## Types of Polygons

A ** polygon** is a figure made from 3 or more line segments. The segments, called

*sides*, intersect at the endpoints of the segments, called

*vertices*(singular:

*vertex*). Congruent polygons can be indicated with the ≅ symbol.

There are many types of polygons.

quadrilateral |
4 sides |

pentagon |
5 sides |

hexagon |
6 sides |

octagon |
8 sides |

n-gon |
polygon with n sides |

equilateral |
all sides equal |

equiangular |
all angles equal |

regular |
all sides equal and all angles equal |

convex |
all angles point “out” |

concave |
has an angle that points “in” |

Triangles and quadrilaterals are the most common polygons on the GRE. Other polygons on the test usually contain triangles and quadrilaterals.

**Parallelograms**

Video Courtesy of (__site-affiliate__) Kaplan GRE prep.

## Perimeter and Area

The ** perimeter** of a polygon is the distance around the outside of the polygon, or the sum of the lengths of the sides of the polygon.

The** area** of a polygon is the amount of surface that is covered by the polygon.

To find the area of triangles and quadrilaterals, you need two measurements: the *base* and the *height*.

Any of the sides of a triangle can be its base. The height of the triangle is the perpendicular distance from the base to the opposite angle.

For the quadrilaterals listed below, the height of the quadrilateral is the perpendicular distance from the base to the opposite side.

These are the formulas for the perimeter, *P*, and area, *A*, of common figures.

### Triangle

*P* = *a* + *b* + *c*

*A* = \dfrac{1}{2} (base *×* height) = \dfrac{1}{2}*bh*

### Right Triangle

*P* = leg + leg + hypotenuse

*A* = \dfrac{1}{2} (leg *×* leg) = \dfrac{1}{2}*bh*

### Rectangle

*P* = 2*l* + 2*w* = 2(*l* + *w*)

*A* = length *×* width = *lw*

*or*

*A* = base *×* height = *bh*

Opposite sides are equal.

All angles are equal and are 90°

### Square

*P* = 4*s*

*A* = *s*^{2}

All sides are equal.

All angles are equal and are 90°.

### Parallelogram

*P* = 2*l* + 2*w* = 2(*l* + *w*)

*A* = base *×* height = *bh*

Opposite sides are equal.

Opposite sides are parallel.

Opposite angles are equal.

Adjacent angles add to 180°.

### Rhombus

*P* = 4*s*

*A* = base *×* height = *bh*

All sides are equal.

Opposite sides are parallel.

Opposite angles are equal.

Adjacent angles add to 180°.

### Trapezoid

*P* = *a* +* b*_{1} + *c* + *b*_{2}

*A* = \dfrac{1}{2}(base 1 + base 2)(height)

= \dfrac{1}{2}(*b*_{1} + *b*_{2})*h*

One pair of opposite sides, the bases, is parallel.

### Circle

*C* = 2*πr* = *πd*

*A = πr*^{2}

*r* = radius, *d* = diameter, and *d* = 2*r*

*π* ≈ 3.14 ≈ \dfrac{22}{7}

Note: A circle is not a polygon. Circles are covered in the next section.

## Example

Find the perimeter and area of the figure.

### Solution

The area of the triangle is *A* = (1/2)(48)(10) = 240 square units

To find the perimeter, use the right triangle formed by the height and half of the base.

That triangle’s side lengths are

10 : 24 : *x*.

This is a multiple of a

5 : 12 : 13 triangle, so *x* = 2(13) = 26.

*P* = 48 + 26 + 26 = 100 units

## Example

Find the area of the shaded region.

### Solution

The shaded region is a rectangle minus a trapezoid.

The area of the rectangle is *A* = (8)(5) = 40 square inches.

The side lengths of the trapezoid are height = 2, base_{1} = 3 and base_{2} = 5.

*A* = (1/2)(3 + 5)(2) = 8 square inches.

The area of the shaded region is

40 – 8 = 32 square inches.

## Maximum and Minimum

The GRE often asks for the maximum or minimum perimeter or area. Maximum area is when a triangle or quadrilateral has perpendicular sides and right angles.

Squares and 45° : 45° : 90° triangles have the maximum area for the smallest perimeter.

## Example

Find the maximum area of a quadrilateral with perimeter 40 feet.

### Solution

A square is the type of quadrilateral that has the maximum area.

If the perimeter of the square is 40 feet, the length of each side is 10 feet. The area is 10^{2} = 100 square feet.

## Example

Find the minimum perimeter of a quadrilateral with an area of 16 square meters.

### Solution

Use the method of plugging in values for the sides of different quadrilaterals that have an area of 16.

The square has the smallest perimeter for the given area.

The minimum perimeter is 16 meters.

## Example

Find the maximum area of a triangle where one side is 5 inches and the other side is 9 inches.

### Solution

This isosceles triangle has sides of 5, 5 and 9.

Using the Pythagorean theorem, the height is about 2.

*A* ≈ (1/2)(9)(2) ≈ 9 in.^{2}

Part of this triangle is a 3 : 4 : 5 right triangle, so the height of the triangle is 3.

*A* = (1/2)(9)(3) = 13.5 in.^{2}

This right triangle has 5 and 9 as the legs.

*A* = (1/2)(9)(5) = 22.5 in.^{2}

The maximum area is 22.5 square inches.

## Interior and Exterior Angles

The sum of the *interior angles* of a polygon is a function of the number of sides. The formula for the sum of the interior angles of a polygon with *n* sides is (*n* – 2)180°.

For a triangle, the sum of the interior angles is 180°.

A quadrilateral has 4 sides, so *n* = 4.

The sum of the interior angles of a quadrilateral is

(4 – 2)180° = 2(180°) = 360°.

For a hexagon, *n* = 6 so the sum of the interior angles is

(6 – 2)180° = 4(180°) = 720°.

For a triangle, the sum of the interior angles is 180°.

If you can’t remember the formula for the sum of the interior angles, count the number of the triangles formed when all the diagonals are drawn from just one vertex. The sum of interior angles is the number of these triangles times 180°. (See more about diagonals below.)

The sum of the *exterior angles* of all polygons is 360°.

You can think of this as going all the way around the polygon, making a complete circle.

## Example

What is the measure of the smallest interior angle of the polygon?

### Solution

The polygon has 5 sides, so the sum of the interior angles is

(5 – 2)(180°) = 540°.

Find the value of *x*.

95 + 70 + *x* + 2*x* + 2*x* = 540

5*x* = 540 – (95 + 70) = 375

*x* = 75

Be careful to remember the question is not asking for *x*, but for the smallest interior angle. Because *x* = 75, the smallest interior angle is 70°.

## Example

Find the measures of all of the angles.

### Solution

The sum of the exterior angles is 360°, so

5*x* + 8*x* + 5*x* = 360

18*x* = 360 *x* = 20

The exterior angle measures are 100°, 100° and 160°.

The interior and exterior angles are supplementary so each pair adds to 180°.

180 – 100 = 80 and 180 – 160 = 20

The measures of the interior angles are 80°, 80° and 20°.

## Symmetry

A figure has ** line symmetry** if the figure can evenly “fold” onto itself. A figure can have more than one line of symmetry.

A figure has ** rotational symmetry** if the figure can be rotated less than 180° and lines up on the original figure.

The point used for the rotation is the

*center of symmetry*.

## Example

Identify the line symmetry and rotational symmetry of the figures.

(a)

(b)

### Solution

(a) The isosceles trapezoid has 1 line of symmetry. It has no rotational symmetry.

(b) The equilateral triangle has 3 lines of symmetry.

Rotating the triangle 120° and -120° matches the original triangle.

## Similar Polygons

Polygons that have the same shape but a different size are called ** similar polygons**.

Polygons are similar if the *corresponding angles* are equal, and the *corresponding sides* have the same ratio.

The ratio of the lengths of the corresponding sides is called the** scale factor**. The scale factor for congruent figures is 1.

Some polygons are similar by definition. For example, all equilateral triangles are similar, all squares are similar, and all circles are similar.

A line parallel to the base of a triangle or a trapezoid also creates similar polygons.

## Example

Segment

ABis parallel to segmentCD. Find the values of the variables.

### Solution

Since *AB* is parallel to *CD*, triangles *AEB* and *CED* are similar.

The top small triangle *CED* is a right triangle. The side lengths are

8 : *x* : 17, so *x* = 15.

Compare 15 and 7.5 to find the ratio: 7.5 is what fraction of 15? Since 7.5 is half of 15, *y* will be half of 8. So *y* = 4.

The side *AB* will not be half of side *CD*. If you compare triangles *AEB* and *CED*, *AEB* is one and a half (1.5) times as big as *CED*.

Since *AB* will be 1.5 times *CD*, *z* = 1.5(17) = 25.5.

## Example

The trapezoids

ABCDandSPENare similar. SegmentABis parallel to segmentCD. SegmentNEis parallel to segmentSP. Find the angle measures and the perimeter of both trapezoids

### Solution

Since *AB* is parallel to *DC*,

∠*C* + ∠*B* = 180°, so ∠*B* = 90°.

∠*N* + ∠*S* = 180°, so ∠*S* = 180° – 65°

= 115°.

Since *NE* is parallel to *SP*,

∠*N* + ∠*S* = 180°, so ∠*S* = 180° – 65°

= 115°.

Use the correspondence between the angles in *ABCD* and *SPEN*.

The angle measures are

*∠C* = ∠*E* = 90°

*∠B* = ∠*P* = 90°

*∠A* = ∠*S* = 115°

*∠D* = ∠*N* = 65°

The longest side in *ABCD* is 12. The longest side in *SPEN* is 30. Find the scale factor.

\dfrac{\textit{ABCD}}{\textit{SPEN}} = \dfrac{12}{30} = \dfrac{2}{5}

\\[3ex]\dfrac{\textit{AB}}{\textit{SP}} = \dfrac{3}{\textit{x}} = \dfrac{2}{5}. SP = 15/2 = 7.5

\dfrac{\textit{BC}}{\textit{PE}} = \dfrac{11}{\textit{y}} = \dfrac{2}{5}. PE = 55/2 = 27.5

The perimeter of *SPEN* is

30 + 7.5 + 27.5 + 20 = 85.

Instead of calculating *CD*, use the scale factor to find the perimeter of *ABCD*.

The perimeter of *ABCD* is

(2/5)(85) = 34.

## Diagonals *(Advanced Topic)*

A ** diagonal** of a polygon is a segment that connects two non-adjacent vertices of the polygon.

Diagonals of a polygon form triangles inside the polygon.

The GRE often uses the triangles formed inside quadrilaterals by the diagonals. The more you know about the quadrilateral, the more you know about the diagonals and the triangles that are formed. Drawing sketches can help you remember common diagonals. Two sketches are drawn for you below.

#### parallelogram

One diagonal of a parallelogram divides the parallelogram into two congruent triangles.

The diagonals of a parallelogram bisect each other.

#### rectangle

One diagonal of a rectangle divides the rectangle into two congruent triangles.

The diagonals of a rectangle are equal and bisect each other.

#### rhombus

One diagonal of a rhombus divides the rhombus into two pairs of two congruent triangles.

Two diagonals of a rhombus divide the rhombus into four congruent triangles.

The diagonals of a rhombus are perpendicular and bisect each other.

The diagonals of a rhombus bisect the angles of the rhombus.

*∠ABC* ≅ ∠*ADC* ∠*BAD* ≅ ∠*BCD*

∠1 = ∠2 = ∠3 = ∠4

∠5 = ∠6 = ∠7 = ∠8

*AE* = *EC* *BE* = *ED*

*∠BEA* = ∠*BEC* = ∠*DEA* = ∠*DEC* = 90°

triangles *BEA* ≅ *BEC* ≅ *DEA* ≅ *DEC*

triangles *BDC* ≅ *BDA*

triangles *BAC* ≅ *DCA*

#### square

One diagonal of a square divides the square into two 45° : 45° : 90° triangles.

The diagonals of a square are equal, perpendicular and bisect each other.

The diagonals of a square bisect the angles of the square.

#### isosceles trapezoid

The diagonals of an isosceles trapezoid are equal.

The diagonals of an isosceles trapezoid divide the trapezoid into two congruent triangles.

Given trapezoid with *AB* = *CD*

*AC* = *BD *

triangles *ABD* ≅ *DCA*