SAT Math Hard Practice Quiz Answers
5.
D
(Estimated Difficulty Level: 5)
Since
ABC
is a right triangle, the length of segment
AC
is
√
6
2
+ 8
2
= 10. (Hint: you will see 3-4-5 and 6-8-10
triangles a lot on the SAT.) The area of triangle
ABC
is (1
/
2)
·
b
·
h
, where
b
is the base and
h
is the height of
triangle
ABC
.
The key thing to remember for this problem is that the
base can be
any
of the three sides of a triangle, not just
the side at the bottom of the diagram. If the base is
AB
, then the height is
BC
and the area of the triangle
is (1
/
2)(6)(8) = 24. If the base is
AC
, then the height is
BD
and the area of the triangle is still 24. This means
that (1
/
2)(
AC
)(
BD
) = (1
/
2)(10)(
BD
) = 24 so that
BD
= 24
/
5.
6.
B
(Estimated Difficulty Level: 4)
Draw a diagram for this problem! With exactly two
parallel lines, the other two lines cannot be parallel to
themselves or to the first two lines. Your diagram may
seem to suggest five points of intersection; however, the
point of intersection of the two non-parallel lines can
overlap with a point of intersection on one of the parallel
lines:
From the figure, the least possible number of intersec-
tion points is then three.
7.
D
(Estimated Difficulty Level: 5)
One formula that a good math student such as yourself
may want to memorize for the SAT is the area of an
equilateral triangle. If the length of each side of the
triangle is
s
, then the area is
√
3
s
2
/
4. The perimeter of
this triangle is 3
s
.
Now, since the perimeter equals the area for this tri-
angle, we have: 3
s
=
√
3
s
2
/
4 so that 3 =
√
3
s/
4 and
s
= 12
/
√
3 = 4
√
3. The perimeter is then 12
√
3, mak-
ing answer D the correct one. (Did you get
s
= 4
√
3
and then choose answer C? Sorry about that.)
8.
A
(Estimated Difficulty Level: 5)
For many difficult SAT questions, it can be very helpful
to know some “extra” math along with the “required”
math. First, when a square is inscribed in a circle,
the diagonals are diameters of the circle. Second, the
diagonals of a square meet at right angles. Third, a
diagonal of a square is
√
2 times as long as the length
of one of the sides. (A diagonal of a square makes a 45-
45-90 triangle with two sides.) For this question, the
length of each side of the square is 6 (since the area is
6
2
= 36), and the length of a diagonal is 6
√
2, so the
radius of the circle is 3
√
2, as shown below:
6
6
6
6
3
√
2
3
√
2
A final piece of needed math: the arc length of a portion
of a circle is the circumference times the central angle of
the arc divided by 360
◦
. Here, the central angle is 90
◦
,
so the needed arc length (shown darkened in the figure
above) is just 1
/
4 times the circle’s circumference. The
arc length is then 2
πr/
4 = 2
π
·
3
√
2
/
4 = 3
π
√
2
/
2 and
the perimeter of the shaded region is 6 + 3
π
√
2
/
2.
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