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
EXAMIANS