Welcome to class!
In today’s class, we will be talking about geometry plane shapes. Enjoy the class!
GEOMETRY PLANE SHAPES
Types of plain shapes and their properties
Similarities and differences between square, rectangles, triangles, trapeziums, parallelograms, and circles
The Circle:
shape
The circle is a shape that can be made by tracing a curve that is always the same distance from a point that we call the centre. The distance around a circle is called the circumference of the circle.
The Triangle:
shape
The triangle is a shape that is formed by 3 straight lines that are called sides. There are different ways of classifying triangles, according to their sides or angles.
According to their angles:
Right triangle: the largest of the 3 angles is a right angle.
Acute Triangle: the largest of the 3 angles is an acute angle (less than 90 degrees).
Obtuse Triangle: the largest of the 3 angles is an obtuse angle (more than 90 degrees).
According to their sides:
Equilateral Triangle: all 3 sides are the same length.
Isosceles Triangle: it has 2 (or more) sides that are of equal length. (An equilateral triangle is also isosceles.)
Scalene Triangle: no 2 sides are of equal measure.
The Rectangle:
shape
The rectangle is a shape that has 4 sides. The distinguishing characteristic of a rectangle is that all 4 angles measure 90 degrees.
The Rhombus:
shape
The rhombus is a shape formed by 4 straight lines. Its 4 sides measure the same length but, unlike the rectangle, any of all 4 angles measure 90 degrees.
The Square:
shape
The square is a type of rectangle, but also a type of rhombus. It has characteristics of both of these. That is to say, all 4 angles are right angles, and all 4 sides are equal in length.
The Trapezoid:
shape
The trapezoid also has 4 sides. It has two sides that are parallel but the other 2 are not.
Trapezium and Its Properties:
A trapezium or a trapezoid is a quadrilateral with a pair of parallel sides. A parallelogram may also be called a trapezoid as it has two parallel sides. The pair of parallel sides is called the base while the non-parallel sides are called the legs of the trapezoid. The line segment that connects the midpoints of the legs of a trapezoid is called the mid-segment.
Parallelograms:
shape
One special kind of polygons is called a parallelogram. It is a quadrilateral where both pairs of opposite sides are parallel.
In our next class, we will be talking more about Geometry Plane Shapes. We hope you enjoyed the class.
The perimeter of a regular polygon, square, rectangle, triangle, trapezium, parallelogram, and circle Area of plain shapes such as square, rectangle, triangle, trapezium, parallelogram, and circle
The different geometrical shapes formula of area and perimeter with examples are discussed below:
Perimeter and Area of Rectangle:
● The perimeter of rectangle = 2(l + b).
Area of rectangle = l × b; (l and b are the length and breadth of the rectangle)
Diagonal of rectangle = √ (l² + b²)
Perimeter and Area of the Square:
The perimeter of square = 4 × S.
Area of square = S × S.
Diagonal of square = S√2; (S is the side of the square)
Perimeter and Area of the Triangle:
Perimeter of triangle = (a + b + c);(a, b, c is 3 sides of a triangle)
Area of triangle = √ (s (s – a) (s – b) (s – c)); where s is the semi-perimeter of triangle
S = 1/2 (a + b + c)
Area of triangle = 1/2 × b × h;(b base, h height)
Area of an equilateral triangle = (a²√3)/4; (a is the side of the triangle)
Perimeter and Area of the Parallelogram:
The perimeter of parallelogram = 2 (sum of adjacent sides)
Area of parallelogram = base × height
Perimeter and Area of the Rhombus:
Area of rhombus = base × height
Area of rhombus = 1/2 × length of one diagonal × length of other diagonal
The perimeter of rhombus = 4 × side
Perimeter and Area of the Trapezium:
Area of trapezium = 1/2 (sum of parallel sides) × (perpendicular distance between them)
= 1/2 (p₁ + p₂) × h (p₁, p₂ are 2 parallel sides)
Circumference and Area of Circle:
Circumference of circle = 2πr
= πd
Where, π = 3.14 or π = 22/7
r is the radius of the circle
d is the diameter of the circle
● Area of circle = πr²
● Area of ring = Area of the outer circle – Area of the inner circle.
In our next class, we will be talking about Three Dimensional Shapes. We hope you enjoyed the class.
Welcome to class!
In today’s class, we will be talking about Wood. Enjoy the class!
THREE DIMENSIONAL SHAPES
THREE DIMENSIONAL SHAPES classnotes.ng
Identification of three-dimensional or 3D shapes Basic properties of cubes and cuboids, and cones Basic properties of cylinders and spheres Volume of cubes and cuboids
Surface Area and Volume of 3D shapes
The two distinct measures used for defining the 3D shapes are:
Surface Area
Volume
Surface Area is defined as the total area of the surface of the three-dimensional object. It is denoted as “SA”. The surface area is measured in terms of square units. The three different classifications of surface area are defined below. They are:
Curved Surface Area (CSA) is the area of all the curved regions
Lateral Surface Area (LSA) is the area of all the curved regions and all the flat surfaces excluding base areas
Total Surface Area (TSA) is the area of all the surfaces including the base of a 3D object
Volume is defined as the total space occupied by the three-dimensional shape or solid object. The volume is denoted as “V”. It is measured in terms of cubic units.
Faces, Edges, and Vertices of 3D Shapes
Three-dimensional shapes have many attributes such as vertices, faces and edges. The flat surfaces of the 3D shapes are called the faces. The line segment where two faces meet is called an edge. A vertex is a point where 3 edges meet.
shape
Faces, Edges, and Vertices
How to Make 3d Shapes for Maths Project
If you know what are three-dimensional shapes, it would ….be easy for you to build a 3d shape project for a house or a building. This would be easy for the students to make as they can measure the rooms easily. Rest all they need is cardboard, glue, scissors and art supplies to make it look exactly like a mini house or building. Here, we are going to discuss the list of different three-dimensional shapes with its properties and the formulas of different 3D shapes
Cube:
shape
A cube is a solid or three-dimensional shape which has 6 square faces. The cube has the following properties.
All edges are equal
8 vertices
12 edges
6 faces
The surface area and the volume of the cube are given below:
The Surface Area of a Cube = 6a2 square units
The volume of a Cube = a3 cubic units
Cuboid:
shape
A cuboid also called a rectangular prism, where the faces of the cuboid are a rectangle in shape. All the angle measures are 90 degree
8 vertices
12 edges
6 faces
The surface area and the volume of the cuboid are given below:
The Surface Area of a Cuboid = 2(lb+bh+lh) Square units
The volume of a Cuboid = lbh Cubic units
Prism:
shape
A prism has a 3D shape which consists of two equal ends, flat surfaces or faces, and also have identical cross-section across its length. Since the cross-section looks like a triangle, the prism is generally called a triangular prism. The prism does not have any curve. Also, a prism has
6 vertices
9 edges
5 faces – 2 triangles and 3 rectangles
The surface area and the volume of the prism are given below:
The Surface Area of a Prism =2(Base Area) + (Base perimeter × length) square units
The volume of a Prism = Base Area × Height Cubic units
Pyramid:
shape
A pyramid a solid shape which has a structure, whose outer faces are triangular and meet to a single point on the top. The pyramid base can be of any shape such as triangular, square, quadrilateral or in the shape of any polygon. The most commonly used type of a pyramid is the square pyramid i.e., it has a square base and four triangular faces. Consider a square pyramid, it has
5 vertices
8 edges
5 faces
The surface area and the volume of the pyramid are given below:
The Surface Area of a Pyramid = (Base area) + (1/2) × (Perimeter) × (Slant height) Square units
The volume of a Pyramid = 1/ 3 × (Base Area) × height Cubic units
Cylinder:
shape
A cylinder is defined as the three-dimensional geometrical figure which has two circular bases connected by a curved surface. A cylinder has
No vertex
2 edges
2 flat faces – circles
1 curved face
The surface area and the volume of the cylinder are given below:
Surface Area of a Cylinder = 2πr(h +r) Square units
The curved surface area of a cylinder = 2πrh
The volume of a Cylinder = πr2 h Cubic units
Cone:
shape
A cone is a three-dimensional object or solid, which as a circular base and has a single vertex. The cone is a geometrical figure that decreases smoothly from the circular flat base to the top point called the apex. A cone has
1 vertex
1 edge
a flat face – circle
1 curved face
The surface area and the volume of the cone are given below:
The Surface Area of a Cone = πr(r +√(r2+h2) Square units
The curved surface area of a cone =πrl
Slant height of a cone = l = √(r2+h2)
The volume of a Cone = ⅓ πr2h Cubic units
Sphere:
shape
A sphere is a three-dimensional solid figure which is perfectly round in shapes and every point on its surface is equidistant from the point is called the centre. The fixed distance from the centre of the sphere is called a radius of the sphere. A sphere has
No vertex
1 curved face
No edges
The surface area and the volume of the sphere are given below:
The Curved Surface Area of a Sphere = 2πr² Square units
The Total Surface Area of a Sphere = 4πr² Square units
The volume of a Sphere = 4/3(πr3) cubic units
In our next class, we will be talking about Construction. We hope you enjoyed the class.
Welcome to class!
In today’s class, we will be talking about construction. Enjoy the class!
CONSTRUCTION
CONSTRUCTION classnotes.ng
Construction of parallel and perpendicular lines
Bisection of a given line segment
Construction of angles 90 and 60 degrees
Construction of Parallel and Perpendicular Lines:
In this section, you will learn how to construct parallel and perpendicular lines.
Parallel Lines:
Parallel lines are the lines which will never intersect and the perpendicular distance between them will be the same at everywhere.
Perpendicular Lines:
The two lines which have the angle of inclination 90° at the point of intersection are called as perpendicular lines.
Construction of Parallel Lines – Example
Using a set square and a ruler draw a line parallel to a given line through a point at a distance of 5cm above it.
Step 1:
(i) Draw a line XY using a ruler and mark a point A on it.
(ii) Draw AM = 5cm with the help of a set square.
Step 2:
Place the set square on the line segment XY.
(i) Place the set scale as shown in the figure.
Step 3:
(i) Pressing tightly the ruler, slide the set square along the ruler till the edge of the set square touches the point M.
(ii) Through M, draw a line MN along the edge.
(iii) MN is the required line parallel to XY through M.
construction
Construction of Perpendicular Lines – Examples
Example 1:
Using a set square and a ruler, draw a line perpendicular to a given line at a point on it.
Solution:
Step 1:
(i) Draw a line AB with the help of a ruler.
(ii) Mark a point P on it.
construction
Step 2:
(i) Place a ruler on the line AB
(ii) Place one edge of a set square containing the right angle along with the given line AB as shown in the figure.
Step 3:
(i) Pressing the ruler tightly with the left hand, slide the set square along the ruler till the edge of the set square touches the point P.
(ii) Through P, draw a line PQ along the edge.
construction
Step 4:
PQ is the required line perpendicular to AB. Measure and check if m 18, 36 and 54.
18 = 2x3x3 = 2 x 32 in index form
Example 2:
Write down all the factors of 22.
State which of these factors are prime numbers
Write the first three multiples of 22
Solution:
Factors of 22 are 1, 2, and 11.
Prime numbers of the factors of 22 are 2 and11
The first three multiples of 22 are 1×22 = 22, 2×22=44, 3 x 22=66 => 22, 44 and 66.
Evaluation
Write down all the factors
State which of these factors are prime numbers
Write the first three multiples of each of the following numbers below
Express each as a product of its prime factors in index form
(A) 12 (B) 30 (C) 39 (D) 48
Welcome to Class !!
We are eager to have you join us !!
In today’s Mathematics class, We will be discussing LCM, HCF and Perfect Squares. We hope you enjoy the class!
mathematics classnotesng
Highest Common Factors
A highest common factor is the greatest number which will divide exactly into two or more numbers. For example, 4 is the highest common factor (HCF) of 20 & 24.
i.e. 20 = 1, 2, (4), 5, 10, 20
24= 1, 2, 3, (4), 6, 8, 12, 24
Example:
Find the H.C.F of 24 & 78
Method 1
Express each number as a product of its prime factors
Workings
2 | 24 2 | 78
2 | 12 2 | 36
2 | 6 2 | 18
3 | 3 3 | 9
3 | 3
24=23x3
78 =(23 x 3) x 3
The H.C.F. is the product of the common prime factors.
HCF=23x3
=8×3=24
Method II
24=2x2x2x3
78=2x2x2x3x3
Common factor=2x2x2x3
HCF=24
hcf of 16 and 24 maths classnotesng
LCM: Lowest Common Multiple
Multiples of 2 are =2,4,6,8,10,12,14,16,18,20,22,24…
Multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40
Notice that 10 is the lowest number which is a multiple of 2 & 5.10 is the lowest common multiple of 2& 5
Find the LCM of 20, 32, and 40
Method 1
Express each number as a product of its prime factors
20=22x5
32=25
40=22x2x5
Content
Fractions and Percentages
Proportion
Ratio
Rate
Fractions and percentages
A fraction can be converted to a decimal by dividing the numerator by its denominator. It can be changed to a percentage by simply multiplying by 100.
Example 5.1
Change 3/8 into a decimal and percentage
Convert 0.145 to a percentage
Solution
1) 3/8 = 0.375 in decimal
3/8 x 100% = 37.5%
Classwork
Change the following to percentage
(a) 0.264 (b) 0.875
Change the following to decimal fractions
(A) 60% (b) 52/3%
APPLICATION OF DECIMAL FRACTIONS AND PERCENTAGES
Consider the following examples.
Find 15% of 2.8kg
Express 3.3 mass a percentage of 7.5
Find 331/3 % of 8.16litres
Solution
15/100 of 2.8kg
15/100 x 2.8 x 1000g
15/100 x 2800
=420g
=420/1000
=0.420kg
2. 3.3/7.5 x 100/1
= 33/75 x 100/1
= 11×4 = 44%
Proportion
Proportion can be solved either by the unitary method or inverse method. When solving by unitary method, always
Write in a sentence, the quantity to be found at the end.
decide whether the problem is either an example of the direct or inverse method
Find the rate for one unit before answering the problem.
Examples
A worker gets N 900 for 10 days of work, find the amount for (a) 3 days (b) 24 days (c) x days
Solution
For 1 day = N 900
1 day = 900/10 = N90
For 3 days =3 x 90 = 270
For 24 days = 24×90 = N 2,160
For x days =X x 90 = N 90 x
Inverse Proportion
Example 1: Seven workers dig a piece of ground in 10 days. How long will five workers take?
Solution
For 7 workers =10 days
For 1 worker =7×10=70 days
For 5 workers=70/5 =14 days
Example 2: 5 people took 8 days to plant 1,200 trees, How long will it take 10 people to plant the same number of trees
Solution
For 5 people =8 days
For 1 person =8×5=40 days
For 10 people =40/10 =4 days
Class Work
A woman is paid N 750 for 5 days, Find her pay for (a) 1 day (b) 22 days
A piece of land has enough grass to feed 15 cows for x days. How long will it last (a) 1 cow (b) y cows
A bag of rice feeds 15 students for 7 days. How long would the same bag feed 10 students
Note on direct proportion: this is an example of direct proportion. The less time worked (3 days) the less money paid (#270) the more time worked (24 days) the more money paid (N N 2,160)
Ratio
Ratio behaves the same way as a fraction. Ratios are often used when sharing quantities.
ratio-to-fraction maths classnotesng
Example
600/800=600/800=3/4
300:400 = 600:800 = 1200:1600 = 3:4
Example
Express the ratio of 96c: 120c as simple as possible
Solution 96c: 120c=96/120=4/5=4.5
Fill in the gap in the ratio of 2:7=28
Solution
let the gap be X
2/7 = X/28
7X =2 x 28
X=2 x 28/7
X=2 x 4
X = 8
Two students shared 36 mangoes in the ratio 2:3 How many mangoes does each student get?
Solution
Total ratio =2+3=5
First share=2/5×35/1=21 mangoes
Rate
Rate is the change in one quantity to the other. Examples are 45km/hr, a km, 1 litre etc
Worked examples
A car goes 160 km in 2 hrs what is the rate in km/hr?
Solution
In 2 hrs the car travels 160 km
In 1 hr the car travels 160/2=80km
Therefore the rate of the car is 80km/hr
A car uses 10 l****s of petrol to travel 74 km. Express its petrol consumption as a rate in km per litre.
Solution
10 l****s =74 km
1 l****s = 74/10 km
=7.4 km
Classwork
A car factory made 375 cars in 5 days, Find its rate in cars per day.
A car travels 126 km in 11/2 hrs. Find the rate in km per hr.
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