Sat Math Hard Practice Quiz Worksheet With Answer Key Page 16
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18SAT Math Hard Practice Quiz Answers
5. D
(Estimated Difficulty Level: 5)
7. D
(Estimated Difficulty Level: 5)
One formula that a good math student such as yourself
Since ABC is a right triangle, the length of segment AC
√
may want to memorize for the SAT is the area of an
2
2
is
6
+ 8
= 10. (Hint: you will see 3-4-5 and 6-8-10
equilateral triangle. If the length of each side of the
triangles a lot on the SAT.) The area of triangle ABC
√
2
triangle is s, then the area is
3s
/4. The perimeter of
is (1/2) · b · h, where b is the base and h is the height of
this triangle is 3s.
triangle ABC.
Now, since the perimeter equals the area for this tri-
The key thing to remember for this problem is that the
√
√
2
angle, we have: 3s =
3s
/4 so that 3 =
3s/4 and
base can be any of the three sides of a triangle, not just
√
√
√
s = 12/
3 = 4
3. The perimeter is then 12
3, mak-
the side at the bottom of the diagram. If the base is
√
ing answer D the correct one. (Did you get s = 4
3
AB, then the height is BC and the area of the triangle
and then choose answer C? Sorry about that.)
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.
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-
6. B
(Estimated Difficulty Level: 4)
45-90 triangle with two sides.) For this question, the
length of each side of the square is 6 (since the area is
Draw a diagram for this problem! With exactly two
√
2
6
= 36), and the length of a diagonal is 6
2, so the
parallel lines, the other two lines cannot be parallel to
√
radius of the circle is 3
2, as shown below:
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
6
lines:
6
6
6
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 2πr/4 = 2π · 3
2/4 = 3π
2/2 and
√
From the figure, the least possible number of intersec-
the perimeter of the shaded region is 6 + 3π
2/2.
tion points is then three.
pg. 16
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