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papgust Community Manager

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Sat Jul 10, 2010 9:59 pm

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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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Sat Jul 10, 2010 10:19 pm

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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]

papgust Community Manager

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Sat Jul 17, 2010 8:40 pm

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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.

papgust Community Manager

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Sat Jul 17, 2010 8:42 pm

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An equiangular quadrilateral DOES NOT have to be equilateral.

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

papgust Community Manager

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Sat Jul 17, 2010 8:44 pm

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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.

papgust Community Manager

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Sat Jul 17, 2010 8:49 pm

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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.

papgust Community Manager

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Sat Jul 17, 2010 8:54 pm

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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)

papgust Community Manager

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Sat Jul 17, 2010 8:59 pm

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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]

papgust Community Manager

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Sat Jul 17, 2010 9:00 pm

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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.

papgust Community Manager

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Sat Jul 17, 2010 9:12 pm

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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

papgust Community Manager

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																		 Sat Jul 17, 2010 9:21 pm
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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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Daily Quote:

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									Thanked by: kumadil2011

surajgarg Rising GMAT Star

Joined: 11 May 2010

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Tue Jul 20, 2010 5:33 am

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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

samark Just gettin' started!

Joined: 05 Sep 2009

Posted:: 8 messages

Sun Aug 01, 2010 12:13 pm

Quote

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!

samark Just gettin' started!

Joined: 05 Sep 2009

Posted:: 8 messages

Tue Aug 03, 2010 1:19 am

Quote

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!

kvcpk GMAT Titan

Joined: 31 May 2010

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Tue Aug 03, 2010 1:38 am

Quote

samark wrote:

papgust wrote:

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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