We can easily count that figure 1 has 1 unit square and figure 2 has 5 unit squares. How many unit squares are present in figure 3?
This problem is part of Arron's set The Cross Investigations - Area .
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I just count it
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FOR every increement 4 units are added on four sides .......thats it
any other idea on this?
1+4=5; 5+4=9
one square = 1 cm square so we can sum the areas of the squares
(x^2)/x
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x
what's the rule man!!!!!!!!!!
every time if one unit is added to each direction, the number increases to 4, as it did in this case, or if we come to know how many are there in just one direction then multiply it by 4 and add 1= 4n + 1, so easy.
just count it ...
4x+1 (i)4(0)+1=1unit, (ii)4(1)+1=5units, (iii)4(2)+1=9
ans is 9
U1 = 1, U2 = 5, U3 = ?
Un = a + (n-1)b
a = 1, b = 5-1 = 4
U3 = 1 + (3-1)4 U3 = 1 + 2.4 U3 = 1 + 8 U3 = 9
ans is 9
9
9
9
9
9 without thinking . . .
4n+1
9
simply 9
simple count..9
9.. no more answers....
9
by counting
9
9
just count it
Simple 9
9
in figure third i am account there is 9 box
4=5=9
just count nd nothing
just 9
jst count it is 9
9
9
wat a simple question????.
so, how it's gonna be the 3rd square?
simply count it
5+4=9
49-49=49
i just counted the box ..
11 is a corret answer
9
9 unit sq.
i know the formula :)
it has 4 side square box which contain 2 small box and it also have one centre box soo 4*2=8 , +1
Just counted it
This was the dumbest question for the level we are rated for - my answer was correct yet directed to the solution...
1,5,9,......
Difference(d) is 4
First term(a) is 1
3rd term = ?
a+(n-1)d= 1+(3-1)4
= 1+8
= 9
same solution,i was thought
We can find here an arithmetic progression: the initial term is 1 and the common difference is 4. So, a 3 = a 1 + 2d = 1 + 2\times 4 = 1 + 8 = 9
X= no. Of squares.
Figure 1. X=1
Figure 2. X= 1+4
Figure 3. X= 1+4(2)
Figure 4. X= 1+4(3)
So on and on.
Suppose, X=figure no(0 to 2) & Law is 4x+1.
So (i) (4*0) +1=1 Squire, (ii) (4*1)+1=5 Squire, (iii) (4*2)+1= 9 Squire.
Just count the unit squares and DON'T count that squares which you have counted earlier....ENJOY.......!!!
4x+1 (1),4(0)=1 (2),4(1)+1=5 (3),4(2)=9
)
very simple.i am expecting some more difficult
arithmetic sequence of difference =4
add 4 unit squares to previous figure
9 number of squares + number of squares-1 =5+4
i just counted it manually :)
start as 1, then add all corners (equivalent of 4). So, the first answer is 5. then add corners again (it means add 4 again) the final answer is 9.
1 box= 1unit....... then 9 boxes=9unitsqr.
In this series 1,5,.......
First term (a) is 1. First term and Secind term differnse is 4. Differense (d) = 4.
Tn=a+(n-1)*d
T1=1+(1-1)*4
T1=1
T2=1+(2-1)*4
T2=5
T3=1+(3-1)*4
T3=9
In 3 squire total 9 units are present
that is stupid just count the squares
9 units
Its so simple..... no of squares is equal to {4(n-1) + 1} where n will be number of squares in any one of the four arms of figure (including center square too)
if Nx = # of rows and Ny-=No columns then Nx X Ny-(Nx-1)X(Ny-1)
nine unit square contains in figure 3...
As easy as drinking wine simply answer is 9 bcs. this figure include 9 blocks
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4x+1 (i)4(0)+1=1unit, (ii)4(1)+1=5units, (iii)4(2)+1=9