Geometry Placement Test Review 1
(Answer Key)
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I hope there are no typos!
1. (See all PROOFS on the last page)
2. a) Conditional Stmt: If each side of a triangle has a length of 10, then the triangle perimeter is 30 (T)
Converse: If the perimeter of a triangle is 30, then each side of the triangle is 10. (F)
Inverse: If each side of a triangle does not have a length of 10, then the triangle perimeter is not 30. (F)
ContraP: If the perimeter of a triangle is not 30, then each side of the triangle is not 10. (T)
b) Conditional Stmt: If an angle is acute, then it has a measure greater than 0 and less than 90. (T)
Converse: If the measure of an angle is greater than 0 and less than 90, then it is acute. (T)
Inverse: If an angle is not acute, then the measure is not greater than zero and less than 90. (T)
ContraP: If an angle’s measure is not greater than zero and less than 90, then it is not acute. (T)
3. (See all PROOFS on the last page)
4. a)
b) ∡JFG = ∡GHJ
5. 75⁰
(let the angle = x = (15): the comp = 90 – x = (75); the supp = 180 – x = (165)
Equation: (180 – x) = 3(90 – x) – 60
6. (See all PROOFS on the last page)
7. a)
||
(yes, because the slopes of both = 1/3)
b)
||
(yes, because the slopes of both = – 3)
c) ∡R is a right angle because the product of the slopes of any pair of adjacent sides = -1:
From a & b above: (1/3)(-3) = -1
8. Quadrilateral to select from: parallelogram, rhombus, rectangle, square, kite, trapezoid, isosceles trapezoid
a) Rhombus (diagonals are ⏊ bisectors -- choices: square or rhombus)
b) Square (only one choice: square – has all the properties of both kite and rectangle)
c) Isosceles Trapezoid (choices: isosceles trapezoid or rectangle)
d) Isosceles Trapezoid (only one choice: isosceles trapezoid)
9.
a) False (planes either intersect or are parallel – skew pertains to lines only!) If two lines are skew, by definition
they must occupy different planes)
b) True (If a line and a plane never meet, they are parallel)
c) False (If two parallel lines lie in different planes -- think of a line on the ceiling being parallel to a line on the
wall – the lines are ||, but the planes are not)
d) True (If a line is perpendicular to two planes, the planes ARE parallel)
e) True (If a plane and a line not in the plane are each perpendicular to the same line, then they are parallel to each
other)
10.
Polygon
Given
a) Pentagon n = 5
b) Hexagon
n=6
c) Dodecagon n = 12
11. x =
Interior
(180 – Exterior)
180 – 72 = 108°
180 – 60 = 120°
180 – 30 = 150°
Equation:
12.
a) If two triangles are similar, they are SOMETIMES congruent
b) If two triangles are congruent, they are ALWAYS similar
c) An obtuse triangle is NEVER similar to an acute triangle
d) Two right triangles are SOMETIMES similar
e) Two equilateral polygons are SOMETIMES similar (think about a rhombus)
f) Two equilateral triangles are ALWAYS similar
g) Two rectangles are SOMETIMES similar if neither is a square
Exterior
360 ÷ n
360 ÷ 5 = 72°
360 ÷ 6 = 60°
360 ÷ 12 = 30°
13.
and
and
and
and
a) From the work above, by a triple use of the distance formula, you should get:
.
Then by the Pythagorean Theorem,
b) The slopes of the three segments are: -1/5, -2/3, and 5. Since (-1/5)(5) = -1 (negative reciprocals), two of the sides
are perpendicular making a right angle.
14. a) x =
b) x = {-1, 6}
15. a) ½
b)
c)
16. Remember: To help you remember the sides to use when writing the trigonometric RATIOS:
SohCahToa
a) sin A =
cos A =
tan A =
b) sin B =
cos B =
tan B =
17. Remember to use the INVERSE trig functions when calculating an angle measure! [
a) 25⁰
b) 73⁰
c) 28⁰
18. diagonal = 17 units
,
, or
Diagonal Formula:
19. Use the special right triangle (30-60-90)for part b!
a) 144
b)
20. Remember
and you are given V = 240, l = 8, h = 10 and you are finding “w”.
21.
(Remember -- the longest side of a triangle is located opposite of the largest angle – that is why the
hypotenuse must always be the longest side of a right triangle!)
22.
23.
24.
25.
26.
27.
28. All triangles have the SAME area because they all have the same base and height measures!
C
Proofs (Questions 1, 3, & 6)
1) Given:
F
D
Prove:
E
Statements
1. ∡CFD
∡EFD
2. FD is an altitude
3. ∡FDC ∡FDE
4.DF DF
5. FDC FDE
6. CD DE
7. DF is a median
Reasons
1. Given
2. Given
3. Alt forms Rt ∡s
4. Reflexive
5. ASA (1, 4, 3)
6. CPCTC
7. If a ray divides a seg into 2
(A)
(A)
(S)
H
3) Given:
segs, then median
M
6
5
Prove:
J
G
K
Statements
1. JH KH
2. HG HM
3. ∡5 ∡6
4. ∡KHG ∡KHG
5. ∡JHG ∡MHK
6. JHG KHM
(S)
(S)
(A)
Reasons
1. Given
2. Given
3. Given
4. Reflexive
5. Addition
6. SAS (1, 5, 2)
6) Given: ⊙ O
O
Prove:
A
Step 1
Step 2
Step 3
Step 4
B
C
Either OB does not bisect ∡ AOC
or
OB bisects ∡AOC
Assumption: OB bisects ∡ AOC
Assuming OB bisects ∡AOC, we now have ∡BOA ∡BOC because the bisector of an angle divides it
into two congruent angles. From ⊙O, we know that OA OC since all radii of a given circle are and
∡A ∡C also, because if two sides of a triangle are , then the angles opposite of those sides are .
We now know that ABO CBO by ASA, and this means that ∡OBA ∡OBC by CPCTC. Since ∡OBA
and ∡OBC are both congruent and supplemental, this means they are right angles and OB is an altitude.
However, this contradicts the given that OB is not an altitude. Our assumption was false and we must
accept the only other conclusion.
∴ OB does not bisect ∡ AOC
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