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Transcript
Geometry Individual Solutions – January 2012 FAMAT State-wide Vero Beach
1)
C) 27 n(n-3)/2=9(6)/2=27
2)
D) 5 2  5 6
By drawing the altitude to side BH, you get a 45-45-90 triangle and a 30-60-90 triangle.
The altitude will have to have a height of 5 2 because of the hypotenuse being 10. That
makes the length from B to the point of intersection with the altitude to be 5 2 as well.
To get the length from H to the point of intersection with the altitude, we can use the 3060-90 rule and see that it is 5 6 . BH = 5 2  5 6
3)
E) NOTA – NONE
I. If the opposite sides of a polygon (if this said both pairs on a quadrilateral, this would
be true) are parallel, this it is a parallelogram. FALSE
II. If the diagonals of a convex quadrilateral are congruent, then it is a rectangle. FALSEIsosceles Trapezoid
III. If two lines are parallel, then their alternate exterior angles are supplementary.
FALSE, they would be congruent.
4)
D)  : 4
If AB  4 and AC  16   2 , that makes BC = π. So the sin A 
5)
D) Opposite rays form a straight line and are opposite in direction:
BA; BC
6)
B)
4  8x  180; x  22
90  22  68
7)
B) space
8)
B) 45
opp 
 .
hyp 4
Geometry Individual Solutions – January 2012 FAMAT State-wide Vero Beach
A  F  90; R  A  180; F  R  180
A  90  F ;( R  90  F  180)  ( F  R  180); 2 R  90  360
R  135; A  45
9)
4
A) 90a(180  )
a
For every regular polygon, the exterior angle is equal to:
360
, so if we set them equal
n
360 4
 , we can solve for n (number of sides) in terms of a.
n
a
n  90a
4
(180  ) = The measure of each interior angle since they are supplementary.
a
4
(180  )90a  Total degrees in a regular polygon with n sides
a
10)
C) 172.5 degrees : 12x-198=180;x=31.5;Angles 2 and 4 are congruent because they are
vertical, so by substituting back into the Angle 2 equation, you get 7(31.5)-48=172.5
11)
B) By 30-60-90 rule ( n : n 3 : 2n ), if the hypotenuse is 15, that makes the height (side
opposite the angle of elevation) equal to
15 3
.
2
12) C) I and III only
I. AB + BC = AC (Segment Addition Postulate)
III. A, B, C are coplanar. (If they are collinear, they must also be coplanar)
13) C) 30
By geometric mean; 62  8BX ;length of BX 
9
; By the Pythagorean Triple 9-12-15,
2
you can get BA = 15/2. Total P = 30
14) B) One pair of opposite sides are congruent and perpendicular. (It has to be one pair
that are congruent and parallel.)
Geometry Individual Solutions – January 2012 FAMAT State-wide Vero Beach
15) A)
2
Distance between  4, 5 and  2, 7  will be the diameter.
(4  2) 2  (5  7) 2 = 4  4  2 2; r  2
16) C) Obtuse
7, 12, 8;c=12
122 ? 7 2  82
144? 49  64
144  113;obtuse
17)
B) 75˚
The sum of the exterior angles is 17x+105 = 360, so x = 17, the largest exterior is
7(15)=105 degrees, so that would make the smallest interior angle equal to 75 degrees.
18) D) If an angle is greater than 90 degrees, then it is not an acute angle. True
19)
B) 48
Isosceles triangle’s vertex angle bisectors are also the altitudes. This allows us to use
Pythagorean Theorem to get the base of the triangle to be 98, half of which is 48, which
is the measure of the midsegment.
24
cm
31
A  0.5( B1  B 2) H
20) B)
48  0.5(50  74) H
24
H
31
21) E) NOTA = 40
SU SA 18

 , so the smallest possible values of the
YU YA 12
missing lengths are SA = 3 and AY = 2, but this would violate the Triangle Inequality
because 3  16  18 , so the next values would be SA = 6 and AY = 4, which works and
makes the perimeter 40.
Since SY is a bisector, the ratio of
Geometry Individual Solutions – January 2012 FAMAT State-wide Vero Beach
22) A) 400 feet
200
120
120
120 so using Pythagorean
Theorem: distance = 400
23) C) 4 6 in.
240
144 3
s2 3
 24 3 for each equilateral triangle with area =
,s  4 6
6
4
24)
D) 13
25)
If a polygon is a parallelogram, then it is a rectangle.
The negation of “If p, then q” is “p and not q”.
B) A polygon is a parallelogram and it is not a rectangle. False
26) Midpoint of (-4, 5) and (-2, 7)= (-3,6). Slope of (-4, 5) and (-2, 7) = 2/2 = 1
Perpendicular Slope is -1.
C) y  6  1( x  3)
27) C) 15
28)
 3x  12    7 x  86  Vertical Angles are congruent
4x=74
37
x=
2
GCO  (3x  12)  90 Exterior Angle is equal to two remote interior angles
.
37
GCO  (3  12)  90  133.5
2
A) 133.5˚
29)
A) 8 x 2 210 x
140x 96x   8x
2
30)
3
2
210 x
Geometry Individual Solutions – January 2012 FAMAT State-wide Vero Beach
A)
3
4
They are in the same ratio.