Download Solutions - FloridaMAO

January Regional Geometry Team Solutions
1. Stewart’s theorem states that (PQ)²(SR)+(QR)²(PS)=(QS)²(PR)+(PS)(PR)(SR). Plugging in and solving gives
QR = 5/
, and applying the theorem again to triangle QST gives ST =
2 3
2.
Each
side
of
the
hexagon
measures
4
meters.
Therefore
the
apothem
=
2 3 .
(
)
Area
of
the
hexagon
=
2 3 i 2 (6 ) = 24 3 3.
(-4) A polygon with 40 sides has 720 diagonals – FALSE
(2)
The area of a trapezoid is equal to the product of its height and the length of its mid-segment - TRUE
(4)
Every square is a rhombus - TRUE
(6)
The measure of an external angle on a regular heptagon is 90 degrees - FALSE
(3)
If the lengths of the sides of a triangle are 50 cm, 120 cm and 130 cm, then the triangle is a right triangle TRUE
(-5)
If the volumes of two similar polygons are in the ratio of 16³ to 49³ then their corresponding surface areas
are in the ratio 4² to 7² - FALSE
(-4) + 6 + (-5) = ‐3
4. Label 3 consecutive vertices of the polygon A, B and C. Now, let BP be the common side to the pentagons
placed on sides AB and BC. Note the measure of angles ABP and PBC is 108°. Since the measure of
angles ABP, PBC and ABC = 360°, angle ABC must equal 144°. A regular polygon with an interior
angle measure of 144° has an exterior angle that measures 36°. There for the polygon must
have 10
sides
.
5. X= 360
Y = 30
Z = 20
A = 11 (orthocenter)
A(X)-Z(Y) = 3960 – 600 = 3360
6.
A = 20°
B= 9m
C = 13 regions
D = 18π cm³
42
+
18π
A+B+C+D=
7.
A
=
504
B
=
10
C
=
60
=
33
(A
+
–
C)B= 4770
8. Let the trapezoid be ABCD with AB = 10 and CD = 15. Let P be the point of intersection of the two diagonals
of ABCD and let
be the segment through P parallel to the bases with X on
Note that ∆PYC ~ ∆ABC, therefore,
two equations together gives
. Also ∆PYB ~ ∆DCB, therefore,
So, PY
same argument will show that PX = 6 as well, so XY = 12
PY
and Y on
.
. Adding these
, hence, PY = 6. The
9. Let A = The sum of two exterior angles of a regular icosagon = 36
Let B = The number of sides that a hendecagon has = 11
Let C = The number of diagonals in a heptagon = 14
Let D = The number of interior angles in a hectagon – 100
=
20
10. a = 105, b = 40, c = 20, d = 75
a + b + c + d = 240
11. Let D = The distance between two points, (-3, 7) and (-12, 3) =
Let H = The length of the hypotenuse in a right triangle with leg measures of 4 and 5 =
Let S = The area of a 135° sector in a circle with area 32π ft² = 12π
Let A = The surface area of a cube with a side length of 5 = 150
Arrange the answers in descending order = 150, 12π,
12.
AD
bisects
BC
since
∆ABC
is
an
equilateral
triangle,
therefore,
CD
=
12.
∆ACD
is
a
30‐60‐90
degree
triangle,
so
AD
=
12
triangle,
so
EC
=
6
and
ED
=
6
.
Likewise,
∆DEC
is
a
30‐60‐90
degree
.
Therefore
AE
=
AC
–
EC
=
24
–
6
=
18.
∆EAF
is
also
a
30‐60‐90
degree
triangle,
so
AF
=
9
and
EF
=
9
30°,
GED
=
60°.
Likewise,
CDE
=
30°
implies
Since
AEG
=
AEF
=
EDG
=
60°.
Therefore
DGE
must
also
measure
60°
and
∆GED
is
equilateral,
so
EG
=
GD
=
ED
=
6
FG
=
EF
–
EG
=
9
–
6
=
3
.
.
FB
=
AB
–
AF
=
24
–
9
=
15
and
BD
=
BC
–
CD
=
12.
Finally,
then,
the
area
of
quadrilateral
BFGD
is
equal
to
the
area
of
∆BFG
+
area
∆BDG
=
B
24
24
F
A
D
G
24
E
C
13. Let’s say the original rectangle has side lengths a and b with a > b. Clearly, we have to cut perpendicular to
a
and b. To be scaled down versions of the
2
a b
original means that the ratio of the side lengths is the same, so we need to have = .
b a
2
the longer side, so the two halves will each have side lengths of
Thus,
a2 2
a
= and =
2
b
1
b
2
1
14. Consider the following diagram, where O is the center of the larger circle. Since the smaller circle has
Q
y
O
P
x
radius of 2, we must have x = 2 and OP = 1. And OQ is a radius of the larger circle, so OQ = 3. Applying the
pythagorean theorem to triangle OPQ we find that 1² + y² = 3². Solving for y gives us y = 2 2 .
The shaded region is a triangle with base 2y and a height of x. Therefore its area is
1
(x )(2 y ) =
2
15.
MN
=
68 + 44
2
=
56
MP
=
12
(triangle
MAP
is
equilateral)
Perimeter
of
ABCD
=
166
Measure
angle
DON
=
90°
4 2