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Kumbalanga 300 gm
Jeerakam – 1 pinch
green chilly -2 psc
Small onion 2 Pcs
Method to PREPARE
Put little oil in a pan and wait for heat well. Put the musteread, jerekam, curry leaf, Onion in small pieces, green chilly cutted in mall round, fry well. after that add the yellow powder. Then we can add the Kumbalanga cutted in small pieces and cook well (20 Min). Finley add the cured (Mixed with little water in the mixi) and off the flame. And it is ready to use.
We already studied about the triangles in previous chapters. And we know that joining 3 noncollinear points in pairs we will get the shape triangle. Same way if we joints such a 4 points in an order we will get shape called quadrilaterals. Quadrilaterals have four side , four angle and four vertices. We can see the figures in page number 135.
Ina quadrilateral ABCD AB, BC, CD and DA are four sides. A,B,C,D are the 4 vertices. And ∠A, ∠B, ∠C, ∠D, are the four angles formed at the vertices.
We can also say that the rectangle, square, parallelogram etc are the special type of quadrilaterals.
angle sum property of a quadrilateral
The sum of the angles of a quadrilaterals is 360°. This can be verify by drawing a diagonal AC in the quadrilateral ABC. This will be divide the quadrilateral in two triangles. fig 8.4 at p-136.
In triangle ADC ∠DAC + ∠ACD + ∠D = 180°
Similarly in triangle ABC ∠CAB + ∠ACB + ∠B = 180° Adding 1 and 2 we will get
∠DAC + ∠ACD + ∠D + ∠CAB + ∠ACB + ∠B = 180 +180 =360° Also we know
∠DAC + ∠CAB = ∠A, ∠ACD + ∠ACB = ∠C
so, ∠A + ∠D + ∠B + ∠C = 360°.
type of quadrilaterals
If one pair of the opposite side of a quadrilateral is parallel, then it is called trapezium. ex. fig 8.5-1.
If both pair of opposite sides of a quadrilateral is parallel then it will called parallelograms. ex. Fig. 8.5 – 2,3,4,5.
If any one of the angle is 90° in a parallelogram, then it will be called rectangle. ex Fig 8.5 – 3.
If all sides of a parallelogram are same, then it will called as rhombus ex Fif 8.5 – 4.
If any one of the angle is 90° and all sides are same in a parallelogram, then it will call as square. ex. Fig 8.5 – 5.
If two pairs of adjacent sides of a quadrilateral is same, then it will call as kite. ex Fig.8.5 – 6.
Properties of parallelogram
Consider a parallelogram ABCD and diagonal AC. Cut the parallelogram through it diagonal AC. as shown in fig 8.7 in p-139, and we can see that it will divide the parallelogram in two triangle. And place that two triangle one over other and can notice that both triangle are congruent.
a diagonal of a parallelogram divide it into two congruent triangle.
Proof : Consider the parallelogram ABCD and AC is the diagonal of it. And the diagonal divide the parallelogram in two triangle ABC AND CDA. Now consider the BC parallel AD and AC transverse
so, ∠BCA=∠DAC (alternative interior angle)
Also AB parallel DC and AC transverse.
so, ∠BAC=∠DCA (alternate interior angle). And AC =CA (commenside)
There for triangle ABC congruent to triangle CDA (ASA rule).
In a parallelogram opposite side are equal.
this can be proved from above proof itself. Because the opposite side of a parallelogram are the corresponding sides / parts of a congruent triangle.
So, AB=DC and AD=BC.
If each pair of opposite side of a quadrilateral is equal, then it is a parallelogram.
This theorem is the converse of the theorem 8.2. Here also we can consider the above figure. From the figure assume that AB = CD and AD =BC. Now we can say the ΔABC and ΔCDA are congruent (SSS). and we can also observe that
∠BCA= ∠DAC (CPCT)
Now ∠BAC + ∠DAC = ∠DCA + ∠BCA
Here the angle BAC + DAC and DCA + BCA are the opposite corners of the quadrilateral . And we know that if the opposite angle are equal then the quadrilateral will be a parallelogram. and this one we can prove in the next theorem.
In a parallelogram , opposite angle are equal.
This can be prove by the same explanation of theorem 8.3. Consider AC is the diagonal of parallelogram ABCD. And by using the laws of parallel lines and transverse we can prove that
And hence opposite angle of parallelogram are equal.
This is the converse of theorem 8.4.
If a quadrilateral ,each pair of opposite angle are equal , then it is a parallelogram.
The diagonals of a parallelogram bisect each other.
This can be check by drawing diagonals of parallelogram and measure the intersecting point. and we can found that the diagonals will intersect at mid of the diagonal each other. This can be also prove by the theorems which we already studied. (parallel and transversal line and congruence of triangle.
This is the converse of the theorem 8.7
If the diagonals of a quadrilateral bisect each other , then it is a parallelogram.
From the fig.8.11 we can see that OA=OC and OB =OD so, Δ AOB congruent to ΔCOD (SAS, vertically opposite angle at O). There for ∠ABO and ∠CDO are same (CPCT). Then we can say that AB parallel to CD (parallel and transversal law). Same way we can proof AD parallel to CB. And hence it is a parallelogram.
See page 141 for examples
A quadrilateral is a parallelogram if a pair of opposite side is equal and parallel.
consider the figure 8.17 (p145) in which AB = CD and AB parallel to CD. and AC is the diagonal of the quadrilateral. ABC and CDA are congruent by SAS rule ( opposite interior angle). So BC and AD are equal and hence it is parallel. (equal line between parallel line will parallel). So this quadrilateral is parallel. see p-145 and 146 for examples and exercises.
The mid-point theorem
Draw a triangle ABC as shown in fig 8.24, mark the midpoint of the sides AB and AC with E and F respectively. Joint E and F and measure EF, BC, angle E and angle B. We can also repeat this experiment with different triangle also. And we can observe that EF = 1/2 BC, and angle AEF= angle ABC. And hence EF parallel to BC. (corresponding angles same).
The line segment joining the mid-point of two side of a triangle is parallel to the third side.
In fig 8.25 E and F are midpoints of AB and AC respectively. And CD Parallel to BA . Δ AEF concourent to Δ CDF (ASA rule.) So EF=DF and BE=AE=DC. Also we know that EF = 1/2 BC.(midpoint theorem). there for each opposite pair are equal and hence BCDE is parallelogram and EF is parallel to BC
This is converse of theorem 8.9, The line drawn through the mid-point of one side of triangle, parallel to another side bysects the third side. See p-149 and 150 for example and exercises.
GRADE-9 NCERT CHAPTERS
GRADE-5 NCERT CHAPTERS
- Perimeter, Area, and volume
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
finding the volume of other shape
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.