Sat Math Medium Practice Quiz With Answer Key Page 14
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16SAT Math Medium Practice Quiz Answers
Geometry
3. B
(Estimated Difficulty Level: 3)
The diagonal of a rectangle divides it into two right tri-
1. D
(Estimated Difficulty Level: 3)
angles; in this case, 30-60-90 triangles. These triangles
are shown on the first page of every SAT math section.
First, let’s try the algebra way: if l is the length and
(Hint: it will help to memorize or be very familiar with
w is the width, then l = w + 4. Since the perimeter is
those formulas and diagrams on the first page.)
16, we know that 2l + 2w = 16, so l + w = 8. We have
two equations and two unknowns: substitute the first
C
equation (l = w + 4) into the second (l + w = 8) to get
60
◦
2
(w + 4) + w = 8 so that 2w = 4, making w = 2. So,
l = 6 and the area is 12.
30
◦
A
D
Another way to go: work with the answers and try to
√
come up easy values for the length and width that mul-
Using that information, CD = 1 and AD =
3 so that
√
tiply to give the area. For example, if the area is 8, the
the area is
3.
length and width could be 4 and 2, respectively, or 8
and 1, but neither one works since we need length to be
4 greater than the width. With an area of 12, length
could be 6 and width could be 2, and this pair works.
4. C
(Estimated Difficulty Level: 3)
Since O is the center of the circle, AO = 6 and OB = 6.
Let’s redraw the diagram, putting in the diameter of the
big circle:
2. D
(Estimated Difficulty Level: 3)
Let x be the area of square A. Remember that when
you increase a number by a percent, you first multiply
the number by the percent (divided by 100), then you
add the original number. So, the equation we need is:
x+0.25x = 80, so 1.25x = 80, which means that x = 64.
6
O
A
B
6
Answer E is tempting at this point, but before you write
down or bubble in your answer to an SAT math ques-
tion, it can be very useful to read the question again
to make sure that you know what the question wants.
In this case, we need the square’s perimeter, not the
area. A square with an area of 64 has a side of 8, so the
perimeter is 32.
The shaded semicircle with diameter OB is equal in
area to the unshaded semicircle with diameter AO. So,
the area of the entire shaded region is just half the area
of the big circle. Since the radius of the big circle is 6,
its area is 36π. The area of the shaded region is then
18π, making answer C the correct one.
pg. 14
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