Question

1.The perimeter of a rectangle is 42cm. And its area is 20cm2

a) What is the sum of length and breadth?

b) Find the length and breadth of rectangle

2. Sum of two numbers is 9 and sum of their squares is 41.

Considering one of the numbers as x,

a) Write a second degree equation

b) Find the numbers

3. How many terms of the Arithmetic Sequence 5,7,9,...must be

added to get 140?​

Answer 1

Step-by-step explanation :

1) The perimeter of a rectangle is 42cm. And its area is 20cm²

•  length = l
•  breadth = b

 we know that,

 Perimeter of the rectangle = 2(length + breadth)

                         42 = 2(l + b)

                        42/2 = l + b

                         l + b = 21

                         l = 21 - b

  Area of the rectangle = length × breadth

                 20 = (21 - b)(b)

                 20 = 21b - b²

                b² - 21b + 20 = 0

                b² - b - 20b + 20 = 0

                b(b - 1) - 20(b - 1) = 0

               (b - 1) (b - 20) = 0

                b = 20,1

 => l = 21 - b = 21 - 1 = 20

a) The sum of length and breadth = 2 cm

b) The length and breadth are 20 cm and 1 cm

2) Sum of two numbers is 9 and sum of their squares is 41.

• Given one of the numbers = x
• Then the other number = 9 - x

       Sum of their squares = 41

          x² + (9-x)² = 41

         x² + 9² + x² - 2(9)(x) = 41

         2x² + 81 - 18x = 41

          2x² - 18x + 81 - 41 = 0

          2x² - 18x + 40 = 0

          2(x² - 9x + 20) = 0

            x² - 9x + 20 = 0

           x² - 4x - 5x + 20 = 0

           x(x - 4) - 5(x - 4) = 0

            (x - 4) (x - 5) = 0

             x = 4 , 5

 a) The required second degree equation is x² - 9x + 20 = 0

 b) The numbers are 4 and 5

3) How many terms of the Arithmetic Sequence 5,7,9,...must be  added to get 140?

•    Given AP sequence : 5,7,9,...
•    first term, a = 5
•    common difference, d = 7 - 5 = 9 - 7 = 2
•    Let there are n terms
•    Sum of n terms = 140

we know that,

            $\boxed{\bf S_n=\frac{n}{2} [2a+(n-1)d]}$

             $140=\frac{n}{2} [2(5)+(n-1)(2)] \\\\ 140 \times 2=n[10+2n-2] \\\\ 280 =n[8+2n] \\\\ 140 \times 2=n[2(4+n)] \\\\ 140=n(4+n) \\\\ 140=4n+n^2 \\\\ n^2+4n-140=0 \\\\ n^2+14n-10n-140=0 \\\\ n(n+14)-10(n+14)=0 \\\\ (n+14)(n-10) = 0\\\\ n=-14,10$

n can't be negative.

So, n = 10

the number of terms of the AP 5,7,9,... to be added to get 140 = 10

Answer 2

Answer:

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