If the length of the diagonal of a square is 20cm,then its perimeter must be

a. 402cm     b. 302cm    c. 10cm    d. 152cm

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.

Let each side of the square be a. Then, area = .a2
New side =125a100=5a4. New area = 5a42 = 25a216

Increase in area = 25a216-a2=9a216
Increase% = 9a216*1a2*100 % = 56.25%.

Let the diagonals of the squares be 2x and 5x respectively.
Ratio of their areas =  12*2x2:12*5x2=4:25

Area of the square =12*diagonal2=12*3.8*3.8=7.22m2

Area to be plastered= [2(l + b) x h] + (l x b)

= [2(25 + 12) x 6] + (25 x 12)

= (444 + 300)

= 744 sq.m

Cost of plastering = Rs.744 x (75/100)

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

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

Let breadth = x metres.
Then, length = (x + 20) metres.

Perimeter =5300/23.50

2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m.

Let original length = x and original breadth = y.

Original area = xy. 

New length = x/2 and New breadth=3y 

New area = 32xy 

Therefore,  Increase in area = New area-original area = 32xy-xy=12xy 

Therefore,  Increase % = increase in area original area*100=12xyxy*100=50 %