A parallelogram is a quadrilateral whose opposite sides are parallel and equal. Specific formulas are known for calculating the areas of various shapes such as rectangle, square, parallelogram, and triangle. Here, we will consolidate the knowledge about these formulas by studying some relationships between the areas of these geometric figures under the condition that they lie on the same base between the same parallel. This article will also clarify some findings on triangle similarity. In this article, we will see how the area of the parallelograms on the same base and between the same parallels are equal.

Figures on the Same Base 

In figure (i) the trapezium \(ABCD\) and the parallelogram \(CDEF\) have a common side \(CD.\) \(CD\) can also be called the common base, or we can say \(ABCD\) and \(CDEF\) have the same base.

Similarly, if we look at figure (ii) we can see both the parallelograms \(PQRS\) and \(MNRS\) share the same base or have the common base \(SR.\) In figure (iii), \(\Delta ABC\) and \(\Delta BCD\) have the same base \(BC.\) In figure (iv), parallelogram \(ABCD\) and \(\Delta DCP\) have the common base \(DC.\)

Figures on the Same Base and Between the Same Parallels

In figure (i) trapezium \(ABCD\) and parallelogram \(EFCD\) are on the same base \(DC.\) In addition to the above, the vertices \(A\) and \(B\) of the parallelogram \(ABCD\) and the vertices \(E\) and \(F\) of the parallelogram \(EFCD\) opposite to the base \(DC\) lie on a line \(AF\) parallel to \(DC.\) We say that trapezium \(ABCD\) and parallelogram \(EFCD\) are on the same base \(DC\) and between the same parallels \(AF\) and \(DC.\)

Similarly, in figure (ii), parallelogram \(PQRS\) and \(MNRS\) are on the same base \(SR\) and between the same parallels \(PN\) and \(SR\) as vertices \(P\) and \(Q\) of \(PQRS\) and vertices \(M\) and \(N\) of \(MNRS\) lie on a line \(PN\) parallel to base \(SR.\)

In the same way, in figure (iii) \(\Delta ABC\) and \(\Delta DBC\) lie on the same base \(BC\) and between the same parallels \(AD\) and \(BC.\) (iv) The parallelogram \(ABCD\) and \(\Delta PCD\) lie on the same base \(DC\) and between the same parallels \(AP\) and \(DC.\)

Therefore, two figures are said to be on the same base and between the same parallels if they have a common base and the vertices opposite to the common base of each figure lie on a line parallel to the base. We have two theorems for figures on the same base and between the same parallels. Let us understand them in detail.

Theorem 1: Parallelogram on the same base and between the same parallels are equal in area.

Proof: Consider two parallelograms \(ABCD\) and \(EFCD,\) on the same base \(DC\) and between the same parallels \(AF\) and \(DC.\)

To prove: area of the parallelogram \(\left( {ABCD} \right)\) is equal to the area of the parallelogram \(\left( {EFCD} \right)\)
Let us consider two triangles \(\Delta ADE\) and \(\Delta BCF.\)
\(\angle DAE = \angle CBF\)……(i)
As these angles are corresponding angles from \(AD\parallel BC\) and transversal \(AF.\)
In the same way, \(\angle AED = \angle BFC\)……(ii)

These are also corresponding angles as \(ED\parallel FC\) and \(AF\) is the transversal.
In \(\Delta ADE\) and \(\Delta BCF,\) by using the angle sum property of a triangle,
\(\angle DAE + \angle ADE + \angle AED = {180^{\rm{o}}}\) and \(\angle BCF + \angle BFC + \angle CBF = {180^{\rm{o}}}\)
\(\angle DAE + \angle ADE + \angle AED = \angle BCF + \angle BFC + \angle CBF\)
Now, substituting \(\angle DAE = \angle CBF,\,\angle AED = \angle BFC\) in above equation we have, \(\angle CBF + \angle ADE + \angle BFC = \angle BCF + \angle BFC + \angle CBF\)
Now, cancelling \(\angle CBF,\,\angle BFC\) from both side we get,
\(\angle ADE = \angle BFC\)….(iii)
We know that the opposite sides of a parallelogram are equal.
So, \(AD=BC\) ….(iv)

From the equations (i), (iii), and (iv)
\(\Delta ADE \cong \Delta BCF\) (by Angle-Side-Angle \(\left( {{\rm{ASA}}} \right)\) rule)
Therefore, the area of \(\Delta ADE\) and the area of \(\Delta BCF\) is the same as the congruent figures have equal area.
Thus, \({\rm{Area}}\,{\rm{of}}\,\Delta ADE = {\rm{Area}}\,{\rm{of}}\,\Delta BCF\)..…(v)

Now,
\({\rm{Area}}\,{\rm{of}}\,\left( {ABCD} \right) = {\rm{Area}}\,{\rm{of}}\,\Delta ADE + {\rm{Area}}\,{\rm{of}}\,\left( {EDCB} \right)\)
\({\rm{Area}}\,{\rm{of}}\,\left( {EFCD} \right) = {\rm{Area}}\,{\rm{of}}\,\Delta BCF + {\rm{Area}}\,{\rm{of}}\,\left( {EDCB} \right)\)
As \({\rm{Area}}\,{\rm{of}}\,\Delta ADE = {\rm{Area}}\,{\rm{of}}\,\Delta BCF,\)
\({\rm{Area}}\,{\rm{of}}\,\left( {ABCD} \right) = {\rm{Area}}\,{\rm{of}}\,\left( {EFCD} \right)\)
Therefore, the area of parallelogram \(ABCD\) is equal to the area of parallelogram \(EFCD.\)
Hence, it is proved that the parallelograms on the same base and between the same parallels are equal in area.

Theorem 2: If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half of the area of the parallelogram.

Let us consider triangle \(\Delta ABP,\) and parallelogram \(ABCD\) be on the same base \(AB\) and between the same parallels \(AB\) and \(PC.\)
You are required to prove that \({\rm{Area}}\,{\rm{of}}\,\Delta ABP = \frac{1}{2}{\rm{of}}\,{\rm{area}}\,{\rm{of}}\,ABCD\)
Construction: Draw \(BQ\parallel AP\) to get another parallelogram \(ABQP.\)
Now, parallelograms \(ABQP\) and \(ABCD\) are on the same base \(AB\) and between the same parallels \(AB\) and \(PC.\)
Therefore, the \({\rm{Area}}\,{\rm{of}}\,\left( {ABQP} \right) = {\rm{Area}}\,{\rm{of}}\,\left( {ABCD} \right)\)
but \(\Delta ABP\) is congruent to \(\Delta BQP\) ( Diagonal \(PB\) divides parallelogram \(ABPQ\) into two congruent triangles)
So, \({\rm{Area}}\,{\rm{of}}\,\left( {\Delta PAB} \right) = {\rm{Area}}\,{\rm{of}}\,\left( {\Delta BQP} \right)\)
Therefore, \({\rm{Area}}\,{\rm{of}}\,\left( {\Delta PAB} \right) = \frac{1}{2}{\rm{Area}}\,{\rm{of}}\,\left( {\Delta ABQP} \right).\)
Thus, the \({\rm{Area}}\,{\rm{of}}\,\left( {PAB} \right) = \frac{1}{2}{\rm{Area}}\,{\rm{of}}\,\left( {ABCD} \right)\)  

Solved Examples – Parallelograms on the Same Base and Between the Same Parallels

Q.1. If \(ABCD\) is a parallelogram and \(CDEF\) is a rectangle, then prove that the area of the parallelogram and the rectangle are the same and area of \(ABCD = DC \times AL.\) Also \(AL \bot DC.\)

Ans:
Given: \(ABCD\) is a parallelogram, and \(CDEF\) is a rectangle having a common base \(DC,\) and they are between the same parallels \(EB\) and \(CD.\)
To prove: \({\rm{Area}}\,{\rm{of}}\,ABCD = {\rm{Area}}\,{\rm{of}}\,CDEF\)
We know that a rectangle is a parallelogram. So, recatangle \(CDEF\) is a parallelogram.
The parallelograms on the same base and between the same parallels are equal in area.
Therefore, \({\rm{Area}}\,{\rm{of}}\,ABCD = {\rm{Area}}\,{\rm{of}}\,CDEF\)
\({\rm{Area}}\,{\rm{of}}\,ABCD = DC \times FC\,\left( {{\rm{Area}}\,{\rm{of}}\,{\rm{the}}\,{\rm{rectangle}} = {\rm{length}} \times {\rm{breadth}}} \right)\)
\(AL \bot DC\)
Therefore, \(AFCL\) also a rectangle
\(AL = FC\)
Hence, the \({\rm{Area}}\,{\rm{of}}\,ABCD = DC \times AL\) (proved).

Q.2. In the figure, \(PQRS\) and \(ABRS\) are parallelograms, and \(X\) is any point on side \(BR.\) Show that 
(i) \({\rm{Area}}\,{\rm{of}}\,\left( {PQRS} \right) = {\rm{Area}}\,{\rm{of}}\,\left( {ABRS} \right)\)
(ii) \({\rm{Area}}\,{\rm{of}}\,\Delta \left( {AXS} \right) = \frac{1}{2}{\rm{Area}}\,{\rm{of}}\,\left( {PQRS} \right)\)

Ans:
Given: \(PQRS\) and \(ABRS\) are parallelograms and \(X\) is any point on side \(BR.\)
To prove: (i) \({\rm{Area}}\,{\rm{of}}\,\left( {PQRS} \right) = {\rm{Area}}\,{\rm{of}}\,\left( {ABRS} \right)\)
(ii) \({\rm{Area}}\,{\rm{of}}\,\Delta \left( {AXS} \right) = \frac{1}{2}{\rm{Area}}\,{\rm{of}}\,\left( {PQRS} \right)\)
Proof:
(i) \(PQ\parallel RS\) as \(PQRS\) is a parallelogram and opposite sides of a parallelogram are equal.
Similarly, \(AB\parallel RS\) as \(ABRS\) is a parallelogram.
Since \(PQ\parallel RS,\,AB\parallel RS\) we can say, \(PB\parallel RS.\)
We can see \(PQRS\) and \(ABRS\) have the same base \(RS\) and they lie between the same parallels \(PB\) and \(RS.\)
Hence, \({\rm{Area}}\,{\rm{of}}\,\left( {PQRS} \right) = {\rm{Area}}\,{\rm{of}}\,\left( {ABRS} \right)\) (proved)
(ii) Since \(ABRS\) is a parallelogram, \(AS\parallel BR\)
Parallelogram \(ABRS\) and \(\Delta AXS\) have the same base \(AS\) and are between the same parallels \(AS\) and \(BR.\)
Therefore, the \({\rm{Area}}\,{\rm{of}}\,\Delta \left( {AXS} \right) = \frac{1}{2}{\rm{Area}}\,{\rm{of}}\,{\rm{parallelogram}}\,ABRS\)
Hence, proved.

Q.3. If the height of a triangle \(\Delta ABD\) is \(10\,{\rm{cm}}\) and the base length of a triangle is \(15\,{\rm{cm}}.\) Find the area of the parallelogram \(ABCD.\)

Ans: In the given figure, \(\Delta ABD\) and the parallelogram \(ABCD\) lie on the same base \(AB\) and between the same parallel lines such as \(AB\) and \(CD.\)
We know that, If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half of the area of the parallelogram.
Area of triangle  \(\Delta ABD = \frac{1}{2} \times {\rm{base}} \times {\rm{height}} = \frac{1}{2} \times 15 \times 10 = 75\,{\rm{c}}{{\rm{m}}^2}.\)
Now, the \({\rm{Area}}\,{\rm{of}}\,\left( {ABD} \right) = \frac{1}{2}{\rm{Area}}\,{\rm{of}}\,\left( {ABCD} \right)\)
Therefore, the area of the parallelogram \(ABCD = 2 \times 75 = 150\,{\rm{c}}{{\rm{m}}^2}.\)

Q.4. Identify the figure in which two figures lie on the same base and between the same parallels.

Ans: In figure (i), \(DC\) is the common side for \(\Delta DPC\) and \(ABCD,\) and they lie between the same parallels that are \(DC\) and \(AB.\)
In figure (ii), \(SR\) is the common base of the quadrilaterals \(MNRS\) and \(PQRS,\) but they do not lie between the same parallels.
In figure (iii), \(QR\) is the common side for \(\Delta QTR\) and \(PQRS,\) and they lie between the same parallels that are \(QR\) and \(PS.\)

Q.5. Show two examples in which two figures lie between the same parallels but do not have the same base.
Ans:

In figure (i), triangles \(\Delta ABC\) and \(\Delta BDE\) lie between the same parallels, but they do not have the same base.
Figure (ii) \(\Delta ABC\) and parallelogram \(PQRS\) lie between the same base, but they do not have the same or common base.

Summary

In this article, we learnt about theorems related to the area of parallelograms on the same base and between the same parallels. We also learnt that if two parallelograms have a common base and they lie between the same parallel lines, then the area of the parallelograms is equal.

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half of the area of the parallelogram.

Learn About Properties of Parallelogram

Frequently Asked Question (FAQs)

Q.1. What can you say about two parallelograms that have the same base and height?
Ans: If two parallelograms have the same base and height, we can say that their area is equal.

Q.2. What is the relation between the parallelogram and triangle standing on the same base between the same parallel lines?
Ans: If a triangle and a parallelogram are on the same base and between the same parallels, then the triangle area is half of the area of the parallelogram.

Q.3. Which parallelograms have the same area?
Ans: The parallelograms on the same base and between the same parallels have the same area.

Q.4. What is the ratio of the areas of two parallelograms that lie on the same base and the same parallels?
Ans: As the parallelogram on the same base and between the same parallels have the same area, the ratio of the area of two parallelograms is \(1:1.\)

Q.5. What is the base and height of a parallelogram?
Ans: The base is the bottom side of the parallelogram. The height is the distance between the base and its opposite side of the parallelogram.

Now you are provided with all the necessary information on the parallelograms on the same base and between the same parallels and we hope this detailed article is helpful to you. If you have any queries regarding this article, please ping us through the comment section below and we will get back to you as soon as possible.