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Ex 6.5, 1 (ii)
Ex 6.5, 1 (iii) Important
Ex 6.5, 1 (iv)
Ex 6.5, 2 (i)
Ex 6.5, 2 (ii) Important
Ex 6.5, 2 (iii)
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Ex 6.5, 3 (i)
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Ex 6.5, 3 (v)
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Ex 6.5, 3 (viii)
Ex 6.5, 4 (i)
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Ex 6.5,22 Important You are here
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Ex 6.5, 27 (MCQ)
Ex 6.5,28 (MCQ) Important
Ex 6.5,29 (MCQ)
Last updated at April 15, 2021 by Teachoo
Ex 6.5, 22 A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum? A wire of length 28m cut into two parts Let π₯ be the length of first part Length of 2nd part = (28βπ₯)π Given that one part is converted into a square Let length of π₯ m be converted into a square Thus, Perimeter of square =π₯ 4(π πππ ππ π ππ’πππ )=π₯ side of square of =π₯/4 Also, other part is converted into Circle β (28βπ₯) m length converted into a Circle of radius π Circumference of circle = (28βπ₯) m 2ππ=(28βπ₯) π=((28 β π₯)/2π) m We need to find length of pieces so that the combined Area of Circle & Square is minimum Let T be the total area of circle & square T = Area of Circle + Area of Square T = ππ^2+(π πππ)^2 T = π((28 β π₯)/2π)^2+π₯^2/16 We have to minimize T Diff T w.r.t π₯ ππ/ππ₯=π/ππ₯ ((28 β π₯)^2/4π+π₯^2/16 " " ) = 2(28 β π₯)/4π . π(28 β π₯)/ππ₯+2π₯/16 = (28 β π₯)/2π (β1)+π₯/8 Putting π π»/π π=π (π₯ β 28)/2π+π₯/8=0 (4(π₯ β 28) + ππ₯)/8π =0 4x β 112 +π π₯=0Γ8 π π₯(4+π)=112 π₯=112/(4 + π) Now finding (π ^π π»)/(π π^π ) ππ/ππ₯=(β28 + π₯)/2π+π₯/8 (π^2 π)/(ππ₯^2 )=1/2π (0+1)+1/8 (π^2 π)/(ππ₯^2 )=1/2π+1/8 > 0 Hence (π^2 π)/(ππ₯^2 )>0 at π₯=112/(4 + π) β΄ Total area is Minimum at π₯ = 112/(4 + π) Finding length of other part Length of other part = 28 β x = 28 β 112/(4 + π) = (28(4 + π)β112)/(4 + π)= (28 π)/(4 + π) β΄ Length of first part is πππ/(π + π ) & another part is πππ /(π + π )