l 2 + b 2 = 41 (or)    l 2 + b 2 = 41  

Also,  l b = 20

l + b 2 = l 2 + b 2 + 2 l b  = 41 + 40 = 81

(l + b) = 9. 

Perimeter = 2(l + b) = 18 cm.

let ABCD be the given parallelogram 

area of parallelogram ABCD = 2 x (area of triangle ABC) 

now a = 30m, b = 14m and c = 40m

 s=1/2 x (30+14+40) = 42 

Area of triangle ABC =  s s - a s - b s - c  

= 42 12 28 2= 168sq m 

area of parallelogram ABCD = 2 x 168 = 336 sq m

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 be x meters
length of two sides = 2x meters
diagonal =  2 x= 1.414x m

saving on 2x meters = .59x m

saving % =  0 . 59 x 2 x * 100%

= 30% (approx)

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

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

Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares
13 . 5 * 10000 m 2 = 135000 m 2 
Let altitude = x metres  and  base = 3x metres.
Then,  1 2 * 3 x * x = 135000 ? x 2 = 90000 ? x = 300 

Base = 900 m and Altitude = 300 m.

Area to be plastered =  2 l + b × h + l × b 

= 2 25 + 12 × 6 + 25 × 12 = 744 s q . m 

Cost of plastering =  744 × 75 100 = R s . 558