Let original length = x metres and original breadth = y metres.

Original area = xy sq.m 

Increased length  =  120 100 and Increased breadth =  120 100 

New area =  120 100 x * 120 100 y = 36 25 x y m 2 

The difference between the Original area  and New area  is:  

36 25 x y - x y

11 25 x y Increase % = 11 25 x y x y * 100= 44%

Let the side of the square(ABCD) be x meters.

Then, AB + BC = 2x metres.

AC =  2 x = (1.41x) m.

Saving on 2x metres = (0.59x) m.

Saving % = 0 . 59 x 2 x * 100 = 30% (approx)

Let the side of the square be x. Then, its diagonal =  2 x 2 = 2 x 

Radius of incircle =  x 2 

Radius of circum circle=  2 × x 2 = x 2 

Required ratio =  ?x 2 4 : ?x 2 2 = 1 4 : 1 2 = 1 : 2

length of wire = 2 ?r= 2 x (22/7 ) x 56 = 352 cm
side of the square = 352/4 = 88cm
area of the square = 88 x 88 = 7744sq cm

ratio =  a 2 2 a 2

The diameter is equal to the shortest side of the rectangle.

 So radius= 14/2 = 7cm. 

Therefore,   a r e a o f c i r c l e = ?r 2 = 22 7 × 49 = 154 cm 2

Other side = [(17 x 17) - (15 x 15)] = (289 - 225) = 8m
Area = 15 x 8 =120 sq. m

let ABC be the isosceles triangle, the AD be the altitude 

Let AB = AC = x then BC= 32-2x       [because parameter = 2 (side) + Base] 

since in an isoceles triange the altitude bisects the base so 

BD = DC = 16-x 

In a triangle ADC,  A C 2 = A D 2 + D C 2 

x 2 = 8 2 + 16 - x 2  ? x = 10 

BC = 32-2x = 32-20 = 12 cm 

Hence, required area =  1 2 * B C * A D=  1 2 * 12 * 10 = 60 sq cm

We know that d=?2s 

Given diagonal = 20 cm 

=> s = 20/ 2 cm

Therefore, perimeter of the square is 4s = 4 x 20/ 2 = 40 2   cm.

Let breadth = x m

Then, length = (x+5)m

Area of a rectangle = Length x Breadth

x(x+5) = 750 

x² + 5x - 750= 0

(x+30)(x-25)= 0

x = 25 or x = -30

Hence, breadth x = 25m

=> Length = x + 5 = 25 + 5 = 30m.