New York City SHSAT 2017

(Marvins-Underground-K-12) #1
a circle is πr^2 , so the area of circle P is π(5)^2 = 25π.

B
Note that the perimeter of the shaded region is in fact equal to the perimeter of the square plus the perimeter—or
circumference—of the circle. The area of a circle is πr^2 , where r is the radius, and since the area of the circle is 25π, its
radius is 5. Circumference is equal to 2πr, or 2 π(5) = 10π. Only choices (B) and (C) contain 10π, so you could eliminate
choices (A), (D), and (E). Since the circle is inscribed within the square, its diameter is equal to a side of the square. The
diameter of the circle is 2r or 10, so a side of the square is 10 and its perimeter is 4(10)= 40. Therefore, the perimeter of the
shaded region is 40 + 10π, choice (B).

33.


F


Slope of a line is defined by the formula where (x 1 , y 1 ) and (x 2 , y 2 ) represent two points on the line. Plug the given
points into the formula (it doesn’t matter which you designate as point 1 or point 2):

34.


D


The circumference of a circle is 2πr, where r is the radius, so a circle whose circumference is 16π has a radius of
The area of a circle is defined by the formula πr^2 , where r is the radius. So in this case the area is π(8)^2 = 64π, choice (D).

35.


J


It might help to sketch a diagram:

Since all sides of a square are equal, notice that the diagonal of the square is also the hypotenuse of an isosceles right
triangle. Use this information to determine the length of a side of the square, marked s in the diagram. The ratio of the sides in
such a triangle is Since the hypotenuse is 6, each leg is (because ). The area of a square is

equal to the length of one side squared, or

36.


37. B

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