- B The length of the garden is p.The width is
half of 44 −p.Therefore, the area is p((44 −p)/2) =
(44p−p^2 )/2. - E You get the maximum number of pieces (11)
by making sure that each cut intersects every
other previous cut in a new spot. Your diagram
should look something like this, with six points of
intersection in the circle. - B Draw the extra line shown here to see that the
shaded region has a perimeter equal to a 10-by-17 rec-
tangle. Therefore its perimeter is 10 + 17 + 10 + 17 =54.
3
10
20
17- A Move the pieces together to see that they
form a right triangle. Since all of the interior an-
gles of an equilateral triangle are 60°and the
bisectors divide them in half, the triangle is a
30 °-60°-90°triangle, so its base must be , and
the area is ()()232 2 23() =.
23
2
2 √ 330 °- E The two right triangles have two pairs of equal
sides (the two radii and the shared hypotenuse),
so they must be congruent triangles. Arc BDis^1 ⁄ 3
of the circle, with circumference 12π. Therefore,
you should be able to determine the measures
shown in this diagram and see that the perimeter
of the shaded region is 12 3 4+ π.
A
B
C
D
° 1230
30 °60 °
60 °66 √ 3 66 √ 3- A If the circle has an area of 4π, its radius is 2. If
the rectangle has an area of 8, its length must be
4. The arc portion of the shaded region is^1 ⁄ 4 of
the circle with circumference 4π, so the perimeter
is 2 + 2 + 4 +π= 8 +π.
P
R
(^22)
4
2 2
π
388 McGRAW-HILL’S SAT
- 81 or 100 Consider the right triangle in the
upper-left corner of the diagram. Notice that it
has a hypotenuse of aand legs of length band
12 −b. The question is asking for the area of the
shaded square, which is a^2. By the Pythagorean
theorem, a^2 =b^2 +(12 −b)^2 = 2 b^2 − 24 b+144. Since
2 b 5, the maximum possible value of a^2 is
2(2)^2 −24(2) + 144 =104, and the minimum pos-
sible value of a^2 is 2(5)^2 −24(5) + 144 =74. Since
amust be an integer, a^2 must be a perfect square,
and the only perfect squares between 74 and 104
are 81 and 100.
12
12 - bab
b