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CSEC Mathematics cheat sheet. complies equations and formulas into one document to assist students om learning
Typology: Cheat Sheet
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Curriculum Planning and Development Division
The booklet highlights some salient points for each topic in the CSEC Mathematics syllabus. At least one basic illustration/example accompanies each salient point. The booklet is meant to be used as a resource for “last minute” revision by students writing CSEC Mathematics.
Produced by: Patrick Ramdath, Curriculum Officer- Mathematics Revised by: Nicole Harris-Knudsen, Acting Curriculum Coordinator- Mathematics
Number Theory Basic Rules Points to Remember Illustration/ Example To find the difference between two numbers when one number is positive and one number is negative the result will be “+” if the larger value is positive or “–“ negative if the larger number is negative.
When multiplying, two positive numbers multiplied together give a positive product; and a negative number multiplied by another negative number gives a positive product. Also, a negative number multiplied by a positive number gives a negative product
e.g. 8 × 5 = 40
Number Theory Positive and Negative Numbers Points to Remember Illustration/ Example The rules for division of directed numbers are similar to multiplication of directed numbers. Use manipulatives- counters (yellow and red)
e.g. 10 ÷ 5 = 2
Natural Numbers (N) : {1,2,3,...} Integers (Z) : {..., - 3, - 2, - 1, 0, 1, 2, 3, ...} Rational Numbers (Q) :. 3/2 (=1.5), 8/4 (=2), 136/ (=1.36), - 1/1000 (=-0.001) Irrational Number : 𝜋, 3.142 (cannot be written as a fraction) Real Numbers (R): 1.5, - 12.3, 99, √2, π
Number Theory Decimals – Rounding Points to Remember Illustration/ Example Rounding up a decimal means increasing the terminating digit by a value of 1 and drop off the digits to the right. Round down if the number to the right of our terminating decimal place is four or less (4,3,2,1,0)
5.47 to the tenths place, it can be can be rounded up to
6.734 to the hundredths place, it can be rounded down to
Number Theory Operations with Decimals Points to Remember Illustration/ Example Find the product of 3.77 x 2.8 =?
Find the product of 3.77 × 2.
3.77 (2 decimal places) 2.8 (1 decimal place) 754 3016 10.556 (3 decimal places)
When dividing, if the divisor has a decimal in it, make it a whole number by moving the decimal point to the appropriate number of places to the right. If the decimal point is shifted to the right in the divisor, also do this for the dividend.
Fractions can always be written as decimals. For example: 2 5 = 0.^
1 2 = 0.^
3 4 = 0.^75 1 4 = 0.^
3 5 = 0.^
3 4 = 0.
Number Theory Computation – Fractions Points to Remember Illustration/ Example When the numerator stays the same, and the denominator increases, the value of the fraction decreases
3
1 , 4
1 , 5
1
4
3 , 5
3 , 6
3 7
3
When the denominator stays the same, and the numerator increases, the value of the fraction increases.
2
7 , 2
8 , 2
9
Equivalent fractions are fractions that may look different, but are equal to each other.^2
1 ,^4
2 , 6
3 8
4
Equivalent fractions can be generated by multiplying or dividing both the numerator and denominator by the same number.
2
1 = 2 2
1 2 x
x = 4
2
5
3 = 5 2
3 2 x
x = 10
6
Fractions can be simplified when the numerator and denominator have a common factor in them^10
6 =^52
3 2 x
x = 5
3
Fractions with different denominators, can be converted to a set of fractions that have the same denominator
4
3 , 3
2 is the same as 12
9 , 12
8
Addition and subtraction of fractions are similar to adding and subtracting whole numbers if the fractions being added or subtracted have the same denominator
12
9
8 = 12
1
When multiplying fractions, multiply the numerators together and then multiply the denominators together and simplify the results.
6
5 × 3
2 = 18
10 = 9
5
Number Theory Prime Numbers Points to Remember Illustration/ Example A prime number is a number that has only two factors: itself and 1e.g. 5 can only be divided evenly by 1 or 5, so it is a prime number. Numbers that are not prime numbers are referred to as composite numbers
Number Theory Computation of Decimals, Fractions and Percentages Points to Remember Illustration/ Example Percent means "per one hundred" 20 % = 20 per 100
To convert from percent to decimal, divide the percent by 100 10 % =^100
10 = 0.
67.5% = 100
To convert from decimal to percent, multiply the decimal by 100
0.10 as a percentage is 0.10 × 100 = 10% 0.675 is 0.675 × 100 = 67.5%
To convert from percentages to fractions, divide the percent by 100 to get a fraction and then simplify the fraction
12 = 100 4
12 4
= 25
3
To convert from fractions to percentages, convert the fraction to a decimal by dividing the numerator by the denominator and then convert the decimal to a percent by multiplying by 100.
25
3 = 0. 0.12 as a percentage is 0.12 × 100 = 12%
Angles formed by a Transversal Crossing two Parallel Lines Vertical Angles are the angles opposite each other when two lines cross. Vertically opposite angles are equal a = d f = g b = c e = h
The angles in matching corners are called Corresponding Angles. Corresponding Angles are equal a = e c = g b = f d = h
The pairs of angles on opposite sides of the transversal but inside the two lines are called Alternate Angles. Alternate Angles are equal d = e c = f
The pairs of angles on one side of the transversal but inside the two lines are called Consecutive Interior Angles. Consecutive Interior Angles are supplementary (add up to 180°) d + f c + e
Illustration of all angles mentioned on a single diagram. The transversal crosses two Parallel Lines
Triangles Pythagoras' Theorem Points to Remember Illustration/ Example Pythagoras' Theorem states that the square of the hypotenuse is equal to the sum of the squares on the other two sides
c^2 = a^2 + b^2 The Hypotenuse is c Find c
c^2 = 5^2 + 12^2 = 25 + 144 = 169 c= √ 169 = 13 units
Triangles Similar Triangles & Congruent Triangles Points to Remember Illustration/ Example Definition: Triangles are similar if they have the same shape, but can be different sizes.
(They are still similar even if one is rotated, or one is a mirror image of the other).
There are three accepted methods of proving that triangles are similar:
If two angles of one triangle are equal to two angles of another triangle, the triangles are similar.
If angle A = angle D and angle B = angle E
Then ∆ABC is similar to ∆DEF
Show that the two triangles given beside are similar and calculate the lengths of sides PQ and PR.
Solution:
∠A - ∠B and ∠R = 180 - ∠P - ∠Q)
Therefore, the two triangles ΔABC and ΔPQR are similar.
Triangles Similar Triangles & Congruent Triangles Points to Remember Illustration/ Example
Therefore: 𝐷𝐸 𝐴𝐵 =^
𝐶𝐷 𝐶𝐴 7 11 =^
15 𝐶𝐴
7CA = 11 × 15
CA = 11 × 7 15
x = CA – CD = 23.57 – 15 = 8.
Mensuration Areas & Perimeters Points to Remember Illustration/ Example The area of a shape is the total number of square units that fill the shape.
Area of Square= a^2 Perimeter of Square= a+ a + a + a
a = length of side
Find the area and perimeter of a square that has a side- length of 4 cm
Area of Square = a × a= a^2 = 4 × 4= 4^2 = 16 cm^2 Perimeter of Square = 4 + 4 + 4 + 4 = 16 cm
a represents the length; b represents the width
Area of Rectangle = a × b Perimeter of Rectangle = a+ a + b + b = 2(a+b)
Find the area of a rectangle of length 5cm, width 3cm
Area of Rectangle = 5 cm x 3cm = 15 cm^2 Perimeter of Rectangle = 5 + 5 + 3 + 3 = 2(5+3) = 16cm
The area of a triangle is : 2
1 x b x h b is the base h is the height
Area = 2
1 x b x h = 2
1 x 20units x 12units = 120 units^2
Area of triangle using "Heron's Formula"- given all three sides:
Step 1: Calculate "s" (half of the triangle’s perimeter):
Step 2: Then calculate the Area :
Example: What is the area and perimeter of a triangle with sides 3cm, 4cm and 5cm respectively?
Step 1: s = 2
3 4 5 = 2
12 = 6
Step 2 : Area of triangle = √ 6 ( 6 − 3 )( 6 − 4 )( 6 − 5 ) = (^) √ 6 ( 3 )( 2 )( 1 ) = 6cm^2
Perimeter of triangle = a + b + c = 3 + 4 + 5 = 12cm
Mensuration Areas & Perimeters Points to Remember Illustration/ Example
Area of Trapezium = ½(a+b) × h = ½(sum of parallel sides) × h
h = vertical height
Perimeter = a + b + c + d
Find the area of the trapezium
A = ½ (a+b) × h = ½(10 + 8) × 4 =½ × (18) × 4 = 36 cm^2
Perimeter = a + b + c + d = 10 + 8 + 4.3 + 4. = 26.4 cm
Area of Parallelogram = base × height
b = base h= vertical height
Find the area of a parallelogram with a base of 12 centimeters and a height of 5 centimeters.
Area of parallelogram= b × h = 12cm × 5cm = 60 cm^2
Perimeter of parallelogram = a + b + a + b = 2 (a + b) = 12cm + 7cm + 12cm + 7cm = 38cm
c d
b
a
Mensuration Surface Area and Volumes Points to Remember Illustration/ Example Volume of Cylinder = Area of Cross Section x Height = π r^2 h
Surface Area of Cylinder = 2π r^2 + 2πrh = 2πr (r + h)
Find the volume and total surface area of a cylinder with a base radius of 5 cm and a height of 7 cm.
Volume = π r^2 h = 7
22 × 52 × 7 =22 × 25 cm^3 = 550cm^3
Conversion to Litres : 1000 cm^3 = 1L
550 cm^3 = 1000
550 L= 0.55 L
Surface Area = 2 π(5)(7) + 2π(5)^2 = 70π + 50π = 120π cm^2 ≈ 376.99 cm^2
Prism- Since a cylinder is closely related to a prism, the formulas for their surface areas are related
Volume of Prism = area of cross section × length = A l
Example: What is the volume of a prism whose ends have an area of 25 m^2 and which is 12 m long
Answer: Volume = 25 m^2 × 12 m = 300 m^3
Volume of irregular prism = A h Surface Area of irregular prism = 2 A + (perimeter of base × h)
Mensuration Surface Area and Volumes Points to Remember Illustration/ Example
Volume of cube = s^3
Surface Area of cube = s^2 + s^2 + s^2 + s^2 + s^2 + s^2 = 6 s^2
Find the volume and surface area of a cube with a side of length 3 cm
Volume of cube = s × s × s = s^3 = 3 × 3 × 3 = 27 cm^3
Surface Area of cube = s^2 + s^2 + s^2 + s^2 + s^2 + s^2 = 6 s^2 = 6(3)^2 = 6 × 9 = 54 cm^2
Volume of cuboid = length x breadth x height = xyz
Surface area = xy + xz + yz + xy + xz + yz = 2 xy + 2 xz + 2 yz = 2( xy + xz + yz)
Find the volume and surface area of a cuboid with length 10cm, breadth 5cm and height 4cm.
Volume of cuboid = length × breadth × height = 10 × 5 × 4 = 200cm^3
Surface Area of cuboid = 2 xy + 2 xz + 2 yz = 2(10)(5) + 2(10)(4) + 2(5)(4) = 100 + 80 + 40 = 220 cm^2
The Volume of a Pyramid
= 3
1 × [Base Area] × Height
Find the volume of a rectangular-based pyramid whose base is 8 cm by 6 cm and height is 5 cm.
Solution:
V = 3
1 × [Base Area] × Height
Mensuration Surface Area and Volumes Points to Remember Illustration/ Example
= 3
1 × [8 × 6] × 5
= 80 cm^3