Parallelogram

Parallelograms are the fundamental polygon of geometry with some unique properties that help in solving various geometrical and mathematical problems related to parallelograms. It is a two-dimensional figure with four sides, in which opposite sides are parallel to one another. A parallelogram is a set of 2D polygons with similar properties (with some variations), such as rectangle, square, or rhombus. 

1.0Definition of Parallelogram 

A parallelogram is a specific quadrilateral in which opposite sides are parallel & equal in length. That is, it is a four-sided figure whose opposite sides are parallel to one another. The parallelogram-shaped figure normally has sides that seem slanted, but the characteristic feature is that the opposite sides are equal & parallel, and the opposite angles are equal. See this Parallelogram diagram to better understand the shape and properties of the quadrilateral. 

2.0Area and Perimeter of Parallelogram 

The area for a parallelogram

The area for a parallelogram can be represented as the product/multiplication of the base and the height of the parallelogram, where the height (h) is the perpendicular distance from its base to its opposite side. Mathematically, the area for a parallelogram can be represented as: 

$\text{ Area of a Parallelogram }=\text{ base }\times \text{ height }$

The area of a parallelogram can also be represented in the vector form. Suppose vectors v and u are two sides of a parallelogram that are next to one another, the parallelogram vector is the cross product of u and v, which can be represented as: 

$\text{ Area }=|u\times v|$

The Perimeter of a Parallelogram 

Perimeter is the sum of all sides of a two-dimensional polygon; hence, the parallelogram perimeter formula can be given as: 

$\text{ Perimeter of Parallelogram }=2(\text{ breath }+\text{ height })$

3.0Parallelogram Properties

1. Opposite sides of a parallelogram are equal & parallel to one another. For example, in the figure, AB = BC and CD = AB. Similarly, AB∥BC and CD∥AB. 
2. Opposite angles of the parallelogram are also equal, with the sum of all the angles equal to 360°. 
3. Diagonals bisect each other, meaning they cut one another in half. 
4. The sum of consecutive angles ∠A and ∠B is equal to 180°. 
5. The diagonals of a parallelogram bisect the figure into two congruent triangles. 

4.0Types of Parallelogram and Its Properties 

A parallelogram represents a quadrilateral whose opposite sides are both parallel and equal. Based on the other properties it may have, a parallelogram may be categorised into various specific forms. Each parallelogram form has distinct characteristics and properties. Here are some general forms of parallelograms and their properties:

Rectangles 

A rectangle is a special type of parallelogram with all four angles equal to 90° or a right angle. The diagonals of any rectangle are equal in length, apart from bisecting each other. 

• The area of a rectangle can be defined as the product of its length and width and can be calculated by the formula: 

$\text{ Area }=\text{ length }\times \text{ width }$

• The perimeter of a rectangle is the sum of all sides of a rectangle and can be calculated using the formula: 

$\text{ Perimeter }=2(\text{ length }+\text{ width })$

Rhombus

A rhombus is a diamond-shaped parallelogram with all equal sides and unequal diagonals, which bisect each other at 90°. The diagonals of any given rhombus also bisect the angles of the rhombus. 

• The area of a rhombus can be expressed as the half of the product of its diagonals, say diagonal 1 and diagonal 2. Mathematically, the area can be written as: 

$\text{ Area of rhombus }=\frac{1}{2}\times \text{ diagonal }1\times \text{ diagonal }2$

• The perimeter of a rhombus is the sum of all the sides of the polygon. The formula for the perimeter of a rhombus is: 

$\text{ Perimeter of rhombus }=4(\text{ Side })$

Square

A square is a perfect mixture of the properties of a rectangle and a rhombus because a square has all the sides equal, and both the diagonals are also equal to each other, with both bisecting at a right angle, or 90°. All the angles of a square are also equal. 

• The area of a square may be calculated by squaring the side of this quadrilateral. Mathematically, it can be written as: 

$\text{ Area of Square }={\text{ side }}^2$

• The perimeter of a square is the same as the perimeter of a rhombus, that is,

$\text{ Perimeter of a square }=4(\text{ side })$

5.0Parallelogram Circumscribing a Circle 

A parallelogram circumscribing a circle is a rhombus. This is because a rhombus possesses the property that all sides are equal, and it can be inscribed with a circle touching all four sides of the parallelogram. The following theorem can be proved as: 

To Prove: ABCD is a Rhombus. 

Given: ABCD is a Parallelogram. 

Solution: By using the theorem, the tangents drawn from a single external point to a circle are equal. Hence, 

SD = DR …..(1)

AS = AP ….(2)

CQ = RC …..(3)

BQ = BP ….. (4)

Adding equations 1, 2, 3, and 4 

SD+AS+CQ+BQ = DR + AP + RC + BP

AD + BC = CD + AB 

Since ABCD is a parallelogram, AB = BC and CD = AB as opposite sides of a rectangle;

2AD = 2CD or simply 

AD = CD 

Similarly, AB = BC

Here, AB = BC = AD = CD

Since all the sides of the rectangle are equal to one another, therefore ABCD is a rhombus, which eventually means the parallelogram circumscribing a circle is always a rhombus. 

6.0Parallelogram Examples

Problem 1: Find the area of a parallelogram with a base of 10 cm and a height of 6 cm.

Solution: Given that Base = 10 cm and height = 6 cm, 

$\begin{array}{c}\text{ Area of a Parallelogram }=\text{ base }\times \text{ height }\\ \text{ Area of a Parallelogram }=10\times 6\\ \text{ Area of a Parallelogram }=60{\mathrm{cm}}^2\end{array}$

Problem 2: Prove that the diagonal of the parallelogram divides it into two congruent triangles. 

Solution: In ABC and CDA 

∠ACB = ∠CAD (alternate interior angle) 

∠BAC = ∠DCA (alternate interior angle) 

AC = AC (common)

ABC ⩭ CDA (ASA) 

Problem 3: A parallelogram has an area of 72 cm² and a base of 12 cm. Find the height of the parallelogram.

Solution: Given that base = 12cm

$\begin{array}{rl}& \text{ Area of a Parallelogram }=\text{ base }\times \text{ height }\\ & 72=12\times \text{ height }\end{array}$

Height = 6cm

Problem 4: Let A(1, 2), B(5, 3), C(4, 7) be three consecutive vertices of a parallelogram ABCD. Find the area of the parallelogram ABCD. 

Solution: Given that A(1, 2), B(5, 3), C(4, 7) are the three consecutive vertices of the parallelogram ABCD. Hence, to find the area of the parallelogram, we will use the vector cross product method, according to which:

$\text{ Area of parallelogram }=|u\times v|$

The coordinates of the consecutive sides, BA and BC, of the parallelogram: 

$\begin{array}{rl}& \overline{BA}=\bar{A}-\bar{B}=(1-5,2-3)=(-4,-1)\\ & \overline{BC}=\bar{C}-\bar{B}=(4-5,7-3)=(-1,4)\end{array}$

Since A, B, and C are consecutive vertices, vectors BA and BC are adjacent sides of the parallelogram.

Hence, 

$BA\times BC=\left|\begin{array}{ccc}i& j& k\\ -4& -1& 0\\ -1& 4& 0\end{array}\right|$

$\begin{array}{rl}& |\hat{i}(-1\times 0-4\times 0)-\hat{j}(-4\times 0-0\times -1)+\hat{k}(-4\times 4-(-1)\times -1)|=\\ & |\hat{k}(-16-1)|=|-17|=17\end{array}$

Hence, the area of the parallelogram is 17 units.

Problem 5: Show that the quadrilateral with vertices A(1, 2), B(4, 6), C(7, 4), and D(4, 0) is a parallelogram.

Solution: Given that A(1, 2), B(4, 6), C(7, 4), and D(4, 0) are the vertices of a quadrilateral. To prove if the vertices belong to a parallelogram, we can check if the opposite sides are equal and parallel using slope and distance.

Using the distance formula: 

$\begin{array}{rl}& \mathrm{AB}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}=\sqrt{(4-1{)}^2+(6-2{)}^2}=\sqrt{9+16}=5\text{ units }\\ & \mathrm{BC}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}=\sqrt{(7-4{)}^2+(4-6{)}^2}=\sqrt{9+4}=\sqrt{13}\text{ units }\\ & \mathrm{CD}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}=\sqrt{(4-7{)}^2+(0-4{)}^2}=\sqrt{9+16}=5\text{ units }\\ & \mathrm{DA}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}=\sqrt{(1-4{)}^2+(2-0{)}^2}=\sqrt{9+4}=\sqrt{13}\text{ units }\end{array}$

Here, AB = CD and BC = DA. Now, using the slope formula to verify the slope of the sides: 

• Slope of $AB=\frac{y_2-y_1}{x_2-x_1}=\frac{6-2}{4-2}=\frac{4}{3}$
• Slope of $CD=\frac{y_2-y_1}{x_2-x_1}=\frac{4-0}{7-4}=\frac{4}{3}$
• Slope of $BC=\frac{y_2-y_1}{x_2-x_1}=\frac{4-6}{7-4}=-\frac{2}{3}$
• Slope of$AD=\frac{y_2-y_1}{x_2-x_1}=\frac{2-0}{1-4}=-\frac{2}{3}$

Since Slope of AB = Slope of CD and Slope of BC = Slope of AD, 

$AB\Vert CD\text{ and }BC\Vert AD$

Therefore, ABCD is a parallelogram.