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Surface area formulas

Right triangles - common lengths and multiples

(3) Bisecting sides --> if you draw a line from the vertex to the midpoint of the opposite side, you have divided the triangle into 2 equal halves. This work regardless of the type of triangle (equilateral, isosceles, irregular, etc.), and regardless of the number of times you draw this line!!!

Cylinders

Coordinate plane/graphs

If I redraw the figure based on what I am told about the figure (e.g. these lines are parallel but these are not), then I can avoid these problems and not get tripped up on bad assumptions.

Quadrilaterals - trapezoid

Quadrilaterals

Quadrilaterals

Examples include:

+ Parallelogram - opposite sides + angles are equal

+ Rectangle - all angles are equal, but only opposite sides are equal

+ Rhombus - all sides are equal, but only opposite angles are equal

+ Square - everything is equal

+Trapezoid - only 1 pair of sides are parallel. That's it!

Terminology:

Corner of a polygon = vertex.

e.g. Vertex has an angle of 65°

Polygons and interior angles

If N is # of sides, then:

(N-2) × 180 = Sum of interior angles of a Polygon

Polygon - # sides - sum of interior angles

Triangle - 3 - 180

Quadrilateral - 4 - 360

Pentagon - 5 - 540

Hexagon - 6 - 720

Area formulas

Most common:

(1) Triangle = (Base × Height) / 2

(2) Rectangle = Length × Width

Least common:

(3) Trapezoid = [(Base1 + Base2) × Height] / 2

--> This is also like taking the average of two bases and then multiplying it by height

(4) Parallelogram = Base × Height

(5) Rhombus = (Diagonal1 × Diagonal2) / 2

--> Diagonals of a rhombus are always perpendicular bisectors(6) Rhombus (alternative) = Base × Height

I will only need to know how to calculate surface area for a rectangular solid and cube.

To do this, I simply sum the areas of 6 faces

GMAT volume trick!!!

Question: How many books, each with a volume of 100 in^3, can be packed into a crate with a volume of 5000 in^3?

The answer is NOT 50!!!

50 is wrong because you don't know the exact dimensions of each book.

When fitting 3D objects into other 3D objects, knowing the volume is NOT enough. You must know the specific dimensions of each object to figure out if the objects fit without leaving gaps.

Triangles

Rules to know are:

(1) The sum of 3 angles is 180

(2) Angles correspond to their opposite sides - largest sides have the largest corresponding angles

(3) The sum of any two sides must be greater than the 3rd side. Note that they can't be equal!

(4) The difference of any two sides must be less than the length of the third side. Note - they can't be equal!

Think about #3 in a different way - you can't have an indirect path to something be shorter (or the same) than the direct path! It's simply illogical!

One consequence of these rules is that, if I have the length of 2 sides of a triangle, I will ALWAYS know the range of possibilities for the length of the 3rd side of the triangle

+ 3/4/5, 6/8/10, 9/12/15, 12/16/20

+ 5/12/13, 10/24/26

+ 7/24/25

+ 8/15/17

Isosceles triangles

45-45-90 triangles are a special type of isosceles triangle - the isosceles right triangle.

The length of the legs of every 45-45-90 triangle is: x, x, x*sqrt(2)

Trick:

The diagonal of a square is x*sqrt(2) since a square is made up of two 45-45-90 triangles

Beware:

The GMAT likes to trick you by e.g. putting sqrt(2) in the legs or hypotenuse. This throws you off when you're calculating areas.

Equilateral triangles and 30/60/90 triangles

+ Equilateral triangles - all sides and angles are equal

+ Two 30/60/90 triangles form one equilateral triangle!

+ Each side length of a 30/60/90 triangle is: x, x*sqrt(3), 2x

+ Therefore, use formulas for 30/60/90 triangles to solve problems with equilateral triangles!

+ Area of equilateral triangle = [S^2 * sqrt (3)] / 4

Trick:

If I know the length of an equilateral triangle, but must figure out the height, how do I solve this problem?

Easy - use the 30/60/90 triangle side length formulas to get the height of the equilateral triangle!

Another trick - if given a right triangle with 2 legs in a 1:2 ratio, then I can automatically say it's a 30-60-90!

Memorize these formulas!

+ The diagonal of a square is x*sqrt(2)

+ The main diagonal of a cube is x*sqrt(3)

If I'm working on a problem in which I am packing something away in a rectangular box, then the main diagonal will set the upper limit of the longest object that I can pack in this box.

I can calculate the main diagonal of a rectangular solid using the "Deluxe Pythagorean Theorem":

a^2 + b^2 + c^2 = d^2

Similar triangles

Triangles are defined as similar if 1 of 2 conditions apply:

(1) all corresponding angles are equal

(2) corresponding sides are in proportion

If two similar triangles have corresponding side lengths in a ratio a / b, then their areas will be in the ratio a^2 / b^2

This rule can be applied more broadly:

(a) for side lengths, heights, or perimeters as well

(b) for any similar figures - quadrilaterals, pentagons, etc.

(c) If solids have sides in ratio a/b, then their volume is in ratio a^3 / b^3

Subtleties to remember about triangles (useful)

(1) Remember that the area of a triangle is the same regardless of which side is chosen as the base.

(1/2) × Base1 × Height1 = (1/2) × Base2 × Height2 = (1/2) × Base3 × Height3

(2) For right triangles, this rule applies in a subtle way since two legs are the same.

(1/2) × Base1 × Base2 = (1/2) * Hypotenuse × Height from hypotenuse

(3) Don't mix up legs and hypotenuse - they are different!

Triangles - calculating areas

The formula for the area of a triangle is (1/2) * base * height.

But finding the height can be tricky -

+ In a simple scenario, the height that I'll need to calculate the area will lie INSIDE the triangle

+ In a complex scenario, the height that I'll need to calculate the area will lie OUTSIDE the triangle.

+ Simple scenarios - you have a simple equilateral triangle in which the base is flat on the ground. Thus the height lies within the triangle.

+ Complex scenario - triangle is still flat on the ground. Now, the vertex is 135-degrees. Thus, the height from the base to the opposite vertex lies OUTSIDE the triangle. To find the height, I have to extend the base horizontally using an imaginary line, then extend that line vertically until it meets the vertex

Tricky triangle tips/tricks (1/4)

(1) Imagine taking a triangle with a right angle on top and base on bottom. Now, draw the height line.

--> All 3 triangles that you see are similar! Think about it - all 3 triangles have right angles, and the 2 smaller triangles share an angle with the bigger triangle.

(2) Imagine taking an isosceles triangle with the base on the bottom. Now draw a line of the triangle which is parallel to the base, so that you have made a smaller triangle.

--> The small triangle is similar to the big triangle!

Tricky triangle tips/tricks (2/4)

Tricky triangle tips/tricks (3/4)

(4) The GMAT WILL use triangle problems to fuck with me. One way they'll do this is to get me to assume right angles so the math is easier. Don't assume right angles!

Example #1 -- if two legs are equal, that doesn't mean you're looking at a 45-45-90 triangle. It could just be that you're looking at a 40-40-100 triangle, or a 23-23-134 triangle

Example #2 -- If isosceles triangle has a perimeter of 12 + 12*sqrt(2), what's the length of one of the legs?

---> you have GOT to think in a structured manner when working problems like this, or else you'll rush through the formulas and fall into trap answers!

The proper way to think about this problem is to realize that either

(A) the hypotenuse is 12, or (B) the two legs sum to 12 (leaving the hypotenuse at 12*sqrt(2))

Tricky triangle tips/tricks (4/4)

Remember that two parallel lines that are cut by a straight line have angles that are related to one another.

The GMAT can hide this relationship, so keep your eyes open when you're working these problems.

When working on problems with these angles, it may help to draw + extend lines so that I can spot relationships.

Circles

Circumference = π × d

Diameter = 2 × r

Area = π × r^2

For DS questions - If I can calculate any of the above, then I can calculate ALL of the above

Central angle = an angle created by 2 radii of a circle

Arc = a curved line that is part of the circumference of a circle

Minor arcs = the shortest arc made by a central angle

Major arcs = the longest arc made by a central angle

Major arc + minor arc = 360 degrees

Circles - arcs with central vs. inscribed angles

Arcs with central angles have their vertex on the center of the circle.

Arcs with inscribed angles have their vertex on the circle itself.

The angle of an inscribed arc is half of the angle of its corresponding central arc!

Since the angle of an arc is equal to a central angle, then its inscribed angle is equal to half of the central angle (and also half of the arc!)

Inscribed triangles

They are triangles whose vertices all lie on the circle itself.

If one of the sides of an inscribed triangle is the diameter of a circle, then the triangle HAS to be a right triangle.

Conversely, any right triangle inscribed in a circle must have the diameter of the circle as one of its sides (thereby splitting the circle in half).

Going 1 step further, I could relate this rule back to the inscribed/central arc angle rule. If the central angle of an arc that covers a semicircle is 180°, then its inscribed arc has to be a right triangle (90°).

Surface area = 2 circles + rectangle = 2(π × r^2) + 2*π*r*h

Volume = π × r^2 × h

-->Note that two cylinders can have the same volume but very different dimensions!!

Triangles - exterior angles

The exterior angle of a triangle is equal to the sum of the two non-adjacent (opposite) interior angles of the triangle.

This is a frequently tested topic!

Make sure I am keeping my eyes open for this. The GMAT can "hide" exterior angle situations

This will be applied in, e.g. triangles where there are two interior angles and an exterior angle.

A SPEEDY shortcut through these sort of problems is to plug in numbers! Specifically, plug in coordinates/points.

For instance, if I need to figure out the quadrants that a line passes through, simply plug in 2 (x,y) coordinates and plot the line quickly.

Geometry trick

Be careful of geometry problems where the figure is drawn to LOOK parallel even though it isn't.

Only go by what you are told. Don't assume away!!

Geometry trick

Redrawing figures is a way to avoid getting mixed up from the problem.

GMAT will try to trick you by drawing figures that are not to scale, aka they will lie about the figure so that I make incorrect assumptions!!

Quadrilaterals - advanced

A common GMAT problem is to MAXIMIZE the area of a quadrilateral, given a fixed perimeter.

SQUARES will always have the largest area of any quadrilateral with fixed perimeter!

--> e.g. if perimeter of a quadrilateral is 25, then the side length for the quadrilateral with biggest area is 6.25

Corollary of this is that: given a quadrilateral with a certain area, the figure that would have the MINIMUM perimeter is the SQUARE!

Polygons - advanced

A regular polgon with all sides equals will maximize area for a given perimeter, and minimize perimeter for a given area

Maximum area of parallelogram or triangle - advanced

If you are given two sides of a triangle or parallelogram, you can maximize the area by placing those 2 sides perpendicular to each other.

To rationalize this rule of thumb, think about the area formulas for triangles (0.5*base*height) and parallelograms (base*height).

The base will be constant while the height changes. Therefore, the height needs to be maximized (since the base will be constant) to get the largest area. Otherwise, you keep getting something smaller, which will involve an angle less than 90 degrees.

Parabolas - advanced

To find where a parabola touches the x-axis, I can plug in y=0 into the quadratic formula to find the solutions for x.

Quadratic formula:

x = [-b ± sqrt(b^2 - 4ac)] / 2a

To quickly know how many solutions the equation has, just look at [b^2 - 4ac].

(1) if >0, then the parabola crosses x-axis twice

(2) if =0, then the parabola touches x-axis once

(3) if <0, then the parabola never touches the x-axis

Perpendicular bisectors - advanced

+ The perpendicular bisector of a line segment has the negative reciprocal slope of the line segment that it's bisecting

+ The product of the 2 slopes of 2 lines that are perpendicular is -1.

+ Exception is when one line is horizontal and one is vertical

Intersecting vs. parallel lines - advanced

With straight lines, there are three possibilities when dealing with these lines and the possibility of them intersecting.

It's especially important to keep these 3 possibilities in mind when dealing with DS questions.

(1) that two lines intersect once and there is one (x,y) that satisfies the equations for both lines

(2) that both lines are parallel and that there are no solutions (x,y) that satisfies the equations for both lines

(3) that both lines are the same and that there are infinitely many solutions (x,y) that satisfies both lines

Trick

Be careful when dealing with >1 triangle. If they share angles or have adjacent angles, or if they have parallel sides, then there's a chance the two triangles are similar.

I need to analyze the triangles to see if they are similar. So, if >1 triangle that are adjoining in some way could possibly be similar (which means the sides are proportional to one another).

Definition = a quadrilateral with one pair of parallel sides

Area = [(b1 + b2) / 2] * h

Perimeter = the sum of all sides - they can all be different lengths so there is no shortcut

Quadrilaterals - Parallelogram

Definition = a quadrilateral with 2 pairs of parallel sides (imagine a stretched out rectangle)

Area = b * h (you'll have to draw the height)

Perimeter = 2b + 2s (s = the other 2 sides)

Diagonals = bisects each other, forms 2 pairs of segments

GMAT tip: look for special triangles when drawing the height to be able to quickly calculate the height

If the diagonals bisect each other AND form right angles, then this parallelogram is really a rhombus, which will also have equal sides. But, if they don't make right angles, you got a parallelogram

Quadrilaterals - Rhombus

Definition = a parallelogram that has 4 equal sides (also called a diamond) (imagine a stretched-out square)

Area = b * h (you'll have to draw the height)

Perimeter = 4s (s is the length of a side)

Diagonals = Perpendicular bisectors of each other. Forms 2 pairs of segments

Other: the interior triangles created by the diagonals are all identical

Quadrilaterals - Rectangle

Definition = a parallelogram that has 4 right angles and 2 sets of congruent opposite sides

Area = l * w

Perimeter = 2l + 2w

Diagonals = bisectors of each other, all segments are equal

Other = the interior triangles created by the diagonals are two sets of identical isosceles triangles

Quadrilaterals - Square

Definition = a rectangle with 4 equal sides, aka a rhombus with right angles

Area = s^2

Perimeter = 4s

Diagonals = Perpendicular bisectors of each other, all segments are equal

Other = the interior triangles created by the diagonals are four congruent isosceles right triangles, otherwise known as 45-45-90

When you go from a parallelogram to a rectangle, the diagonals that bisect each other form equal segments (instead of 2 pairs of equal segments)

When you go from a rhombus to a square, the same thing happens

About this deck

Author: Kunal K.

Created: 2014-02-23

Updated: 2014-06-15

Size: 38 flashcards

Views: 37

Created: 2014-02-23

Updated: 2014-06-15

Size: 38 flashcards

Views: 37

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