Area of a parallelogram
The total space inside the boundary of the parallelogram is called as the area of the
parallelogram. The area of a parallelogram is twice the area of a triangle created by one of its
diagonals.
In a parallelogram, the opposite sides are equal to each other and the opposite angles are also
equal to each other. The sum of all the interior angles of a parallelogram is 3600.
The area of a parallelogram can be found by using the following formula,
Area of a parallelogram, A = bh
where,
b = Base of the parallelogram
h = Height of the parallelogram
The base and the height of the parallelogram are always perpendicular to each other.
We can derive the height of the parallelogram with the help of the area of a parallelogram
formula.
Area of a Parallelogram without Height
There is another formula to find the area of the parallelogram:
For two sides a and b and angle θ, is
Area = ab sin θ.
Area of Parallelogram Vectors
The parallelogram formed by the a ⃗ and b ⃗ with an angle of θ between them. The equation for area of a
parallelogram is given below:
Find the Area of the Parallelogram with Vertices
To find the area of the parallelogram, we need to find 2 vectors that represents two sides of the
parallelogram that share a common vertex. We can either find the angle between the two sides or can
evaluate the cross product directly.
Let us find the area of a parallelogram whose vertices are (0, 0), (3, 7), (4, -2) and (7, 8).
Let parallelogram ABCD has vertices: A(0, 0), B(3, 7), C(4, -2) and D(7, 8).
a ⃗ = AB = (3 - 0, 7 - 0) = (3, 7)
b ⃗ = AC = (4 - 0, -2 - 0) = (4, -2)
Area of parallelogram cross product
Now, Area of parallelogram = |a ⃗ ×
a ⃗ | = |(3 * -2) - (7 * 4)|
= |-6 - 28|
= 34
∴ The area of the parallelogram is 34 sq units.
Question 3: The area of a parallelogram is 34 m2. If the height of that parallelogram is 8 m, then
find the base of the parallelogram.
Solution:
Given, area A = 34 m2 and height h = 8 m.
We know that the area of parallelogram is,
A=bh
Substitute the value of A and h,
34 = b (8)
b = 348
b = 4.25 m
So, the base of the given parallelogram is 4.25 m.
Question 4: What is the parallelogram height if the area of the parallelogram is 400 cm2 and base
of the parallelogram is 20 cm?
Solution:
Given, Area of the parallelogram is 400 cm2 and base of the parallelogram = 20 cm.
Parallelogram height = (Area of the parallelogram) /( Base of the parallelogram).
= 40020
= 20 cm.
The height of the parallelogram is 20 cm.
Parallelogram Theorem
Quadrilateral means a closed figure formed by four line segments and a parallelogram is a quadrilateral
in which the opposite sides are parallel to each other. A point is used to represent a position in space. A
plane means a surface increasing infinitely in every directions such that all points lying on the line joining
any two points on the surface. The parallelogram theorems are given below.
Parallelogram Proof
Below you could see three theorems of parallelogram
Theorem 1:
Parallelograms are the same base and between the same parallel lines are equal in area.
Activity:
Draw a line segment AB. Draw a line l parallel to AB. Mark a point C on l. Draw AL
perpendicular to l. Measure the length of AL.
We find the area of the triangle ABC as 1/2 × base × height
= 1/2 * AB * AL
Mark another point P on l. We find the area of ?ABP as 1/2 × base × height = 1/2 * Base *
Height
= 1/2 * AB * AL
thus, we seen that the area of the triangle remains the same for all positions of the vertex C on
the line l.
Theorem 2:
A parallelogram is a rhombus if its diagonals are 90 degree.
Given: ABCD is a parallelogram where the diagonals AC and BD are perpendicular.
To prove: ABCD is a rhombus.
Construction: Draw the diagonals AC and BD. Let M represent the point of intersection of AC
and BD (see Figure).
Proof: In triangles AMB and BMC,
(i) ?AMB = ?BMC = 90°
(ii) AM = MC
(iii) BM is common.
? By SSA criterion, ?AMB = ?BMC.
In particular, AB = BC.
Since ABCD is a parallelogram, AB = CD, BC = AD.
? AB = BC = CD = AD.
Hence ABCD is a rhombus. The theorem is proved.
Theorem 3:
The quadrilateral is a parallelogram, then the one pair of opposite sides are parallel and equal.
Given: ABCD is a quadrilateral, where AB || CD and AB = CD.
To prove: ABCD is a parallelogram.
Construction: Draw the diagonal AC (see Figure).
Proof: In triangles ABC and ADC,
(i) AB = CD (given)
(ii) AC is common
(iii) m?BAC = m?ACD
By SAS criterion, ?ABC = ?ADC. sides are equal
AD = BC, m?DAC = m?ACB.
AD || BC.
Hence ABCD is a parallelogram. The theorem is proved
Theorem:
The diagonals of a parallelogram bisect each other.
Proof:
ABCD is a parallelogram. AC and BD are diagonals.
By ASA criterion, ?AMB = CMD (see Figure).
? AM = CM, BM = DM.? The diagonals bisect each other.
Eucledian Proof
According to scientist Adler, in his book on modern geometry he mention “One of the main
defects in traditional Euclidean proof is its almost complete disregard of such notions as the two
sides of a line and the interior of an angle. Without clarification of these ideas, absurd
consequences result.”
Here is an illustration to show traditional proof of a geometrical figure parallelogram PQRS
Let PQRS parallelogram where, PQ||RS and QR||PS and O be the intersection of the diagonals
PR and QS. Prove PO = RO
The traditional proof (Euclidean Proof):
To prove :
Δ PRQ = RSP (hence PQ = RS) and Δ POQ = Δ SOR (hence PO = RO)
Proof :
By congruency of triangle ∠ RPQ = ∠ PRS
because alternative angles of parallels PO and RS.
As we assumed the trivial fact that point S and Q are either sides of the line PR .
Draw back lies in harder situation to prove the last point.