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TALLAHASSEE STATEWIDE
1/14/2017
GEOMETRY TEAM SOLUTIONS
1.
a. perpendicular bisector
b. HL
c. HK
d. altitude
2.
a. 2x + 4x + 6x = 180, so x = 15. 2(15) + 6(15) = 120
b. 360/22.5 = 16
c. Let x = the angle and 5x + 8 equal the supplement of the angle. x + 5x + 8 = 180, so x = 86/3 and
the supplement is 454/3. 454/3 – 86/3 = 368/3.
d. Area of a rhombus is half the product of its diagonals. 288 = 0.5(2x)(x), so x = 12 2 . The
diagonals are 12 2 and 24 2 .
3.
a. sometimes – Only if the statement itself is true.
b. sometimes – Three points will always lie in exactly one plane. The fourth point could be in the
same plane as the other three, but it may be in a different plane.
c. sometimes – Supplementary angles simply have a sum of 180…they do not necessarily have to
share a vertex and a side.
d. sometimes – A scalene triangle could also be right or acute.
4.
a. Diagonals of a rhombus bisect each other and are perpendicular, forming four right triangles with
legs of length 4 and 6. Use the Pythagorean theorem to find that each hypotenuse (side of the
rhombus) is 2 13 . The perimeter is the sum of all four sides: 8 13 .
b. Area of a rhombus is half the product of its diagonals. A = 0.5(8)(12) = 48.
c. If x is half the length of one side of the equilateral triangle: 16 3  0.5(2 x)( x 3 ) , so x = 4. One
side = 8, so the perimeter = 24.
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TALLAHASSEE STATEWIDE
1/14/2017
GEOMETRY TEAM SOLUTIONS
d. The height is the long leg of a 30-60-90 triangle where the hypotenuse is one side of the
equilateral triangle. The height is 4 3 .
5.
a. 9 + 7 = 16
b.
180n  2
 156 , so n = 15
n
c.
360 40
, so n = 27

n
3
d. x = 360 and y 
360
 36 , so x – y = 324
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6. m3  35 because it forms a linear pair with m5 . By the triangle exterior angle theorem
m3  m1  m6 , so 35 + x + 20 = 3x – 5 and x = 30.
a. m6 = 3(30) – 5 = 85
b. m4 = 180 – m1 = 180 – (30 + 20) = 130
c. 2x – 11 + x – 10 = 90, so x = 37. mDEB = 90 + (37 – 10) = 117
d. mAEC = 90 + (2(37) – 11) = 153
7.
a. The sum of two sides of a triangle must be greater than the third, so treat the missing side as
either the longest side (x + x + 9 > 3rd side…2x + 9 > 3rd side), or the shortest side (x + 9 – x < 3rd
side…9 < 3rd side). The length of the third side must be greater than 9 and less than 2x + 9.
b. 2 feet = 24 inches, so x + 2x = 24 and x = 8. The longer piece is 2(8) = 16.
c. The altitude to the hypotenuse is equal to the geometric mean of the segments of the hypotenuse:
4  18  6 7
d. Because the lines are parallel, alternate interior angles are congruent: AAS.
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TALLAHASSEE STATEWIDE
1/14/2017
GEOMETRY TEAM SOLUTIONS
8. Use properties of parallel lines and transversals to find the missing measures:
a. 49 + 76 = 125
b. 153 – 104 = 49
c. 153 – 27 = 126
d. 76 + 153 = 229
9.
a. The circumcenter is equidistant from the vertices of a triangle. In a right triangle, the circumcenter
is located at the midpoint of the hypotenuse. (–1, 1)
b. The incenter is equidistant from the sides of a triangle; it is formed by the intersection of the angle
bistectors of a triangle. Bisect the right angle by drawing a line with a slope of 1 that passes through
point C: y = x + 3. The incenter must lie on this line. Since the distance from a point to a line is
measured as the length of the segment from the point perpendicular to the line, the distances from
sides CB and AC would have to be measured horizontally and vertically, respectively. The point (–3,
0) lies on y = x + 3 and is exactly 2 units from both sides CB and AC. The formula for the distance
Axo  By 0  C
from a point (xo, yo) to a line Ax + By + C = 0 is
. Substitute (–3, 0) for the point and
A2  B 2
 3 3  40  1
–3x – 4y + 1 = 0 for the line (the equation of the line that forms side AB) and get
 32   42

9 1
5
 2 . The coordinates of the incenter are (–3, 0).
c. F is the centroid…the point formed by the intersection of the medians of the triangle. Draw the
median from C to D (found in part a). Draw the median from A to M, the midpoint of BC. The
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centroid is the intersection of these two lines: y  1  x  1 and y  1   x  5 . Solve the system
4
8
 7 
to find that they intersect at the centroid:   ,0  .
 3 
d. The orthocenter of a right triangle is located at the right angle: (–5, –2).
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TALLAHASSEE STATEWIDE
1/14/2017
GEOMETRY TEAM SOLUTIONS
10. The sum of the angles in a triangle = 180 and the sum of the angles in a quadrilateral is 360:
a. mSFU = 21
b. mSUO = 21 + 120 = 141
c. mW = 120
d. mWFS = 47 + 21 = 68
11.
a. Use the Pythagorean Theorem to find that the side adjacent to Q is 8. cosQ = 8/17
b. See the diagram below. x is half the length of one side of the triangle. Since the hypotenuse of
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the 30-60-90 triangle is 6, y = 3, and x = 3 3 . The area = 6 3 (9) = 27 3
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 
c. - Draw a nonagon with the vertices labeled A, B, C, D, E, F, G, H, and I to
represent each of the businessmen. Draw lines to connect A to each of the
eight other points, then lines from B to each of the seven remaining points,
lines from C to each of the six remaining points, and so in. In all there are
35 lines that connect the nine vertices, each line representing two handshakes
each. 35(2) = 70.
d. The possible dimensions are: 1 & 120, 2 & 60, 3 & 40, 4 & 30, 5 & 24, 6 & 20, 8 & 15, and 10 &
12. The dimensions with the smallest perimeter that is greater than 60 is 4 & 30 (perimeter is 68).
12.
a. Supplementary angles have a sum of 180, so 5x – 28 + 11x – 64 = 180. Solve to find that x = 17.
Substitute 17 for x to find that the angles measures are 57 and 123. The smaller angle is 57.
b. The sum of the measures of the interior angles in a pentagon is 180(5 – 2) = 540. If three of the
angles have an average of 139, 3(139) = 417 degrees have already been accounted for. 540 – 417 =
123 to divide evenly between the remaining two angles. 123/2 = 61.5.
c. An interior angle and its adjacent exterior angle form a linear pair. Let the exterior angle equal x:
x + (x + 36) = 180. Solve to find the exterior angle x = 72. The number of sides can be found by
360/72 = 5.
d. The sides of EFB form the three midsegments of ADC . Midsegments of a triangle are ½ the
side they are parallel to, so the lengths of the sides in EFB are 6, 11, and 14. The perimeter is 31.
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TALLAHASSEE STATEWIDE
1/14/2017
GEOMETRY TEAM SOLUTIONS
13.
a. Each face of a cube is a square with diagonal 8. The diagonal cuts the square into two 45-45-90
triangles with the diagonal as the hypotenuse, thus, each side is 4 2 . The area of one face is
4 2 
2
 32 . There are six congruent faces on a cube, so the total surface area is 32(6) = 192.
b. Since the sixth and seventeenth stones are directly opposite each other, stones 7-16 occur
between them (a total of ten). There must be exactly ten more stones on the other half, so in all there
are ten stones on one side, ten stones on the other, plus stones 6 and 17: 22.
c. The circumcenter is equidistant from the vertices of the triangle, so XZ = 9. The circumcenter is
formed by the intersection of the perpendicular bisectors of the sides of a triangle, so YX has been
bisected, making YX = 14. XZ + YX = 9 + 14 = 23.
d. Half the diagonal of the square is equal to the radius of the circle. If the perimeter of the square is
20, each side has a length of 5. The diagonal cuts the square into two 45-45-90 triangles with the
5 2
diagonal as the hypotenuse, so the diagonal has length 5 2 . Thus the radius of the circle is
.
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14.
a. Use the properties of 45-45-90 and 30-60-90 triangles to find the missing lengths as shown in the
diagram. The perimeter of ADEF = 8 + 4 + 4 2 + 4 + 4 + 4 3 = 20  4 2  4 3 .
b. There are two pairs of congruent adjacent sides given, so it is a kite.
c. The wasted area will be the area of the square minus the area of each
of the nine circles. The square has area 30(30) = 900. Each circle has a radius of 5, and thus an
area of  5 2  25 . The total area of all circles is 925   225 . The wasted area is 900  225 .
d. The figure shown is a kite. Diagonals of a kite are perpendicular and the angles between the noncongruent sides are bisected. So, m1  90 and m3  38 . The acute angles in a right triangle have
a sum of 90, so m2  52 and m4  37 . m4  m5 so m5  37 . 2(37) + 38 + 3(90) – 2(52) – 37
= 241.
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