Area Problems
A circular road runs rounds a circular ground. If the difference between the circumference of the outer circle and the inner circle is 66 meters, the width of the road is?
Distance covered in one revolution = total distance travelled / total number of revolution. = ( 88 x 1000) / 1000 m = 88 mWe know that the distance covered in one revolution = circumference of the wheel.? ?d = 88? 22d / 7 = 88? d = 28 m
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 parallel sides be 3a and 5a.So Area of trapezium = 1/2 x sum of parallel side x perpendicular distance between them.? 1/2 (3a +5a) x 12 = 384? 8a = 64? a =8? Smaller side = 3x = 3 x 8 = 24 cm.
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
EXAMIANS