? ( 22 x r2) / 7 = 38.5
? r2 =(38.5 x 7) /22
? r = 3.5 cm
? Circumference = 2 x (22/7) x 3.5 cm
= 22 cm
? 22/7 x r2 = 13.86 x 10000
? r2 = (13.86 x 10000 x 7) / 22
? r = 210 m
? Circumference = [2 x (22/7) x 210 ] m
= 1320 m
Cost of fencing = Rs. (1320 x 20)/100
= Rs . 264
? 2?r - r = 37
? [(2 x 22/7) -1]r= 37
? 37r / 7= 37
? r = 7
So, area of the circle =(22/7) x 7 x 7 cm2
= 154 cm2
Original area = (22/7) x 9 x 9 cm2
New area = (22/7) x 7 x 7 cm2
? Decrease = 22/7 x [(9)2 -(7)2] cm2
=(22/7) x 16 x 2 cm2
Decrease percent = [(22/7 x 16 x 2) /( 7/22 x 9 x 9)] x 100 %
= 39.5 %
Original area = ? x (r/2)2 = ?r2/4
Reduction in area = ? r2 - 3? r2/4
? Reduction per cent = [ 3?r2/4 x 4/(?r2) x 100 ] %
= 75%
Let the side of the square = y cm
Then, breadth of the rectangle = 3y/2 cm
? Area of rectangle = (40 x 3y/2) cm2
= 60y cm2
? 60y = 3y2
? y = 20
Hence, the side of the square = 20 cm
circumference = 2?r = 2 x (22 / 7) x r = 352
? r = (352 x 7) / (22 x 2) =56 cm
Now area = ?r2 = (22 / 7) x 56 x 56 m2
= 9856 m2
Let h be the altitude of triangle.
So area of triangle = (1/2)xh
area of square = x2
From question area of square = area of triangle
x2 = (1/2)xh
? h = (2x2)/x = 2x
? s= (13 + 5 + 12) / 2 cm
= 15cm
s-a = 2 cm,
s-b = 10 cm and
s-c = 3 cm
? Area = ? (15 x 2 x 10 x 3 ) cm2
= 30 cm2
? (12 x h) / 2 = 30
? h = 5 cm
Let lateral side = (5y) cm and base = (4y) cm
? perimeter = 5y + 5y + 4y = 14
?y = 1
So, the sides are 5 cm , 5 cm and 4 cm
Now s= 1/2 (5 + 5 + 4) cm = 7 cm
(s-a) = 2 cm
(s-b) = 2 cm and
(s-c) = 3 cm
? Required Area = ? (7 x 2 x 2 x 3) cm2
=2?21 cm2
s = (13 + 14 + 15 ) / 2 = 21,
s-a = 8 ,
s-b = 7,
s-c= 6
? Area to be painted = ? [s(s-a) (s-b) (s-c)]
=? [21 x 8 x 7 x 6] m2
= 84 m2
? Cost of painting = Rs. (84 x 8.75) = Rs. 735
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