QUADRILATERALS

TRAPEZIOD

One pair of opposite sides is parallel, but not equal
Area = 1/2 * (a+b)*h

Isosceles Trapezoid
What is the area of the trapezoid as show?
Angle A = 120 degrees
The perimeter of the trapezoid ABCD = 36

The area of a trapezoid is equal to the average of the bases multiplied by the height. In this problem, you are given the top base (AB = 6), but not the bottom base (CD) or the height. (Note: 8 is NOT the height!) In order to find the area, you will need a way to figure out this missing data. Drop 2 perpendicular lines from points A and B to the horizontal base CD, and label the points at which the lines meet the base E and F.


missing data. Drop 2 perpendicular lines from points A and B to the horizontal base CD, and label the points at which the lines meet the base E and F.

EF = AB = 6 cm. The congruent symbols in the drawing tell you that Angle A and Angle B are congruent, and that Angle C and Angle D are congruent. This tells you that AC = BD and CE = FD.
Statement (1) tells us that Angle A = 120. Therefore, since the sum of all 4 angles must yield 360 (which is the total number of degrees in any four-sided polygon), we know that Angle B = 120, Angle C = 60, and Angle D = 60. This means that triangle ACE and triangle BDF are both 30-60-90 triangles. The relationship among the sides of a 30-60-90 triangle is in the ratio of x: x √3: 2x , where x is the shortest side. For triangle ACE, since the longest side AC = 8 , CE = 4 and AE = 4 √3 . The same measurements hold for triangle BFD. Thus we have the length of the bottom base (4 + 6 + 4) and the height and we can calculate the area of the trapezoid


asurements hold for triangle BFD. Thus we have the length of the bottom base (4 + 6 + 4) and the height and we can calculate the area of the trapezoid

Statement (2) tells us that the perimeter of trapezoid ABCD is 36. We already know that the lengths of sides AB (6), AC (8), and BD (8) sum to 22. We can deduce that CD = 14. Further, since EF = 6, we can determine that CE = FD = 4. From this information, we can work with either Triangle ACE or Triangle BDF, and use the Pythagorean theorem to figure out the height of the trapezoid. Now, knowing the lengths of both bases, and the height, we can calculate the area of the trapezoid

EACH STATEMENT ALONE IS SUFFICIENT


POLYGONS – 3 DIMENSIONAL


CIRCLES

Circumference of a circle= 2πr
Circumference of a semi-circle= πr
Perimeter of a semi-circle = πr + D (Diameter of the circle)
Area of a circle = πr²
Area of a semi-circle = πr² / 2
Length of the arc = (θ/360) × (2 π r)
Area of the sector = (θ/360) × πr²
Distance travelled by a wheel in n revolutions = n x circumference
Perimeter of a sector = (θ/360) × (2 π r) + 2r
Circles, when graphed on the coordinate plane, have an equation of x2 + y2 = r2 where r is the radius (standard form) when the center of the circle is the origin.  When the center of the circle is (h, k) and the radius is of length r, the equation of a circle (standard form) is (x - h)2 + (y - k)2 = r2


CIRCLES, ARCS AND SECTORS
A circle is a set of points that are equidistant from a fixed point called the centre. It comprises 360 degrees.
RADIUS: distance from the centre of the circle to any point on the circle; can be shown as a line segment connecting the centre to a point on the circle – OC
DIAMETER: is a line segment that connects two points on the circle and goes through the centre of the circle - AE
Diameter = 2r
CHORD: is any line segment whose endpoints are any two points on the circle – BD
The tangent line and the radius of the circle that has an endpoint at the point of tangency are perpendicular to each other – AO IS PERPENDICULAR TO XY
ARC – a portion/distance on the circle

A
C
B
E
O
D
Y
X
A
B
O
C


INSCRIBED CIRCLES
When a circle is inscribed inside a square, the side equals the diameter


CO – ORDINATE GEOMETRY


CO-ORDINATE GEOMETRY – GENERAL CONCEPTS
CO-ORDINATE POINTS

Points in the plane are identified by using an ordered pair of numbers, (x , y), where x is the x co-ordinate and y is the y co-ordinate. For example, in (4, -2), the x co-ordinate is 4 and the y co-ordinate is -2
Any point on the X axis can be taken as (a,0)
Any point on the Y axis can be taken as (0,b)
The point (0, 0), where the axes cross, is called the origin
Along the X axis, the y co-ordinate is 0 and along the Y axis, the x co-ordinate is 0

INTERCEPTS: The point where a line intersects a co-ordinate axis is called an intercept
X intercept: the point at which the line intersects the x axis
The x-intercept is expressed using the ordered pair (x, 0), where x is the point where the line intersects the x-axis. The x-intercept is the point on the line at which y = O
Y intercept: the point at which the line intersects the y axis
The y-intercept is expressed using the ordered pair (0, y), where y is the point where the line intersects the y-axis. The y-intercept is the point on the line at which x = O


ssed using the ordered pair (0, y), where y is the point where the line intersects the y-axis. The y-intercept is the point on the line at which x = O

LINE
A line can be formed in any of the three ways
By knowing the co-ordinates of any two points on the line
By knowing the co-ordinates of one point on the line and the slope of the line
By knowing the equation of the line

SLOPE OF A LINE: Y2 – Y1 / X2 – X1
A line in the plane is formed by the connection of two or more points. Therefore, the slope is defined as the “rise” over the “run” – that is, how much the line rises vertically divided by how much the line runs horizontally
For example, what is the slope of the line for the following co-ordinates: (2,-3) and (6, 5)
Slope = 5 – (-3) / (6 – 2) = 8/4, which is equal to 2

The slope is always the co-efficient of x in the equation (point-intercept) form y = mx + b. Therefore, in the equation y = 4x + 8; the slope is 8
If the equation of the line is given in the general form ax + by + c = 0; then the slope is –a/b and the intercept is –c/b


y = 4x + 8; the slope is 8
If the equation of the line is given in the general form ax + by + c = 0; then the slope is –a/b and the intercept is –c/b

The slope of all points/ co-ordinates on a line will ALWAYS be the same. However, just because two points have the same slope, doesn’t necessarily mean that the lie on the same line and vice versa. For instance, parallel lines have the same slope; conversely, if two lines have the same slope they could either be parallel or lie on each other
If the slope is 1, the angle formed by the slope is 45 degrees
Parallel lines have equal slopes
Perpendicular lines have negative reciprocal slopes
SLOPE DIRECTION - There are four types of slope an equation can have
Positive – rises upward from left to right
Negative – falls downward from left to right
Zero – parallel to the x axis
Undefined – parallel to the y axis
Every line (but the one that crosses the origin OR the one that is parallel to X or Y axis OR the X and Y axis themselves) crosses three quadrants


CO-ORDINATE GEOMETRY – EQUATION OF A LINE/ SLOPE-INTERCEPT LINE
What is the slope-intercept form for a line with the equation 6x + 3y = 18?
Rewrite the equation by solving for y as follows:
6x + 3y = 18
2x + y = 6
y = – 2x + 6
Thus, the slope is -2 and the y-intercept (b) is 6

Horizontal and vertical lines are not expressed in the y = mx +b form. Instead, they are expressed as simple, one-variable equations. Horizontal lines are expressed in the form: y = some number, such as y = 3 or y = 5 (i.e., equation of the line parallel to the x axis is 3 or 5). Vertical lines are expressed in the form: x = some number, such as x = 4 or x = 7 (i.e., equation of the line parallel to the y axis is 4 or 7)
Equation of the X axis is Y= 0
Equation of the Y axis is X= 0


CO-ORDINATE GEOMETRY – 3
DISTANCE BETWEEN TWO POINTS
The distance between any two points in the coordinate plane can be calculated by using the Pythagorean Theorem
Example: what is the distance between the points (1, 3) and (7, -5)?
Draw a right triangle connecting the points
Find the lengths of the two legs of the triangle by calculating the rise and the run
The y-coordinate changes from 3 to -5, a difference of 8 (the vertical leg)
The x-coordinate changes from 1 to 7, a difference of 6 (the horizontal leg)
Use the Pythagorean Theorem to calculate the length of the diagonal, which is the distance between the points
62 + 82 = c2
36 + 64 = c2
100 = c2
c = 10

POSITIVE AND NEGATIVE QUADRANTS
Quadrant I: x and y are both positive
Quadrant II: x is negative, y is positive
Quadrant III: x is negative, y is negative
Quadrant IV: x is positive, y is negative


nd y are both positive
Quadrant II: x is negative, y is positive
Quadrant III: x is negative, y is negative
Quadrant IV: x is positive, y is negative

The GMAT sometimes asks you to determine which quadrants a given line passes through. For example: Which quadrants does the line 2x + Y = 5 pass through?
2x + y = 5
y = 5 – 2x
Since b = 5, the y-intercept is the point (0, 5). The slope is -2, so the line slopes downward steeply to the right from the y-intercept. Although we do not know exactly where the line intersects the x-axis, we can now see that the line passes through quadrants I, II, and IV

PARALLEL LINES
Two lines are said to be parallel when their slopes are equal and the y – intercept is different
For example: If Point A is (5,20) and Point B is (30,7), define a line through point C (12,10) that is parallel to the line that passes through both A and B
Slope of AB = (7-20)/(30-5)  -13/25  -0.52
Substituting in equation y-y1 = m(x-x1)  y = -0.52(x – 12) + 10
y = -0.52x + 16.24
Distance between two parallel lines y = mx + b and y= mx + c can be found by the formula D = | b – c |/ √m2 + 1


CO-ORDINATE GEOMETRY – 4
PERPENDICULAR BISECTORS
The perpendicular bisector of a line segment forms a 90° angle with the segment and divides the segment exactly in HALF. The key to solving perpendicular bisector problems is remembering this property: the perpendicular bisector has the negative reciprocal slope of the line segment it bisects. That is, the product of the two slopes is -1 (the only exception occurs when one line is horizontal and the other line is vertical, since vertical lines have undefined slopes)
Example: If the coordinates of point A are (2, 2) and the coordinates of point B are (0, -2), what is the equation of the perpendicular bisector of line segment AB?
Slope of AB is 2
Slope of the perpendicular bisector is – (1/2)
Now we know that the equation of the perpendicular bisector has the following form: y = (-1/2)x + b. However, we still need to find the value of b To do this, we will need to find one point on the perpendicular bisector
The perpendicular bisector passes through the midpoint of AB. Thus, if we find the midpoint of AB, we will have found a point on the perpendicular bisector. The x co-ordinate of the perpendicular bisector will be the average of the x co-ordinates of A and B. Similarly, the y co-ordinate of the perpendicular bisector of AB will be the average of the y co-ordinates of A and B.
Plug in the values of x (1) and y (0) to get b = 1/2


dinate of the perpendicular bisector of AB will be the average of the y co-ordinates of A and B.
Plug in the values of x (1) and y (0) to get b = 1/2

INTERSECTION OF TWO LINES
Recall that a line in the coordinate plane is defined by a linear equation relating x and y. That is, if a point (x, y) lies on the line, then those values of x and y satisfy the equation. For instance, the point (3, 2) lies on the line defined by the equation y = 4x - 10, since the equation is true when we plug in x = 3 and y = 2. On the other hand, the point (7, 5) does not lie on that line, because the equation is false when we plug in x = 7 and y = 5
So, what does it mean when two lines intersect in the coordinate plane? It means that at the point of intersection, BOTH equations representing the lines are true. That is, the pair of numbers (x, y) that represents the point of intersection solves BOTH equations. Finding this point of intersection is equivalent to solving a system of two linear equations
Example: At what point does the line represented by y = 4x - 10 intersect the line represented by 2x + 3y = 26?
Solving the equations simultaneously, we get x=4 and y=6
Therefore, the point of intersection is (4,6)
If two lines in a plane do not intersect, then the lines are parallel. If this is the case, there is NO pair of numbers (x, y) that satisfies both equations at the same time
The two equations/lines in a plane may also represent the same line. In this case, infinitely many points (x, y) along the line satisfy the two equations (which must actually be the same equation in two disguises)


line. In this case, infinitely many points (x, y) along the line satisfy the two equations (which must actually be the same equation in two disguises)

The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2), internally, in the ratio m:n are (nx1 + mx2)/m+n and (ny1 + my2)/m+n


ADVANCED CONCEPTS
EQUATION OF A CIRCLE
The equation of a circle with centre (h, k) and radius r is (x – h)2 + (y – k)2 = r2
The equation of a circle with centre (0, 0) and radius r is x2 + y2 = r2

In the coordinate plane, a circle has center (2, -3) and passes through the point (5, 0). What is the area of the circle?
(5 – 2)2 + (0 – (-3))2 = r2
(3)2 + (3)2 = r2
18 = r2
Therefore, area = 18 π

Find the equation of a circle whose center is at (4, 2) and is tangent to y-axis
Since the circle is tangent to y-axis, the radius of the circle is perpendicular to y-axis. It also means that the length of the radius is also the length of the perpendicular segment from the center of the circle to y –axis. The point of tangency is at (0, 2). To find the length of the radius, use the distance formula:
(0 – 4)2 + (2 – 2)2 = r2
r = 4


the circle to y –axis. The point of tangency is at (0, 2). To find the length of the radius, use the distance formula:
(0 – 4)2 + (2 – 2)2 = r2
r = 4

Write the equation of the circle with the given condition: (10, 8) and (4, -2) are the endpoints of the diameter
In a circle, the radius is one-half of the diameter. Since the given are the endpoints of the diameter, the center of the circle is the midpoint of the diameter
(h,k) = [(10+4)/2, (8-2)/2)]  (7,3)
The next step is to get the length of the radius. Since radius is one-half of the circle, it will equal the distance from the center to any one end point of the diameter: (10 – 7)2 + (8 – 3)2 = r2  r = root 34
Equation = (x – 7)2 + (y – 3)2 = 34


equal the distance from the center to any one end point of the diameter: (10 – 7)2 + (8 – 3)2 = r2  r = root 34
Equation = (x – 7)2 + (y – 3)2 = 34

What is the radius for the following – Circle tangent to the line 3x – 4y = 24 with center at (1, 0)
The radius of the circle is equal to the distance of the center (1, 0) from the line 3x – 4y = 24
To find the distance/radius from a point to a line, we use the formula: d = Ax + By + C / - √(A2 + B2 ), where A and B are the coefficients of x and y, and C is the constant in the equation of line. X and y are co-ordinates of the point
Therefore, A = 3, B = -4 and C = -24, x = 1, y = 0
Radius = 21/5


ADVANCED CONCEPTS
PARABOLAS

The equation y = Ax2 + Bx + C is the equation of the quadratic graph which is a parabola with axis parallel to y axis
C is the y intercept (i.e., the value of y when x is 0)

If A > 0, the parabola opens upwards. If A < 0, the parabola opens downwards

If B2 > 4AC, the parabola cuts the x-axis at 2 different points
If B2 = 4AC, the parabola touches the x-axis at one point (the two points become co-incident)
If B2 < 4AC, the parabola does not cut the x-axis at all

The vertex of a parabola represents the maximum or minimum value of the function. The vertex is located at point (-b/2a, c – (b2/4a))


MINOR PROBLEM TYPES


MINOR PROBLEM TYPES
OPTIMIZATION
SCHEDULING
GROUPING
Scheduling problems, which require you to determine possible schedules satisfying a variety of constraints, can usually be tackled by careful consideration of extreme possibilities, usually the earliest and latest possible time slots for the events to be scheduled

Example
How many days after the purchase of Product X does its standard warranty expire? (1997 is not a leap year
When Mark purchased Product X in January 1997, the warranty did not expire until March 1997
When Santos purchased Product X in May 1997, the warranty expired in May 1997

Rephrase the two statements in terms of extreme possibilities:
Statement 1
Shortest possible warranty period: Jan. 31 to Mar. 1 (29 days later)
Longest possible warranty period: Jan. 1 to Mar. 31 (89 days later)
Statement 2
Shortest possible warranty period: May 1 to May 2, or similar (1 day later)
Longest possible warranty period: May 1 to May 31 (30 days later)


ement 2
Shortest possible warranty period: May 1 to May 2, or similar (1 day later)
Longest possible warranty period: May 1 to May 31 (30 days later)

Even taking both statements together, there are still two possibilities-29 days and 30 days -so both statements together are still insufficient
Note that, had the given year been a leap year, the two statements together would have become sufficient!
In general optimization problems, the objective is to maximize or minimize some quantity, given constraints on other quantities. These quantities are all related through some equation


the objective is to maximize or minimize some quantity, given constraints on other quantities. These quantities are all related through some equation

Example
The guests at a football banquet consumed a total of 401 pounds of food. If no individual guest consumed more than 2.5 pounds of food, what is the minimum number of guests that could have attended the banquet?
Begin by considering the extreme case in which each guest eats as much food as possible, or 2.5 pounds apiece. The corresponding number of guests at the banquet works out to 401/2.5 = 160.4 people
You obviously cannot have a fractional number of guests at the banquet. Thus the answer must be rounded. To determine whether to round up or down, consider the explicit constraint: the amount of food per guest is a maximum of 2.5 pounds per guest. Therefore, the minimum number of guests is 160.4 (if guests could be fractional), and we must round up to make the number of guests an integer: 161
Note the careful reasoning required! Although the phrase "minimum number of guests" may tempt you to round down, you will get an incorrect answer if you do so. In general, as you solve this sort of problem, put the extreme case into the underlying equation, and solve. Then round appropriately


er if you do so. In general, as you solve this sort of problem, put the extreme case into the underlying equation, and solve. Then round appropriately

In grouping problems, you make complete groups of items, drawing these items out of a larger pool. The goal is to maximize or minimize some quantity, such as the number of complete groups or the number of leftover items that do not fit into complete groups. As such, these problems are really a special case of optimization. One approach is to determine the limiting factor on the number of complete groups


. As such, these problems are really a special case of optimization. One approach is to determine the limiting factor on the number of complete groups

Example
Orange Computers is breaking up its conference attendees into groups. Each group must have exactly one person from Division A, two people from Division B, and three people from Division C. There are 20 people from Division A, 30 people from Division B, and 40 people from Division C at the conference. What is the smallest number of people who will not be able to be assigned to a group?
The first step is to find out how many groups you can make with the people from each division separately, ignoring the other divisions. There are enough Division A people for 20 groups, but only enough Division B people for 15 groups and only enough Division C people for 13 groups. So the limiting factor is Division C. Therefore the total number of people left over will be 7 from A + 4 from B + 1 from C = 12
COMPUTATION
Very occasionally, the GMAT features problems centered on computation problems that contain no variables at all.
Example
Five identical pieces of wire are soldered together to form a longer wire, with the pieces overlapping by 4 cm at each join. If the wire thus made is exactly 1 meter long, how long is each of the identical pieces? (1 meter = 100 cm)
It is easy to assume that, because there are 5 pieces, there must be 5 joins. Instead, there are only 4 joins. Each join includes 4 cm of both wires joined, but is only counted once in the total length of 100 cm. Therefore, the total length of all the original wires is 100 + 4(4) = 116 cm. Because there are 5 wires, each wire is (116/5) = 23.2 cms long


MENTAL MATH / SHORT CUTS