Let x and y be the length and breadth of the rectangle respectively.
Then, x - 4 = y + 3 or x - y = 7 ----(i)
Area of the rectangle =xy; Area of the square = (x - 4) (y + 3)
(x - 4) (y + 3) =xy <=> 3x - 4y = 12 ----(ii)
Solving (i) and (ii), we get x = 16 and y = 9.
Perimeter of the rectangle = 2 (x + y) = [2 (16 + 9)] cm = 50 cm.

distance covered in 1 revolution = 2 ?r = 2 x (22/7) x 20 = 880/7 cm
required no of revolutions = 17600 x (7/880) = 140

let the radius of the pool be Rft

Radius of the pool including the wall = (R+4)ft

Area of the concrete wall = ? R + 4 2 - R 2= ? R + 4 + R R + 4 - R = 8 ? R + 2 sq feet

=> 8 ? R + 2 = 11 25 ?R 2 = > 11 R 2 = 200 R + 2

Radius of the pool R = 20ft

Let length = x meters, then breadth = 0.6x 

Given that  perimeter = 800 meters 

=> 2[ x + 0.6x] = 800 

=> x = 250 m 

Length = 250m and breadth = 0.6 x  250 = 150m 

Area = 250 x 150 = 37500 sq.m

Area of the sheet =  20 × 30 c m 2  

Area used for typing =  20 - 4 × 30 - 6 c m 2   

required % = 384 600 × 100=64%

circumference of a circle = 2 ?r

=> 2 × 22/7 × r ? r = 37

=> 37/7 × r = 37

=> r = 7 cm.

Diameter D = 2r = 7×2 = 14 cm.

Distance covered in one revolution =  88 × 1000 1000= 88m. 

2 ?R 2 = 88 ? 2 × 22 7 × R = 88 ? R = 88 × 7 44

The triangle with sides 26 cm, 24 cm and 10 cm is right angled, where the hypotenuse is 26 cm.

Area of the triangle =1/2 x base x height => 1/2 x 24 x 10 = 120 sq.cm

Given length of the rectangle = 3 cm

Breadth of the rectangle = 4 cm

Then, the diagonal of the rectangle  D = 3 2 + 4 2 = 25 = 5

Then, it implies side of square = 5 cm

We know that Area of square = S x S = 5 x 5 = 25 sq.cm.

We have: l = 20 ft and lb = 680 sq. ft.So, b = 34 ft.

Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.