ratio =  1 / 2 × 2 d 2 1 2 d 2 = 4 : 1

Area of the room=(544 * 374)

size of largest square tile= H.C.F of 544 & 374 = 34 cm

Area of 1 tile = (34  x  34)  c m 2

Number of tiles required== [(544 x 374) / (34 x 34)] = 176

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.

Circumference = No.of revolutions * Distance covered

Distance to be covered in 1 min. = (66 X1000)/60 m = 1100 m.
Circumference of the wheel = 2 x (22/7) x 0.70 m = 4.4 m.
Number of revolutions per min. =(1100/4.4) = 250.

area of the room = 544 x 374 sq.cm

size of largest square tile = H.C.F of 544cm and 374 cm= 34 cm

area of 1 tile = 34x34 sq cm

no. of tiles required = (544 x 374) / (34 x 34) = 176

Let a = 13, b = 14 and c = 15. Then,  s = 1 2 a + b + c=21

 (s- a) = 8, (s - b) = 7 and (s - c) = 6.

Area = s s - a s - b s - c = 21 × 8 × 7 × 6 = 84 sq.cm   

Let the breadth of floor be 'b' m.

Then, length of the floor is 'l = (b + 25)'

Area of the rectangular floor = l x b = (b + 25) × b

According to the question,

(b + 15) (b + 8) = (b + 25) × b

b 2 + 8 b + 15 b + 120 = b 2 + 25 b

2b = 120

b = 60 m.

l = b + 25 = 60 + 25 = 85 m.

Area of the floor = 85 × 60 = 5100 sq.m.

Let original length = x and original breadth = y.

Original area = xy. 

New length = x/2 and New breadth=3y 

New area =  3 2 x y 

Therefore,  Increase in area = New area-original area =  3 2 x y - x y = 1 2 x y 

Therefore,  Increase % =  i n c r e a s e i n a r e a o r i g i n a l a r e a * 100 = 1 2 x y x y * 100 = 50 %

100 cm is read as 102 cm. 

A1 = (100*100)Sq.cm 

A2 = (102*102)Sq.cm

(A2 - A1) = 102 2 - 100 2 = (102 + 100) x (102 - 100) = 404 sq.cm.