Area of Parallelogram
Area of parallelogram is measured by the product of its base
and height. i.e. Area of parallelogram = base×height = b×h. This can be understood by the following activities.
1. Let’s take a piece of paper of a parallelogram
shape.
2. Fold along PQ and QA which are perpendicular to
each other. Cut it to get ΔPQA.
3. Place ΔPQA along SR such a way that PQ = RS,
thus the rectangle APSA’ is formed by this arrangement.
Now, the area of parallelogram PQRS = area of rectangle
APSA’
= length × breadth
= PS × PA
= base × height
∴ Area of parallelogram = base×height = b×h
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Some theorems related to area of parallelogram
1. Diagonal of a parallelogram bisects the
parallelogram. Or, The area of each triangle formed by a diagonal of
parallelogram is half of the area of the parallelogram.
2. Area of Parallelograms on same base and between same
parallel lines are equal.
3. The area of a triangle is equal to half of the
area of a parallelogram standing on same base and between same parallel lines.
Proofs:
1. Diagonal of a parallelogram bisects the
parallelogram. Or, The area of each triangle formed by a diagonal of
parallelogram is half of the area of the parallelogram.
Given: ABCD is a parallelogram in which BD is its
diagonal.
To prove: ΔABD = ΔBCD
Proof:
Statements
Reasons
1. In ΔABD and ΔBCD
i. ∠ABD = ∠BDC (A) -------------->
Alternate angles
ii. BD = BD (S)
----------------------> Common side
iii. ∠ADB = ∠CBD (A) -------------->
Alternate angles
2. ΔABD ≅ ΔBCD
--------------------------> By A.S.A. axiom
3. ΔABD = ΔBCD
----------------> Congruent triangles are equal in area
Proved.
2. Parallelograms on a same base and between
same parallel lines are equal in area.
Given: Parallelograms ABCD and ABEF are on same base and between same parallel lines.
To prove: ⎕ ABCD =
⎕ ABEF
Construction: CX perpendicular to AB is drawn from the point C.
Proof:
Statements
Reasons
1. ⎕ ABCD = AB × CX-------------------> Area of
parallelogram = base × height
2. ⎕ ABEF = AB × CX--------------------> Area of
parallelogram = base × height
3. ⎕ ABCD = ⎕ ABEF ------------------->
From statement 1 and 2.
Proved.
3. Area of a triangle is ½ of area of a parallelogram on the same base and between same
parallel lines.
Given: ΔABC and ⎕ ABCD
are on the same base and between same parallels.
To prove: ΔABC = ½ ⎕ BCDE
Construction: A
line CF is drawn parallel to BA.
Proof:
Statements
Reasons
1. ABCF is a parallelogram ------------->
Being BC∥AD(given)
and BA∥CF (construction)
2. ΔABC = ½ ⎕ ABCF -----------------> Diagonal
bisect the parallelogram
3. ⎕ ABCF = ⎕ BCDE
-------> Parallelograms on same base and between same parallel lines
4. ΔABC = ½ ⎕ BCDE -----------------> From statements 2 and 3.
Proved.
Workout Examples
Example 1: Find the area of a parallelogram
whose base is 10cm and height is 6cm.
Solution:
From the figure,
Base
of the parallelogram (b) = 10cm
Height
of the parallelogram (h) = 6cm
Now,
Area
of parallelogram (A) = base (b) × height (h)
= 10cm × 6cm
= 60 cm2
Example 2: In the given figure, ABCD is a
parallelogram. AE ⊥ BC and AN ⊥ CD. If BC = 12cm, AE =
5cm and AN = 8cm, find the length of CD.
Solution:
Here,
BC
= 12cm
AE
= 5cm
AN
= 8cm
Area
of ⎕ ABCD = BC × AE
=
12cm × 5cm
=
60 cm2
Again,
Area
of ⎕ ABCD = CD × AN
or, 60cm2 = CD × 8cm
or, CD = 60cm2/8cm
or, CD = 7.5cm
Example 3: In the given figure are of trapezium
AECD is 80cm2 and area of ΔDEC is 27cm2. Find the area of
ΔBCE.
Solution:
Here,
Area
of ⎕ ABCD = 2 × ΔDEC ----------> Area of ⎕ is double the area of Δ on same base and
between same parallel lines.
= 2 × 27cm2
= 54cm2
Area
of ΔACE = Area of trapezium AECD – Area of ⎕ ABCD
= 80cm2 – 54cm2
= 26cm2
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regarding the area of parallelogram here.
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