Area Problems
The area of a trapezium is 384 sq.cm 2. If its parallel sides are in ratio 3 : 5 and the perpendicular distance between them be 12 cm, The smaller of parallel sides is?
Let the sides of trapezium be 5k and 3k, respectively According to the question, (1/2) x [(5k + 3k) x 12] = 384? 8k = (384 x 2)/12 = 64 ? k = 64/8 = 8 cmLength of smaller of the parallel sides = 8 x 3 = 24 cm
Distance travelled in 1 revolution = circumference of the wheel = ?d= ( 22/7 ) x 63= 198 cmSo the distance travelled in 100 revolutions = 100 x distance travelled in 1 revolution = 100 x 198 cm= 19800 cm = 198 m
Let area 100 m2Then, side = 10 m New side = 125 % of 10= (125/100) x 10= 12.5 m New area = 12.5 x 12.5 m2=(12.5)2 sq. m? Increase in area = (12.5)2 - (10)2 m2= 22.5 x 2.5 m2=56.25 m2% Increase = 56.25 %
Original area = (22/7) x 9 x 9 cm2New area = (22/7) x 7 x 7 cm2? Decrease = 22/7 x [(9)2 -(7)2] cm2=(22/7) x 16 x 2 cm2Decrease percent = [(22/7 x 16 x 2) /( 7/22 x 9 x 9)] x 100 %= 39.5 %
EXAMIANS