Rectangle Tiling

The area of the rectangle is $1^2+2^2+2^2+2^2+3^2+4^2+5^2 = 63$ Since 63 factors as $1 \times 63, 3 \times 21, 7 \times 9$ , and the length and breath of the rectangle must each be more than 5, hence the only possibility is $7 \times 9$ . In this case, the rectangle has perimeter $2\times(7 + 9 )= 32$ .

It remains to check that this can actually be done:

Hence, the perimeter is 32.

Place the square with side length 4 directly to the left of the square with side length 5. Place the square with side length 3 to the left of the square with side length 1, and both on top of the square with side length 4. Place the 2 squares of side length 2 in the top right corner. This gives a 7 by 9 rectangle. Hence, the perimeter is $2(7+9)=32$ .

The areas of the individual squares are: 1, 4, 4, 4, 9, 16 and 25. Their sum is 63. This has to be the product of the sides of the big rectangle. The smallest side of that rectangle has to be 5 or larger, since the largest small rectangle has the side 5. We try sides 5 and up. 63 is not divisible by 5 or 6. 63 is divisible by 7, which makes the big rectangle 7*9. Its perimeter is then 2(7+9) = 32.

The area of the 7 squares combined is 63 units square, and the side lengths of the rectangle must be integers i.e. 7x9, 3x21 or 1x63. Since there is a square with side length 5, 3x21 and 1x63 don't work and it is easy to construct a 7x9 rectangle

1^2+3 2^2+3^2+4^2+5^2=9 7

so 2(9+7)=2.16=32

At first we have calculated the area that is covered by those 7 squares .

Now note that the side length of rectangle is integer . So factorize the are . There are two possible factorization 63=3 21=7 9

Note if a sidelength of rectangle is 3 then its impossible to put the last two square in the rectangle .

So sidelengths are 7,9 and the perimeter is 2(7+9)=32

Clearly, the area of the rectangle is 1+4+4+4+9+16+25=63 63 = 3^2*7. Note that since the largest square is of length 5, both the length and the breadth of the rectangle must be at least 5. The sides of the rectangle must clearly be integer values as well. This leaves the possible rectangles to be 1x63, 3x21, or 7x9. Only the last has both dimensions at least length 5, so the required answer is 2(7+9)=32

The total area of the rectangle is 1^2+3(2^2)+3^2+4^2+5^2=63. Since we could fit a 5 by 5 square inside it, each of the dimensions must be at least 5. The only possible set of side lengths of the rectangle that has a product of 63 and each side at least 5 is to have the length 9 and width 7. Thus, the perimeter is 2(7+9)=32

Since the side lengths of the 7 squares are 1, 2, 2, 2, 3, 4 and 5, the areas of the individual squares are 1, 4, 4, 4, 9, 16 and 25. The total area is then 1+4+4+4+9+16+25 = 63. The question says that there is no overlap or gaps in-between, so the area of the rectangle is 63. The sides of the rectangle have to be integers, because it is not possible to line squares up to cover a decimal length. We then check the possible dimensions for a rectangle with area 63:

63 x 1 is not possible because squares of side length 2, 3, 4 and 5 do not fit.

21 x 3 is not possible because squares of side length 4 and 5 do not fit.

7 x 9 is possible because both the length and width are large enough for each square to individually fit in the rectangle.

These are the three cases. At this point we still do not know whether the squares fit together in a 7 x 9 rectangle, but that is the only case that could potentially work. Therefore, the answer is 7+7+9+9 = 32.

place square with biggest side, and adjacent to it place square of side 4 on one side and other side place 2square with side 2 and now complete square by other sides.

since the squares are to be fitted tin rectange with no gaps remaining that means area of both has to be same so we calculate area of individual square and sum up which comes up 63...this has to be area of rectangle and 63 can be broken into 7 9 or 21 3 but the maximum side of square is 5 which can't be fitted if the sid eof rectangle is 3 so the dimension of rectangle has to be 7*9 and perimeter 32

total area=63 square unit. so sides of the rectangle are 7,9 or 3,21. but we have two squares which has a side larger than 3. so sides of the rectangle are 7 and 9. therefore perimeter of the rectangle is 2*(7+9)=32

The total area of the rectangle=63 We have to find two number let say x and y,so that xy=63 since 5 can also fit into the rectangle...x,y must be greater than 5 63=7*9 therefore,the perimeter of the rectangle=2(7+9)=32

This is what I did:

1. I squared all numbers and sum them up. Got 63.

2. Since the rectangle cannot have width of 1, then the width must be 7 or 9. (Factors of 63)

3. Perimeter: 2 times (7+9) = 32.

the pmt is 2 ( l + b ) 2 ( 9 + 7 ) 2 ( 16 ) 32. ans