Question

It takes 5 toothpicks to build the top trapezoid shown at below. You need 9 toothpicks to build 2 a trapezoids and 13 toothpicks for 3 trapezoids. (6) (ii) If 1000 toothpicks are available, how many trapezoids will be in the last complete row? How many toothpicks will you use to construct these rows?​

Answer 1

Answer:

As given below

Step-by-step explanation:

Given data

Number of toothpicks to built the top trapezoid = 5

Number of toothpicks to built 2nd row trapezoid = 9

Number of toothpicks to built 3rd row trapezoid = 13

here we need to find how many trapezoids will be built by using 1000 toothpicks

from given data

toothpicks in 1st row = 5

toothpicks in 2nd row = 9

toothpicks in 3rd row  = 13

⇒ the number series of toothpicks

= 5, 9, 13... are in AP    here first term a = 5 , common difference d = 4

here we will find at which term the sum of toothpicks will be equals to 1000 or less than 1000 or greater than 1000.

We know that sum terms in AP  = $\frac{n}{2} [ 2a + (n - 1) d]$  

                                              = n/2 [ 2(5)+ (n-1)4]  

                                              = n/2 [ 10 + 4n - 4 ]    

                                              = $\frac{n}{2}$(2)[ 3 + 2n]  

                                              = n [ 3 + 2n ]  

here  at n = 21    ⇒  21 [ 3 + 2(21)]

                           ⇒  21 [ 45] = 945  

         at n = 22   ⇒ 22 [ 3 + 2(22)]  

                           ⇒ 22 [ 47]  = 1034 > 1000

⇒ we can built 21 trapezoids with help of 945 and there will be 55 toothpicks still

⇒ to built 22 trapezoid we will need 89 toothpicks

⇒ with help of 1000 toothpicks we can built 21 rows of trapezoids

Answer 2

Answer:

If 1000 toothpicks are available, how many trapezoids will be in the last complete row?It takes 5 toothpicks to build the top trapezoid. You need 9...

Let   the top single trapezoid which has

5

So the 1st row   has 5 tooth pics.

a1

= 4*1+1.

In the 2nd row , there would be 2 adjoined trapezoids and

has a2 = 5+4= 4*2+1 = 9 tooth pics.

In the 3rd row the re are 3

reapezoids with 3*4+1 = 13.

Like that in the nth row there are an

=4n+1  toothpicks

 Therefore the sum Sn of the tooth picks of n rows

is given by:

Sn = a1+a2+a3+...+an

Sn =

(1*4+1)+(2*4+1)+(3*4+1) +....(n*4+1)

Sn =

(1+2+3+...+n)4+(1+1+1....+1), there are n 1's in the 2nd bracket.

Sn

= n(n+1)4/2 +n = 2n(n+1)+n = 2n^2+3n = 1000.

2n^2+3n

=1000.

2n^2+3n > 1000.

For n =

21,  n(2n+3) = 945.

fOr n =22, n(2n+3) =

1034.

So there are 21 rows.

The last

complete row has 21 trapezoids with 21*4 +1 = 85 tooth picks.

The

last 23rd incomplete row has 13 trapezoids with 13*4+1 = 53

sticks

Totatal tooth pickes used = 21(21*4+1) = 945+13*4+1  =

998.