matha namo namaha


ĀDYAMĀDYENĀNTYA - MANTYENA
The Sutra ' adyamadyenantya-mantyena' means 'the first by the first and the last by the last'.
Suppose we are asked to find out the area of a rectangular card board whose length and breadth are respectively 6ft . 4 inches and 5 ft. 8 inches. Generally we continue the problem like this.
        Area = Length X Breath

                    = 6’ 4" X 5’ 8" Since 1’ = 12", conversion

                    = ( 6 X 12 + 4) ( 5 X 12 + 8) in to single unit

                    = 76" 68" = 5168 Sq. inches.
Since 1 sq. ft. =12 X 12 = 144sq.inches we have area

            5168    =    144) 5168 (35
            ¯¯¯¯
             144                   432
                                    ¯¯¯¯
                                     848
                                     720    i.e., 35 Sq. ft 128 Sq. inches
                                    ¯¯¯¯¯
                                     128
By Vedic principles we proceed in the way "the first by first and the last by last"

        i.e. 6’ 4" can be treated as 6x + 4 and 5’ 8" as 5x + 8,

            Where x= 1ft. = 12 in; x2 is sq. ft.
Now ( 6x + 4 )(5x + 8 )

            = 30x2 + 6.8.x + 4.5.x + 32
            = 30x2 + 48x + 20x + 32
            = 30x2 + 68. x + 32
            = 30x2 + ( 5x + 8 ). x + 32 Writing 68 = 5 x 12 + 8
            = 35x2 + 8. x + 32
            = 35 Sq. ft. + 8 x 12 Sq. in + 32 Sq. in
            = 35 Sq. ft. + 96 Sq. in + 32 Sq. in
            = 35 Sq. ft. + 128 Sq. in
It is interesting to know that a mathematically untrained and even uneducated carpenter simply works in this way by mental argumentation. It goes in his mind like this

                        6’    4"

                        5’    8"

        First by first i.e. 6’ X 5’ = 30 sq. ft.

        Last by last i.e. 4" X 8" = 32 sq. in.

        Now cross wise 6 X 8 + 5 x 4 = 48 +20 = 68.
Adjust as many '12' s as possible towards left as 'units' i.e. 68 = 5 X 12 +8 , 5 twelve's as 5 square feet make the first 30+5 = 35 sq. ft ; 8 left becomes 8 x 12 square inches and go towards right i.e. 8 x 12 = 96 sq. in. towards right ives 96+32 = 128sq.in.
Thus he got area in some sort of 35 squints and another sort of 128 sq. units. i.e. 35 sq. ft 128 sq. in

Another Example:

                   

        Now 12 + 2 = 14, 10 x 12 + 24 = 120 + 24 = 144

        Thus 4′ 6″ x 3′ 4″ = 14 Sq. ft. 144 Sq. inches.

        Since 144 sq. in = 12 X 12 = 1 sq. ft The answer is 15 sq. ft.

        We can extend the same principle to find volumes of parallelepiped also.
I. Find the area of the rectangles in each of the following situations.

        1).   l = 3’ 8" , b = 2’ 4 "             2).   l = 12’ 5" ,  b = 5’ 7"

        3).   l = 4 yard 3 ft. b = 2 yards 5 ft.(1yard =3ft)

        4).   l = 6 yard 6 ft. b = 5 yards 2 ft.


    II. Find the area of the trapezium in each of the following cases. Recall area = ½ h (a + b) where a, b are parallel sides and h is the distance between them.

        1).   a = 3’ 7", b = 2’ 4", h = 1’ 5"

        2).   a = 5’ 6", b = 4’ 4", h = 3’ 2"

        3).   a = 8’ 4", b = 4’ 6", h = 5’ 1".
Factorization of quadratics:

        The usual procedure of factorizing a quadratic is as follows:

                       3x2 + 8x + 4
                    = 3x2 + 6x + 2x + 4
                    = 3x ( x + 2 ) + 2 ( x + 2 )
                    = ( x + 2 ) ( 3x + 2 )
But by mental process, we can get the result immediately. The steps are as follows.
i). Split the middle coefficient in to two such parts that the ratio of the first coefficient to the first part is the same as the ratio of the second part to the last coefficient. Thus we split the coefficient of middle term of 3x2 + 8x + 4 i.e. 8 in to two such parts 6 and 2 such that the ratio of the first coefficient to the first part of the middle coefficient i.e. 3:6 and the ratio of the second pat to the last coefficient, i.e. 2: 4 are the same. It is clear that 3:6 = 2:4. Hence such split is valid. Now the ratio 3: 6 = 2: 4 = 1:2 gives one factor x+2.

ii). Second factor is obtained by dividing the first coefficient of the quadratic by the fist coefficient of the factor already found and the last coefficient of the quadratic by the last coefficient of the factor.
        i.e. the second factor is

                      3x2          4
                     ____  +  ___  =  3x + 2
                       x          2
        Hence 3x2 + 8x + 4 = ( x + 2 ) ( 3x + 2 )
Eg.1: 4x2 + 12x + 5
i) Split 12 into 2 and 10 so that as per rule 4 : 2 = 10 : 5 = 2 : 1 i.e.,, 2x + 1 is first factor.

ii) Now
            4x2       5
            ___  +  __  =  2x + 5   is second factor.
            2x         1
Eg.2: 15x2 – 14xy – 8y2
i) Split –14 into –20, 6 so that 15 : - 20 = 3 : - 4 and 6 : - 8 = 3 : - 4. Both are same. i.e., ( 3x – 4y ) is one factor.

ii) Now
            15x2        8y2
            ____   +  ___   =  5x + 2y is second factor.
              3x         -4y
Thus 15x2 – 14xy – 8y2 = ( 3x – 4y ) ( 5x + 2y ).
It is evident that we have applied two sub-sutras ‘anurupyena’ i.e.‘proportionality’ and ‘adyamadyenantyamantyena’ i.e. ‘the first by the first and the last by the last’ to obtain the above results.
Factorise the following quadratics applying appropriate vedic maths sutras:

            1).   3x2 + 14x + 15

            2).   6x2 – 23x + 7

            3).   8x2 – 22x + 5

            4).   12x2 – 23xy + 10y2

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