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 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.
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.
In a triangleSum of two sides is always greater than 3rd side i.e., x < 25 + 15 = 40 .....(i)Difference of two sides is always less than 3rd side i.e., 25 - 15 = 10 < x ...(ii) From Eqs. (i) and (ii) , we get 10 < x < 40
Original area = ?(d/2)2= (?d2) / 4New area = ?(2d/2)2= ?d2Increase in area = (?d2 - ?d2/4)= 3?d2/4? Required increase percent = [(3?d2)/4 x 4/(?d2) x 100]%= 300%
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