**Area of parallelogram and area of triangles-**

Parallelogram is a special type of quadrilateral in which both pair of opposite sides is parallel. Area of parallelograms and triangles between the same parallel lines can be deduced using the properties of parallelogram and triangle.

**Area axiom:**

Every polygonal region has an area, measured in standard units. The area of a square of side 1 unit is 1 square unit and area is always a positive quantity.

**Congruent axiom:** Two shapes or figures are said to be congruent, if they have the same shape and same size. Area of congruent triangles is always equal but any triangles of equal areas need not to be congruent.

If ∆ABC and ∆CDE are two congruent triangles, then

Area of (∆ABC) = Area (∆CDE)

**Points to be remembered:**

- A triangle’s median divides it into two triangles of equal area.
- Area of an object (or figure) is a number (with some standard unit) associated with the part of the plane enclosed by that object (or figure).
- Any planar region formed by a figure Z is made of two non-overlapping planar region A and B, then

Area of (Z) = Area (A) + Area (B)

**Theorem:** Parallelograms on the same base and between the same parallel sides are equal in area.

**Proof:** Two parallelograms ABCD and ABEF, on the same base DC and between the same parallel line AB and FC.

To prove that area (ABCD) = area (ABEF).

In ∆ADF and ∆BCE,

AD=BC (∴ABCD is a parallelogram ∴ AD=BC)

AF=BE (∴ABEF is a parallelogram ∴AF=BE)

∠ADF=∠BCE (Corresponding Angles)

∠AFD=∠BEC (Corresponding Angles)

∠DAF =∠CBE (Angle Sum Property)

∆ADE ≅ ∆BCF (From SAS-rule)

Area(ADF) = Area(BCE) (By congruence area axiom)

Area(ABCD)=Area(ABED) + Area(BCE)

Area(ABCD)=Area(ABED)+Area(ADF)

Area(ABCD)=Area(ABEF)

Hence, the area of parallelograms on the same base and between the same parallel sides is equal.

**Corollary**: A parallelogram and a rectangle on the same base and between the same parallels are equal in area.

**Proof:** Since a rectangle is also a parallelogram so, the result is a direct consequence of the above theorem.

**Theorem:** The area of a parallelogram is the product of its base and the corresponding altitude.

**Given:** In a parallelogram ABCD, AB is the base.

**To prove** that Area(\(||^{gm}\) ABCD) = AB×AL

**Construction:** Complete the rectangle ALMB by Drawing BM perpendicular to CD.

**Proof:** Since (\(||^{gm} \)ABCD) and rectangle ALMB are on the same base and between the same parallels.

∴ Area(\(||^{gm}\) ABCD) =Area(rect.ALMB)

= AB×AL (Area of rectangle= base ×X height)

Hence, Area(\(||^{gm}\) ABCD)= AB×AL

**Theorem:** Two triangles on the same base (or equal bases) and between the same parallel sides are equal in area.

**Proof:** Consider a parallelogram ABCD with diagonal AC.

**Construction:** AN perpendicular to DC.

∆ADC≅ ∆CBA (A diagonal of a parallelogram divides it into two congruent triangles)

Therefore, Area(ADC)= Area(CBA) (Area of congruent triangles are equal)

Now, Area(∆ADC) = \(\frac{1}{2}\) area(ABCD) =\(\frac{1}{2}\) (DC×AN)

Area of ∆ADC= \(\frac{1}{2}\)× Base(DC) ×Altitude (AN).

The area of triangle is always half the product of its base and height.

Area (Triangle) = \(\frac{1}{2}\) Area ( parallelogram)

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