| Geometry |
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(A) Statement (1) ALONE
is sufficient, but statement (2) is not sufficient. |
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(B) Statement (2) ALONE
is sufficient, but statement (1) is not sufficient. |
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(C) BOTH statements
TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. |
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(D) EACH statement ALONE
is sufficient. |
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(E) Statements (1) and
(2) TOGETHER are NOT sufficient. |
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| Basics |
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| 1 |
Land for a pasture is
enclosed in the shape of a 6-sided figure; all sides are the same length and
all angles have the same measure. What is the area of the enclosed land? |
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(1) |
Each side is 8 meter long. |
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(2) |
The distance from the center of the land to the
midpoint of one of the sides is 4√3 meters. |
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The best answer choice is (D); Obvious answer was
(A) |
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. M 4 P |
Facts to remember |
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. 30º |
1. |
If all 6
sides length are equal and all 6 angles are equal, then the hexagone is made
out of 6 equalateral triangles |
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. 8 |
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2. |
all angles within the exagone = 60º |
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. O |
3. |
The line OP
makes the equalateral triangle into 2 true triangles where the inner angles
are: 90º, 60º and 30º |
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60º |
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4. |
Then if as
said in (1) each side is 8 meter long, then OM = 8 meters as well; then MP =
4 meters; then OP = 4√3 meters; then it is possible to calculate the
area of each triangle (area of a triangle = 1/2 base x high) and then of the
hexagone itself. |
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Then the possible answers should be A or D |
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5. |
From (2)
you can find the same measures - if OP = 4√3 meters then you can
theoritically find the measure of OM
and of MP. |
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The the best possible answer is D |
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| Geometry |
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| Triangles |
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| 2 |
If x, y and z are the lenghs of the three sides of
a trinagle, is y > 4 |
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(1) |
z = x + 4 |
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(2) |
x = 3 and z = 7 |
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The best answer choice is (D); Obvious answer was
(E) |
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Facts to remember |
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x |
z |
1. |
The sum of
any two side length of a triangle is always > the the lengh of the 3rd
side |
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2. |
Therefore; for instance x + y > z |
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3. |
Statement (1) implies that x + y > x + 4; then y
> 4 |
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y |
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Then the possible answers should be A or D |
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4. |
Statement (2) implies that 3 + y > 7; then y
> 4 |
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The the best possible answer is D |
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| 3 |
In the
figure shown, QRS is a straight line and line TR biscets < PRS. Is it true
that lines TR and PQ are parallel? |
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(1) |
PQ = PR |
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(2) |
QR = PR |
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The best answer choice is (B); Obvious answer was
(C) |
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. P |
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. pº |
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xº |
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. qº |
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rº |
xº |
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Q |
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R |
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S |
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Facts to remember |
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1. |
Since QRS is a straight line, then rº + xº + xº =
180º (or rº + 2xº = 180º) |
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2. |
For PQ ║ TR, xº must be = pº |
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3. |
From statement (1) you know that qº = rº, but you cannot know if pº = xº |
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Then the possible answers should be B, C or E |
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4. |
From statement (2) you know that qº = pº and then
rº + 2pº = 180º |
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5. |
If rº + 2xº = 180, then xº = pº, therefore PQ
║ TR |
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The the best possible answer is B |
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| Geometry |
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| Triangles |
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| 4 |
If each side of the Δ ACD shown has lengh 3
and if AB has lengh 1, what is the area of rgion BCDE? |
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A. |
9 |
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B. |
7 |
√3 |
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C. |
9 |
√3 |
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D. |
7 |
√3 |
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E. |
6 + √3 |
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4 |
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4 |
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4 |
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2 |
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C |
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The best answer choice is (B); Obvious answer was
(D) |
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3 |
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Facts to remember |
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B |
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1 |
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1. |
Region BCDE = area of
Δ ACD - area of Δ ABE |
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60º |
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2. |
Δ ACD is a
equilateral |
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A |
F |
E D |
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3. |
The area of Δ ACD
can be calculated as follow: |
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AF = 3/2 |
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AC |
√3 |
x |
AC |
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3 |
√3 |
x |
3 |
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9 |
√3 |
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CF = 3/2√3 |
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2 |
2 |
2 |
2 |
4 |
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4. |
Δ ABE is a right
triangle with 30º and 60º angles (since < BAE is 60º) |
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5. |
The area of Δ ABE
can be calculated as follow: |
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AB |
√3 |
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1 |
√3 |
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√3 |
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2 |
2 |
2 |
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6. |
Therefore, the area of
BCDE is: |
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9√3 |
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√3 |
= |
9√3 |
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2√3 |
= |
7 |
√3 |
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4 |
2 |
4 |
4 |
4 |
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The the best possible
answer is B |
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| 5 |
A ladder 25
feet long is leaning against a wall that is perpendicular to level ground.
The bottom of the ladder is 7 feet from the base of the wall. If the top of
the ladder slips down 4 feet, how many feet will the bottom of the ladder
slip? |
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A. |
4 |
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B. |
5 |
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C. |
8 |
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D. |
9 |
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E. |
# |
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. A |
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The best answer choice is (C); Obvious answer was
(E) |
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. E |
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Facts to remember |
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1. |
AB² = AC² + BC² / 25² = AC² + 7² / AC² = 625 - 49 =
576 / AC = 24 |
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2. |
EF² = CE² +
CF² / 25² = 21² + CF² / CF² = 625 - (24-4)² = 625 - 400 = 225
CF = 15 |
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. C |
B |
F |
3. |
So the answer should be CF - CB / 15 - 7 = 8 |
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AB = EF = 25 |
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The the best possible
answer is C |
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CB = 7 |
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CE = AC - 4 |
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| Geometry |
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| Triangles |
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| 6 |
Is ΔMNP isosceles? |
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(1) |
Exactly two of the
angles, <M and <N, have the same measure |
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(2) |
<N and <P do not
have the same measure |
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P |
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The best answer choice
is (C); Obvious answer was (A) |
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Facts to remember |
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1. |
If <M =
<N, then the triangle is either an isosceles or an equilateral. Since
every equilateral is an isosceles, then the answer is yes |
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M |
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N |
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Then the possible answers should be A or D |
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2. |
If <N =
<P, then it could be any type of triangle, therefore the information is
not enough to determine if the triangle is an isosceles |
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The the best possible
answer is A |
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In the figure shown, D
is a point on side AC of ΔABC. Is ΔABC isosceles? |
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(1) |
The area of triangular
region ABD is equal to the area of triangular region DBC |
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(2) |
BD ┴ AC and AD = DC |
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The best answer choice
is (B); Obvious answer was (E) |
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Facts to remember |
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1. |
As noted in
statement (1), you can figure that AD = DC but not if ΔABC is isosceles |
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. A |
D C |
Then the possible answers should be B, C or E |
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2. |
As noted in
statement (2), we can figure that ΔABD and ΔBDC are right triangles
and since AD = DC and BD is the common side, so AB = BC and ΔABC is an
isosceles |
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The the best possible
answer is B |
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| Geometry |
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| Triangles |
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| 7 |
If the area of
triangular region RST is 25, what is the perimeter of RST? |
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(1) |
The lengh of one side of
RST is 5√2 |
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(2) |
RST is a right isosceles
triangle |
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R |
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The best answer choice
is (B); Obvious answer was (C) |
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Facts to remember |
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1. |
If for
instance ST = 5√2, then ½(5√2)h = 25, if it could be proven that
it would be the same formula applies on RS so we could know that ΔRST is
a right triangle and then the perimetter could be calculated, but the data is
not sufficient to determine that |
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S |
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Then the possible answers should be B, C or E |
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2. |
Statement
(2) shows that for instance RS = ST and that < SRT = < RTS = 45,
therefore ½x² = 25 / x = 5√2, therefore based on pythagorean theorem,
the perimetter can be calculated |
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The the best possible
answer is B |
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| 8 |
Dan and
Karen, who live 10 miles apart, meet at a café that is directly north of
Dan's house and directly east of Karen's house. If the café is 2 miles closer
to Dan's house than to Karen's house, how many miles is the café from Karen's
house? |
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Karen's house |
Café |
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6 |
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B. |
7 |
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C. |
8 |
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D. |
9 |
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E. |
# |
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x |
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The best answer choice
is (C); Obvious answer was (A) |
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x-2 |
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10 miles |
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1 |
Using the pythagorean
theorem, the calculation should be as follow: |
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Dan's house |
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x² + (x -2)² = 10² |
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x² + x² - 4x + 4 = 100 |
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2x² - 4x -96 = 0 |
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2(x² - 2x - 48) = 0 |
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2(a - 8)(x + 6) = 0 |
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a = 8, or a = -6 |
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2 |
Since it is a distance
to calculate, therefore the answer is 8 |
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The the best possible
answer is C |
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| Geometry |
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| Quadrilaterals |
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| 9 |
A rectangle
is defined to be "silver" if and only if the ratio of its length to
its width is 2 to 1. If rectangle S is silver, is rectangle R silver? |
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(1) |
R has the same area as S |
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(2) |
The ratio of one side of
R to one side of S is 2 to 1. |
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The best answer choice is (E); Obvious answer was
(C) |
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Facts to remember |
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1. |
Statement 1
is not sufficient because is rectangle S has one side of 2 and the other of 4
(8m2 area), the other
could have one side of 1 and the other of 8 (8m2 area as well). |
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Then the possible answers should be B, C or E |
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2. |
Statement 2
alone is not sufficient because it gives the relations of one side of the
rectangle to one side of the other rectangle, not the other side of one to
the other. |
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Then the possible answers should be C or E |
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3. |
Because
statement 1 and 2 alone are not sufficient for the same reason - one
rectangle can be of 4 x 2 and the other 8 x 1, the area is the same (8m2) and
the ratio between one side to the other is 1:2 or 2:1, but yet, the second
triangle is not silver, but could be if the data is changed with the same
logical path. |
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The the best possible answer is B |
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| 10 |
In the
figure shown, two rectangles with the same dimensions overlap to form a
shaded region. If each rectangle has perimeter 12 and the shaded region has
perimeter 3, what is the total length of the heavy line segments? |
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c e |
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A. |
# |
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B. |
# |
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C. |
# |
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D. |
22 |
E. |
# |
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f |
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h |
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d |
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g |
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The best answer choice is (C); Obvious answer was
(D) |
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1. |
[(ab + bd + dc + ca) +
(ef + fg + gh + he)] = 24 |
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2. |
id + dh + hi = 3 |
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3. |
We need to find what is
the length of: |
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4. |
ac + ci + bd + ab + ie +
ef + fd - Let's call it L |
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5. |
L = [(ab + bd + dc + ca)
+ (ef + fg + gh + he)] - [id + dh + hi] |
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L = 24 - 3 = 21 |
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The the best possible answer is C |
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| Geometry |
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| Quadrilaterals |
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| 11 |
Is quadrilateral Q a
square? |
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(1) |
The sides of Q have the
same length |
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(2) |
The diagonals of Q have
the same length |
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The best answer choice is (C); Obvious answer was
(A) |
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Facts to remember |
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1. |
If the
sides of Q have the same length it could be either a square or either a rhombus - so statement 1 is not
sufficient |
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Then the possible answers should be B, C or E |
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2. |
If the
diagonals of Q have the same length it could be either a square or either a
rectangle |
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Then the possible answers should be C or E |
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3. |
From the two statements,
we can conclude that Q is a square |
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The the best possible answer is C |
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| 12 |
The surface
area of a square tabletop was changed so that one of the dimansions was
reduced by 1 inch and the other dimension was increased by 2 inchs. What was
the surface area before these changes were made? |
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(1) |
After the changes were
made, the surface area was 70 square inches |
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(2) |
There was 25 percent
increase in one of the dimensions |
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The best answer choice is (D); Obvious answer was
(E) |
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Facts to remember |
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1. |
We know
before the change it was a square with each side = s. fo after the change,
the area of the rectangle = (s - 1)(s + 2). |
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2. |
From
statement 1, we understand that the area of the rectangle after the change =
70 square inches. So (s - 1)(s + 2) = 70. The result will be of one positive
number and one negative number. The positive number will be the length of one
side of the before change square and therefore of the area of the before
change square. |
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Then the possible answers should be A or D |
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3. |
From
statement 2, we know that the increase was only on one side (increase of 2
inches), and that this increase was by 25%. So 0.25s = 2, therefore s = 8 and
the area of the square can be determined |
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The the best possible answer is D |
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| Geometry |
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| Quadrilaterals |
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| 13 |
Is quadrilateral RSTV
a rectangle? |
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(1) |
The measure of <RST is
90° |
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(2) |
The measure of <TVR is
90° |
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The best answer choice is (E); Obvious answer was
(C) |
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Facts to remember |
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1. |
In manner for
quadrileteral RSTV to be a rectangle, all its angles have to be = 90°. |
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2. |
Statement 1
is not sufficient because it states that one angle (RST) equals 90°. The
shape could be either of a rectangle, either a trapezoid with a right angle,
either any other quadrilateral with one right angle, either a rectangle |
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S |
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S |
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S |
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R |
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V |
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T |
V |
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T |
V |
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T |
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Then the possible answers should be B, C or E |
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3. |
Statement 2 alone is the
same as statement 1. |
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Then the possible answers should be C or E |
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4. |
Statement 1
+ 2 are not sufficient because with 2 right angles, the shape could either
look like that: |
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S |
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S |
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R |
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V |
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V |
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or either a rectangle,
so therefore, cannot be determined. |
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The the best possible answer is E |
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| 14 |
The figure
shown is composed of 6 squares, each with side s centimeters. If the number
of centimeters in the perimeter of the figure is equal to the number of
square centimeters in its area, what is the value of s |
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A. |
1 |
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B. |
5 |
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C. |
2 |
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D. |
5 |
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E. |
7 |
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3 |
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2 |
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3 |
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The best answer choice is (E); Obvious answer was
(B) |
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| Do not make the error of turning the figure into a rectangle for
perimeter calculation sakes |
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| 1. |
The area =
6s2. The perimeter = 14s, since the perimeter = the area, therefore 6s2 =
14s. Divide all by s (since s = 0), so 6s = 14 - s = 7/3 |
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| The the best
possible answer is E |
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| Geometry |
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| Circles |
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| 15 |
The outline
of a sign for an ice-cream store is made by placing ¾ of the circumference of
a circle with radius 2 feet on top of an isosceles triangle with height 5
feet, as shown. What is the perimeter, in feet, of the sign? |
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A. |
3π + 3√3 |
B. |
3π + 6√3 |
C. |
3π + 2√33 |
D. |
4π + 3√3 |
E. 4π + 6√3 |
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The best answer choice is (B); Obvious answer was
(?) |
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O |
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2 |
2 |
1. |
In the
shown figure, adding the extra material in red, O is the center of the circle
= 2. The measure of <AOB = 90° because it cuts off an arc that is ¼ of the
circumference (placing ¾ of the circumference of a circle). |
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D |
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A |
B |
2. |
So, the right ∆AOB
- AB = √22 + 22 = √8 = 2√2 |
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3. |
Therefore, AD = DB = ½AB
= ½ 2√2 = √2 |
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4. |
In right ∆ CDB, BC
= AC = √52 + (√2)2 = √27 = 3√3 |
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5. |
The length of the curved
line = ¾ x (circumference) = ¾ x 2πr = ¾ x 2π x 2 = 3π |
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5 ft |
6. |
Therefore, the perimeter
of the shape is: |
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3π + AC + BC =
3π + 3√3 + 3√3 = 3π + 6√3 |
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The the best possible answer is B |
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