## Wednesday, May 17, 2017

### 2016/2017 - GRADE 9 - STUDENT'S ACTIVITY - CLINOMETER

Teacher : Nuri Martini, S.Si, M.Pd

TRIGONOMETRY
Learning Objective
Students will be able to APPLY the concept of TRIGONOMETRY in real - life problems

Student's Activity of mathematics that deals with the relationship between the side lengths and angle of triangles. We
1. Brief Explanation of Trigonometry Concepts
TRIGONOMETRY  is a branch of mathematics that deals with the relationship between the side lengths and angles of triangles

We can use trigonometric ratios to find unknown side lengths and angles in right angled triangles.
FINDING SIDE LENGTHS
Suppose we are given the angles of a right angled triangle, and the length of a side. We can use the trigonometric ratios to find the other side lengths.
Step 1 : Redraw the figure and mark on it HYP, OPP, and ADJ relative to a given angle.
Step 2 : Choose an appropriate trigonometric ratio, and construct an equation.
Step 3 : Solve the equation to find the unknown side length

FINDING ANGLES
If we know two side lengths of a right angled triangle, we can use trigonometry to find the angles. If In the right angled triangle shown sin A = 3/5, we say that A is the INVERSE SINE of 3/5. We can use a calculator to evaluate inverse sines.

PROBLEM SOLVING USING TRIGONOMETRY
The trigonometric ratios can be used to solve a wide variety of problems involving right angled triangles. When solving these problems it is important to follow the steps below:
* Draw a DIAGRAM to illustrate the situation
* Mark on the diagram the UNKNOWN angle or side that needs to be calculated
* Locate a RIGHT ANGLED TRIANGLE in your diagram
* Write an EQUATION connecting an angle and two sides of the triangle using an appropriate trigonometric ratio
* SOLVE the equation to find the unknown

ANGLES OF ELEVATION AND DEPRESSION
When an object is HIGHER than an observer, the ANGLE OF ELEVATION is the angle from the horizontal UP to the object.
When an object is LOWER than an observer, the ANGLE OF DEPRESSION is the angle from the horizontal DOWN to the object.

THE AREA OF A TRIANGLE
We can use trigonometry to find the area of a triangle if we are given the lengths of TWO SIDES, as well as the INCLUDED ANGLE between the sides.
The area of a triangle is a half of the product of two sides and the sine of the included angle.
Area = 1/2 x ab sin C

THE SINE RULE
The sine rule is a set of equations which connects the lengths of the sides of any triangle with the sines of the opposite angles.
sin A/a = sin B/b = sin C/c
If we are given two angles and one side of a triangle, we can use the sine rule to find another side length. Finding angles using the sine rule is more complicated than finding sides because there may be two possible answers.

THE COSINE RULE
The cosine rule relates the three sides of a triangle and one of its angles. If we are given two sides of a triangle and the included angle, we can use the cosine rule to find the third side.

PROBLEM SOLVING USING THE SINE AND COSINE RULES
When using trigonometry to solve problems, you should draw a diagram of the situation. The diagram should be reasonably accurate, and all important information should be clearly marked on it

2. Making a Clinometer (please read or search from the internet how to make it)
Material Needed
1. Protractor
2. Straw
3. Sellotape
4. String
5. Glue Tac

* Take data for measuring height of object (Outdoor and Indoor ) Min 1 Object and Max 3 Object with 3 different data
Wednesday, 10th May 2017 : Outdoor Objects
Thursday, 18th May 2017 10.45 - 11.30 : Indoor Objects (Min 1 objects and Max 3 objects) Every objects should take 3 times

* Make a written report (Every Thursday, 18th & 25th May 2017 --> 1 Period)
- Title (Created by students)
- Introduction (Explain about concepts use in this investigation such as Pythagora's Theorem , basic trigonometry)
- Purpose of Investigation
- Procedure (include pictures)
1. Making Clinometer
2. Measuring Objects (Indoor and Outdoor)
- Table Data
1. Indoor Objects
2. Outdoor Objects
- Calculation
- Analysing and Conclusion
- Resources
Deadline : Wednesday, 31st May 2017

3. Practice to apply the concepts of basic trigonometry from their textbook
Exercise 12.A1 No. 1bf, 2behk, 3c, 4 --> Monday, 15th May 2017
Exercise 12.A2 No. 1efhi, 2c, 3 --> Monday, 15th May 2017
Exercise 12B No. 1 - 16 --> Wednesday, 17th May 2017
Exercise 12F No. 1bc, 3ac, 4ab, 5 - 8 --> Monday, 22nd May 2017
Exercise 12G1 No. 1ab, 2bc, 3, 4bc --> Wednesday, 24th May 2017
Exercise 12G2 No. 2, 3cd, 4c --> Wednesday, 24th May 2017
Exercise 12I No. 1 - 4, 10 - 11 --> Monday, 29th May 2017

Teacher : Nuri Martini, S.Si, M.Pd

LEARNING OBJECTIVE
Students will be able to solving quadratic equations algebraically use variety methods (Factorising, Completing Square, and General Formula)

INTRODUCTION
The study of algebra is vital for many areas of mathematics. We need it to manipulate equations, solve problems from unknown variables, and also to develop higher level mathematical theories.
QUADRATIC EQUATION is an equation which can be written in the form ax2 + bx + c = 0 where a, b, and c are constants and a is not equal to zero.
A quadratic equation may have two, one or zero real solutions.

SOLUTION BY FACTORISATION
For quadratic equations which are not of the form x2 = k, we need an alternative method of solution. One method is to FACTORISE the quadratic and then apply the NULL FACTOR LAW.
The Null Factor Law states that :
When the product of two or more numbers is zero, then at least one of them must be zero.
so if ab = 0 then a = 0 or b = 0

Step 1 : If necessary, rearrange the equation so one side is ZERO
Step 2 : Fully Factorise the other side (usually the LHS/Left Hand Side)
Step 3 : Apply the Null Factor Law
Step 4 : Solve the resulting linear equations
Example
 Solution by Factorisation

COMPLETING THE SQUARE
Some quadratic equations, such as x^2+4x-7=0, cannot be solved using the factorisation methods already practised. This is because the solutions are irrational.
Instead, we use a method called completing the square.
 Completing the Square
The process of creating a perfect square on the left hand side is called completing the squareTo complete a perfect square, the number we must add to both sides is found by halving the coefficient of x, then squaring this value.

Many quadratic equations cannot be solved easily by factorisation, and completing the square is rather tedious. Consequently, the quadratic formula has been developed.

## Thursday, February 2, 2017

### 2016/2017 - GRADE 9 - PROVING PYTHAGORAS' THEOREM

Teacher : Nuri Martini, S.Si, M.Pd

PROVING PYTHAGORAS' THEOREM

Learning Objective
Students will be able to do investigation of Pythagoras' Theorem from variety methods

Explanation
A right angled triangle is a triangle which has a right angle as one of its angles. The side opposite the right angle is called the hypotenuse, and is the longest side of the triangle. The other two sides are called the leg of the triangle.
Around 500 BC, the Greek mathematician Pythagoras discovered a rule which connects the lengths of the sides of right angled triangles.
In geometric form, Pythagoras' theorem states that :
In any right angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides.
A theorem is a fact that has been proven to be true.

Student's Activity
1. Students do investigation of proofing Pythagoras' theorems with the following steps
a. Research about methods of proofing Pythagoras' theorem and choose minimum 4 methods to be investigated by students
b. Learning the method choose and try to make a model of it use colour paper with random size then stick on A4 paper.
c. Showing all the proofing on written report

2. Format of written report
a. Cover title
b. Introduction (Describe your knowledge about : Who is Pythagoras? What is the Pythagoras' theorem, When do you use the Pythagoras' theorem)
c. What to do (Brief Description of each method that you choose include with algebraic explanation and picture/model of the methods)
d. Conclusion
e. Attachment (All models of each methods need to be stick on A4 paper separately for each methods)
f. Resources

MATHEMATICS_INVESTIGATION_R...

## Sunday, December 13, 2015

### 2015/2016 - GRADE 12 SCIENCE UN - SUKUBANYAK

Teacher : Nuri Martini, S.Si, M.Pd

Resources
Sartono Wirodikromo, Matematika untuk SMA kelas XI Program Ilmu Alam, Bab 5 Sukubanyak, hal 140 - 175, Penerbit Erlangga

SUKUBANYAK

Standar Kompetensi
Menggunakan aturan sukubanyak dalam penyelesaian masalah

Kompetensi Dasar
• Menggunakan algoritma pembagian sukubanyak untuk menentukan hasil bagi dan sisa pembagian
• Mengunakan  teorema sisa dan teorema faktor dalam pemecahan masalah

PENGERTIAN SUKUBANYAK
Sukubanyak atau polinom dalam  variabel x yang berderajat n secara umum dapat ditulis sebagai berikut :

Derajat dari suatu sukubanyak dalam variabel x ditentukan oleh pangkat yang paling tinggi bagi variabel x yang ada dalam sukubanyak itu.

OPERASI ANTAR - SUKUBANYAK
Penjumlahan atau pengurangan sukubanyak f(x) dengan sukubanyak g(x) dapat ditentukan dengan cara menjumlahkan atau mengurangkan suku - suku yang sejenis dari kedua sukubanyak itu.
Sedangkan perkalian sukubanyak f(x) dengan sukubanyak g(x) dapat ditentukan dengan cara mengalikan suku - suku dari kedua sukubanyak itu. Dalam mengalikan suku - suku dari kedua buah sukubanyak itu digunakan sifat distributif perkalian baik distributif perkalian terhadap penjumlahan maupun distributif perkalian terhadap pengurangan

Misalkan f(x) dan g(x) masing - masing merupakan sukubanyak berderajat m dan n, maka :
• f(x) +/- g(x) adalah sukubanyak berderajat maksimum m atau n
• f(x)  . g(x) adalah sukubanyak berderajat ( m + n )
KESAMAAN SUKUBANYAK
Sukubanyak f(x) dikatakan memiliki kesamaan dengan sukubanyak g(x), jika kedua sukubanyak itu mempunyai nilai yang sama untuk semua variabel x bilangan real.
Kesamaan dua sukubanyak f(x) dan g(x) itu ditulis sebagai

PEMBAGIAN SUKUBANYAK
Yang dibagi = pembagi x hasil bagi + sisa pembagian

Metode Horner
Persamaan yang menghubungkan sukubanyak yang dibagi f(x) dengan sukubanyak pembagi (x - k), sukubanyak hasil bagi H(x), dan sisa pembagian S adalah
f(x) = (x - k)  . H(x) + S

Jika sebuah sukubanyak f(x) dibagi dengan (x - k) menghasilkan sisa S = 0 maka dikatakan f(x) habis dibagi oleh (x - k) atau (x - k) adalah faktor dari f(x).

TEOREMA SISA
Misalkan sukubanyak f(x) dibagi dengan P(x) memberikan hasil bagi H(x) dengan sisa pembagian S(x). Persamaan yang menyatakan hubungan antara f(x) dengan P(x), H(x) dan S(x) adalah :
f(x) = P(x) . H(x) + S(x)
dengan
• f(x) sebagai sukubanyak yang dibagi, misalnya diketahui berderajat n
• P(x) sebagai sukubanyak pembagi, misalnya diketahui berderajat m dan m lebih kecil atau sama dengan n
• H(x) sebagai sukubanyak hasil bagi, berderajat (n - m) yaitu derajat sukubanyak yang dibagi dikurangi dengan derajat sukubanyak pembagi
• S(x) sebagai sukubanyak sisa pembagian, berderajat paling tinggi atau maksimum (m - 1) yaitu berderajat maksimum satu kurangnya dari derajat sukubanyak pembagi.
Teorema 1
Jika sukubanyak f(x) berderajat n dibagi dengan (x - k) maka sisanya ditentukan oleh
S= f(k)
Teorema 2
Jika sukubanyak f(x) berderajat n dibagi dengan (ax+b) maka sisanya ditentukan oleh
S = f(-b/a)

TEOREMA FAKTOR
Teorema 3
Misalkan f(x) adalah sebuah sukubanyak, (x - k) adalah faktor dari f(x) jika dan hanya jika f(k) = 0. k disebut akar atau nilai nol dari persamaan sukubanyak f(x) = 0.

AKTIVITAS SISWA
Kegiatan
Menentukan rumus keliling dan luas bangun data segi - n beraturan serta luas dan volume prisma segi-n beraturan

Tujuan
Memahami penggunaan sukubanyak dalam menentukan rumus untuk bangun datar segi-n beraturan dan prisma segi-n beraturan

 Siswa membuat bangun datar segi - n beraturan menggunakan batang korek api yang dilem pada kertas untuk n = 4, 5, - - -, 10
 Siswa menghitung luas bangun datar segi-n beraturan dan menentukan rumus umus untuk setiap n

 Siswa membuat Laporan Hasil Percobaan