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What is the area of the inscribed quadrilateral ABCD? A 28 B 36 C 42 D 48 E 56 Solution: AThe graph cuts the y-axis at points where 0=x, and hence where 1242=−yy, that is, where 01242=−−yy, that is, 0)6)(2(=−+yy, giving 2−=yand 6=y. Hence A is the point (0,6), Cis the point )2,0(−and so AChas length 8. Similarly, B and D are points where 0=yand hence 0122=−+xx, that is, 0)3)(4(=−+xx, giving 4−=xand 3=x. So B is the point )0,3(, D is the point )0,4(−, and so BD has length 7. The area of the inscribed quadrilateral ABCD is therefore BDAC.2128)78(21=×=. Extension Problems 16.1 The solution above takes it for granted that the area of the quadrilateral ABCD is half the product of the lengths of its diagonals. Give an example to show that this is not truefor all quadrilaterals. 16.2 What is the special property of ABCD that makes it correct that in this case the area is half the product of the length of the diagonals? Give a proof to show that your answer is correct. 17. The diagram shows a pattern found on a floor tile in the cathedral in Spoleto, Umbria. A circle of radius 1 surrounds four quarter circles, also of radius 1, which enclose a square. The pattern has four axes of symmetry. What is the side length of the square? A 21B 22−C 31D 21E 12−Solution: B Let O be the centre of the circle, let P be one of the points where two of the quarter circles meet, let Q be the centre of one of these quarter circles, and let R be the vertex of the square that lies on OQ, as shown. Then in the triangle OPQthere is a right angle at Pand 1==QPOP, and therefore, by Pythagoras’ Theorem, 2=OQ. Since 1=QR, it follows that 12−=OR. It follows that the diagonal of the square has length)12(2−. Using, Pythagoras’ Theorem again, it follows that the side length of the square is())12(221−22−=. Extension Problem 17.1 In the proof above we claim, by Pythagoras’ Theorem, that if the diagonal of a square has length x, then its side length isx21. Show that this is indeed a consequence of Pythagoras’ Theorem.
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