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Papgust's GMAT MATH FLASHCARDS directory

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papgust Community Manager
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Post Sat Jul 10, 2010 8:59 pm
Really sorry for a long break. I will be posting flashcards on "Geometry", the last topic of these flashcards.

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papgust Community Manager
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Post Sat Jul 10, 2010 9:19 pm
Types of angles:

Figure 1:



1. Adjacent angles: Any 2 angles that share a common side separating the 2 angles and that share a common vertex.
E.g.: | 1 and | 2 are adjacent angles. [See Figure 1]


2. Vertical angles: Any 2 angles that are not adjacent angles. Vertical angles are EQUAL in measure.
E.g.: | 1 and | 3 are vertical angles. [See Figure 1]


3. Complementary angles: Any 2 angles whose sum is 90 degrees. Complementary angles NEED NOT be adjacent to each other.


4. Supplementary angles: Any 2 angles whose sum is 180 degrees.[img][/img]



Figure 2:


L and M are parallel lines. T is a traversal.

5. Corresponding Angles: Angles that appear to be in the same relative position in each group of four angles. Corresponding angles are EQUAL when two parallel lines are cut by a traversal.
E.g.: | 1 and | 5 are corresponding angles. [See Figure 2]


6. Alternate Interior Angles: Angles within the lines being intersected, on opposite sides of the traversal, and are not adjacent. Alternate interior angles are EQUAL when two parallel lines are cut by a traversal.
E.g.: | 4 and | 6 are alternate interior angles. [See Figure 2]


7. Alternate Exterior Angles: Angles outside the lines being intersected, on opposite sides of the traversal, and are not adjacent. Alternate exterior angles are EQUAL when two parallel lines are cut by a traversal.
E.g.: | 1 and | 7 are alternate interior angles. [See Figure 2]


8. Consecutive Interior Angles: Angles are same-side interior angles. Consecutive interior angles are SUPPLEMENTARY when two parallel lines are cut by a traversal.
E.g.: | 4 and | 5 are consecutive interior angles. [See Figure 2]


9. Consecutive Exterior Angles: Angles are same-side exterior angles. Consecutive exterior angles are SUPPLEMENTARY when two parallel lines are cut by a traversal.
E.g.: | 1 and | 8 are alternate interior angles. [See Figure 2]

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papgust Community Manager
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Post Sat Jul 17, 2010 7:40 pm
Lines segments in Triangles:

* ALTITUDES:

Altitudes are the perpendicular segments from a vertex to the opposite sides. The three lines containing the altitudes intersect in a single point, which may or may not be inside the triangle.


* MEDIANS:

A Median is the line segment drawn from a vertex to the mid-point of its opposite side. The three medians meet in one point inside the triangle.


* ANGLE BISECTOR:

Angle bisector is a segment drawn from a vertex that bisects the vertex angle. The three angle bisectors meet in one point inside the triangle.


TAKEAWAY:

Altitude drawn from the vertex angle can be proven to be a median as well as an angle bisector in an ISOSCELES triangle.

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papgust Community Manager
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Post Sat Jul 17, 2010 7:42 pm
An equiangular quadrilateral DOES NOT have to be equilateral.

An equilateral quadrilateral DOES NOT have to be equiangular. [--- Unlike Equilateral Triangle --]

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Post Sat Jul 17, 2010 7:44 pm
Regular Polygons:


* When a polygon is BOTH equilateral and equiangular.

* Sum of INTERIOR angles of a convex polygon with 'n' sides = (n-2) * 180

* Sum of EXTERIOR angles of a convex polygon = 360 degrees.

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Post Sat Jul 17, 2010 7:49 pm
How to prove a figure as PARALLELOGRAM?

5 ways:

* If both sides of opposite sides of a quadrilateral are EQUAL.

* If both pairs of opposite angles of a quadrilateral are EQUAL.

* If all pairs of consecutive angles of a quadrilateral are SUPPLEMENTARY.

* If one pair of opposite sides of a quadrilateral is both EQUAL and PARALLEL.

* If the diagonals of a quadrilateral BISECT each other.


TAKEAWAY:
A diagonal of a Parallelogram DIVIDES it into 2 congruent triangles.

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Post Sat Jul 17, 2010 7:54 pm
Isosceles Trapezoids:


* If the legs are EQUAL.

* Base Angles are EQUAL.

* Diagonals are EQUAL.

* Median of any trapezoid:
1. Is parallel to both bases.
2. Has length = 1/2 * (Sum of bases)

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Post Sat Jul 17, 2010 7:59 pm
Regular Polygons:

* One point in its interior that is equidistant from its vertices is called the center of the regular polygon.

* An Apothegm is a line segment that goes from the center and is perpendicular to one of the polygon's sides.


Perimeter (regular n-gon) = n * s [n--> no of sides and s--> length of a side]

Area (regular n-gon) = 1/2 * a * p [a--> Apothegm length and p--> perimeter of regular n-gon]

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Post Sat Jul 17, 2010 8:00 pm
Similar Polygons:

* Two polygons with the same shape.

* When two polygons are similar, then the following MUST be true.
i. Corresponding angles are EQUAL.
ii. The ratios of pairs of corresponding sides must all be EQUAL.

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Post Sat Jul 17, 2010 8:12 pm
Similar Triangles:

If two triangles are similar, then the ratio of any two corresponding segments (such as altitudes, medians, angle bisectors) EQUALS the ratio of any two corresponding sides.

Example:



If ∆ QRS ~ ∆ TUV,
then QR/TU = RS/UV = QS/TV.

According to the theorem,

Length of altitude RA / Length of altitude UD = QR / TU

Length of median QB / Length of median TE = QR / TU

Length of bisector CS / Length of bisector FV = QR / TU

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Post Sat Jul 17, 2010 8:21 pm
Perimeters and Areas of Similar Triangles:


When two triangles are similar, the reduced ratio of any two corresponding sides is called the scale factor of the similar triangle.




i. If two similar triangles have a scale factor of a:b, the the ratio of their perimeters is a:b.

Example:

6/3 = 8/4 = 10/5 ==> 2/1 (or) 2:1 (Perimeter)


ii. If two similar triangles have a scale factor of a:b, then the ratio of their areas is a^2 : b^2.

Example:

Area of ∆ ABC / Area of ∆ DEF = 24 / 6 = 4 / 1 (or) 4:1

4:1 is nothing but a^2:b^2 (or) 2^2 : 1^2.

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surajgarg Senior | Next Rank: 100 Posts
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Post Tue Jul 20, 2010 4:33 am
papgust wrote:
Perimeters and Areas of Similar Triangles:


When two triangles are similar, the reduced ratio of any two corresponding sides is called the scale factor of the similar triangle.




i. If two similar triangles have a scale factor of a:b, the the ratio of their perimeters is a:b.

Example:

6/3 = 8/4 = 10/5 ==> 2/1 (or) 2:1 (Perimeter)


ii. If two similar triangles have a scale factor of a:b, then the ratio of their areas is a^2 : b^2.

Example:

Area of ∆ ABC / Area of ∆ DEF = 24 / 6 = 4 / 1 (or) 4:1

4:1 is nothing but a^2:b^2 (or) 2^2 : 1^2.
Pls don't end it, keep posting more Smile

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samark Newbie | Next Rank: 10 Posts
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Post Sun Aug 01, 2010 11:13 am
papgust wrote:
How to find REMAINDER for LARGE POWERS of numbers:


There are 2 ways to do so:

1. Pattern Method:

Example:

What is the remainder when 2^56 / 7 ?

Solution:
Remainder when 2^1 is divided by 7 is 2
Remainder when 2^2 is divided by 7 is 4
Remainder when 2^3 is divided by 7 is 1
Remainder when 2^4 is divided by 7 is 2 --> Repeats again.

The remainder repeats after 3 steps i.e. in the 4th step.

Now, Divide the power (or index) by 3 (no of steps after which remainder repeats) and compute a new remainder.

56 % 3 --> 2 (remainder)

Now, raise the base (2) to the power 2 (new remainder). 2^2 % 7 --> 4.

Thus, 4 is the remainder when 2^56 / 7.



2. Remainder Theorem Method: (NOT RECOMMENDED unless clear)

Example:

What is the remainder when 2^51 / 7 ?

Solution:
2^51 can be changed to (2^3)^17.
7 can be changed to (8-1) OR (2^3 - 1)

Substitute 'x' in place of 2^3,

x^17 / (x-1)

Remainder is f(1). Substitute 1 in 'x',

Remainder is 1.

Thus, 1 is the remainder when 2^51 / 7.
Is this rule only applicable for powers of 2? Thanks!

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samark Newbie | Next Rank: 10 Posts
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Post Tue Aug 03, 2010 12:19 am
papgust wrote:
If ax^2+bx+c > 0, where a > 0,
Then
x DO NOT LIE between xL and xU

(xL and xU are lower and upper limits of x respectively)


If ax^2+bx+c < 0, where a > 0,
Then
x LIES between xL and xU

(xL and xU are lower and upper limits of x respectively)
Can someone give an example? It's not completely clear to me..Thanks!

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kvcpk Legendary Member
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Post Tue Aug 03, 2010 12:38 am
samark wrote:
papgust wrote:
If ax^2+bx+c > 0, where a > 0,
Then
x DO NOT LIE between xL and xU

(xL and xU are lower and upper limits of x respectively)


If ax^2+bx+c < 0, where a > 0,
Then
x LIES between xL and xU

(xL and xU are lower and upper limits of x respectively)
Can someone give an example? It's not completely clear to me..Thanks!
Hi Samark,
Its simple. Let us take the equation as
x^2 -1 >0
Which values of x will satisfy this?
The lower limit of x is -1 and the upper limit of x is 1

Now the above rule says that x does not lie between -1 and 1.

Example take x=0 -> 0-1>0
-1 is not greater than 0.

if you take x=2, 4-1>0
3 is greater than 0.

Hence x should not lie between -1 and 1

Similarly, lets take x^2-1<0
Again upper and lower limits are -1 and 1
Now as per the rule above, x should lie between -1 and 1

Example take x=0 -> 0-1<0
-1 is less than 0.

if you take x=2, 4-1<0
3 is not less than 0.

Hope this helps!!

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