Sat Math Hard Practice Quiz Worksheet With Answer Key Page 17
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18SAT Math Hard Practice Quiz Answers
9. 4
(Estimated Difficulty Level: 5)
11. C
(Estimated Difficulty Level: 5)
Make a diagram, and fill it in with the information that
You need to know half of the “third-side rule” for trian-
is given. (You should do this for any difficult geome-
gles to solve this question: The length of the third side
try question without a figure.) Since the perimeter of
of a triangle is less than the sum of the lengths of the
square ABCD is x, each side of the square has length
other two sides. For this question, we will make AC the
x/4, so your figure should look something like this:
third side.
x
Now, suppose that the length of each of the other two
4
sides of the triangle is x, so that AB = BC = x. Then,
B
C
F
the third-side rule says that AC is less than the sum of
AB and BC: 7 < x + x. Simplifying gives: 2x > 7 so
x
x
x
x
that x > 3.5. The smallest possible integer value for x
4
4
4
4
is 4.
x
A
D
E
G
4
Now, use the third-side rule for triangles: The length
of the third side of a triangle is less than the sum of
the lengths of the other two sides and greater than the
positive difference of the lengths of the other two sides.
When the rule is applied to EG as the third side, we
10. B
(Estimated Difficulty Level: 5)
get: 0 < EG < x/2. If y is the perimeter of the triangle,
then y = x/4 + x/4 + EG = x/2 + EG. Solving for EG
One way to do this question is to use the fact that the
gives EG = y
x/2. Substituting into the inequality
product of the slopes of two perpendicular lines (or line
gives 0 < y
x/2 < x/2 so that x/2 < y < x, mak-
segments) is
1. The slope of the line segment on the
ing answer C the correct one. To make this problem
left is (a
0)/(2
0) = a/2. The slope of the line
less abstract, it may help to make up a number for the
segment on the right is (0
a)/(10
2) =
a/8. The
perimeter of the square. (A good choice might be 4 so
two slopes multiply to give
1:
that x = 1. You’ll find 1/2 < y < 1, the same as answer
C when x = 1.)
2
a
a
a
·
=
=
1.
2
8
16
2
Solving for a gives a
= 16 so that a = 4. A messier
way to do this problem is to use the distance formula
12. 2 < x < 3
(Estimated Difficulty Level: 5)
and the Pythagorean theorem. The length of the line
√
2
2
segment on the left is
2
+ a
, and the length of line
In order to determine at what point two lines intersect,
segment on the right is
(10
2)
2
+ (0
a)
2
. Then,
set the equations of the lines equal to one another. In
the Pythagorean theorem says that:
this case, we have: 2x
1 = x + c so that x = c + 1.
In other words, x = c + 1 is the x-coordinate of P , the
2 2
2 2
2
2
2
2
+ a
+
(10
2)
+ (0
a)
= 10
.
point where the lines intersect. Now, if c is between 1
and 2, then c + 1 is between 2 and 3. Any value for the
2
Simplifying the left-hand side gives: 2a
+ 68 = 100 so
x-coordinate of P between 2 and 3 is correct.
2
2
that 2a
= 32. Then, a
= 16, making a = 4.
pg. 17
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