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Tampa Bay Tech Invitational
Geometry Individual
For all questions, E. NOTA means “none of
the above answers is correct”. Diagrams are
NOT drawn to scale.
1. Points A, B, and C are collinear, but they do not
necessarily lie on a line in the order named. If
AB  5, BC  3, what is the length of segment AC ?
A. either 2 or 8
C. 2
January 11, 2014
4. Which proportion is not equivalent to
a c
 ?
b d
(no denominator is equal to 0)
a b

c d
a c
D. 
d b
A.
B.
b
d

ab cd
C.
b d

a c
E. NOTA
B. either 2 or 4
D. 8
E. NOTA
5. In ABC with right angle at B, D is a point on
AC such that BD  AC , AD  6 ,
2. Which statement would not guarantee that A
and B are congruent?
A. A and B are adjacent angles formed by
perpendicular lines.
BD  4 3 . Find the length of BC .
A. 4 7
B. 6 3
D. 2 21
E. NOTA
C. 6 7
B. A and B are vertical angles.
C. A and B are both complements of C.
D. A and B are same side interior angles formed
by two lines and a transversal.
6. The median to the hypotenuse of a right triangle
divides the triangle into two triangles that are both
E. NOTA
A. similar
D. isosceles
3. In ABC , point D is on AB , point E is on AC
B. right
E. NOTA
C. scalene
AB  2 x  1, AE  2 x  2, EC  x  1 , find the
7. Given rectangle ABCE and right isosceles
triangle DEC with common side EC.
DE  DC  EA  BC  x . If the perimeter of
length of AC .
ABCDE = 16  4 2 , find the value of x.
such that DE || BC . If AD  x  2
A. 7
D. 15
B. 8
E. NOTA
C. 12
A. 1
B. 2
D. 4
E. NOTA
C. 3
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Geometry Individual
8. Given: m1   x 2  3 y  , m2   20 y  3  ,
m3   3 y  4 x  , x  0. Find the m1 in degrees.
1
12. The perimeter of ABCD is 85. If
AB  3x  4, BC  4 x, CD  6 x  11, and
AD  5x  7, find the length of AB .
2
3
A. 5.5
D. 42.5
A. 37
D. 67
B. 40
E. NOTA
January 11, 2014
B. 20.5
E. NOTA
C. 21.5
C. 53
13. Given parallelogram ABCD with
mA  x, mD   3x  4  . Find mB  mD.
9. AB  CD at point E.
mAEC  ( x  2 y), mAED   x  2 y   ,
mDEB  50 . By how much does x exceed y in
degrees?
A. 70
B. 90
D. cannot be determined
C. 110
E. NOTA
A. 46
D. 268
B. 134
E. NOTA
C. 180
14. The perimeter of equilateral triangle ABC is
36 3 . The bisectors of  A and  B meet at D.
Find the length of AD .
10. Which group of numbers can be the lengths of
the sides of an obtuse triangle?
A. 3,5,7
3 5
D. ,1,
4 4
B. 2,4,6
A. 6
B. 6 3
D. 12 3
E. NOTA
C. 12
C. 0.5,0.6,0.7
E. NOTA
15. Point X is equidistant from vertices T and N of
TEN . Point X must lie on which of the following?
A. bisector of E
11. Which information does not prove that
quadrilateral ABCD is a parallelogram?
A. AC and BD bisect each other
B. median to TN
C. perpendicular bisector of TN
D. altitude to TN
E. NOTA
B. AD || BC ; AD  BC
C. AB || CD; AD  BC
D. A  C; B  D
E. NOTA
16. One of two complementary angles is twice the
other, Find the supplement of the smaller angle.
A. 30
D. 150
B. 60
E. NOTA
C. 120
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Geometry Individual
January 11, 2014
17. ABCD is an isosceles trapezoid with upper base
22. In ZYW , M is a point on WZ and N is a point
AD and diagonals intersecting at point E. Given
BE  x  7, CE  y  3, AE  x  5, BD  y  4. Find
on YZ such that NM || YW . If
the numerical length of AC .
A. 10
D. 16
B. 12
E. NOTA
C. 14
18. If the sum of the measures of the interior angles
of a polygon is increased by 900, how many sides
will have been added to the polygon?
A. 4
D. 10
B. 5
E. NOTA
C. 9
19. The lengths of the diagonals of a rhombus are in
the ratio 3:4 and the perimeter is 40. Find the sum
of the lengths of the diagonals.
A. 14
D. 28
B. 16
E. NOTA
C. 24
ZY  2 x  9, ZM  10, ZN  x  3, and MW  x, then
find the value of x.
A. 12
D. 12
B. 5
E. NOTA
C. 2  34
23. The inverse of the contrapositive of the
converse of a conditional statement is equivalent to
the __?__
A. inverse
B. converse
D. conditional
C. contrapositive
E. NOTA
24. The average lengths of the sides of ABC is 14.
If AB  x  5, BC  2 x  1, and AC  4 x  6, how
much longer than the average is the longest side?
A. 2
D. 24
B. 4
E. NOTA
C. 22
25. HGF is equilateral. If the perimeter is 6 y  24
20. In PRT , S is a point on PT such that RS is an
angle bisector. If PS  6, ST  8, and the perimeter
of PTR is 42, find the length of PR .
A. 12
D. 29
B. 12.75
E. NOTA
perimeter is 4 18,
B. 3 12
E. NOTA
HGF.
A. 38
D. 120
B. 72
E. NOTA
C. 114
C. 18
21. Find the length of the diagonal of a square if the
A. 2 18
D. 72
and HG  3 y  7, find the numerical perimeter of
C. 36
26. One side of a triangle is 4 cm shorter than a
second side. The ray bisecting the angle formed by
these sides divides the opposite side into 4 cm and 6
cm segments. Find the perimeter of the triangle.
A. 30
B. 32
D. cannot be determined
C. 44
E. NOTA
Tampa Bay Tech Invitational
Geometry Individual
27. Each interior angle of a regular polygon
measures 160 and each side has length 5. Find the
perimeter of the polygon.
A. 20
D. 90
B. 55
E. NOTA
C. 85
28. In a trapezoid, the length of the bases are 4 and
16, and the lower base angles have measures 30
and 60 . Find the distance between the midpoints
of the bases.
A. 2 3
B. 3 3
D. 12 3
E. NOTA
C. 6
29. Regular hexagon ABCDEF has side AF in
common with regular hexagon AFGHIJ, and side
BC in common with regular hexagon BCKLMN.
All three hexagons are coplanar and nonoverlapping. If AB = 64, find the length of JN .
A. 64
D. 128
B. 80
E. NOTA
C. 96
30. In right triangle ABC, the altitude to the
hypotenuse AB is drawn intersecting the
hypotenuse at point E. If BC  3 5, BE  5 , find
the length of AC .
A. 3
D. 9
B. 4
E. NOTA
C. 6
January 11, 2014
Tampa Bay Tech Invitational
Geometry Individual
SOLUTIONS
1. A
2. D
3. E
4. D
5. A
6. D
7.
D
Let the legs of the triangle be x
which makes the width of the rectangle x. EC has
7. D
8. A
9. C
10. A
11. C
12. B
13. D
14. C
15. C
16. D
17. D
18. B
19. D
20. A
21. E
22. D
23. D
24. B
25. C
26. A
27. D
28. C
29. D
30. C
1.
A
There are two possibilities. With AB-C, B is between A and C and the length of AC
would be 8. With A-C-B, C is between A and B and
the length of AC is 2.
2.
3.
D
January 11, 2014
A. Both would be right angles and 
B. Vertical, congruent.
C. Both complements of 90 are
congruent.
D. Same side interior angles are
supplementary.
E
The triangles are similar by AA so
2x  2 x  2

, x  7,0 .
the sides are proportional.
x 1
x 1
Reject the 0. This makes AC equal to 20.
4.
D
Check all the rules of proportions or
put numbers in and see if the cross products are the
same.
5.
A
The triangles are similar by AA so
the geometric mean formulas can be used.
DC  6  4 3, DC  8, BC  8 8  6  4 7 .
6.
D
The median to the hypotenuse is ½
the length of the hypotenuse so the triangles both
have 2 sides congruent making the triangles
isosceles.
length x 2 . So the perimeter is 4 x  x 2. Which
now gives 4 x  x 2  16  4 2 ,


x 4  2  16  4 2, solving this gives x  4.
8.
A
Use systems:
x 2  3 y  20 y  3  180, x 2  23 y  177.
20 y  3  3 y  4 x  180, 4 x  23 y  177.
x 2  23 y  4 x  23 y, x 2  4 x  0, x  4,0.
Reject the 0 since the question says x  0.
Substituting x  4 in x 2  23 y  177 gives y  7.
m1  x 2  3 y  16  21  37.
mAEC  mAED  180, so we
9.
C
have x  2 y  x  2 y  180, x  90. And angles
AEC and DEB are vertical so x  2 y  50,
substituting for x, 90  2 y  50, y  20. x exceeds
y by 110, 90   20  110.
10. A
Use c 2  a 2  b 2 which makes the
triangle isosceles.
A. 49  9  25 , obtuse
B. This is not a triangle. 2  4  6 .
C. 0.49  0.25  0.36 , acute
25 16 9
  , right
D.
16 16 16
11. C
Follow the rules for proving a quad
is a parallelogram.
A. Yes, diagonals bisect each other.
B. Yes, one pair of sides is both parallel and
congruent.
C. No.
D. Yes, opposite angles are congruent.
12. B
Find the value of x by adding the
sides and setting equal to the perimeter 85.
41
 11 
18x  11  85, x  5.5. AB  3    4 
 20.5 .
2
2
Tampa Bay Tech Invitational
Geometry Individual
13. D
Consecutive
angles
in
a
parallelogram
are
supplementary.
4 x  4  180, x  46.mD  134, mD  mB  268.
14. C
After drawing the diagram, ABD is
an isosceles triangle with base angles of 30 .Draw
the altitude from D to AB forming 2 30  60  90
triangles. The length of a side of the original
January 11, 2014
20. A
Since the perimeter is 42 and the
sum of the lengths of PS and ST is 14, the sum of
the other two sides is 28. Use the Triangle Angle
Bisector Theorem to get the proportion:
x 28  x

, x  12 .
6
8
21.
E
Simplifying the perimeter is 12 2 .
One side of the square is 3 2 . The diagonal would
be
2 times the length of the side 3 2 which is 6.
triangle is 12 3 . That makes the base of each of
the 30-60-90 triangles have a length of 6 3 . Since
it is opposite a 60 angle, divide by 3 to get the
length of the altitude DE which is 6. Double that to
get the length of AD – 12.
22. D
Triangles are similar by AA.
Sides are proportional.
x3
10

, x  12, 5 .
2 x  9 x  10
15.
23.
C
By definition, this point would be on
D
q  p,~ p ~ q, p  q, conditional
the perpendicular bisector of TN .
16.
Let the angles be x and 90  x .
D
x  2  90  x  , x  60 and the complement will be
30. The supplement of the smaller angle is 150.
17. D
The diagonals are congruent.
BD  AE  CE ; y  4  x  5  y  3 . Solving this
24. B
The average length of the sides is 14
so the perimeter is 42. x  5  2 x  1  4 x  6  42 ,
x  6. That makes AB  11, BC  13, AC  18. So
18 – 14 is 4.
25.
C
3 3 y  7   6 y  24, y  15. The
perimeter would be 6 15  24  114.
gives x  2. Now we have x  7  y  3,
substituting 2 for x, we have y  12. This makes
AC  16.
18.
(n  2)180  900   n  2  x 180,
B
180n  540  180n  180 x  360, x  5.
19. D
Since the diagonals of a rhombus are
perpendicular, the half diagonals would be 3x and
4x. Using Pythagorean theorem,
 3x 
2
  4 x   100, x  2. This makes the
2
3x  6,4 x  8 and the sum of the diagonals would
be twice (6+8)=28.
26. A
Two possibilities for the 4 and 6 on
the side of the triangle. Use the triangle angle
x4 x
 , x  12 or
bisector theorem to get
4
6
x4 x
 , x  8 . Reject this solution as the side
6
4
of the triangle can’t be negative. The lengths of the
sides of the triangle are 12+8+10 = 30.
27. D
One exterior angle would be 20  .
The sum of the exterior angles of a polygon, one at
each vertex is 360, so divide 360 by 20 to get 18
angles/sides. The perimeter is 18  5  90.
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Geometry Individual
28. C
Draw the altitudes to get a rectangle
and two 30-60-90 triangles. In the triangle with
30  base angle, let the side opposite the 60 angle
x
be x, the side opposite the 30 angle will be
.
3
In the triangle with the 60 base angle, the side
opposite the 30 angle will be 12  x (the length of
the base minus [length of the upper base 4 and x]),
the side opposite the 60 angle will be 12  x  3 .
x
and 12  x  3 are both altitudes of the
3
trapezoid so they are equal. Set them equal and
solve for x gives 9. So the segments of the base are
3 – 4 – 9. We want the distance between the
midpoints of the bases. ½ the longer base is 8. So
the midpoint of the base is 8 units from either
vertex of the lower base. Using the triangle with
the 30 base angle, go in 8 units – this is one
vertex of the right triangle we need to find the
distance between the midpoints. The other vertex is
the midpoint of the upper base. The third vertex is
the perpendicular from the midpoint of the upper
The
base. The altitude of the trapezoid is 3 3 , the
length of the side on the lower base is 3. Using
Pythag, the distance we want is 6. Whew, that was
a long solution.
29. D
Draw the diagram and you will see
that JNBA is a trapezoid. AB is the shorter base
and JN is the longer base. The measure of angle
AJN and JNB is 60 and the upper base angles are
120. Draw an altitude and you have a 30-60-90
triangle. The legs of the trapezoid are 64 making
the part of the lower base opposite the 30 degree
angle having a length of 32. So the 3 lengths of the
base JN are 32 + 64 + 32 = 128.
30.
C
The 3 triangles are similar. Using the
geometric mean to find the length of AB :
3 5  5  AB , AB  9 . Then to find AC:
AC  4  9, AC  6 .
January 11, 2014