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TALLAHASSEE STATEWIDE
GEOMETRY INDIVIDUAL ANSWERS
1/14/2017
1. C - Since the angles are complementary, their sum is 90. 4x + 17 + 2x + 1 = 90, so x =
12. The angles therefore have measures of 65 and 25. Their positive difference is 65 – 25 =
40.
2. C - Because the sides of the rectangle are in ratio 9 : 2, we can solve 2(9x + 2x) = 44
since 44 is the perimeter. x = 2, so the sides of the rectangle have lengths 18 and 4. The area
of the rectangle is 18(4) = 72, so the area of the square is 36. Each side of the square must be
6, making the perimeter 24.
3. E - Choices A and C don’t prove any lines are parallel. Choices B and D prove that lines c
and d are parallel.
4. D - Choice A allows you to prove the triangles are congruent by ASA. Choice B allows
you to prove the triangles are congruent by AAS. Choice C allows you to prove the triangles
are congruent by SAS. Choice D creates SSA, which is not a valid means of proving triangles
are congruent.
5. B 52 is
52
128 forms a linear pair with the angle now labeled as 52.
corresponding with x + y, so they have equal measures.
52 + 86 + (x – y) = 180 since these angles form a triangle.
Solve the system: x + y = 52
x – y = 42
x = 47 and y = 5
6. B
7. D
8. D - ii is true because a statement and its contrapositive always have the same truth value.
iii is true because the vertical angles could be formed by perpendicular lines.
9. B - Because the triangle is equiangular, it is also equilateral. Solve the system: 2x – 3y =
14
x + 2y =
14
x = 10 and y = 2 so x –
y=8
10. B
11. A - Choices A and B give the correct statement. It is false because a rhombus is not a
kite but its diagonals are perpendicular.
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TALLAHASSEE STATEWIDE
GEOMETRY INDIVIDUAL ANSWERS
1/14/2017
12. B - slope of KJ = 1/6, slope of JI = -2, slope of HI = 1/2, slope of KH = -4. KJ = 37 , JI =
5 , HI = 3 5 , and KH = 17 . No sides are parallel and no sides are congruent, therefore the
figure is simply a quadrilateral.
18012  2 
360
. Substituting these values into the answer choices
 150 and b =
12
7
360
1020
gives A =
,B=
, C = 180, and D = 90.
7
7
13. D - a =
14. C - The sum of two sides must be greater than the third. Choice C is the only one that
satisfies this requirement.
15. E - The sum of the angles in a triangle is 180, therefore:
80 
In a triangle, the shortest side is across from the smallest angle
and the longest side is across from the largest angle. So, in PAH ,
from shortest to longest the sides are PH, PA, and AH. In ATH ,
85
from shortest to longest the sides are HT, AH, and AT. AH is the
longest in PAH , but the mid-length side in ATH , so AT must be the longest segment
overall.
16. C - Use the midpoint formula and solve:
4 x
8 y
 2 and
 2 . x = -8 and y = 12.
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17. D - Diagonals of a rectangle are congruent, so x² – 48 = 2x. So, x² – 2x – 48 = 0,
(x – 8)(x + 6) = 0, and x = 8 or -6. -6 is extraneous since it would result in a negative length.
5
(16) = 20.
KA = 2(8) = 16 and
4
18. B - The angle bisector splits the sides and segments of AC proportionally:
2 x  12
x

. Solve to get x = 30. Substitute to find that AD = 56, AB = 48, DB = 47, so the
2x  4 x  5
perimeter is 151.
19. D - ii is false because a rhombus doesn’t necessarily have four right angles. All other
statements are true.
20. D - The altitude from the vertex angle of the isosceles triangle to the base bisects the
base. Use the Pythagorean theorem to find that the height is 15. The area of the triangle is
0.5(16)(15) = 120. So, the volume of the prism is 120(7) = 840.
21. D - Since AL is an angle bisector, m1  m2 . Solve 2x + 5 = x + 22 to find x = 17.
m1  m2  39  . mSAL  180   m2  180   39   141 .
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TALLAHASSEE STATEWIDE
GEOMETRY INDIVIDUAL ANSWERS
1/14/2017
22. B - Draw a heptagon with the vertices labeled A, B, C, D, E, F, and G to represent each
of the teams. Draw lines to connect A to each of the six other teams, then lines from B to each
of the five remaining teams, lines from C to each of the four remaining teams, and so in. In all
there are 21 lines that connect all of the seven vertices, each line representing three games
each. 21(3) = 63.
23. C - DE || AB since they are both perpendicular to CA . CED  B since corresponding
angles are congruent. The triangles are congruent by AA.
24. B - See the diagram at right. The diagonals of the square are 18 inches
long, so each side of the square (which is the radius of the table) is 9 2 inches.
The farthest the table reaches from the corner of the wall is the diagonal of the
square plus one radius of the table: 18  9 2 .
25. A - See the diagram at right. a = 25 and b = 23. Solve the right
triangle that is formed by using the Pythagorean theorem. h = 4 6
a
26. C - The altitude to the hypotenuse is equal to the geometric mean
of the segments of the hypotenuse. 18  22  6 11
27. B - Because the distances between lines l, m, n, and p are congruent
on the transversal on the left, AB = BC = CD as well. Solve the system:
2x + 4 y = 32
3x + 10 y = 64
x = 8 and y = 4
28. A - Diagonal of a rhombus are perpendicular, so mJTA  90  . Diagonals of a rhombus
bisect opposite angles, so 1  TJA  TCA . Acute angles of a right triangle are
complementary, so m1  m3  90  : 9 x  22  x 2  90 , x 2  9 x  112  0 , x  7x  16  0 , so x
= 7 or -16. -16 is extraneous since it would result in a negative angle measure.
mTCA  m1  9(7)  22  41 .
29. E - The sum of two sides of a triangle must be greater than the third. 8 + 11 = 19, so no
triangle can be formed.
30. B - See the diagram at right. 90 is the long leg of a 30 -60 -90 triangle, so the short leg is
30 3 . The total height of the bridge is 30 3  3  31 3 .
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