GRE gives special attention to the section of quadrilaterals in geometry. It is important for every GRE aspirant to understand the properties of quadrilaterals as mere properties can help you solve the problems in split second otherwise you would be banging your head still won’t get any clue of how to solve the problem.

Let us understand what quadrilaterals are and how many types of quadrilaterals are there.

Any four sided, closed figure can be termed as a quadrilateral. There are various forms of quadrilateral, likely, square, rhombus, trapezium, rectangle, parallelogram, kite etc.

## Quadrilaterals: Type and Properties

**Square:** If all the sides of the quadrilateral is equal and perpendicular to one another then it is a square. The diagonals of a square are perpendicular to each other as well as they bisect one another. It is a special type of rhombus.

**Parallelogram:** It is the most favorite figure of GRE test designers. You will find majority question from this topic. The opposite sides of parallelogram are equal and parallel to one another but they won’t be perpendicular in nature. The area of parallelogram is equal to base × height. The diagonals of parallelogram bisect one another.

**Rectangle:** If the angles of a parallelogram makes right angle then it will become a rectangle. The diagonals of a rectangle are equal in length and they bisect each other at an angle of 90°.

**Rhombus:** All sides of a rhombus are equal but only the opposite angles are equal. In a rhombus, adjacent angles sum up to form supplementary angle. The diagonals of a rhombus bisect each other at complementary angles.

**Trapezoid:** Trapezoid is a four sided figure in which only one pair of sides will parallel. The sum of the adjacent sides of a trapezium is equal to 180°.

**Kite:** The length of the adjacent sides of a kite is equal. The diagonals bisect one another at angle of 90°. OX = OY (Where O is the point of intersection)

Let’s solve a question to understand these concepts:

* Question:* In a parallelogram, the base is \(7\frac{6}{5} \; mm\) and height is \(2\frac{3}{41} \; mm\). Find its area.

* Solution: *Base = \(7\frac{6}{5} \; mm\) ; Height = \(2\frac{3}{41} \; mm\)

Area = \(\frac{41}{5} \times \frac{85}{41} = 17 \; mm^{2}\)

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