# Quadrilaterals - GMAT Math Study Guide

A quadrilateral is a four-sided polygon. The five main quadrilateral types (listed in order from most to least symmetric) are: squares, rectangles, parallelograms, rhombuses, and trapezoids. Each type of quadrilateral has its own defining features and equations to determine the area and the perimeter. The interior angles for all quadrilaterals must sum to 360°.

## Square • All sides are equal in length
• All angles are right (i.e., 90°)
• Opposite sides are parallel
Perimeter = 4(side)
Area = (side)2

## Rectangle • Opposite sides are equal in length
• Opposite sides are parallel
• All angles are right (i.e., 90°)
Perimeter = 2(width) + 2(length)
Area = (length)(width)

## Parallelogram • Opposite sides are equal in length
• Opposite sides are parallel
• Opposite angles are equal (i.e., bottom-right = upper-left)
Perimeter = 2(width) + 2(length)
Area = (width)(height)

## Rhombus • All sides are equal in length
• Opposite angles are equal
• Diagonals form right angles
• Diagonals bisect each other
• Diagonals bisect end angles

Note: In a rhombus, while all sides are equal, all angles are not necessarily equal.

Perimeter = 4(side)
Area = 1/2(diagonal1 * diagonal2)

## Trapezoid • One pair of sides is parallel
Perimeter = A + B + C + D
Area = 1/2(height)(base1 + base2)

## Types of GMAT Problems

Many times, GMAT questions that appear to involve only one quadrilateral actually involve several. Often, the various shapes will be labeled by the points that make up their corners. In the picture below, ABCD would refer to the square while ABE would refer to the triangle. In the diagram below, if ABCD is a square whose area is 100, CDEF is a parallelogram, and the perimeter of ABCFED is 80, what is the length of CF? A) 10 B) 20 C) 15 D) 22 E) 30
1. Since ABCD is a square, AB = CD = BC = DA.
2. Since CDEF is a parallelogram, CD = EF and CF = DE. Thus, AB = EF.
3. For the perimeter of ABCDEF, there are only two different lengths:
(1) AB = AD = BC = CD = EF
(2) CF = DE
4. In calculating the perimeter of ABCFED, AB is used four times while CF is used twice.
5. P = 80 = 4*AB + 2*CF.
6. The area of square ABCD is (AB)2 where AB is the length of the side AB.
Since the area of square ABCD is 100, the length of side AB is the square root of 100, which is 10.
7. Substitute 10 in for AB in the equation P = 4*AB + 2*CF.
P = 4*10 + 2*CF = 80.
8. Solve for CF.
P = 40 + 2CF = 80
2CF = 80 - 40 = 40
CF = 40/2
CF = 20