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62 Theyarh70 8 T17rh2 3r4r0rh7 56They are 708 T718 T728 T738 T748 T758 T768 T778 T788 T798, hence (A). 96 h TT3e T\b7 TT 3h (T 4) T\b 49 T3h 16 \b 1, hence (E). ,6 (Also I8) The perimeters of P, Q, R and S are 4 e 2, 8, 2e 2 and 4 respectively, hence (B). n6 Solving, 7 yy 194, and so yy The y \b 27 a y. So the smallest ycan be is 28, and any value yy 28 is also a solution, hence (C). c6 The table top is a circle with radius 3 metres, so its area in square metres is a(3) T\b 9 a . Because ais slightly greater than 3, 9a is slightly greater than 27. Of the given choices, 30 is closest, hence (B). (6 (Also I11) Theyar7e08y 1 The sum of the exterior angles of the pentagon is 360 \b 90 + 4 h(180 Tr), so that 180 Tr\b 270 74 \b 670 5 andr\b 180 T6705\b112 05, hence (E). Theyar7e08y 2 The sum of the interior angles of the pentagon is 3 h180 \b 90 + 4 r, so thatr\b 450 a4 \b 1120 5, hence (E). A6 (Also I12) If8 is the point directly below 1and to the left of 2, then the right triangle 12 8 has sides 16 and 12, and hypotenuse r. Thenr T\b 16 T+ 12 T\b 400 and r\b 20, hence (A). )6 Square both sides of the equation to obtain: e rT+1 \b r+2 \b rr T+1 \b r T+4r+4 \b r4r \b T30 Hence, the only possible solution is r\b T r e. We substitute this value into the original equation and verify that both sides are indeed equal to a e, hence (B). Th They ar7 0 812345 8469,342n 2016 AMC – Senior Solutions

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