A certain triangle has sides of lengths 5, 12, and x.Which of the following could be the value of x? (Choose all that apply.)
Answer(s): B,C,D
For any triangle, the sum of the lengths of any two sides must be greater than the length of the third side.With sides 5, 12, x:5 + 12 > x x < 175 + x > 12 x > 712 + x > 5 (always true for positive x)Thus, x must satisfy 7 < x < 17
When rounded to the nearest 0.01, both numbers x and y are equal to 65.28.Quantity A|y x|Quantity B0.001
Answer(s): D
Because a number rounds to 65.28 (to the nearest 0.01) precisely when it lies in the interval [65.275, 65.285),both x and y must be in this interval. Hence|y - x| < 65.285 - 65.275 = 0.01.If x = y, then |y - x| = 0 < 0.001.If x = 65.275 and y = 65.284, then |y - x| = 0.009 > 0.001.If x = 65.275 and y = 65.276, then |y - x| = 0.001.Thus |y - x| can be less than, equal to, or greater than 0.001.Quantity A and Quantity B cannot be definitively compared.
The area of each building lot in a future housing development will be at least acre and at most acre. The combined area of all of the building lots in the development will be x acres, where x is a positive integer.Quantity AThe greatest possible number of building lots in the development minus the least possible number of building lots in the developmentQuantity Bx
Answer(s): A
Each lot area satisfiesFor a total of x acres:Greatest possible number of lots: use the minimum area per lot acreLeast possible number of lots: use the maximum area per lot acreQuantity A = Nmax - Nmin = 4x - 2x = 2x.Quantity B = xSince 2x > x for every positive integer x, Quantity A > Quantity B.
Triangle ABC is inscribed in the circle with center O and diameter AC.Quantity A3 times the area of triangular region ABCQuantity BThe area of the circular region
Answer(s): B
Let the circle have radius r; then the diameter AC = 2r.Because AC is a diameter, ABC is right-angled at B (Thales' theorem). Thuswith AB2 + BC2 = AC2 = (2r)2 = 4r2.For fixed AB2 + BC2 = 4r2, the product AB BC (and hence the area) is maximized when AB = BC:Hence 3(area of ABC) 3r2.Since r2 > 3r2, even the greatest possible value of Quantity A is less than Quantity B, and for all other positions of B Quantity A is still smaller.
Lines l and m are parallel.90 < c < 180Quantity AcQuantity Ba + b
Because l and m are parallel, the acute or obtuse size of each of the three labeled angles depends on how steeply the two oblique lines happen to intersect the parallels.If the oblique lines meet at a very narrow angle, a and b will be small (so a + b will be small), whereas will be very obtuse (greater than a + b).If the oblique lines meet at a very wide angle, one of the angles a or b becomes obtuse, making a + b exceed c.It is also possible to choose the slopes of the oblique lines so that a + b equals c.Hence a + b can be less than, equal to, or greater than c.
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