Difficulty:
Medium
NCTM
Standard:
Geometry
Standard
for
Grades
9–12:
explore
relationships
(including
congruence and similarity) among classes of two- and three-dimensional geometric objects, make
and test conjectures about them, and solve problems involving them.
The number
n
satisfies the equation
√
80 +
√
125 =
√
n
. What is the value of
n
?
(A)
205
(B)
305
(C)
405
(D)
500
(E)
1000
2010 AMC 10 B, Problem #6—2010 AMC 12 B, Problem #4—“Note that√80 = 4√5and√125 = 5√5.”
Solution
Answer (C):
Simplifying gives
√
80 +
√
125 = 4
√
5 + 5
√
5 = 9
√
5 =
√
405
. Hence
n
= 405
.
Difficulty:
Medium-easy
NCTM Standard:
Number and Operations Standard for Grades 9–12: understand numbers, ways
of representing numbers, relationships among numbers, and number systems.
The square of
x
is 16 larger than the square of
y
, and
x
is 2 larger than
y
. What is the product
xy
?
(A)4(B)8(C)15(D)24(E)32
2010 AMC 10 B, Problem #7—“Note thatx2−y2= (x−y)(x+y) = 16.”
Solution
Answer (C):
Because
x
2
−
y
2
= (
x
−
y
)(
x
+
y
) = 16
, and
x
−
y
= 2
,
we have
x
+
y
=
16
2
= 8
. Solving the simultaneous equations
x
−
y
= 2
and
x
+
y
= 8
yields
x
= 5
and
y
= 3
. Thus the product
xy
= 5
·
3 = 15
.
Difficulty:
Medium-easy
NCTM
Standard:
Algebra Standard for Grades 9–12:
write equivalent forms of equations,
inequalities, and systems of equations and solve them with fluency-mentally or with paper and
pencil in simple cases and using technology in all cases.
The area of triangle
EBD
is one third of the area of
3
–
4
–
5
triangle
ABC
. Segment
DE
is perpendicular
to segment
AB
. What is
BD
?
A
B
C
D
E
3
4
5
2010 AMC 10 B, Problem #8—“The area of△ABCis12·3·4 = 6.”
Solution
Answer (D):
The area of
△
ABC
is
1
2
·
3
·
4 = 6
, so the area of
△
EBD
is
1
3
·
6 = 2
. Note that triangles
ABC
and
EBD
are right triangles with
an angle in common, so they are similar. Therefore
BD
and
DE
are in
the ratio 4 to 3. Let
BD
=
x
and
DE
=
3
4
x
. Then the area of
△
EBD
can be expressed as
1
2
·
x
·
3
4
x
=
3
8
x
2
. Because
△
EBD
has area 2, solving
yields
BD
=
4
√
3
3
.
OR
Because
△
EBD
and
△
ABC
are similar triangles, their areas are in the
ratio of the squares of their corresponding linear parts. Therefore
(
BD
4
)
2
=
1
3
and
BD
=
4
√
3
3
.
Difficulty:
Medium
NCTM
Standard:
Geometry Standard for Grades 9–12:
analyze properties and determine
attributes of two- and three-dimensional objects.
If a rectangle’s width is increased by 3, the rectangle
becomes a square and the area of the rectangle is
increased by 54.
What is the area of the original
rectangle?
2010 AMC 10 B, Problem #9—“Letxandybe the width and length of the originalrectangle, respectively. Form an equation to show thedifference in areas.”
Solution
Answer (E):
Let
x
and
y
be the width and length of the original rectangle,
respectively. Increasing the width by 3 is equivalent to attaching a rectangle
with dimensions 3 and
y
, as shown. The extra area is 54, so
3
y
= 54
and
y
= 18
. Then
x
= 15
and the area of the original rectangle is
15
·
18 = 270
.
x
y
3
Difficulty:
Medium
NCTM
Standard:
Geometry Standard for Grades 9–12:
analyze properties and determine
attributes of two- and three-dimensional objects.
On the way to the fair, Paul, Sue, and Ellen compare
how much money they have.
Paul has 20 percent
more than Sue and 25 percent more than Ellen. Ellen
has
x
percent less than Sue. What is
x
?
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- Winter '13
- Kramer
- Platonic solid, United States of America Mathematical Olympiad
