Download m B xm D x ∠ = + ∠ = +

GEOMETRY TEAM #1
FAMAT January Regional
A = the m∠C in parallelogram ABCD with m∠B = (4 x + 15), m∠D = (6 x + 2)°
B = the degree measure of the smallest angle in triangle ABC with
m∠A = ( x + 20)°, m∠B = ( x + 7)°, m∠C = (2 x − 15)°
Find the value of A − B.
GEOMETRY TEAM #2
FAMAT January Regional
A = the number of diagonals for a regular polygon with 24 sides
B = the measure, in degrees, of one interior angle in a regular dodecagon
C = the sum of the exterior angles of a regular octagon, one at each vertex
D = the length of the apothem in a regular hexagon having a side of 6
Find the value of A + B − C − D2 .
GEOMETRY TEAM #3
FAMAT January Regional
A = the degree measure of the smaller angle formed by the hands of a clock at 4:36
B = the maximum number of unique scalene triangles having all sides of integral lengths and perimeter less than 13
C = the sum of the possible integral lengths of the third side of a triangle having the other two sides of 3 and 8
Find the value of A − (C + B ) .
GEOMETRY TEAM #4
FAMAT January Regional
A = the complement of the smaller angle when the same side interior angles of two parallel lines have
measures of x° and ( 3x − 20 ) °
B = the length of EF when E is the midpoint of DF and DE = 2 x − 3, DF = 7 x − 27
suur suur
C = the m∠RST in the diagram when RX || TY , m∠R = 118°, m∠T = 159°
D = the m∠BAC in ΔABC when m∠B = 26°, the exterior angle at ∠A is x°
and the exterior angle at ∠C is 138°
Find the value of ( A + D ) − ( B + C ) .
GEOMETRY TEAM #5
FAMAT January Regional
In the diagram, m∠R = 30°, RS = 5, m∠TSP = 45°
and PT is an altitude of ΔRSP .
Let the perimeter of ΔRPT be A + B 3 .
Find the value of ( A − B ) .
2
FAMAT January Regional
GEOMETRY TEAM #6
The degree measure of the larger angles of a rhombus are double the degree measures of the smaller angles. The shorter
diagonal has a length of 10.
A = the length of a side of the rhombus
B = the length of the longer diagonal
C = the degree measure of the larger interior angle
D = the perimeter of the rhombus
Find the value of
C B2
.
+
D A
GEOMETRY TEAM #7
FAMAT January Regional
A = the length of the hypotenuse of a right triangle when the sum of the squares of the lengths of the three sides
of the triangle is 200
B = the perimeter of a rectangle when the width is 12 and the diagonal is 6 more than the length
Find the value of A + B .
GEOMETRY TEAM #8
FAMAT January Regional
Consider the following statements.
If the statement is true, assign it a value of 2.
If the statement is false, assign it a value of −1 .
Find the sum of the values.
I.
Through any two points there is exactly one plane.
II.
Two intersecting lines are coplanar.
III.
At least one line can be drawn through any three points.
suur
IV.
If points P and Q lie in a plane, then PQ lies in that plane.
V.
Two triangles are congruent if the measure of the three angles of one are equal to the
measure of the corresponding parts of the other.
GEOMETRY TEAM #9
FAMAT January Regional
Given regular octagon ABCDEFGH.
A = the length of HC if the side of the octagon is 1
B = the measure of one interior angle of the octagon
C = the complement of one exterior angle of the octagon
D = the area of trapezoid ABCH
Find the value of A + B − C − 2 • D .
GEOMETRY TEAM #10
FAMAT January Regional
Given ΔFHJ , point K is on FJ and G is a point on HF so that GK || HJ .
FG = x − 2, HJ = y, GH = 9, KJ = x + 3, FK = 4, GK = 5 . Find the perimeter of ΔHJF expressed as a simplified improper fraction.
GEOMETRY TEAM #11
FAMAT January Regional
A = the length of an angle bisector in an equilateral triangle with a side having a length of 8
B = the least value of diagonal AC in rectangle ABCD with a perimeter of 20
Find the value of
A2 + B 2
.
2
GEOMETRY TEAM #12
FAMAT January Regional
A side of one equilateral triangle is congruent to the altitude of a second equilateral triangle.
Find the ratio of the perimeter of the larger triangle to the perimeter of the smaller triangle.
GEOMETRY TEAM #13
In the diagram, m∠AEC = x 2 − 6 x, m∠BED =
FAMAT January Regional
1
x + 42 .
2
Find the m∠AEB expressed as a simplified mixed number.
GEOMETRY TEAM #14
FAMAT January Regional
Which of the following has the largest perimeter?
Your answer will be the name of the figure, either square, or triangle, or hexagon, spelled correctly of course.
1. A square with a diagonal of 4 2
2. An isosceles triangle with legs having length 2 3 and base angles of 30°
3. A regular hexagon with apothem 3
GEOMETRY TEAM SOLUTIONS
#1
#2
#3
90
15
35
#8
#9
#10
#4
#5
58
25
#11
#12
#6
33
#7
52
FAMAT January Regional
1
90
169
4
7
2 3
3
#13
3
132
4
#14 Hexagon
1. 90
A: 139 B: 49
A − B = 139 − 49 = 90
B and D are opposite angles so they are congruent. 4 x + 15 = 6 x + 2, x =
A:
13
, m∠B = m∠D = 41 .
2
Angle C is the supplement of B and D so its measure is 139 ° .
B: The sum of the angles is 180° . 4 x + 12 = 180, x = 42, m∠A = 62, m∠B = 40, m∠C = 69 . The smallest angle is 49° .
2. 15
A:
A: 252
B: 150 C: 360
( n − 3) n = 24 • 21 = 252
2
2
( )
2
A + B − C − D2 = 252 + 150 − 360 − 3 3 = 15
D: 3 3
B: One exterior angle of a 12 sided polygon is
360
= 30 . One interior angle would be the
12
supplement of 30, which is 150.
C: The sum of the exterior angles of any polygon, one at each vertex, is 360 ° .
D: Draw a triangle from a vertex of the hexagon to the center of the hexagon and the foot of the apothem on the side of
1
the triangle. This is a 30-60-90 triangle. The hypotenuse is 6, the apothem is opposite a 60° angle and is
3 times the
2
hypotenuse, 3 3 .
3. 35
A: 78
B: 3
C: 40
A − (C + B) = 78 − ( 40 + 3) = 35.
60h − 11m 60 • 4 − 36 • 11
=
= 78 B: Triangles with side lengths of : 2,5,4;3,4,5;2,3,4.
2
2
C: The third side must solve 5 < x < 11 which would be 6,7,8,9,10 and the sum is 40.
A:
4. 58
( A + D ) − ( B + C ) = (40 + 112) − (11 + 83) = 58
A: These two angles are supplementary so 4 x = 200, x = 50 and the smaller angle is 50. The complement is 40.
B: Double DE and set it equal to DF. 4 x − 6 = 7 x − 27, x = 7, DE = EF = 11
suur
suur
C: Draw a line through S parallel to RX and TY . Then use same side interior angles.
The angle on the upper side of the parallel is 62 and the angle on the under side is 21.
The sum is 83.
D: Exterior + interior angle of a triangle = 180. m∠BCA = 42 , and the sum of the remote interior
angles equals the exterior angle so 42 + 26 = x, x = 68 , m∠BAC = 180 − 68 = 112.
Let PT = ST = x, SP = x 2, RP = 2 x.
5. 25
Using Pythagorean Theorem in ΔRPT , we have
x 2 + ( x + 5) = ( 2 x ) , and solving this gives x =
2
2
5+5 3
. The perimeter is x + x + 5 + 2 x which is 4 x + 5 .
2
⎛5+5 3 ⎞
20 + 20 3 10 30 + 20 3
4⎜
+ =
⎟+5=
, which simplifies to 15 + 10 3 .
2
2
2
2
⎝
⎠
( A − B)
2
6. 33
2
B: 10 3
A: 10
7. 52
= (15 − 10) = 25.
A: 10
C: 120
D: 40
C B 2 120 300
=
+
= 3 + 30 = 33
+
40 10
D A
A + B = 52
B: 42
A: a + b + c = 200, a + b = c2 . Substituting c 2 for a 2 + b2 in the first equation gives
2
2
2
2
2
2c 2 = 200, c = ±10, reject the −10 .
B: Let the length of the rectangle be x and the diagonal x + 6 . Using Pythag,
122 + x 2 = ( x + 6) , x = 9, −8 , reject −8 . Length is 9, width is 12, perimeter is 42.
2
8. 1
I. is false (there is exactly one line)
II is true
III. is false (At least one line can be drawn through any 2 points)
IV is true
V is false (these triangles are similar)
9. 90
After drawing the octagon and HC , draw the altitudes from A and B to HC . This gives us a trapezoid
with upper base and legs all equal to 1. Since the triangles formed by drawing the altitudes are 45-45-90,
the length of the altitude is
A: The length of HC is
2
.
2
2
2
+1+
= 1+ 2 .
2
2
B: One interior angle of the octagon is 135° . The exterior angle is
360
= 45° and the interior angle is the
8
supplement of that.
C: The complement of an exterior angle, which is 45, is 45.
D: The area is
1
2
2 +1
.
•
1+ 2 +1 =
2 2
2
(
)
⎛ 2 +1⎞
The value of A + B − C − 2 • D = 2 + 1 + 135 − 45 − 2 ⎜
⎟ = 90 .
⎝ 2 ⎠
10.
169
4
ΔFKG ~ ΔFJH by AA therefore sides are proportional.
FG FK x − 2
4
=
,
=
, solving gives x = 6, −7 ,reject the −7 . This makes
GH KJ
9
x−3
FG = 4, FH = 13, KJ = 9, FJ = 13 .
To find JH,
11. 7
KG 4 5 4
65
65 169
.
= , = , y = . The perimeter of ΔHJF = 13 + 13 + =
4
4
JH 13 y 13
4
A2 + B 2
48 + 50
=
=7
2
2
A: The angle bisector in an equilateral triangle is also a median and an altitude.
A: 4 3
B: 5 2
Draw the segment and find it is opposite a 60 degree angle which would be 4 3.
B: The perimeter is 20 which makes the sum of a width and a length 10. By trial and error, using
different lengths for the width and length, such as 1 and 9, the diagonal would be
the diagonal would be
82 using 2 and 8,
68 so you can see the diagonal is getting smaller. So, trying 5 and 5 makes the
diagonal 5 2 . 12.
2 3
3
Let the length of a side of one triangle be x and the altitude of the other x . In the 2nd triangle, the altitude
is opposite a 60 degree angle so to find the side of the triangle divide by
the side
3 and multiply by 2 making
2x 3
. This is the larger side and we want the ratio of the larger to the smaller. Since the ratio
3
2x 3
3 =2 3.
of the sides is the same as the ratio of the perimeters, we have
x
3
13. 132
3
4
These two angles are vertical so they are congruent. x 2 − 6 x =
2 x 2 − 13x − 84 = 0, ( 2 x − 21)( x + 4 ) = 0, x = −4 or
14. Hexagon
1
13
x + 42, x 2 − x − 42 = 0,
2
2
21
2
Square – The diagonal is 4 2 making the side 4 and the perimeter is 16.
Triangle – The legs are 2 3 , drawing the altitude and it is opposite a 30 degree angle,
1
the altitude is the hypotenuse which is 3 . Half of the base, since it is opposite a 60 degree angle, is
2
3. And the base is 6 making the perimeter 6 + 4 3 .
Hexagon – Apothem is 3 is opposite the 60° angle, divide by
the hexagon that has a perimeter of 12 3 .
Which is largest? hexagon
3 to get
3 . Double this to get the side of