Area of a parallelograms and triangle grade-9 -01

Introduction

If we have a triangular field and want to divide it in to 3 equal parts. Can we divide it with out finding the area. Suppose we divide only one side of the triangular field in three equal parts and joint the points to the opposite vertex.  is it equal in size?. We are handling this type of situation in this chapter. to check this we have to find the area of the each plane. The area of the plane is defined as the magnitude of  the planar region. And the planar region is the plane enclosed by the simple closed figure. We are also know about the congruent of figures. Two figures are congruent if they have in same shape and same size. then we can keep one figure over other. And if two figure are congruent then they are in same size. But the converse of the statement is not possible. Area of afigure can be write as some number of unit ex 12 cm sq.

In the figure 9.3-p-153. The figure T formed by two planar region formed by figure P and Q. wecan also denote the area of the figure A as ar(A) . The properties of the area of the figure is as follows

  1. If A and B are two congruent figures then ar(A) = ar(B)
  2. If a planar region is formed by a figure T is  made up of two non-overlapping planar regions formed by figure P and Q then ar(T) = ar(P) + ar(Q).
Figures on same base and between same parallel

See figure 9.4 and 9.5 on p-154.

Figures are said to be on the same base and between the same parallels, if they have common base (side) and the vertices (or vertex)opposite to the common base of each figure lies on a line parallel to the base.

see figure 9.6 and 9.7 on p-155 see exercise also.

Fraction grade-5-01

INTRODUCTION

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.

Like fractions:  fractions with same denominator is called like fractions. ex. 3/7, 5/7, 1/7.

Unlike fractions : Fractions with different denominator is called unlike fraction ex. 4/7, 5/10, 5/9.

Proper fraction : if the numerator less than denominator in a fraction, then it is called proper fraction. The value of proper fraction is always less than 1. ex. 1/8, 4/9,6/11.

Improper fraction : If the numerator is equal to or greater than denominator, then it is called improper fraction. The value of improper fraction is always equal or greater than 1. ex. 5/2, 12/7, 8/8.

Mixed fractions:   This is the combination of a whole number and fraction. For example 3 apple and a half apple. ex 3½.

Comparing the like fraction : In Order to compare the like fractions we can just compare the numerator only and can decide which one is smaller, bigger or equal.  Do some exercise. p-79

Changing in to mixed fractions : we can change an improper fraction into mixed fraction by divide the numerator with denominator. The whole part will be the quotient, numerator will be the remainder and divisor will be the denominator of the mixed fraction. do exercise in p-79.

Changing in to Improper fractions : We can change a mixed fraction into improper fraction. The numerator of the improper fraction can obtain by multiplying the whole part and denominator and add the numerator with the product. The denominator of the improper fraction is same as the denominator of the mixed fraction see exercise in p-79. See video

Equivalent fractions

Same fraction of a whole can represent with different fractional numbers. that means different fractional number can be same value. ex. 1/2 of an apple and 2/4 an apple. See activity on p-80. we can find the number of equivalent fraction of a fractional number by multiplying the numerator and denominator by same number. We can also find the equivalent fractions by dividing the numerator and denominator by same number. see p-80

checking the equivalent fraction 

In order to check whether  to fraction are same or not we can cross multiply them. Cross multiplication can be done by multiply the numerator of the first number and denominator of the second number. Then multiply the numerator of the second number and denominator of the first number. If both product are same then the two fractions are same. The product is the value of the fraction whose denominator is multiplied. By this we can find which one is big or small also. see p-81 for exercise

lowest form of the fraction

As we know same fraction can be write with different numerator and different denominator. ex 2/4, 4/8,  6/ 12. So in this section we will study how we can check whether the given fraction is in its lowest form. If the given fraction number is in its lowest form then there will not be any common factor of numerator and denominator expect 1. (or the hcf will be 1).  For ex. 6/12, the HCF of 6 na12 is 6so it is not 1. There for 6/12 is not in its lowest form. If we consider 1/6, then the hcf is 1 so it is in it lowest form.

To change a fraction in its lowest form, we can find the HCF of numerator and denominator and then we have to divide both numerator and denominator with HCF  and we will get the lowest form of the fraction. ex 6/12 HCF is 6 so divide both numerator and denominator we will get 1/6 this is the lowest form. see exercise 6B in p-83.

Comparing the fractions

Comparing the like fractions:  as the denominator of the like fractions are same just compare the numerator. The biggest numerator fraction is the bigger one.

comparing the unlike fraction : In this case all denominator will be different. In this case if the numerator is same for all  the biggest denominator fraction is the smallest one. If denominator and numerator are different for fractions then we can find the LCM if the denominators and make all fraction with same denominator as LCM, then we can compare the numerator only. To do this we have to divide the LCM with denominator of each fraction. And multiply  the quotient with numerator and denominator of each fraction respectively. See exercise 6c p-86

Adding the fractions

Adding the like fractions : As the like fraction have the same denominator we have to add only the numerators of the fractions and the denominator is the same.ex. 2/12 + 4/12 = 6/12.

Adding the unlike fractions: In order to add two or more unlike fractions we have to make them  like fractions. To do this we have to find the LCM of the all denominators, and convert the denominator of all the fractions to the LCM. Now all the fraction is in like fractions and we can add them. see examples and exercise in page 88.

Adding the mixed numbers with like fractions : For do this we have to add the whole part together first, and this will be the new whole part. Now add the two like fraction part. If the result is in improper fraction convert that in to mixed fraction and now we will get the proper fraction in the fraction part of the mixed fraction. Now we can add the previously obtained new whole part along with the whole part of the mixed fraction and keep the fraction part as it is. See ex. i page-89.

Adding the mixed numbers with unlike fractions : This one also we can do as per the above steps. But before doing that we have to covert the unlike fractions in like fractions. see ex.  and exercise in page-89.

Subtracting the fractions

To subtract any faction, we have to make them first in to the like fraction. To do this we can find  LCM of all denominators, and make all the fractions denominator to the LCM. Then we can subtract the numerators and keep the denominator same. Like this we can subtract like fractions and unlike fractions. even it is proper or improper. see the exercise i page-90.

Subtracting the mixed fractions:

Heron’s formula grade 9- NCERT-01

introduction

We all ready know about how to find out the area, perimeter of different shapes like rectangle and square. In this chapter we will study details about how we can find the area of various type of triangle.

area of the triangle

Area of the triangle = ½ * base * height

We can see that if the triangle is right angle triangle, we can directly take two side, which contain the right angle, as base and height. see the the fig.12.1 in p-197. So in this triangle ABC, BC and CA are the height and base ( any one can select as base or height).  There for the

Area of the triangle ABC = ½ * 5 * 12  =30 cm².

Now we can consider an equilateral triangle PQR as shown in fig 12.2 p-198. In this we can know all the side of the triangle. But how we can find the height. We can find the midpoint of QR and mark as M and join it to P. Now we can see that ΔPMQ is aright angle triangle. And we can find the PQ using the pythagorean theorem.

PQ ² = PM² + QM²

10²=PM²+5²

PM²= 75

PM= √75=5√3

Area of the triangle = ½ * 10 * 5√3 =25√3 cm².

Now we can consider an isosceles triangle. In this two sides is same. How we can find the height. We can consider the triangle XYZ in the fig 12.3 p-198. Find and mark the midpoint P on YZ and joint it with point x. Now we can see the triangle XYP ia right angle triangle.  So

XP² = XY² – YP² = 5² -4² = 25-16=9

XP= √9 = 3

Area of the triangle = ½ * 8 * 3 =12 cm².

Now we can consider the scalene triangle. In this type of triangle all sides are different. So we cannot find the height of the triangle as described in the above method. Her we can use the Heron’s formula. Heron is the famous mathematician (p-199). The formula given by him can use to find the area of any type of triangle. And it is called hero’s  formula

Area of a triangle = √(s(s-a)(s-b)(s-c))

Where a,b,c are the three side of the triangle. ANd s is the semi perimeter of the triangle. So  s= (a+b+c)/2.

Now consider the triangle ABC in the fig 12.5 p-200. Here a= 40, b=24,c=32. so

s= (40+24+32)/2=  48

s-a= 48-40=8

s-b=48-24=24

s-c=48-32=16

Area of the triangle = √(48*8*24*16) =384 m² see page 200,201,203 for exercise

we can also use this formula for finding the area of quadrilateral by dividing it in to two triangle, see page 206 for exercises.

 

Triangle and its property grade-7 NCERT-1

Triangle is a simple closed curve made up with 3 lines segment. it have 3 vertex, 3 side and 3 angle. In a triangle ABC, the side are AB, BC, CA. Angles are ∠BAC, ∠ABC, ∠BCA. Vertices  are A,B,C. In the above triangle the side opposite to the vertex A is BC. please try to say the opposite side of other vertex.

Based on side triangle can classifieds in to Scalene ( all three sides are unequal), Isosceles (2 sides are same), And Equilateral triangle (3 side are same).

Based on Angles Triangle can classifieds in to Acute-angle, Obtuse-angle, and Right angle triangle. see page 113.

Medians of triangle

A median is the line segment between a vertex and the midpoint of the opposite side. A triangle can be 3 medians. If we cut a triangle in a sheet of paper we can fold each of its sides and can be find the midpoint of each side.  see (P114) for think and discuss.

Altitude of a triangle

It is nothing but the height of the triangle. It is the line segment between the base and opposite vertex. See (P115) for think and discuss. See (P116) for exercise.

LCHF diet plan, food which can be eat or not eat

Food can be eat

Egg, All fish, Read meat, chicken, Olive Oil, Coconut oil, Broccoli, Cauliflower, Mushrooms, cabbage, tomato, ladies finger, curry leaf, coriander leaf, Green leafs vegetable, chilli, Bitter Gourd, Full fat milk and yogurt, butter, cheese, Dry fruits like, almond, walnut, pumpkin seeds. Drink at least 3.5 liters water, use lemon water with salt. Coconut

food CANNOT eat

Sugar, sugar product, juses , fresh fruits, Rice, chapati , perotas, beverages, bakery items. Potato, carrot, all items growing under the ground  . dal, and all type payer items,

Morukari with kumbalanga

Item required

Kumbalanga 300 gm

Curd 500g

mustard

Jeerakam – 1 pinch

green chilly -2 psc

Curry leaf

Small onion 2 Pcs

Yellow Powder.

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.

Quadrilaterals -01

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 

trapezium

If one pair of the opposite side of a quadrilateral is parallel, then it is called trapezium.  ex. fig 8.5-1.

Parallelograms

If both pair of opposite sides of a quadrilateral is parallel then it will called parallelograms. ex. Fig. 8.5 – 2,3,4,5.

rectangle

If any one of the angle is 90° in a parallelogram, then it will be called rectangle. ex Fig 8.5 – 3.

rhombus

If all sides of a parallelogram are same, then it will called as rhombus ex Fif 8.5 – 4.

square

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.

Kite

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.

 Theorem 8.1.

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).

Theorem 8.2.

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.

Theorem 8.3.

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

∠BAC= ∠DCA

∠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.

Theorem 8.4.

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

∠BAC= ∠DCA

∠BCA= ∠DAC

And hence opposite angle of parallelogram are equal.

Theorem 8.5.

This is the converse of theorem 8.4.

If a quadrilateral ,each pair of opposite angle are equal , then it is a parallelogram.

Theorem 8.6.

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.

Theorem 8.7.

This is the converse of the theorem 8.7

If the diagonals of a quadrilateral bisect each other , then it is a parallelogram.

Proof :

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

Theorem 8.8

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).

Theorem 8.9

The line segment joining the mid-point of two side of a triangle is parallel to the third side.

Proof :    

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

Theorem 8.10

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.