Geometry Chapter 1: Angles and Lines
  
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1: Introduction

Geometry is always a logical puzzle. The GMAT gives you a tiny piece of information about a figure, and you must logically deduce the information the question wants. Geometry questions require you to follow a series of “if…then” statements, constantly solving the puzzle, piece by piece, to the correct answer.


II. Lines and Angles (Beginner)

The most basic element in geometry is the point:

 

 

If we have two points, we can connect them to form a line
When two lines originate from the same point, they form an angle:



The point of intersection is the vertex and the two lines form the sides of the angle. The angle is designated in a number of ways. It is BAC where the middle letter is the vertex, A where A is the vertex, or x where x is located inside the angle.

Angles are measured in degrees, denoted by this symbol: °. For example, a 30 degree angle would be written as 30°. When placed next to one another, angles may be added up.

  The sum of x + y here is 90°.
  The sum of x + y here is 180°. Notice that x and y create a straight line. All straight lines are 180°.

 

Example 1


 


A beam of light is shown in the figure to the left reflecting off a mirror. What is the value of x?

 

 

Solution

Because these three angles form a straight line, they must add up to 180°. Therefore:
x + 3x + x = 180°
5x = 180°
x = 36°

 


   Geometry Chapter 2: Intersecting Angles
 


Vertical Angles

When two lines intersect each other, they create four angles:



In this case, the angles opposite one another are called vertical angles. As you can see, vertical angles are by definition always equal to each other. Therefore, in this case, 1 = 2, and 3 = 4. Another way to say this is that both "little angles" will equal each other and both "big angles" will equal each other. Also, note that a "little angle" plus a "big angle" will always equal 180°.


     

 

 

Parallel Lines

Two lines that never intersect and never get closer or farther away from one another are called parallel lines. || is the symbol for parallel lines. In the figure to the right, A || B, or line A is parallel to line B.


When two parallel lines are cut by a third line, they form a system of Vertical Angles.

In this figure, we can see that 1 =4 and 5 =8. But, because they are formed by two parallel lines, they are all equal to each other, so: 2 = 3 = 6 = 7.

There are many terms to describe these angles, such as “alternate interior” or “alternate exterior,” but these terms are not used on the test. For the GMAT, it is simply enough to know that all the little angles will always be equal to each other and all the big angles will always be equal to each other.

Additionally, you should realize that any little angle added to any big angle will always equal 180°.


Perpendicular Lines

When two lines intersect each other at a 90° angle, they are said to be perpendicular to one another. is the symbol for perpendicular, and mn means that m and n are perpendicular to each other.


Midpoint and Bisect

Two more vocabulary words you need before you can finish this section. Midpoint is the center point of any line. In the example below, point B is the midpoint of line AC.

Bisect means “to cut in half”. Anything may be bisected: a line, an angle, a circle, a square. In the figure above, point B “bisects” line AC.

 

Example 1

A complement of an angle the second of a pair that ads to 90°. A supplement is the second of a pair that adds to 180°. If the complement of an angle is one quarter of its supplement, what is the angle?



Solution

Let x be the angle.
Its complement y is y = 90 - x
Its supplement z is z = 180 - x

90 - x = (180 - x)/4
If y = z/4, we can set this.
360 - 4x = 180 - x
Multiply both sides by 4.
180 = 3x
Subtract 180 from both sides and add 4x to both sides.
x = 60

 


   Geometry Chapter 3: Triangles
 

While there are many different kinds of triangles, there are some rules that are specific to all triangles, and we will start with these. Figure 1, below, is a generic triangle. If you draw different kinds of triangles, these rules will always hold true.

     
1.

The angles of any triangle will ALWAYS add up to 180°. For the three internal angles of a triangle: x° + y° + z° = 180°.

 
   
2. The biggest side is ALWAYS opposite the biggest angle and the smallest side is ALWAYS opposite the smallest angle. In this case, we can therefore see x must be the smallest angle, because it is opposite side A, which is the smallest side.  
     
3. Any side of a triangle will always be less than the sum of the other two sides but greater than the difference of the other two sides. In Figure 1, if we take side C, for example, we can say
(B – A) < C < (B + A).
         
4. If we draw an external line, as in Figure 2, the angle formed will always be equal to the sum of the other two angles in the triangle.
In this case, n = x + y. That is because x + y + z = 180 and n + z = 180. Therefore n must equal x + y. This property will be true for any triangle
   
         

Perimeter and Area

The perimeter of any figure is the distance around the outside of the figure, or the sum of the sides of that figure. The perimeter in Figure 1 is A + B + C.

The area of any figure is the amount of space that is inside that figure. Each figure has a different formula for finding its area.

To find the area of a triangle, we always need two elements: the base and the height.

The base of any triangle can be any of its sides. The height of the triangle is the perpendicular distance from the base to the opposite angle. Here are some examples of triangles with different bases and heights.

Every triangle has three bases and three corresponding heights. To find the area, use the following formula:

Triangle Types

There are three important triangle types on the GMAT.

Isosceles Triangle

 

An isosceles triangle has two equal sides and two equal angles. In the figure to the right: A = C, and sides AB = BC. This is in line with rule 2 above: if the two sides are equal, their opposite angles must be equal as well.

NOTE: The height of an isosceles triangle always bisects the triangle, creating two equal right triangles.
     

Equilateral Triangle

 

An equilateral triangle has three equal sides and three equal angles. Because a triangle’s angles must add up to 180°, each angle of an equilateral triangle is always 60°.

     

Right Triangle

 



A right triangle is any triangle with a 90° angle. The two perpendicular sides are called legs and the side opposite the right angle is called the hypotenuse.

Right Triangles

There is a constant relationship between the legs and the hypotenuse called the Pythagorean Theorem. The Pythagorean Theorem states that the square of the hypotenuse will equal the sum of the squares of the legs.

a2 + b2 = c2

Try using the Pythagorean Theorem yourself on a few triangles:


32 + 42 = c2
9 + 16 = c2
25 = c2
5 = c

122 + b2 = 132
144 + b2 = 169
b2 = 25
b = 5
42 + 82 = c2
16 + 64 = c2
80 = c2


Special Right Triangles based on Sides

There are several common right triangles on the GMAT, and if you are familiar with them, you will be able to recognize them easily on the exam.

3 – 4 – 5 (see above)
5 – 12 – 13 ( see above)
7 – 24 – 25
8 – 15 – 17

Note two important things. First, the largest side is ALWAYS the hypotenuse. If two legs of a triangle are 3 and 5, the hypotenuse WILL NOT be 4. Try it with the Pythagorean Theorem and you will see why. Second, these values are ratios. This means that multiples of these triangles are also special right triangles. The 3 – 4 – 5 is also the 6 – 8 – 10 and the 9 – 12 – 15. The 5 – 12 – 13 is also the 10 – 24 – 26.


Special Right Triangles based on Angles

There are also two special right triangles that we identify by the measurements of their angles. The first is the 45° – 45° – 90° triangle.

As you can see from the figure, the 45° – 45° – 90° triangle is an isosceles triangle, so all the rules we know for isosceles triangles apply here. There are two equal sides and two equal angles. For that reason, there is a constant relationship between the legs and the hypotenuse. Whatever the legs are, the hypotenuse is always that times .

 

 


The 30° – 60° – 90° right triangle works in the exact same way as the 45° – 45° – 90° triangle, but with different dimensions.

As you can see from the diagram, the dimensions of the 30° – 60° – 90° triangle are x - x - 2x. It is imperative that you memorize this triangle and these dimensions.

One easy way is to remember our rule from above – the smallest side is opposite the smallest angle and vice versa. Since x is the smallest, it is always opposite the 30° angle, and 2x, which is the largest, must always be opposite the 90° angle (meaning it is always the hypotenuse). Since x is between 1x and 2x ( it is approximately 1.7 - between 1 and 2), it is always opposite the middle angle, 60°.


An equilateral triangle may be divided into two 30° – 60° – 90° right triangles. The height of an equilateral triangle always bisects the triangle, creating two equal right triangles that are both 30-60-90 right triangles.

Because of this property, from the side of an equilateral triangle we may always figure out the height. That means we do not need to be given the height of an equilateral triangle to find the area.

Look at the example in the figure on the left. The sides are all length 10, but with the height drawn, the triangle has been bisected. The base is now cut in half, creating two 30° – 60° – 90° triangles, each with x = 5 and 2x = 10. The height of the triangle, then, must be x, or, in this case, 5. The area, , would be .

Note: Since the sides of an equilateral triangle are always proportional, there is a special formula for its area. If the side of the triangle is s, the formula is. On the above equilateral triangle it would be 100 / 4 = 25 .


 
800score.com Tip:
Keep in mind that Mr. GMAT knows that you have memorized common triangle structures, whether they be 3-4-5 triangles or 45-45 right triangles. The GMAT isn't supposed to be "beatable" by preparation, so Mr. GMAT gets around this by setting traps designed to fool people who have studied from second-rate GMAT prep books. As a result, if you see 3 and 4 as sides of a triangle DO NOT ASSUME that the hypotenuse is 5, because Mr. GMAT isn't always so nice.

Similar Triangles

Triangles with the same angles are always proportional to each other. These triangles are called similar triangles because we can relate them to each other. In each set of similar triangles, the same sides opposite the same angles are proportional. There are three ways similar triangles can appear on the exam.

1.

Figure 1: Two triangles, same angles.
Notice that the second triangle’s sides are half the first triangle. We can make this comparison because they have the same angles.

2.

Figure 2: One triangle, with a parallel line cutting through its middle.
This figure is composed of two triangles. They each share angle y, and their bases are parallel. For that reason, they both have angle x and angle z, and the smaller, inner triangle is proportional to the larger triangle.

3.

Figure 3: Two triangles, connected at one vertex, forming vertical angles, with parallel bases.
In this figure, the two triangles are similar because they share angle z, and their bases are parallel. In this case, angles x and y are on opposite sides from one another (because of alternate interior angles).



Example 1 (easy)

 

 

For the triangle shown, find L.

Solution

The small box in the corner signifies a right triangle. The ratio of the two legs is 12/16 = 3/4. It is a 3-4-5 triangle (4 times the 3-4-5 triangle); consequently, its hypotenuse L = 4 × 5 = 20.

Or we could have used the Pythagorean Theorem to obtain:
L2 = 122 + 162
L2 = 400
L = 20

 

Example 2 (easy)


 

 

  Calculate the length L for the triangle shown.

Solution

This is a right triangle, a 45°- 45°- 90° triangle. The length of a leg of such a triangle is:
1 / times the hypotenuse.

This gives:



 

Example 3 (medium)


 

A given isosceles triangle has two equal angles of 30°. The side common to the 30° angles has a length of 4. How long are the equal sides?

Solution

A sketch of the triangle is always helpful. Let x be the unknown length. You know how to use the properties of a right triangle to solve for the sides of a triangle, so if you have to solve for the side of a different kind of triangle, you can use a right triangle within the given triangle. Can you see how one of the triangles we've just discussed could be helpful in solving the problem? By dividing the isosceles triangle into 2 right triangles, we get two 30° - 60° - 90° triangles.

The ratio of the side adjacent to the 30° angle and the hypotenuse is : 2. Hence,


Example 4 (hard)


A triangle has angles of 45° and 75°. The side opposite the 45° angle has a length of 6. What is the length of the side opposite the 75° angle?


Solution

Sketch the triangle. The remaining angle is 180 - (75 + 45) = 60°. Again, see if you can solve the problem by creating right triangles. Form two right triangles and label the unknowns x, y, z. The side adjacent to the 60° angle is 1/2 the hypotenuse.

Hence, y = 3. The side opposite the 60° angle is x = 3 (the triangle is 3 times as big as the base 30° - 60° - 90° triangle shown previously). Since the legs of a 45° - 45° - 90° triangle are equal, z = x = 3. The length is then

 


   Geometry Chapter 4: Circles
  

The diameter (d) of a circle is twice the radius (r). A circle's circumference is d or 2r ( = 3.14 or 22/7- which is approximately 3.14).

A central angle has its vertex at the center of a circle, and its measure equals the measure of the arc it intercepts (in degrees). For example, if AOB = 60, then the measure of arc AB is 60°, or 60/360 = 1/6 of the circle's circumference.

Circumference = 2r =d

AOB = arc AB

An inscribed angle has its vertex on the circle itself, and its measure is 1/2 of the measure of the arc it intercepts:

ACB = 1/2 arc AB.

A line that just touches a circle is called a tangent. It is perpendicular to the radius drawn to the point of touching.

ABC is a right triangle if CB is the diameter. A triangle inscribed in a circle is a right triangle if one of its sides is a diameter. Obviously, A has its vertex on the circle, and it intercepts half of the circle so that A = 180°/ 2 = 90°.

Note: if an inscribed triangle has a leg as the diameter, that leg is the hypotenuse.


Example 1 (medium)

What arc length is intercepted by an inscribed angle of 42° on a circle with r = 12 (where = 3.14 = 22/7)?




Solution

The 42° inscribed angle intercepts 1/2(arc°) or arc° = 84°; that is, 84/360 of the circle is intercepted by the angle. The circumference is 2r = 24 so that the arc length is, using = 22/7,

arc length = 84/360 × 24 =
factor out the 12s in 84 and 360, factor 24 into 6 and 4, and convert into 22/7.

(7 × 12)/(30 × 12) × (6 × 4) 22/7 = 88/5 = 17.6



Example 2 (easy)

A triangle is inscribed in a circle with shorter sides 6 and 8 units long. If the longer side is a diameter, find the length of the diameter.

 


Solution

A triangle so inscribed (with one side a diameter) is a right triangle. Consequently,

d = 6 + 8 = 36 + 64 = 100; therefore d = 10. Or, you could have just seen that 6, 8 is double 3, 4.


Example 3 (hard)

A certain clock has a minute hand that is exactly 3 times as long as it's hour hand. Point C is at the tip of the minute hand, and point D is at the tip of the hour hand. What is the ratio of the distance that point C travels to the distance that point D travels in 6 hours?

A. 3:1
B. 6:1
C. 12:1
D. 18:1
E. 36:1

Solution

In 6 hours, the point C on the minute hand travels 6 circumferences (where point C to the middle of the clock is the radius of its circle). The point D on the hour hand only travels half way round the clock, half a circumference (where point D to the middle of the clock is the radius of it's circle).

Since the minute hand is 3 times as long as the hour hand, let the distance between point C and the center of the clock be 3r and the distance from point D to the center be r.

Point C travels 6 × 2(3r) = 36r
Point D travels 0.5 × 2r = r

36r
r

Thus, the ratio of the distance that point C travels to the distance that point D travels in 6 hours is 36:1. The correct answer is E.

 

 


   Geometry Chapter 5: Perimeters and Areas
 
 

The perimeter of a figure is the distance around the figure. The perimeter, P, and area, A, of common figures are shown.


 Circle



P= 2r
A=r

 


 Rectangle



P= 2h + 2b

A=bh


 Square




P = 4h
A = h

 


 Right triangle



A = bh/2


Triangle

A = bh/2


Parallelogram



A = bh

 

 

Example 1 (easy)

What is the radius of a circle if its circumference is numerically equal to twice its area?

 


Solution

The perimeter is the same as the circumference = 2r. The area is r, so that
2r = 2(r ); therefore, r must equal 1.





Example 2 (medium)

An automobile travels 2 miles. How many rotations does a 14-inch radius tire make (use 22/7 for )?


 

Solution

The circumference of a tire is 2r = 2 × 14 = 28 inches. First, make the units commensurate by converting miles to inches (12 inches in a foot, 5280 feet in a mile).

No. of rotations =
(2 miles × 5280 ft/mile × 12 inches/ft)
28
Substitute 22/7 for (we put 7 on top and 22 on bottom)
(2 × 5280 × 12 × 7 )
28 × 22
Cancel out 2 into 22, 7 into 28 to make 4 and then that 4 into 12.
(2 × 5280 × 312 × 7 )
28 × 22 11

(5280 × 3)
= 1440
11

In the above, we simplified by canceling out common factors and then multiplied and divided. It is important to first simplify to save time in the final step.



Example 3

A square is inscribed in a circle of radius 10. Determine the ratio of the area of the circle to the area of the square.




Solution



First, sketch the figure. The area of the circle is r = 100. That's easy.

Now let's go over the area of the circle. The diameter is 20, which is also equal to the diagonal of the square. The diagonal of the square is also the hypotenuse of a right triangle inside of the square, a 45-45 triangle.

The legs of the triangle are equal so that b = 20/. Since it is a 45/45 right triangle, the legs are equal to the hypotenuse /. The area of the square is the leg squared (20 / ) = 200. The ratio of the areas is:

Area circle/Area square = 100/200 =/2


 


   Geometry Chapter 6: Solids
 

There are two solids that interest us in preparing for the test. They are the rectangular solid (a box) and a circular cylinder. A cube is a special rectangular box whose sides are all equal. The volume of a box is the product of its three sides: V = bwh (base × width × height) . The volume of a circular cylinder is the area of the base times the height: V = rh (base circle × height).

The surface area of a cube is 6 × (area of one side) since all six sides have the same area. The surface area of a cylinder is composed of the top and bottom circular areas and the area around the cylinder. The top and bottom are simply 2 × × r . The area of the side is 2 × × radius × height.

Box



V = bhw

 

Cylinder



V =rh



Example 1 (easy)

How many liters does it take to fill a box that is 2m by 20 cm by 20 mm?




Solution

There are 1000 liters in a cubic meter. Hence, we find the volume in cubic meters and multiply by 1000. The volume is

V = bwh
   = 2 × 0.2 × 0.02 = 0.008m

1000 liters/m ×  0.008m = 8 liters


Example 2 (easy)

It takes about 7.5 gallons to fill a volume of one cubic foot. How many gallons are needed to fill a cylinder 2 ft high and 28 inches in radius (=22/7)?




Solution

The volume of a cylinder is the area of its circular base times its height:

V = rh
= 22/7
r = 28 inches, or 28/12 feet (divide 28 by 12 to convert) or 7/3 (factor out the 4), r = 49/9
h = 2, height equals 2.

Now let's plug the numbers into the equation: V =rh
= 22/7 × (49/9) × 2 =
cancel out the 7 under 22 and the 49 above 9 and multiply

= 308 / 9ft

Since it takes 7.5 gallons to fill one cubic foot, multiply the cubic area by 7.5.

= 308/9 × 7.5 gallons = 256 2/3 gallons





Example 3 .

A gallon of paint covers 400 ft of wall area. How many gallons are required to paint the walls of a building with perimeter 200 ft and height 10 ft (assuming there are no windows)?



Solution


The perimeter is the distance around the building, that is, the length of the rectangles that make up its sides. Since each rectangular side is assumed to be 10 ft high, the total area is:

A = bh = 200 × 10 = 2000ft

Number of gallons required = 2000/400 = 5

Note: the shape of the building makes no difference here. The building could be circular (think of taking the wrapper off of a bottle to make a rectangle).

 


   Geometry Chapter 7: Coordinate Geometry
\


Rectangular Coordinates

A point P is positioned relative to two perpendicular lines, called the coordinate axes. The perpendicular distance from the y-axis to point P is the x-coordinate; the perpendicular distance from the x-axis to point P is the y-coordinate. The coordinates x and y form an ordered pair (x, y).

Often, a grid is used to display points relative to the coordinate axes.

   

The point (4, 3) is located 4 units from the y-axis to the right and 3 units above the x-axis; the point (-2, 1) is 2 units to the left of the y-axis and 1 unit above the x-axis. The distance, d, between the two points can be found by the Pythagorean Theorem. The horizontal leg is the total distance in the x-direction: 4 - (-2) = 6; the vertical leg is the distance in the y-direction: 3 - 1 = 2. The distance is then


d = 6 + 2=

= 2

Example 1 (easy)

A square has two corners of a diagonal at (6, 8) and (2, 4). What is its area?




Solution

Compare the x1 and x2 values and the y1 and y2 values. The difference in the x-direction is 6 - 2 = 4 and in the y-direction 8 - 4 = 4. The sides are both of length 4, so that the area is A = 4 × 4 = 16.


Slope Formula

How do you measure the slant of a line? By definition, it is the ratio of the vertical change to the horizontal change (see figure below).


Forming the vertical change over the horizontal change (above) figure results in slope formula (where m is the slope).

Use this formula to calculate slopes of lines.

 

Example 2 (easy)

What is the slope of this line?



 

Solution


To solve this problem, plug the digits in the line into the slope intercept equation.

y = 4, b = 2

x = 5, a = 1


The slope is 1/2.


 

Slope Intercept Formula

If you have a formula, such as x - 2y = 4, how do you calculate the slope of the line?

If you want to graph a line, the formula to use is:

y = mx + b

In this equation, m is the slope of the line and b is the y-intercept.

The y-intercept is when x = 0 in an equation.
The x-intercept is when y = 0 in an equation.

The x-intercept is the point where the line crosses the x-axis. It is found by setting y = 0 and solving the resulting equation. The y-intercept is the point where the line crosses the y-axis. It is found by setting x = 0 and solving the resulting equation.

Example 3 (easy)

Graph the equation 4x - y = 5.


Solution


Try to convert the equation 4x - y = 5 into the format

y = mx + b

4x - y = 5 can be converted into

y = 4x - 5

This means that m = 4 (the slope is 4) and b = -5 (the y-intercept is -5)


In the above graph, the slope is 4 and the y-intercept is -5.

In the above graph, the dot is the y-intercept at (0, -5).

The line slopes up at a rate of 4 up for every 1 across (slope of 4). It intersects lines at (1,-1) and (2,3).


Distance Formula

The distance formula is an adaptation of the Pythagorean Theorem which is used to find the distance between two points on the coordinate plane. The formula states:

d = √(X2 - X1)2 + √(Y2 - Y1)2


Example 4 (medium)

What is the distance between the points (3, 6) and (4, 7) on the coordinate plane?

 

 

Solution

In this problem, there are two ways to figure out the distance.

One is to draw a triangle using these points where the line from (3,6) to (4,7) is the hypotenuse, a line from (3,6) to (4,6) is one leg, and a line from (4,6) to (4,7) is another leg. From this, find out the distance of each leg, plug it into the Pythagorean theorem and you get that the hypotenuse is equal to √2.

The other way to approach this problem is to use the distance formula on the two points given.

d = √(4 - 3)2 + √(7 - 6)2

d = √(1)2 + √(1)2

d = √(2)

Both ways work to give the same answer, but sometimes the distance formula is a much more direct way to figure out the distance between two points.

 



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