document.write( "Question 1204506: A square piece of cloth of maximum size is cut from a circular piece, and then a circular piece of maximum size is cut from the square piece. If the radius of the original piece of cloth was 72 cm, what is the amount of cloth, in cm² that was wasted?\r
\n" );
document.write( "\n" );
document.write( "A) 2486 pi B) 2576 pi C) 2592 pi D) 1296 pi E) 2660 pi \n" );
document.write( "
Algebra.Com's Answer #840788 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Diagram \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Radius of larger circle = 72 cm \n" ); document.write( "Diagonal of square = 2*72 = 144 cm \n" ); document.write( "Side length of square = 144/sqrt(2) = 72*sqrt(2) cm \n" ); document.write( "Radius of smaller circle = 72*sqrt(2)/2 = 36*sqrt(2) cm\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Area of large circle = pi*r^2 = pi*(72)^2 = 5184pi \n" ); document.write( "Area of smaller circle = pi*r^2 = pi*(36*sqrt(2))^2 = 2592pi\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Area of wasted material = (Area of large circle) - (Area of small circle) \n" ); document.write( "Area of wasted material = 5184pi - 2592pi \n" ); document.write( "Area of wasted material = 2592pi\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "It's not a coincidence that the smaller circle area and the amount of wasted material is the same. \n" ); document.write( "In other words, the area of the smaller circle is half that of the larger circle. \n" ); document.write( "Here's scratch work to prove this claim. \n" ); document.write( "r = large circle radius \n" ); document.write( "2r = square diagonal \n" ); document.write( "2r/sqrt(2) = r*sqrt(2) = square side length \n" ); document.write( "r*sqrt(2)/2 = smaller circle radius \n" ); document.write( "smaller circle area = pi*(smaller radius)^2 \n" ); document.write( "smaller circle area = pi*(r*sqrt(2)/2)^2 \n" ); document.write( "smaller circle area = (1/2)*pi*r^2 \n" ); document.write( "smaller circle area = (1/2)*(larger circle area)\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Answer: Choice C) 2592pi \n" ); document.write( " \n" ); document.write( " |