Find the number of rectangles in the adjoining figure(a square is also considered a rectangle): Find the number of rectangles in the adjoining figure(a square is also considered a rectangle):
figure is in attachments
anyone please !!! help me
You are right..its 6*12
no its answer is not 6*12 as you can see there 2square also make a rectangle so we have to calculate allof them
So, its a bit logical kind..okay, I'll check it
6*12 means there are 6 row and 12 column
Its 12! x 6!
i can give you a way ...becoz i solved a question in which there was a rect with 4*4 dimention ..so i solved it by no of 4*4 square=1 no of 3*3 square=4 we can write it 2^2 no of 2*2 square=9 ........3^2 no of 1*1 square= 16 .....4^2 it become a series 1^2+2^2+3^2+4^2 and solved it by formula n(2n+1)(n+1)/6
opctions are a..864 b.3276 c. 1638 d.. None of these
The answer I got was too large in number..Just a sec
i got it how to solve first no of rectangle =6*12 then 6*(12-1) similarly 6*(12-2) ,6*(12-3),6*(12-4) ........6*(12-1) is that right
No no that's cant be right...It gives the number of boxes and you are repeating many of them
take an example of 2*6 rectangle and calculate no of boxes
Calculate the answer you got...
468 but now i think its was some how not correct
i take rectangle of 1*1,1*2,1*3,1*2 but not take 2*3,2*4.....we will have to take this too
Its not 12! x 6!..its a typo sorry..its \[\sum_{1}^{12}n \times \sum_{1}^{6}n = 1638\]
wts the logic u apply here
Take one row at first...how many rectangles can you select? 12boxes at a time in 1 way, 11boxes at a time in 2 ways, 10boxes at a time in 3 ways......1box at a time in 12 ways..So, the sum is sigma12..there are six such rows..so, sigma12 x 6 ways...now, select two rows at a time and you can do that in 5 ways..the total no. of rectangles are sigma12 x 5...now three rows and you get sigma12 x 4...Likewise, the total number of rectangles are : sigma12(6+5+4+3+2+1) = sigma12 x sigma6 = 1638
Try it and tell me when you understand it...
confused but trying
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