The greatest product possible for one vertex in c is 8 4 3 96 4 5 3 2 8 1 3 1 7

The greatest product possible for one vertex in c is

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The greatest product possible for one vertex in c) is 8 ! 4 ! 3 = 96 . 4 5 3 2 8 1 3 1 7
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E XTRA C HALLENGES This resource may be copied in its entirety, but is not to be used for commercial purpose s without permission from the Centre for Education in Mathematics and Computing, University of Waterloo. For more activities and resources from the University of Waterloo’s Faculty of Mathematics, please visit . 2 Centre for Education in Mathematics and Computing Faculty of Mathematics University of Waterloo Waterloo, ON Canada N2L 3G1 Thus, c. will yield the greatest product at one vertex. 4. Let OJ represent the volume of orange juice, PJ represent the volume of pineapple juice, CJ represent the volume of cranberry juice, W represent the volume of water, and V represent the total volume in the pitcher. We know OJ + PJ + CJ + W = V . Now, 2 CJ = PJ , OJ = PJ and W = 1 8 V . So, PJ + 2 CJ + CJ + 1 8 V = V 2 CJ + 2 CJ + CJ + 1 8 V = V 5 CJ = 7 8 V CJ = 7 40 V 7 40 of the total volume is cranberry juice, so in 320 mL there is 7 40 ! 320 or 56 mL of cranberry juice. 5. The area of the large square is 36 m 2 . The diameter of the circle is equal to the length of one side of the large square, so the radius of the circle is 3 m. Thus, the area of the circle is 9 ! m 2 . The diagonal of the small square is equal to the diameter of the circle. The diagonal divides the small square into two congruent right-angled isosceles triangles. Using the Pythagorean theorem, we can determine the length of each side of the small square. If each side of the smaller square is of length a , then a 2 + a 2 = 6 2 2 a 2 = 36 a 2 = 18 Thus, the area of the small square is 18 m 2 . The area of the shaded region is 36 ! 9 " + 18 = 54 ! 9 " ! 25.73 m 2 .
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