# Area of Parallelogram

## 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’
= 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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