PSAT Math
Issue 1
Super Mathter
May 9, 2003
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AREA AND PERIMETER .............................................................................................................. 2
COORDINATES ............................................................................................................................. 7
–1–
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PSAT Math
3. A square is formed by connecting two
equal rectangles as below. The area of
the square is 36 in2, what is the
perimeter of one rectangle?
Area and Perimeter
length
width
width
rectangle
Issue 1
length
perimeter = 2×(width + length)
area = width × length
Area of a trapezoid = 1(height)(top+base)
top
height
base
A
1. A picture frame has inner dimensions
of 20 in by 10 in, the width of the
frame is 1 in, find the area of the frame.
4.
E
B
C
D
(a) Find the area of ∆BDE if the area of
£ACDE is 16.
1 in.
20 in.
10 in.
2. A rectangle is formed by connecting
three equal squares. The rectangle has a
perimeter of 24 in. What is the area of
the rectangle?
(b) Find the area of £ACDE if the area
of ∆BDE is 16.
–2–
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PSAT Math
5. When a certain rectangle is divided into
two halves, two squares are formed. If
each of these squares has perimeter 48,
what is the perimeter of the original
rectangle?
6. Two squares are shown below. Square
B has a side 20 in and Square A has a
smaller perimeter than Square B by 20
in. What is the difference of the areas of
the two squares?
9. ABCD is a trapezoid. AD = 8 cm, AB
= 12 cm.
20 in
B
A
Issue 1
8. Given right triangles, I, II, III, IV, and
V. The following are lengths of legs of
each triangle:
I) 7 and 4
II) 12 and 2
III) 8 and 3
IV) 6 and 4
V) 24 and 1.
Which of these have the same area?
D
C
A
M
B
(a) If M is a midpoint of AB, then
area(ABCD) =
7. In the rectangle ABCD several triangles
are drawn.
A
D
E
B
C
All the following triangles have
identical areas EXCEPT
A) ADC
B) EDC
C) ACB
D) AED
E) BCD
(b) If area(∆MBC) = area(AMCD),
then area(ABCD) =
–3–
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PSAT Math
10. The cross shape in the figure is formed
by putting together 5 identical squares.
A
D
Issue 1
B
(d) The area of the cross shape is 80 sq.
inches, then the perimeter of the square
ABCD=_________.
C
(a) The area of the cross shape is 80.
Find the perimeter of the cross =
_________.
11. As the figure below, sides AD and BC
of rectangle ABCD pass through the
cents of circles I and III. If the
circumference of each circle is 7π, what
is the area of the rectangle ABCD?
A
(b) The perimeter of the shape is 36 in.,
then the area = _________.
B
I
II
III
C
D
12. Two identical rectangles with width =
5 and length = 25 are crossed to form a
cross.
(c) The area of the cross shape is 50 sq.
inches, then the area of the square
ABCD=_________.
25 cm
5 cm
5 cm
–4–
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PSAT Math
(a) What is the area of the figure?
Issue 1
(b) What is the area of the uncovered
square?
14. Use the figure below to answer the
following questions.
(b) What is the perimeter of the figure?
20
5
15
25
5
(a) What is the area of the shape?
13. A square is to be decorated by covering
smaller squares around its sides. The
dimensions of the two squares are
shown in the figure.
5 in
40 in
(b) What is the perimeter of the shape?
(a) How many such smaller squares are
needed in total?
15. A square has an area of 100 sq. in. If
each side is increase by 5 in, what is the
increase of the area?
–5–
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PSAT Math
16. A rectangle has a perimeter of 120 in
and a length of 40 in.
Issue 1
17. Find the shaded area of each of the
following figures.
16 in.
(a) What is the original width?
12 in.
16 in.
(a)
(b) If each side has the same increase,
and the perimeter increases 40 in. What
is the increase of each side?
Each circle with radius = 1 cm.
(b)
(c) What are the new length and new
width?
5 in.
5 in.
(d) What is the increase of area?
10 in.
(c)
–6–
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PSAT Math
18. Square A has an area of 36 in2. Square B
has a perimeter shorter than Square A
by 12 in. What is the area of Square B?
A
Issue 1
Coordinates
21. On the number line shown below, the
length of YZ is how times the length of
XY?
X
Y
Z
B
22. If AB = BC in the figure below, what
is the x-coordinate of point B?
19. A kitchen has the following
dimensions. If the kitchen to be
covered 2×2 in2 tiles, how many pieces
are needed?
(Hint: 1 ft = 12 in)
(1, 4) (x, y) (11, 4)
A
B
C
2 in
2 in
3 ft
4 ft
23. In the figure below, the slope of the
line that passes through points A and B
is -3/5. What is the value of r?
20. Find the area of the shaded polygon.
Y
A( 2,r )
12 cm
3 cm
B( 7,1 )
10 cm
3 cm
X
O
10 cm
–7–
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PSAT Math
24. What is the area of the region bounded
by the five darkened line segments in
the figure below?
Issue 1
27. What is the area of a triangle with
vertices (2, 3), (8, 3), and (13, 6) in the
XY-plane?
(3, 5)
(0, 3)
(5, 3)
O
(5, 0)
28. Five points A, B, C, D, and E, lie on a
line, not necessarily in that order.
Segment AB has a length of 24. Point C
is the midpoint of AB, and point D is
the midpoint of segment AC. If the
distance between D and E is 5, what is
the one possible distance between A
and E?
25. If line segment RT above has length 5,
what is the value of k?
6k
4k
R
S
T
O
26. In the figure below, the slope of line L
is J. What is the value of b?
x
y
p
t
Y
S(4, 9)
29.
L
R(-2, b)
X
In the circle with center O above, the
two triangles have legs of lengths x, y,
p, and t, as shown. If x2 + y2 + p2 + t2
= 72, what is the circumference of the
circle?
A) 8π
B) 9π
C) 12π
D) 24π
E) 36π
–8–
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PSAT Math
P
30.
Issue 1
D) -1
E) -1/4
T
L
In the figure above, points P and T lie
on line L. How many different points
on L are twice as far from point T as
from point P?
A) None
B) One
C) Two
D) Four
E) More than four
A(4.5, 3.5)
33.
31. Points A, B, C and D lie on a number
line in that order. Given the lengths
BD =11, AC = 8, and CD = 6, what is
the length of AB?
A) 3
B) 4
C) 5
D) 6
E) 7
In the coordinate graph system above, a
circle is to be drawn with its center at
point A. What is the least possible
radius circle A can have if circle A
includes parts of at least three of the
quadrants of the system?
A) 2
B) 4
C) 5
D) 6
E) 7
Y
II
P
32.
0
I
O
1/8
III
34.
On the number line above, the marks
are equally spaced. What is the
coordinate of point P?
A) -3/2
B) -11/8
C) -5/4
X
IV
In the above figure, which quadrants
contain pairs (x, y) so that xy is
negative?
A) II only
B) III only
–9–
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PSAT Math
C) I and III only
D) II and IV only
E) I, II, III and IV
Issue 1
D) 17
E) 389
x3
37.
C
B
A
35.
D
On the graph above, a circle is to be
drawn with an area of 25π. The center
of the circle is at (2, -1). Which of the
four points shown on the graph could
be located within the circle?
A) A and B only
B) B and D only
C) C and D only
D) A, B, and D
E) A, B, C and D
x2
x
If x, x2, x3 lie on a number line in the
order shown above, which of the
following could be the value of x?
A) -2
B) - 1
C) H
D) 1
E) G
38. In a coordinate graph system, a circle is
drawn whose center is at the origin and
whose radius is 4. All of the points
described by the following coordinates
will fall within the circle EXCEPT
A) (0, 3)
B) (-2, -2)
C) (3, 4)
D) (-1, 2)
E) (-3, 0)
36. In a coordinate graph system, a triangle
is drawn with vertices located at points
with coordinates (-2, -1), (10, 4), and
(-2, 4). What is the length of the longest
side of the triangle?
A) 12
B) 13
C) 5 20
– 10 –
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PSAT Math
Y
Issue 1
(2, 8)
(2, 3)
39.
X
In figure above, the line segment
joining the points (2, 3) and (2, 8) forms
one side of a square. Which of the
following could be coordinates of
another vertex of that square?
A) (-2, 5)
B) (-2, 3)
C) (5, 2)
D) (7, 2)
E) (7, 8)
– 11 –
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PSAT Math
Issue 1
Super Solver
Given a ∆ADC, sliding the vertex A
along a parallel line AB (AB//C
D) will not change the area of the
triangle. That is, the area of
∆ADC=area of ∆EDC=area of ∆BDC.
(See the figure below.)
Area and Perimeter
1. 64 in2
22×12 - 20×10 = 64
2. 27 in2
The perimeter of the rectangle is 8
times that of a square. The length of
each side of a square is 3. Thus, the area
of a square is 32 = 9, the area of the
rectangle is 3 times the square area,
which equals to 9×3 = 27 in2.
A
D
C
9. (a) 72 cm2
The area of a trapezoid is
1(top+base)×height. The area of the
ABCD is 1(6+12)×8=72.
(b) 64 cm2
Since area(∆MBC) = area(AMCD), AM
: MB = 1 : 2, therefore, AM = 4 and
MB = 8. The area of the trapezoid =
1(8)(4 + 12) = 64
4. (a) 8
area(∆BDE) = 1area(£ACDE) = 1(16)
=8
(b) 32
16×2 = 32
10. (a) 48
4×3×4 = 48
(b) 45
36÷4÷3 = 3 ⇒ 32×5 = 45
(c) 90
5. 72
12
12
12
12
B
8. II, III, IV, and V
3. 18 in
36= 62, so each rectangle has a length of
6 in and a width of 4 in. The perimeter
of a rectangle is 2(3+6) = 18 in.
12
E
12
9
50× 5 = 90
(d) 48 in
80÷5 = 16 = 42 ⇒ Each side of ABCD
is 4×3 = 12, the perimeter is 12×4 = 48
in.
6. 175 in2
Square A has a side of 20- 20÷4 = 15,
202 - 152 = 400 - 225 = 175 in2.
7. D
This problem is important since it
requires the notion of invariance.
11. 98
The diameter of each circle = 7, thus
– 12 –
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PSAT Math
the rectangle has a width of 7 and a
length of 14. The area of the rectangle
= 7×14=98.
12. (a) 5×25 = 125, 125 + 125 = 250, The
overlapping part is 5×5=25, so 250-25
= 225.
(b) 25 - 5 = 20, the perimeter is 4(10 +
5 + 10) = 100 cm.
Issue 1
area = 8×4=32. There are 8 circles.
Thus, the shaded area = 32 8×3.14=32-25.12=6.88 cm2.
(c) Let cut a quarter circle to fill the
missing corner of the square. The area
of the semicircle is 1π(52) = 39.25 in2.
The area of the square is 102 = 100 in2.
The total area of the shaded part
becomes 139.25 in2.
13. (a) 40÷5 = 8, 8×4 = 32, 32- 4 = 28
(smaller squares)
(b) The side of the uncovered square is
40 - 2×5 = 30, 30×30 = 900 in2
5 in.
5 in.
5 in.
5 in.
10 in.
14. (a) The width of the letter is 5 (20-15).
The total area = 5×25 + 5×15 +
5×15=5×55 = 275.
18. Each side of Square A is 36 = 6×6, its
perimeter is 4×6 = 24. The perimeter
of Square B is 24 - 12 = 12, each side is
12 ÷ 4 = 3 in, the area is 3×3 = 9 in2.
5
15
25
5 in.
5 in.
5
19. 4 ft = 48 in, 48 ÷ 2 = 24, 3 ft = 36 in,
36÷2 = 18, 24×18 = 432 (pieces)
(b) The perimeter is 2×(5 + 25) +
2×(5+15)×2 - 10×2 = 60 + 80 - 20 =
120
20. Complete the trapezoid by adding a
square to the corner, which has an area
of 9 cm2. The area of the trapezoid is
1(10+15)×13=162.5 cm2. Thus, the
shaded part has an area = 162.5 - 9 =
153.5 cm2.
15. Since 10×10 = 100, each side is
increased by 5, then it becomes 5+10 =
15 in, the new area is 15×15 = 225 sq.
in, so the difference is 225 - 100 = 125
sq. in.
16. (a) 120÷2 = 60, 60 - 40= 20 (width)
(b) Since each has the same increase,
40÷4 = 10.
(c) 40+10 = 50 (new length), 20+10 =
30 (new width)
(d) 40×20 = 800 (old), 50×30 = 1500
(new), the increase = 1500 - 800 = 700
sq. in.
12 cm
10 cm
3 cm
10 cm
3 cm
Coordinates
21. 2.5
22. 6
17. (a) The inner square = 162 = 256. The
four radiating triangles =
4×(1×16×12)=384. The shaded area =
256 + 384 = 640 in2.
(b) The diameter = 2 cm. The rectangle
23. 4
24. 20
– 13 –
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PSAT Math
25. 1/2 or .5
10k=5, so k=0.5.
-10×7 = -
26. 1
The slope formula is
∆y
∆x
=
9− b
4+ 2
=
27. 9
1(base×hieght) = 1(6×3) = 9
28. 1 or 11
See the diagram below.
34. D
E
A
D
C
5
5
6
B
6
35. E
Each of four points is located within a
distance less than 5. (Use Pythagorean
theorem to verify.)
12
29. C
Knowing that x2+y2=p2+t2 = r2, So,
2r2 = 72, r2=36, so r = 6, thus the
perimeter is 12π.
36. B
This is a right triangle. The longest side
is the hypotenuse, the side between (-2,
-1) and (10, 4) so the length is
30. C
There are two desired points S and S'.
As the figure below shows, S is an ideal
point since PS:ST=1:2. S' is another
ideal point since PS':S'T=1:2.
122 + 52 = 13
37. C
38. C
The length of (3, 4) = 5, which exceeds
the radius, 4.
2
1
P
T
L
S
S'
5
4
33. C
Key idea of this problem is the distance
of A to the origin ≈ 5.7. When radius =
2, or 4 is too short, it can only include
at most 2 quadrants because the circle
cannot reach the Y-axis. When radius 6
and 7 are too long, it will include four
quadrants since the circle will include
the origin. The only choice is C.
9− b
6
= J ⇒ 3(9 - b) = 24 ⇒ 27 - 3b = 24 ⇒
3b = 3 ⇒ 1
E
Issue 1
32. C
1
39. E
2
31. A
AB=3. See the figure below.
3
11
B
A
C
8
6
D
– 14 –
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