Let the side of the square be 's' cm

length of rectangle = (s+5) cm

breadth of rectangle = (s-3)cm

(s+5) (s-3) = 

s 2 - 5s - 3s - 15 =  s 2 

2s = 15

Perimeter of rectangle = 2(L+B) = 2(s+5 + s?3) = 2(2s + 2)

= 2(15 + 2) = 34 cm 

From statement (A),

20b = (1/2) × b × h

h = 40 cm.

Let breadth = x metres, length = 3x metres, height = H metres. 

Area of the floor=(Total cost of carpeting)/(Rate) = (270/5) sq.m = 54 sq.m 

x × 3 x 2 = 54 ? x 2 = 54 × 2 ? x = 6  

So, breadth = 6 m and length = 3 6 2 = 9 m.

Now, papered area = (1720/10) =  172 sq.m 

Area of 1 door and 2 windows = 8 sq.m 

Total area of 4 walls = (172 + 8) sq.m = 180 sq.m 

2 × 9 + 6 H = 180 ? H = 6

speed = 12 km/h =  12 × 5 18 = 10 3 m / s 

distance covered =  20 × 2 × 22 7 × 50 = 44000 7 m 

time taken  = distance /speed =   44000 7 × 3 10 s e c = 4400 × 3 7 × 1 60 m i n = 220 7 m i n 

area of cross roads = (55 x 4) + (35 x 4)- (4 x 4) = 344sq m

cost of graveling = 344 x  (75/100) = Rs. 258

perimeter = total cost / cost per m = 10080 /20 = 504m
side of the square = 504/4 = 126m
breadth of the pavement = 3m
side of inner square = 126 - 6 = 120m
area of the pavement = (126 x126) - (120 x 120) = 246 x 6 sq m
cost of pavement = 246*6*50 = Rs. 73800

Area = 5.5 × 3.75 sq. metre.

Cost for 1 sq. metre. = Rs. 800

Hence total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = Rs. 16500

We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
Area = (l x b) = (63 x 40) = 2520 sq.m

Let inner radius be r metres. Then, 2 ?r = 440 ; r =  440 × 7 44= 70 m.

Radius of outer circle = (70 + 14) m = 84 m.

Let the breadth of the given rectangle be x then length is 2x.
thus area of the given rect is  2 x 2

after dec 5cm from length and inc 5cm breadth , new lenght becomes 2x-5 and breadth is x+5.thus new area =(2x-5)(x+5)= 2 x 2 + 5 x - 25

since new area is 75 units greater than original area thus
2 x 2 + 75 = 2 x + 5 x - 25
5x=75+25
5x=100
therefore x=20
hence length of the rectangle is 40 cm.