Area of Parallelogram:
A quadrilateral having two pairs of parallel sides is called a parallelogram.
In a parallelogram,
* Opposite sides are equal in length
* Opposite angles are equal.
* All parallelograms are rectangles.
A parallelogram is a quadrilateral with the opposite sides parallel and equal to each other.
The angles of a parallelogram need not be equal to 90⁰. All rectangles are parallelograms but all parallelograms are not rectangles.
In parallelogram ABCD, AB = DC, AD = BC and AB ⏸DC.
AD ॥ BC. None of the angles are equal to 90⁰.
However, notice that ∠DAB = ∠BCD and ∠ABC = ∠ADC, or the opposite angles of a parallelogram are equal to each other.
Theorem: The opposite sides of a parallelogram are equal; the opposite angles are equal; and a diagonal bisects the parallelogram.
Given : A parallelogram ABCD.
Construction: Join BD.
To prove: i) AD = CB and DC = AB.
ii) Â = Ĉ and AĎC = CBA
iii) Δ ADB = Δ CBD
Proof: In the triangles ADB and CBD,
DB is common.
a = a₁ (alternate; AD parallel to BC).
b = b₁ (alternate; DC parallel to AB).
Therefore the triangles ADB and CBD are congruent
In particular, AD = CB; AB = CD and A = C.
Area of a parallelogram = base x height
= bh
Q.1. The figure shows a parallelogram OQRS where PQ = 9 cm and PS = 6 cm. QU is perpendicular to PS and QT is perpendicular to SR. If QU = 8 cm. Calculate the length of QT.
Solution:
Area of the parallelogram = base x height
PQ x QT = PS x QU
9 x QT = 6 x 8
QT = 16/3
Length of QT = 5 1/3 cm
Q.2. The figure shows a parallelogram ABCD where AB = 14 cm and BC = 10 cm. If DE = 8 cm, calculate
i) the area of the parallelogram
ii) the perimeter of the parallelogram
Solution:
I) Area of the parallelogram = base x height
= 14 x 8
= 112 cm2
ii) Since opposite sides of a parallelogram are equal in length
AB = DC and BC = AD
Perimeter of the parallelogram = 14 x 2 + 10 x 2
= 48 cm Ans.
Ex. 42 B
1. In the parallelogram ABCD, AB = 8 cm and AD = 6 cm. The distance between AB and CD is 3 cm. Find the distance between AD and BC.
Solution:
In the parallelogram ABCD we are given that AB = 8 cm and AD = 6 cm. The distance between AB and CD is EF = 3 cm.
Area of the parallelogram = AB X EF
= 8 X 3 cm²
= 24 cm²
Let, the distance between AD and BC is GH.
The area of the parallelogram is = AD X GH
= 6 cm X GH
6 cm X GH = 24 cm²
or, GH = 4 cm² Ans.
3. The area of a trapezium is 14 cm². The parallel sides are 3cm and 4 cm long. Find the distance between them.
Solution:
ABCD is the trapezium and AB and CD are the two parallel sides where AB = 4 cm and CD = 3 cm.
The area of the trapezium ABCD is = 1/2 ( AB + CD ) - h
According to the question,
1/2 (AB + CD) h = 14 cm²
or, 1/2 ( 3 +4 ) h cm = 14 cm²
or, 7/2 cm h = 14 cm²
or, h = 14 X 2 cm²/ 7 cm
or, h = 4 cm Ans.
Q. In the given quadrilateral, the three angles ∠1, ∠2 and ∠3 are 60⁰ , 80⁰ and 100⁰ respectively. Find the fourth angle (∠4).
∠1 + ∠2 + ∠3 = 60⁰ + 80⁰ + 100⁰ = 240⁰
Now, in the quadrilateral,
∠1 + ∠2 + ∠3 = 360⁰
Therefore, ∠4 = 360⁰ - (sum of three angles)
=360⁰ - 240⁰ = 120⁰
