Area Problems
The area of a trapezium is 384 sq.cm. If its parallel sides in ratio 3 : 5 and the perpendicular distance between them be 12 cm. The smaller of parallel sides is?
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.
According to the question, Area of semi- circle = 77 m(1/2) x ? x r2 = 77? r2 = (77 x 2 x 7)/22? r = 7m ?Circumference of semi- circle =?r + 2r= ( ? + 2)r = [(22/7) + 2] x 7 = 36 m
Let the length, breadth and height of the room be l, b and h respectively As per question Cost of 2(l + b) x h = Rs. 48 ? Required cost = cost of 2 (2l + 2b) x 2h= cost of 4 [2(l + b) x h ]= 4 x Rs. 48= Rs. 192
Area of park = 100 x 100 = 10000 m2Area of circular lawn = Area of park - area of park excluding circular lawn= 10000 - 8614 = 1386Now again area of circular lawn = (22/7) x r2 = 1386 m2? r2 = (1386 x 7) / 22= 63 x 7= 3 x 3 x 7 x 7? r = 21 m
Let original length = x metres and original breadth = y metres. Original area = (xy) m2. New length = ❨ 120 x ❩m = ❨ 6 x ❩m. 100 5 New breadth = ❨ 120 y ❩m = ❨ 6 y ❩m. 100 5 New Area = ❨ 6 x x 6 y ❩m2 = ❨ 36 xy ❩m2. 5 5 25 The difference between the original area = xy and new-area 36/25 xy is = (36/25)xy - xy = xy(36/25 - 1) = xy(11/25) or (11/25)xy ∴ Increase % = ❨ 11 xy x 1 x 100 ❩% = 44%
EXAMIANS