

The Train of thought when preparing to answer a question that looks complicated

Solving an Expression that involves triangles that are similar to each other.

Chapter 2: The best way to learn Euclidean Geometry is by doing Euclidean Geometry

EXERCISE on proving a proportionality theorem

EXERCISE after analysing a Sketch

EXERCISE on identifying theorems within a sketch

Highlighting the IMPORTANCE of analysing a sketch properly.

Maximising on the time spent analyzing a sketch

What to do when you get stuck in between a question

Proving that a line is a tangent to a circle passing through 3 points

When trigonometry gets mixed up with Euclidean Geometry

What to do when the Area Rule Shows up in Euclidean Geometry

Chapter 3: Proportionality and Similarity Questions Galore

Analysing a sketch before diving into questions

Getting 3 other angles that are equal to x

The MAIN! way to show that triangles are similar

Giving a number as an answer in a question that has ZERO lengths or angles given

Using a proportionality theorem to answer n exam style question.

Free marks from a proportionality of similar triangles question.

Banking time by properly analysing a sketch

Answering a PROPORTIONALITY exam style question which has 4 triangles

Chapter 4: Sketches without circles

Analysing a sketch euclidean geometry sketch that has NO circle.

Showing that a point is the MIDPOINT of a line thats part of a triangle

Using a given ratio to get the length of a side in a triangle

Getting a value for the ratio between the 2 areas of 2 triangles

Working with Ratios of Triangles and their Areas

Using the RATIOS between the sides of triangles to prove similarity

Getting the length of a line using a proportionality theorem

EXERCISE on proving similarity and a tangent to a circle

Pulling out all the stops in a proportionality theorems question

James Carter

I worked through almost all of the booklet before starting the course. I had a few questions leftover that I did not do and Levy showed me that the only reas...

I worked through almost all of the booklet before starting the course. I had a few questions leftover that I did not do and Levy showed me that the only reason I could not do them was because I was thinking about them wrong. He has shown me an innovative way of doing geometry and now I feel I am ready to tackle any problem that comes my way. More so, I have applied this style of thinking to the rest of the maths topics and find all other problems easier too. Thank you Levy for teaching me innovative thinking, and for showing me the power of colour!!

Euclidean Geo

Promise langa.

The course simplified everything that i thought was hard. The course was a confidence booster and it opened up a channel of understanding geometry in depth.

Impressed!!

Very helpful! Everything is thoroughly explained. I would recommend this to other Grade 12 learners.

Wonderful Course

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This is genuinely a great course. The rigorous lecture lessons are delivered with a bit of humor and thus making the course more enjoyable. Quite convinced t...

This is genuinely a great course. The rigorous lecture lessons are delivered with a bit of humor and thus making the course more enjoyable. Quite convinced that every individual who undertakes this course will be able to sit in an exam room and go through the Euclidean Geometry questions with ease and efficiency. This course is very helpful and necessary, 10/10 would recommend.

Rethabile Matsile

This course is wonderful and takes you deep into the topic

Euclidean geometry

Benedict phanzu, concise and informative, roche erasmus.

very helpful, need some more clarity on the very last question though

Tshiamo Sefolo

Absolutely love the way that this course is set out and it most definitely pushed my limits and I realised that I can actually do it. Euclidean Geometry and ...

Absolutely love the way that this course is set out and it most definitely pushed my limits and I realised that I can actually do it. Euclidean Geometry and I are at a better place now because of this course. Absolutely worth the investment. Thanks a mil Levy!!

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I Really never enjoyed Euclidean geometry. It has always been my achilleas' heel when it came to paper 2, however Mr Levi's famous words " just because you d...

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Benefitted a lot from this. Your explanations are impeccable.

Course Description

As you go through this Euclidean Geometry Grade 12 course, you will learn how to answer difficult euclidean geometry questions. 

The content you will have covered when you complete this course in about a week or so is all at Grade 12 maths past papers level. The sketches we go through, are analyzed in detail explaining the thought process that you need to follow in order to equip yourself with enough information from a sketch before answering questions. 

Levy From Bright Young Brains

Levy Nyamariwata

30% of your ENTIRE maths paper 2 exam is Euclidean Geometry

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Mathematics Grade 12

Euclidean geometry grade 12 questions and answers.

Euclidean geometry Grade 12 Questions and Answers:Euclidean geometry is a mathematical system attributed to Alexandrian Greek mathematician Euclid, which he described in his textbook on geometry: the Elements. Euclid’s method consists in assuming a small set of intuitively appealing axioms, and deducing many other propositions from these.

• 1.1 Congruent triangles (EMCJ2)
• 2 Euclidean Geometry Grade 12 Questions and Answers (Downloadable pdf)
• 3 Questions and Answers
• 4 Have you made your career choice yet?

Types of triangles 

Congruent triangles (emcj2).

Source:https://intl.siyavula.com/read/maths/grade-12/euclidean-geometry/08-euclidean-geometry-01

Euclidean Geometry Grade 12 Questions and Answers (Downloadable pdf)

Hey, Grade 12 Learners. My Courses portal has everything you needed to ace your matric exams, tests, assessments, research tasks, and assignments. Feel free to explore all resources for grade 12 learners, such as Study Guides, Department of Basic Education Past Exam Papers with Memos , and Speech Topics.

Questions and Answers

My Courses has a large Questions and Answers repository for the most popular High School and Tertiary Schools subjects. This comes in handy when doing your revision or preparing for exams, tests, research tasks, and assignments.

Have you made your career choice yet?

Now that you have made it all the way from your 1st grade to the final (Grade 12) grade, have you made up your mind about what career you want to pursue after matric? Have a look at the below questions and click on the links for more information:

Do you know:

• If you want to study at a University or College ?
• What career or course/qualification do you plan to pursue after school ?
• What subjects you will need to study for this career ?
• Which country do you want to study at, Canada maybe?
• How will you pay for your studies?

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Mathematics Grade 12 November 2020 Question Papers and Memos for study revision

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Mathematics grade 11, cubic functions grade 12 questions and answers pdf.

Cubic Functions Grade 12 Questions and Answers pdf: In mathematics, a cubic function is a function of the form f(x)=ax^3+bx^2+cx+d where the coefficients a, b, c, and d are real numbers, and the variable x takes real values, and a ≠ 0. In other words, it is both a polynomial function of degree three, and a real function.

• 1 Downloadable Cubic Functions Grade 12 Questions and Answers pdf
• 2 Questions and Answers

Downloadable Cubic Functions Grade 12 Questions and Answers pdf

General solution trigonometric grade 12 pdf.

General Solution Trigonometric Grade 12 pdf: This page contains a guide for Mathematics Grade 12 Students with the following Trigonometric notes: Trigonometric Functions, 3D Trigonometry, Identities Problems with Solutions, and Challenging Exam Practices.

• 1 Concepts and skills
• 2 Prior knowledge:
• 3 Solutions Trigonometric Equations
• 4 How to solve general trigonometric equations formula and find solutions
• 5 Questions and Answers
• 6 Download General Solution Trigonometric Grade 12 pdf
• 7 Identities Problems with Solutions for exam Practice

Concepts and skills

• Simplify trigonometric equations.
• Find the reference angle.
• Use reduction formulae to find other angles within each quadrant.
• Find the general solution of a given trigonometric equation.

Prior knowledge:

• Trigonometric ratios.
• Trigonometric identities.
• Solve problems in the Cartesian plane.
• Co-functions.
• Factorisation.
• Revise the following trigonometric ratios for right-angled triangles, which you learnt in Grade 10.

Will start with the basic equation, sin x = 0 . The principal solution for this case will be x = 0, π, 2π as these values satisfy the given equation lying in the interval [0, 2π]. But, we know that if sin x = 0, then x = 0, π, 2π, π, -2π, -6π, etc. are solutions of the given equation. Hence, the general solution for sin x = 0 will be, x = nπ, where n∈I.

Similarly, general solution for cos x = 0 will be x = (2n+1)π/2, n∈I, as cos x has a value equal to 0 at π/2, 3π/2, 5π/2, -7π/2, -11π/2 etc. Below here is the table defining the general solutions of the given trigonometric functions involved equations.

Solutions Trigonometric Equations

Watch: Video

How to solve general trigonometric equations formula and find solutions

The periodicity of the trigonometric functions means that there are an infinite number of positive and negative angles that satisfy an equation. If we do not restrict the solution, then we need to determine the general solution to the equation. We know that the sine and cosine functions have a period of 360360° and the tangent function has a period of 180180°.

Method for finding the solution:

• Simplify the equation using algebraic methods and trigonometric identities.
• Determine the reference angle (use a positive value).
• Use the CAST diagram to determine where the function is positive or negative (depending on the given equation/information).
• Restricted values: find the angles that lie within a specified interval by adding/subtracting multiples of the appropriate period.
• General solution: find the angles in the interval [0°;360°][0°;360°] that satisfy the equation and add multiples of the period to each answer.
• Check answers using a calculator.

Download General Solution Trigonometric Grade 12 pdf

Identities problems with solutions for exam practice.

1.  (1 – sin A)/(1 + sin A) = (sec A – tan A) 2

L.H.S = (1 – sin A)/(1 + sin A)

= (1 – sin A) 2 /(1 – sin A) (1 + sin A),[Multiply both numerator and denominator by (1 – sin A)

= (1 – sin A) 2 /(1 – sin 2  A)

= (1 – sin A) 2 /(cos 2  A), [Since sin 2  θ + cos 2  θ = 1 ⇒ cos 2  θ = 1 – sin 2  θ]

= {(1 – sin A)/cos A} 2

= (1/cos A – sin A/cos A) 2

= (sec A – tan A) 2  = R.H.S.  Proved.

2.  Prove that, √{(sec θ – 1)/(sec θ + 1)} = cosec θ – cot θ.

L.H.S.= √{(sec θ – 1)/(sec θ + 1)}

= √[{(sec θ – 1) (sec θ – 1)}/{(sec θ + 1) (sec θ – 1)}]; [multiplying numerator and denominator by (sec θ – l) under radical sign]

= √{(sec θ – 1) 2 /(sec 2  θ – 1)}

=√{(sec θ -1) 2 /tan 2  θ}; [since, sec 2  θ = 1 + tan 2  θ ⇒ sec 2  θ – 1 = tan 2  θ]

= (sec θ – 1)/tan θ

= (sec θ/tan θ) – (1/tan θ)

= {(1/cos θ)/(sin θ/cos θ)} – cot θ

= {(1/cos θ) × (cos θ/sin θ)} – cot θ

= (1/sin θ) – cot θ

= cosec θ – cot θ = R.H.S.  Proved.

3.  tan 4  θ + tan 2  θ = sec 4  θ – sec 2  θ

L.H.S = tan 4  θ + tan 2  θ

= tan 2  θ (tan 2  θ + 1)

= (sec 2  θ – 1) (tan 2  θ + 1) [since, tan 2  θ = sec 2  θ – 1]

= (sec 2  θ – 1) sec 2  θ [since, tan 2  θ + 1 = sec 2  θ]

= sec 4  θ – sec 2  θ = R.H.S.  Proved.

More problems on trigonometric identities are shown where one side of the identity ends up with the other side.

4.  . cos θ/(1 – tan θ) + sin θ/(1 – cot θ) = sin θ + cos θ

L.H.S = cos θ/(1 – tan θ) + sin θ/(1 – cot θ)

= cos θ/{1 – (sin θ/cos θ)} + sin θ/{1 – (cos θ/sin θ)}

= cos θ/{(cos θ – sin θ)/cos θ} + sin θ/{(sin θ – cos θ/sin θ)}

= cos 2  θ/(cos θ – sin θ) + sin 2  θ/(cos θ – sin θ)

= (cos 2  θ – sin 2  θ)/(cos θ – sin θ)

= [(cos θ + sin θ)(cos θ – sin θ)]/(cos θ – sin θ)

= (cos θ + sin θ) = R.H.S.  Proved.

5.  Show that, 1/(csc A – cot A) – 1/sin A = 1/sin A – 1/(csc A + cot A)

1/(csc A – cot A) + 1/(csc A + cot A)

= (csc A + cot A + csc A – cot A)/(csc 2  A – cot 2  A)

= (2 csc A)/1; [since, csc 2  A = 1 + cot 2  A ⇒ csc 2 A – cot 2  A = 1]

= 2/sin A; [since, csc A = 1/sin A]

1/(csc A – cot A) + 1/(csc A + cot A) = 2/sin A

⇒ 1/(csc A – cot A) + 1/(csc A + cot A) = 1/sin A + 1/sin A

Therefore, 1/(csc A – cot A) – 1/sin A = 1/sin A – 1/(csc A + cot A)  Proved.

6.  (tan θ + sec θ – 1)/(tan θ – sec θ + 1) = (1 + sin θ)/cos θ

L.H.S = (tan θ + sec θ – 1)/(tan θ – sec θ + 1)

= [(tan θ + sec θ) – (sec 2  θ – tan 2  θ)]/(tan θ – sec θ + 1), [Since, sec 2  θ – tan 2  θ = 1]

= {(tan θ + sec θ) – (sec θ + tan θ) (sec θ – tan θ)}/(tan θ – sec θ + 1)

= {(tan θ + sec θ) (1 – sec θ + tan θ)}/(tan θ – sec θ + 1)

= {(tan θ + sec θ) (tan θ – sec θ + 1)}/(tan θ – sec θ + 1)

= tan θ + sec θ

= (sin θ/cos θ) + (1/cos θ)

= (sin θ + 1)/cos θ

= (1 + sin θ)/cos θ = R.H.S.  Proved.

https://www.math-only-math.com/problems-on-trigonometric-identities.html

https://intl.siyavula.com/read/maths/grade-12/trigonometry/04-trigonometry-04

https://byjus.com/maths/trigonometric-equations/

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