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SAT模拟考试题:SAT数学练习题(14)

2016-08-11 16:53:56来源:网络

  Since DEB angle is 90o and equal to angle BCD, EBCD is a rectangle and ED = BC = 3

  The area is (1/2)·4·3 = 6.

  Question #7: The side AB of the ABCD square is 10, AE = ED and BF = FC. What is the area of parallelogram BFDE?

SAT模拟考试题:SAT数学练习题(14)

  (a) 20

  (b) 25

  (c) 50

  (d) 60

  (e) 75

  Answer: The easiest way to solve this problem is to notice that we can calculate the area of the parallelogram BFDE by subtracting twice the area of triangle ABE from the area of the square ABCD.

  The area of the square ABCD is 10·10 = 100.

  The area of triangle ABE is (1/2)·AB·AE = (1/2)·10·5 = 25.

  In conclusion, the area of parallelogram BFDE is 100 - 2·25 = 50.

  Question #8: The side of the cube C1 is equal to the height of a cylinder C2 and twice the radius of the same cylinder C2.

  Column AColumn B

  Volume of the cube C1Volume of the cylinder C2

  (a) The quantity in Column A is greater then the quantity in Column B.

  (b) The quantity in Column B is greater then the quantity in Column A.

  (c) The two quantities are equal.

  (d) The relationship cannot be determined from the information given.

  Answer: The volume of the cube will be a·a·a = a3

  The volume of the cylinder will be h·¶·r2 = a·¶·(a/2)2 = (¶/4)·a3

  Since (¶/4) < 1, the volume of the cube will be greater than the volume of the cylinder, so tthe quantity in Column A is greater then the quantity in Column B.

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本文关键字: SAT模拟考试 SAT数学

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