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
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
Area to be plastered = [2(l + b) x h] + (l x b) = {[2(25 + 12) x 6] + (25 x 12)} m2 = (444 + 300) m2 = 744 m2. ∴ Cost of plastering = Rs. ❨ 744 x 75 ❩ = Rs. 558
Let length of rectangle = 5kand breadth of rectangle = 3kAccording to the quecation,5k - 3k = 8 ? 2k =8? k = 4? Lenght = 5k = 5 x 4 = 20 mBreadth = 3k = 3 x 4 = 12 m? Required area = Lenght x Breadth = 20 x 12 = 240 sq m
We know that in any triangle "the sum of two sides is always greater than its third side" and "the difference of two sides is always less than its third side".(i) 2 + 3 is not greater than 5 (ii) |5 - 2| not less than 3
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