Download Condensed Test

GEOMETRY TEAM #1
A:
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Given parallelogram ABCD, AB  x  5, AD  x  9, CD  2 x  1.
Find the perimeter of ABCD.
B:
Given parallelogram ABCD, mA  x, mD   3x  4  .
Find the sum of the mD  mB .
C:
If 78 is divided into three parts in the ratio 3:5:7,
find the sum of the smallest and largest parts.
D:
Find the number of diagonals of a regular nonagon.
GEOMETRY TEAM #2
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A:
Find the length of FD in a regular hexagon ABCDEF with side length 8
B:
Find the mAFD in regular hexagon ABCDEF.
C:
Find the length of the apothem in a regular hexagon with a perimeter of 24 6 .
D:
Find the perimeter of a regular hexagon when the longest diagonal has a length of 11.
GEOMETRY TEAM #3
A:
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Write the hypothesis of the contrapositive of the statement: “If an animal is a dog, then the
animal is not a cat”.
B:
In parallelogram PRST, the perimeter is 10 more than 5  RS  . If PR  26, find the length of
segment RS .
C:
The measure of  A is 6 greater than twice the measure of  B . If the sum of the two angles is
42 , find the measure of A .
D:
Given: BD bisects ABC and mABC  24, mABD  2 x  y , mABC  x  2 y .
Find the value of x  y expressed as a decimal.
GEOMETRY TEAM #3
A:
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Write the hypothesis of the contrapositive of the statement: “If an animal is a dog, then the
animal is not a cat”.
B:
In parallelogram PRST, the perimeter is 10 more than 5  RS  . If PR  26, find the length of
segment RS .
C:
The measure of  A is 6 greater than twice the measure of  B . If the sum of the two angles is
42 , find the measure of A .
D:
Given: BD bisects ABC and mABC  24, mABD  2 x  y , mABC  x  2 y .
Find the value of x  y expressed as a decimal.
GEOMETRY TEAM #4
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Use the diagram to name the result. Be sure to use the correct symbols when necessary.
A:
AB  BC
B:
CD  EB
C:
AC  EC
D:
BA  BC
D
C
E
AA
GEOMETRY TEAM #5
B
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A: The complement of an angle is ½ the sum of twice the angle and 20.
Find the measure of the supplement of the angle.
B: The supplement of an angle is four times the complement of the angle.
Find the measure of the complement.
C: The measure of the supplement of an angle is 30 less than five times the measure of the
complement. Find two-fifths the measure of the complement.
D: The supplement of an angle is 60 less than twice the supplement of the complement of the
angle. Find the measure of the complement.
GEOMETRY TEAM #6
A:
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Given parallelogram ABCD, mB   4 x  y  , mC   x  y  , mD   2 y  x   .
Find the mA.
B:
In parallelogram ABCD, the ratio of AB to BC is 5:3. If the perimeter of ABCD is 32,
find the length of AD .
C:
In ABC , D is a point on AB and E is a point on AC such that DE || BC . F is a point on DE such
that E is between D and F. If mB  3x  y , mC  3x  5 y , mADE  2 x  6 y  5 , mA  3x  4 y ,
find mAEF .
D:
Given: ABC is isosceles with base BC , E is the midpoint of AC , D is the midpoint of AB ,
AD  2 x  3, CE  5x  9, BC  4 x. Find the perimeter of ABC.
GEOMETRY TEAM #7
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Write the most descriptive name for each figure. Must be spelled correctly.
A:
A four-sided figure in which the diagonals are perpendicular bisectors of each other.
B:
A four-sided figure in which the diagonals bisect each other.
C:
A triangle in which the sides have different lengths.
D:
A four-sided figure in which the diagonals are congruent and all sides are congruent.
GEOMETRY TEAM #8
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A:
AB  CD at point E. mAEC  2 x  40, mCEB  2 y  40, mDEB  x  2 y . Find the mAED.
B:
If AB  16, BC  8, AC  24, which point is between the other two?
C:
What are the restriction on y when the measure of an obtuse angle is 5 y  45 ?
D:
In equilateral triangle ABC, point D is in the interior and is the point of intersection of the
perpendicular bisectors. Find the perimeter of DBC when the perimeter of ABC is 17.
GEOMETRY TEAM #9
A:
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Given a rhombus with a side length of 25 and the longer diagonal with a length of 40,
find the length of the shorter diagonal.
B:
Given the same rhombus as A above, find the perimeter of a similar rhombus with a
longer diagonal with a length of 32.
C:
Given square ABCD with equilateral triangle ABE having AB as one side and point E in the
exterior of the square. If the apothem of the equilateral triangle is 9, find the perimeter of the
figure AEBCD.
D:
In a rhombus, each side has a length of 4. Find the sum of the squares of the diagonals.
GEOMETRY TEAM #10
A:
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In ABC the lengths of the sides are AB  15, BC  10, AC  14. The angle bisector from point B
is drawn dividing AC into two parts. Find the length of the shorter of the two parts.
B:
In right triangle ABC with right angle at C, cos A  7 . Find sin A  tan A .
10
C:
The lengths of the sides of ABC are BC  12, CA  13, and AB  14. If M is the midpoint of CA ,
and P is the point where CA is cut by the bisector of B, find the length of MP .
D:
The perimeter of a rectangle is 20 inches. Find the least value of a diagonal of the rectangle, in
inches.
GEOMETRY TEAM #11
A:
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The sum of the degree measures of the interior angles of a regular polygon is 5040.
Find the number of sides of this polygon.
B:
The degree measure of one interior angle of a regular polygon is eight times the measure of an
exterior angle of the regular polygon. Find the number of diagonals of this polygon.
C:
The measure, in degrees, of an exterior angle of a regular polygon is 15 .
Find the perimeter of the polygon when one side has a measure of 5 ½ .
D:
The sum of the measures of the interior angles of a polygon is 1620 . When the measure of one
interior angle is expressed as a simplified mixed number, give the whole number part of the
mixed number.
GEOMETRY TEAM #12
A:
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ABC is equilateral and the length of a side is 8. D is a point on BC such that AD  BC .
E is the midpoint of AD . Find the length of BE .
B:
Given regular octagon ABCDEFGH and square BCRS – sharing a common side BC . Points R and S
are in the exterior of the octagon. Find the number of degrees in ABS.
C:
Give the name, spelled correctly, of the center of gravity of a triangle.
D:
On a recent walk, Sam traveled due north for 2 miles, then due west for 3 miles and finally due
south for 7 miles. Find the shortest distance, in miles, that he must travel to reach his
starting position.
GEOMETRY TEAM #13
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For each description below, choose the most descriptive name for each quadrilateral.
Must be spelled correctly.
Choices are: kite, parallelogram, rectangle, rhombus, square, trapezoid
A:
A quadrilateral whose consecutive sides measure 15, 18, 15, 18.
B:
A quadrilateral whose consective sides measure 15, 18, 18, 15.
C:
A quadrilateral with consecutive angles of 30,150,110,70 .
D:
A quadrilateral whose diagonals are perpendicular bisectors of each other.
GEOMETRY TEAM #14
A:
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Find the degree measure of the complement of the smaller angle when two supplementary
angles are in the ratio 11:7.
B:
C:
D:
Find the degree measure of the supplement of an angle when three times the degree measure
of the supplement less five times the degree measure of the complement is 11 times the measure
of the angle.
The exterior angles of a triangle, one at each vertex, are in the ratio 2:3:4.
Find the measure of the largest interior angle.
The sides of one of two similar figures have lengths of 3, 3, 4, and 5. If the longest side of the
other quadrilateral is 8, find its perimeter.
ANSWERS
GEOMETRY TEAM
A
44
#1
#2
B
268
90
8 3
“If an animal is a cat”
or “an animal is a cat”
B or point B
EC
140
72
rhombus(all must be spelled
correctly)
100
30
6
parallelogram
12
135
scalene
B
9  y  27
30
28
3
or 5 or 5.6
5
5
80
#10
Null set or empty set
30
2 7
parallelogram (all must be
spelled correctly)
20
#13
#14
D
27
33
 or { }or
#9
#11
#12
C
52
14
#4
#8
January 11, 2014
6 2
30
#3
#5
#6
#7
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16.8
or equivalent
90 3
½ or .5
ABC or
CBA
70
60
square
34 3  51
9
64
5 2
17 51
70
135
135
132
centroid
kite
trapezoid
34
rhombus
170
100
24
147
SOLUTIONS
#1
A: 44
B: 268
C: 52
D: 27
A: Opposite sides are congruent. x  5  2 x  1, x  4 . AB  CD  9, AD  BC  13 . Perim. is 2  9  13  44 .
B: Consecutive angles are suppl. 4 x  4  180, x  46.mD  mB  134 ,
C: 15 x  78, x 
largest
#2
26
. So the parts are
5
 26  78  26 
 26  182
3    ,5    26,7   
. The sum of the smallest and
5
 5  5  5 
 5 
78 182 260


 52.
5
5
5
A: 8 3
B: 90
mD  mB  2 134   268 .
D: A nonagon has 9 sides.
C: 6 2
n  n  3 9  6 

 27.
2
2
D: 33
A: Draw FD . FED an isosceles triangle with base FD. mE  120 . Draw the altitude of this triangle
forming 2 30-60-90 triangles. Using either one of them, the hypotenuse is 8, the side opposite the 30
(which is the altitude) is 4, and the side opposite the 60 is 4 3 . Double 4 3 to get the length of
FD which will be 8 3 .
B: Using the same triangle as in A above, since mEFD.  30, mAFD  mAFE  mEFD  90 .
C: Draw the apothem forming an equilateral tri. A side of the triangle is
1
of 24 6 which is 4 6 . The
6
1
3 making the apothem 6 2 .
2
11
D: Draw the longest diagonal. This has a length of 11 and this has a length of
since it is two side of
2
 11 
the equilateral tri. formed by drawing all of the longest diags. That makes the perimeter 6    33.
2
apothem is opposite a 60  angle so multiply 4 6 by
#3
A: If an animal is a cat or an animal is a cat
B. 14
C. 30
D. 16.8 or 16
4
5
A: The contrapositive would be “If an animal is a cat, then it is not a dog.” The hypothesis is the “if”
part. The hypothesis is “If an animal is a cat” or “an animal is a cat”.
B:
Opposites sides are congruent so p  10  5(RS), 2 RS 52  10  5 RS, RS  14.
C: A  6  2 B, A  B  42,42  B  6  2 B, B  12, A  6  24  30.
D: 2 x  y  12, x  2 y  24, Solving this system gives x 
#4
A. B or point B
#5
A. 140
48
, y  7.2 .
5
B.  or { } or empty set or null set
B. 30
C. 12
x  y  16.8.
C. EC
D. ABC or CBA
D. 70
For all parts of the solutions of these problems, the angle will be represented by x, the complement will
be 90  x , and the supplement will be 180  x .
A: 90  x 
1
 2 x  20  , x  40, supplement will be 140.
2
B: 180  x  4  90  x  , x  60 , the complement will be 30.
C: 180  x  5 90  x   30, x  60 , complement is 30,


2
of that is 12.
5
D: 180  x  2 180   90  x   60, x  20 , complement is 70.
#6
A. 72
B. 6
C. 135
D. 60
A: Opposite angles are congruent, and consecutive angles are supplementary. Set up a system for
equations. 4 x  y  x  y  180,5x  2 y  180 and 4 x  y  2 y  x,5 x  y  0. Solving gives
y  60, x  12, mA  mD  x  y  72.
B: Since the ratio of consecutives sides is 5:3 and the semiperimeter is 16, 8x  16, x  2 making those
sides have lengths of 10 and 6. AD  BC  6.
C: The lines are parallel making corresponding angles congruent, and the sum of the angles of the
triangle is 180. Use these facts to set up a system of equations. 3x  y  2 x  6 y  5, x  5 y  5.
And the sum of the 3 angles is 9 x  180, x  20. Substituting in the other equation gives y  3.
mAEF  mADE  mA. mADE  63, mA  72, mAEF  135.
D: This triangle is isosceles and the congruent sides are AB and AC. The midpoints of these two sides
makes all 4 segments congruent, thus AC=CE. 2 x  3  5x  9, x  4. The sides have lengths of 16, 22, and
22 making the perimeter 60.
#7 A. rhombus
#8
B. parallelogram
A. 100
B. B
C. scalene
C. 9  y  27
D. square ALL MUST BE SPELLED CORRECTLY.
34 3  51
,
9
D.
A: Vertical angles are congruent and adjacent angles are supplementary. Using the system of equations,
solve for x and y. 2 x  40  x  2 y, x  2 y  40. Solve this w/ 2 x  2 y  100, x  20, y  30.
mAED  mCEB  100.
B: AB  16, BC  8, AC  24. B is between A and C.
C: 90  5 y  45  180,9  y  27.
D: The point D is where the perpendicular bisectors, medians, altitudes, and angle bisectors intersect.
The point where the medians intersect is 2/3 of the length of the median or altitude. The altitude is
opposite a 60  angle so its length is
1  17 
17 3
.
  3
2 3 
6
DB =
2  17 3  17 3
. The perimeter of DBC


3 6 
9
 17 3  17 34 3  51
.


9
9
3


is wanted so double the length of DB & add BC. 2 
#9
A. 30
B. 80
C. 90 3
D. 64
A: Draw the two diagonals. This is a rhombus so the diagonals are perpendicular and bisect each other
forming a right triangle with hypotenuse 25 and one leg of 20. Using Pythagorean Theorem, find the
other let to be 15 which makes the diagonal 30.
B: The ratio of the longer diagonals is 32:40 which simplifies to 4:5. The perimeter of the original
rhombus is 100 so the prop. to solve for the perim. of the smaller rhombus is
4
x

, x  80.
5 100
C: The apothem is 1/3 of the length of the median which is also the altitude. The length of the altitude
is 27. To find a side of the equilateral triangle, divide by
3 and multiply by 2 which will be 18 3 . The
perimeter will be 5 times 18 3 which will be 90 3 .
D: Draw the diagonals and let one be represented by d1 and the other as d 2 . Since the diagonals
perpendicular they form right angles with the legs
2
are
1
1
d1 and d 2 . Using the Pythagorean theorem,
2
2
2
1 2 1 2
1  1 
2
2
 d1    d 2   16, d1  d 2  16, d1  d 2  64.
2
2
4
4

 

#10
A.
28
3
or 5 or 5.6
5
5
B.
17 51
70
C. ½ or .5
D. 5 2
A: When the angle bisector is drawn, use the triangle angle bisector theorem. Let the smaller part of
AC be x, and the longer part is 14  x . The proportion is
15
10
28
3
 ,x 
or 5 or 5.6.
14  x x
5
5
B: Find the length of the third side of the triangle using Pythagorean Theorem. The length is
sin A  tan A 
51
51 17 51


.
10
7
70
51 .
13
. To find the length of AM, letting AM = x, use the triangle
2
14
12
13 1

, x  7. So, AP  AM  7   or .5.
angle bisector theorem. The prop. will be
x 13  x
2 2
C: P is the midpoint so both AM and MC =
D: The smaller diagonal would be in a square. s  5, d  5 2 .
#11
A. 30
B. 135
C. 132
D. 147
A:  n  2 180  5040, n  30 .
B: Let n be the number of sides of the polygon. One interior angle is 8 times one exterior angle.
 n  2 180  8  360  .

 n  18 so the number of diagonals is
 n 
n
C:
n  n  3 18 15

 135.
2
2
360
 24 sides.
15
D:  n  2 180  1620, n  11 . To find one interior angle, divide the sum by 11 giving
147
1620
which is
11
3
. The whole number part is 147.
11
#12
A. 2 7
B. 135

A: AD  4 3 , ED  2 3, 2 3
C. centroid

2
D.
34
 4 2   BE  , BE  2 7 .
2
B:An interior angle of an octagon is 135. An angle of the square is 90. At the vertex B, where the square
meets the octagon, the sum of the three angles is 360. Subtract the sum of 135 and 90 from 360 to get
135.
C: centroid
D: Drawing the diagram and connecting the segment from the starting point to the ending point gives and
triangle with legs of 3 and 5 making the hypotenuse
34 .
#13
A. parallelogram B. kite C. trapezoid D. rhombus ALL MUST BE SPELLED CORRECTLY.
#14
A. 20
B. 170
C. 100
A: Angle is x, supplement is 180  x .
D. 24
x
11
 , x  70. The complement of 70 is 20.
180  x 7
B: Angle is x, supplement is 180  x , complement is 90  x . 3180  x   5  90  x   11x, x  10 . So the
supplement of the angle is 170.
C: 9 x  360, x  40. Ratio is 2:3:4 so exterior angles are 80, 120, 160 making the interior angles 100, 60,
20 and the largest is 100.
D:
8 x
 , x  24.
5 15