A parallelogram is a quadrilateral with opposite sides parallel and equal. The opposite angles of a parallelogram are congruent. Any side of a parallelogram is called its base (in general, the bottom horizontal line is considered to be base). The perpendicular distance from the base to the opposite side is called the altitude or height of a parallelogram.

• A parallelogram with all equal sides and equal angles of 90° is a square.
• A parallelogram with all right angles is a rectangle.
• A parallelogram with all sides equal is a rhombus.

## Parallelogram Theorems

Theorem 1: Parallelograms with the same base and between the same parallel lines are equal in area.

Proof: $\square$ABCD and $\square$BCEF are parallelograms with the same base and between the same parallel lines AE and BC.

The theorem is proved by two column proof using the figure below.

 Statement Reason 1) ABCD and BCEF are two parallelograms with the same base BC and between the same parallels AE and BC. 1) Given 2) h is the altitude of the parallelogram ABCD and also of the parallelogram BCED. 2) By the definition of altitude. 3) Area of Parallelogram ABCD = $\frac{1}{2}$ X BC X h. 3) By the definition of area of parallelogram. 4) Area of Parallelogram BCEF = $\frac{1}{2}$ X BC X h. 4) By the definition of area of parallelogram. 5) Area of parallelogram ABCD = Area of parallelogram BCEF 5) From 3 and 4.

Theorem 2: A parallelogram is a rhombus if its diagonals are perpendicular.

Given: $\square$ABCD is a parallelogram where the diagonals AC and BD are at 90° to each other.

Construction: Join AC and BD to form diagonals. Let M be the point of intersection of diagonals.

Proof: In triangles AMD and DMC,

(i) $\angle$AMD = $\angle$DMC = 90°

(ii) AM = MC

(iii) DM is common.

Therefore, By SSA Postulate, $\Delta$AMD = $\Delta$DMC.

In particular, AD = CD by CPCT.

Since ABCD is a parallelogram, the opposite sides are congruent and so AB = CD, AD = BC.

Therefore, AB = BC = CD = AD.

Hence ABCD is a rhombus. The theorem is proved.

Theorem 3: A quadrilateral, with one pair of opposite sides parallel and equal is a parallelogram.

To prove: ABCD is a parallelogram.

Construction: Draw the diagonal BD.

Proof: In triangles ABD and BDC,

(ii) BD is common

(iii) m$\angle$ABD = m$\angle$CDB (opposite interior angles)

By SAS postulate, $\Delta$BAD $\equiv$ $\Delta$BCD.

By CPCT, sides are equal and so AB = CD, m$\angle$ADB = m$\angle$CBD.

Thus we have, AB || CD.

Hence ABCD is a parallelogram. The theorem is proved.

Theorem 4: The diagonals of a parallelogram bisect each other.

Given: ABCD is a parallelogram. AC and BD are diagonals.

Proof: In $\Delta$AMB and $\Delta$CMD,

Since ABCD is a parallelogram, AB = BC

Also AB || CD, thus $\angle$ABM = $\angle$CDM

And $\angle$MAB = $\angle$DCM

So, by ASA criterion, $\Delta$AMB = $\Delta$CMD.

Thus, AM = CM, BM = DM.

Hence the diagonals of a parallelogram bisect each other.

## Parallelogram Diagonals

The diagonals of a parallelogram bisect each other and divide the parallelogram into four congruent triangles.

## Area of Parallelogram

Like for most quadrilaterals, the area of a quadrilateral is the length of one side times its perpendicular height or altitude. The area of a parallelogram is given by the formula
Area = $bh$, where b is the length of any base and h is the corresponding altitude

## Perimeter of Parallelogram

In Geometry, the Perimeter of any polygon is the total distance around the outside of a 2-dimensional shape. In particular, in a parallelogram, each pair of opposite sides is the same length, so the perimeter of a parallelogram is twice the base plus twice the side length. And the formula can be written as
Perimeter = 2(b + a),where b is the length of the base and a is the side length

## Properties of Parallelograms

1. Two sets of sides are parallel.
2. Two sets of sides are congruent.
3. The opposite angles are congruent.
4. The consecutive angles are supplementary.
5. The diagonals bisect each other.
6. The diagonals form two congruent triangles.
7. The sum of the squares of the sides equals the sum of the squares of the diagonals. (This is the parallelogram law.)

## Parallelogram Problems

### Solved Examples

Question 1: A Parallelogram has base 5cm and height 4cm. What is the area of parallelogram?
Solution:

Given that Base = 5cm, Height = 4cm

Area of the parallelogram (A) = b x h square units

Thus the Area of the given parallelogram = 5 x 4 = 20 sq.cm.

Question 2: A Parallelogram has an area of 32 sq cm and height 4 cm. What is the length of the base of parallelogram?
Solution:

Given that area = 32 sq cm, Height = 4 cm.

Let the length of the base = b.

Area of the parallelogram (A) = b x h square units.

Thus the Area of the given parallelogram = b x 4 = 4b sq.cm.

But the area is given as 32 sq.cm.  Thus, 4b = 32

b = $\frac{32}{4}$ = 8.

Thus the base length of the given parallelogram is 8 cm.

Question 3: A parallelogram has a base of 2x, altitude of x, and the other side of the parallelogram (not the base) is 3x. If the area of this parallelogram is 18, what is its perimeter?
Solution:

Given the base of the parallelogram, b = 2x

Altitude, h = x

Area of the parallelogram (A) = b * h = 2x * x = 2x$^2$.

But the area is 18.

Thus 2x$^2$ = 18 $\rightarrow$ x$^2$ = 9 and so x = 3.

The perimeter would be 2(b + l) where l = 3x.

Thus the perimeter = 2(2 * 3 + 3 * 3) = 2(6 + 9) = 2(15) = 30.