Daniel Sadoc Posted November 12, 2005 Share Posted November 12, 2005 Hi People, I have two data comparison questions: 1 - Given a square and a rectangle inscribed in that square: (a) perimeter of the rectangle (b) perimeter of the square Correct answer: (b) Why? Is it easy to prove that (b) is the correct answer? 2 - Given a square and a quadrilateral inscribed in that square: (a) perimeter of the quadrilateral (b) perimeter of the square Best regards, DAniel Quote Link to comment Share on other sites More sharing options...
coolsamurai_83 Posted November 12, 2005 Share Posted November 12, 2005 1. If a rectangle is inscibed in a square, then the perimeter of the square is obviously greater than the perimeter of the rectangle See for yourself - draw a square and put a rectangle in the square. Length or breadth of the rectangle would be equal to the length of the side of the square and the other side of the rectangle would be less than the side of the square therefore the perimeter of the square would be greater than the perimeter of the rectangle Quote Link to comment Share on other sites More sharing options...
coolsamurai_83 Posted November 12, 2005 Share Posted November 12, 2005 2. Quadrilateral is any four sided figure. Take the example of the rectangle itself ( in 1) Therefore the answer is again b, that is the perimeter of the square is greater than the perimeter of the quadrilateral Quote Link to comment Share on other sites More sharing options...
Daniel Sadoc Posted November 12, 2005 Author Share Posted November 12, 2005 1. If a rectangle is inscibed in a square, then the perimeter of the square is obviously greater than the perimeter of the rectangle See for yourself - draw a square and put a rectangle in the square. Length or breadth of the rectangle would be equal to the length of the side of the square and the other side of the rectangle would be less than the side of the square therefore the perimeter of the square would be greater than the perimeter of the rectangle But one of the sides of the rectangle may be greater than one of the sides of the square! Quote Link to comment Share on other sites More sharing options...
Daniel Sadoc Posted November 12, 2005 Author Share Posted November 12, 2005 But one of the sides of the rectangle may be greater than one of the sides of the square! Ok. The result may be proved using the triangular inequality (considering the four triangles that are formed between the quadrilateral and the square). Quote Link to comment Share on other sites More sharing options...
reemagre Posted November 12, 2005 Share Posted November 12, 2005 need help with this question(quantitative comparison from barrons coordanate geometry the distance from (b,5)to (c,-3) is 10. b c-b =6 Quote Link to comment Share on other sites More sharing options...
Daniel Sadoc Posted November 13, 2005 Author Share Posted November 13, 2005 need help with this question(quantitative comparison from barrons coordanate geometry the distance from (b,5)to (c,-3) is 10. b c-b =6 dist = sqrt ( (b - c)^2 + 64 ) = 10 therefore (b-c)^2 = 36 |b-c|=6 so c-b=6 Quote Link to comment Share on other sites More sharing options...
denizen06 Posted November 13, 2005 Share Posted November 13, 2005 1 - Given a square and a rectangle inscribed in that square: (a) perimeter of the rectangle (b) perimeter of the square Correct answer: (b) 2 - Given a square and a quadrilateral inscribed in that square: (a) perimeter of the quadrilateral (b) perimeter of the square Best regards, DAniel 1.b 2.b Quote Link to comment Share on other sites More sharing options...
kronique Posted November 14, 2005 Share Posted November 14, 2005 dist = sqrt ( (b - c)^2 + 64 ) = 10 therefore (b-c)^2 = 36 |b-c|=6 so c-b=6 err.... i did not get the soln.:mad: ...can u plz explain...it is a comparison right?? Quote Link to comment Share on other sites More sharing options...
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