GRADE-9 NCERT CHAPTERS
Perimeter,Area And volume-3
Volume is the space that occupy by an object. It is used to compare the size of the solid object. We can also understand the volume of an object like breadbox by measuring how much i can hold.
consider Shweta and Swati have bread boxes. One box is look like little more length than other. the second one box is look like little more height than first one. How we can compare which one is exactly bigger than other. In this type of cases we can find volumes of both and can find which one is bigger.
As the volume is the space occupied by the object, we can check the volume of the bread boxes by filling it. Consider the equal size of sandwiches. In the first box the first layer contain 2 rows of 4 sandwiches (that is 8 psc). And this first bos can holt sucha 2 layers. so the first bos can hold total 2*8 = 16 psc of sandwiches. Where the second box can have only 2 rows of 3 sandwiches (6 psc). But it can hold such a 3 layer. So the second box can hold 3*6 = 18 pcs. So the second box have more volume, because it can occupy more things.
The volume of an object is the amount of space it occupies.
Different type of unit of measurement of volume.
See figure in page 187
A Millimeter cube (MM cube) is about the size of grain of sugar.
A Centimeter cube (C.M cube) is about the size of a die
A Meter cube (M cube) is about the size of a very large carton.
A MM cube is used to measure the volume of very small object.
A meter cube is used to measure the volume of large object.
Now we can find the volume of a bread box by filling it with CM cubes
see the picture in page 188. Here the base of the bread box is fill with 10 rows of 4 cubes. So 10*4 = 40 cubes at one layer. And this box is fill up with four such a layer. so there are 40*4 = 160 Cm cubes in this box. there for the volume of the box is 160 CM CUBES.
We can also find the volume of container with different size of layers in CM cubes. In such a case we have to calculate the Size of each layer separately and can add the size of each layer so that we can get the volume of the things. See the figure in page 188
As we seen above we can find the volume by finding the number of cubes in a layer and number of layers.
volume = number of cubes in a layer * number of layer
Number of cubes in a layer = length * breadth
Number of layer = height
There for volume = Length * breadth * Height
This can be used as short cut
V = L * B * H
See p- 189 for examples and exercises
This is the one of the method to find the volume of any shapes. In this method we will use a measuring glass. and fill water in this glass up to 200 ml. Now we will use 15 nos 1 cm cube interlock, and will make a cuboid with 5 cm length , 5 cm height and 2cm width. So that we can calculate the volume of the cuboid which we made. Volume = l *b*h = 5*5*2= 50 cubic cm. as shown in the picture in page 191. Now we can put the cuboid in the water on the measuring glass. And see how much water is rise in the glass. it will rise up to 250ml, that means the water rise 50l after placing the cuboide. That means the cuboide of 50 cm. cube will occupy the same space of 50 ml. Or we can say 50 ml = 50 cm cube. Please do the all examples and exercise in page 192,193.
Perimeter, Area and Volume-2
In order to find the method for finding the area of the triangle, first we will consider a right angle triangle. Also see the pictures in page No. 180. Now we can draw two lines from non adjacent vertices of the triangle, so that they will meet together and form a rectangle as shown in the figure. Now we can observe that the area of the given triangle is exactly half of the rectangle formed. And we can also know how to find out the area rectangle. which is length * breadth. So now the
area of the triangle is 1/2( length * breadth).
The above method is can use for right angle triangle (90° between base and height) only. So if we have find the area of other type of triangle( less than 90° between base and height) we can do the following things.
First we have to draw a line from opposite vertex from base to the base, in such a way it will make the given triangle in two right angle triangle. as shown in the fig -on page 180. Now we can find the area of that two right angle triangle spritely as described above. And we can add both area to get the area of the triangle.
Now please do the all exercise in page 181.
Now consider one different type of triangle (>90° between base and height) as shown in the figure
In this case we can consider the whole rectangle R1, which cover completely the given triangle T1. And now we can find the area of the R1 rectangle and half of this area will be equal to the area of the triangle T1 and T2 and can be considered area 1. Now consider the rectangle R2 and find the area of it. We can see that the half of the area will be equal to the triangle T2. Now we can subtract the area of the triangle T2 from area 1 and will get the area of the given triangle.
Area have different units to mark it. This different type of unit is using for more convenient. Because if we mention the area of a small switchboard, room and a city all are in on unit like sq. cm, then it will be difficult to understand. so we will express the area of a small think like switch board in sq. cm. the area of the room in sq.meter. And the area of a city in sq. km. Smaller area will be in smaller unit (CM) and bigger area in biger unit (KM). see the exercise in page 182.
We can find out the area of irregular shape in a squared paper as described below . Also see the figures in page 183.
step-1 – under estimate by counting the whole square. There are 20 squares.
step-2 – overestimate by counting whole and part square. There are 48 such a square.
step-3 – Add two figures and divide by 2 and ignore remainder if any.
(20+48)/2 = 68/2 = 34 square unit. (apx).
Do all the exercise on page 184,185and 186.
Perimeter is the distance around the edge of the figure. Or length of the total edge. Or all the points in the edge only. Area is the amount of the surface a figure cover. or area is the amount of all the points inside the figure. Please see the below picture for better understanding.
see the picture in the textbook, there the shapes are draw in the squared paper. And each square having the size of 1cm. So from that fig. we can find out the perimeter of the shape by counting the outside squares which touched on the boundary of the shape. then can get the perimeter = 16cm. since each sure side is 1cm.
In the same way we can find out the area also, By counting all the square inside the shape. which will give 12 sqcm. Since each square have 1 cm in sides, the area of the square is 1sqcm.
By repeating this experiment we can understand the perimeter of the rectangle= length+ breath+length+breath. Further we can solve this in to an equation as shown below
Perimeter of the rectangle = 2(length + breadth)
Same way for the area we can understand we will get the area of a rectangle by multiplying the length*breadth. and we can write the equation as shown.
Area of the rectangle = Length*breadth
As we studied the perimeter of the rectangle is length+breadth + Length + breadth . But in the case of square the length and breadth of the square is same, that means all the side of the square is same. So the perimeter of the square can be found by side + side + side + side. And it can be write as 4* length of the side or 4*side
Perimeter of the square is 4 * Side
As we know the area of the rectangle is Length * breadth. But in the case of the square, all the side is same . So we can find the Area of the square by the following equation.
Area of square is Side * Side.
Now please practice the exierce in the page No. 177,178,179.
A pentomino is a figure made up of 5 equal square joint together to share sides.
please see the page number 179 for example of pentomino. We can also make 5 different kinds of pentomino.
fraction is used to represent the part of the whole thing. Fraction can be write in the form of (Numerator /denominator).
ex. ½, ¾, ¼.
½ Means, we are cutting an apple in 2 equal part and taking only one part. The denominator will mention the whole thing is cutted in how much equal part. Here the apple is cutted in 2 equal part so the denominator become 2. And we are only taking one pice from that. So the numerator become 1 in this example. ½ can also be called as half.
Please also remember that the half (any fraction) of the different thing will not be same in amount.
ex. Half of a big apple and half of the small apple is not same in amount.
So for the fraction to be equal they should be from the same whole, or equal sized whole.
Fraction that have same denominator called like fraction.
ex. 1/3, 2/3.
fractions that have different denominator is called unlike fraction.
ex 2/5. 3/7, 3/8. all are unlike fractions.
Fraction that name the same part are called equivalent fractions. That means, if two fraction are representing the same amount is called equivalent fractions.
ex. 2/4, 4/8 are equivalent fractions. that means 2/4 = 4/8.
see the example. Mr.Singh has to write 15 certificate for distribution on school day. so far he has written 2/5 of the certificate. here we are going to be find out how many certificate he has completed.
2/5 of 15 certificate he has completed means, the 15 certificate is equally divided in 5 parts and from that 5 parts 2 parts he completed.
So the sequel 5 part of the 15 is 15/5 =3. each 5 part of the certificate have 3 certificates. 2 part from that 5 part= 3*2=6.
So 2/5 of 15=6. and singh has completed 6 certificates.
Find the 3/4 0f 20
divided the number with the denominator 20/4=5
multiply the quotient we get by the numerator 5×3=15.
So 3/4 of 20 = 15.
Number 1 is divisible with all numbers.
If a number is an even number then it is divisible by 2. Or if the last digit (ones place) of the number is 0,2,4,6,8, then it is divisible by 2.
The given number is divisible by 3, then the sum of the all digits if the number will be the multiple of 3.
A number with 3 or more digits is divisible with 4, then the number formed by it last 2 digits (once and tens place) will be divisible wit 4.
The number which has either 0 or 5 in its once place is divisible with 5.
If a number is divisible with 2 and 3 both then the number is divisible with 6 also.
A number with 4 or more digit is divisible by 8, if the number formed by its last 3 digit is divisible by 8.
If the sum of the digits of a number is divisible by 9 then the number itself is divisible by 9.
If a number is factor of another number x, then the number should be divisible with that number x. That means when this number is divided with number x then the remainder should get zero, then the the number is factor of the number x.
In this method we should divide number x with 1, 2, 3, …. upto half of the given number x. Then write all the numbers which make the reminder zero while division. All that number will be the factors of that given number.
Before starting explain this method we will see two words. First one is prime number. If any number have only two factors (1 and That number itself) then that number is called prime number. Second one is prime factor, If any of the factors of a number is a prime number then that factor is called prime factor.
As shown in the figure we can find the factorization of a number in 2 different way. As in the figure-1a divide the number with least possible prime number in each step until it reach to 1. So the prime factors of 8= 2x2x2
As in the figure-1b divide the number with least possible prime factor and write it as shown in the figure. Repeat the steps until we get all the factors are prime factor. So the prime factor of 42= 2x7x3
After finding the prime factors of two numbers by any one of the above method, we can multiply all the common prime factors together to get the HCF.
so the HCF of number 8 and 42 is 2. Because 2 is the only one common prime factors.
we can also find the HCF BY following method also.
When a line l cut two lines m and n at a distant point, then the line l is called transverse to line m and n. This can be see in the fig1. If a line l transverse to the line m and n then then the line l cut the line m and n at the point P and Q . and it will also create four angles at each points P and Q. This angles are shown as 1 to 8 in the fig1.
Here the angle 1,2,7,8 are called exterior angle. The angles 3,4,5,6 are called interior angle. Angle 1&5, 4&8, 2&6, 3&7 are called corresponding angle. The angles 4&6 ,3&5 are called alternate interior angles. Angle 1&7, 2&8 are called alternate exterior angles. The angles 4&5, 3&6 are called interior angle on thee same side of the transverse. It can also be called as consecutive interior angles, allied angles or co-interior angles.