This article aims to discuss parallelograms that are between the same parallels and on the same base. Students are expected to understand figures and should be complete knowledge of the properties of different figures like triangles and parallelogram. Parallelograms on the same base and between the same parallels have same area. This article will explain this theorem in details. It will be easier for students to understand the concepts associated with figures that share the same base and are within the same parallels.

Introduction

In the above figure, the parallelogram \(ABCD\) and \(△BCE\) lie on the same base \(BC\) and between the same parallels \(BC\) and \(AD.\) Thus, we can say that when two figures or shapes are said to be on the same base if they share a common base or equal base and lie on the same line parallel to the base.

Figures Between Same Parallels: Theorems

There are many related theorems on the concept of figures between the same parallels. Let’s go through these theorems.

Theorem 1: Parallelograms on the same base and between the same parallel lines are equal in area.

Given: Two parallelograms\(\left( {\parallel \,{\rm{gms}}} \right)\) \(ABCD\) and \(ABEF\) on the same base \(AB\) and between the same parallel lines \(AB\) and \(DE.\)

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To prove: Area of \({\parallel \,{\rm{gm}}}\) \(ABCD=\) area of \({\parallel \,{\rm{gm}}}\) \(ABEF.\)

The area addition postulate says that if we have two shapes that do not overlap, their total area equals the sum of the areas of the individual shapes. Hence, two parallelograms on the same base and between the same parallels lines are equal in area.

Theorem 2: Parallelogram with equal bases and between the same parallels are equal in area.

Proof: Let \(ABCD\) and \(EFGH\) be two parallelograms with equal bases, i.e., \(AB=EF\) and between the same parallel lines \(EB\) and \(HC.\)

Now, from \(B\) draw \(BM⊥DC\) and from \(E,\) draw \(EN⊥HG,\) then \(BM=EN.\)

Therefore, \(EBMN\) is a rectangle.

Now, area of a parallelogram \({\rm{ = base \times height}}\)

\(=AB×BM\)

Now, since, \(AB=EF\) and \(BM=EN\)

Thus, \(AB×BM=EF×EN\)

\(=\) area of parallelogram \(EFGH\)

Hence, parallelograms with equal bases and between the same parallel lines are equal in area.

Theorem 3: Parallelograms on the same base and having equal areas lie between same parallel lines.

Given: Two parallelograms \(ABCD\) and \(ABEF\) on the same base \(AB\) and area of ∥ gm \(ABCD=\) area of \(\parallel \,{\rm{gm}}\,ABEF.\)

To prove: \(AB∥FC\)

Construction: From \(D,\) draw \(DM⊥AB\) and from \(E,\) draw \(EN⊥AB.\)

Proof:

Hence, parallelograms on the same base and having equal areas lie between the same parallel lines.

Theorem 4: Area of a Triangle is Half that of a Parallelogram if they are on the Same Base And Between the Same Parallel Lines.

Given: A triangle \(ABE\) and a parallelogram \(ABCD\) on the same base \(AB\) and between the same parallel lines \(AB\) and \(DC\)

To prove: Area of \(\Delta ABE = \frac{1}{2}\) area of \(\parallel \,{\rm{gm}}\,ABCD\)

Construction: Through \(B,\) draw \(BF∥AE\) to meet \(DC\) produced at \(F.\)

Proof:

Hence, the area of a triangle is half that of a parallelogram if they are on the same base and between the same parallel lines.

Theorem 5: Triangles on the same base and between the same parallel lines are equal in area.

Given: Two triangles \(ABC\) and \(BCD\) on the same base \(BC\) and between the same parallel lines \(BC\) and \(DI.\)

To prove: Area of \(△ABC=\)Area of \(△BCD\)

Construction: Through \(C,\) draw \(CE∥BA\) and \(CF∥BD\) to meet line \(l\) at \(E\) and \(F\) respectively.

Proof:

Hence, triangles on the same base and between the same parallel lines are equal in area.

Solved Examples – Figures Between Same Parallels and Related Theorems

Q.1. In the below-given figure, \(ABCD\) is a rectangle with sides \(AB = 8\,{\rm{cm}}\) and \(AD = 5\,{\rm{cm}}{\rm{.}}\) Compute the area of a parallelogram \(ABEF.\)

Ans: In the figure, we can see that the rectangle and the parallelogram lie on the same base \(AB\) and between the same parallel lines \(AB\) and \(DE.\)
Therefore,
Area of a \(\parallel \,{\rm{gm}}\,ABEF = \)Area rectangle \(ABCD\)
Area of a \(\parallel {\rm{gm}}\,ABEF = 8 \times 5 = 40\;{\rm{c}}{{\rm{m}}^2}\)
Hence, the area of \(\parallel {\rm{gm}}\,ABEF = 40\;{\rm{c}}{{\rm{m}}^2}\)

Q.2. \(P\) and \(Q\) are any two points lying on the sides \(DC\) and \(AD\) respectively of a parallelogram \(ABCD.\) Show that the area of \(△APB=\)area of \(△BQC.\)
Ans: Given a parallelogram \(ABCD,\) and \(P\) and \(Q\) are points lying on the sides \(DC\) and \(AD\) respectively, as shown in the figure.

As \(APB\) and ∥ gm ABCD are on the same base and between the same parallel lines \(AB\) and \(DC,\)
Area of \(\Delta APB = \frac{1}{2}\) area of \(\parallel {\rm{gm}}\,ABCD \ldots (1)\)
Also, \(\Delta BQC\) and \(\parallel {\rm{gm}}\,ABCD\)are on the same base and between the same parallel lines \(AD\) and \(BC,\)
Area of \(\Delta BQC = \frac{1}{2}\) area of \(\parallel {\rm{gm}}\,ABCD \ldots (2)\)
Now, from the equation \((1)\) and \((2)\)
Area of \(△APB=\) Area of \(△BQC\)

Q.3. In the below-given figure, \(ABCD\) and \(ABEF\) are parallelograms, and \(P\) is any point on \(DC.\) If the area of \(\parallel {\rm{gm}}\,ABEF = {\rm{90}}\,{\rm{c}}{{\rm{m}}^{\rm{2}}},\), find the area of \(△ABP\).

Ans: As we can see that, \(△ABP\) and \(\parallel {\rm{gm}}\,ABEF\) are on the same base \(AB\) and between the same parallel lines \(AB\) and \(DC,\)
Therefore, the area of \(\Delta ABP = \frac{1}{2}\) area of \(\parallel {\rm{gm}}\,ABCD\)
\( = \frac{1}{2} \times 90\;{\rm{c}}{{\rm{m}}^2}\)
\( = 45\;{\rm{c}}{{\rm{m}}^2}\)
Hence, the area of \(△ABP\) is \(45\;{\rm{c}}{{\rm{m}}^2}.\)

Q.4. A point \(E\) is taken on the side \(BC\) of a parallelogram \(ABCD. AE\) and \(DC\) are produced to meet at \(F.\) Prove that the area of \(△ADF=\) area of the quadrilateral \(ABFC.\)

Ans: Join \(AC\) and \(BF\)

As triangles \(△ACF\) and \(△BCF\) have the same base \(CF\) and are between the same parallels \(AB\) and \(CF.\)
Therefore, the area of \(△ACF=\) area of \(△BCF…(1)\)
The diagonal, \(AC\) divides the parallelogram \(ABCD\) into two triangles of equal area.
Thus, the area of \(△DAC=\)area of \(△ABC…(2)\)
Now, on adding the equations \((1)\) and \((2,)\) we get,
Area of \(△ACF+\) area of \(△DAC=\) area of \(△BCF+\)area of \(△ABC\)
Area of \(△ADF=\) area of the quadrilateral \(ABFC.\)

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Q.5. In the below-given figure, \(ABCD\) is a parallelogram. Point \(P\) and \(Q\) on \(BC\) trisect \(BC\) in \(3\) equal parts. Prove that area of \(△APQ=\) area of \(\Delta DPQ = \frac{1}{6}\) area of \(\parallel {\rm{gm}}\,ABCD\).

Ans: Given points, \(P\) and \(Q\) on \(BC\) trisect \(BC,\) so \(BP=PQ=QC\)
Therefore, \(PQ = \frac{1}{3}BC\)
From A, draw \(AM⊥BC\) and from \(D,\) draw \(DN⊥BC\) (when produced)

As \(AM\) and \(DN\) are both perpendicular to \(BC,\) so \(AM∥DN.\) Also, \(AD∥BC,\) i.e., \(AD∥MN,\) therefore, \(AMND\) is a parallelogram.
Thus, \(AM=DN,\) as the opposite sides of a parallelogram are equal.
Now, the area of \(\Delta APQ = \frac{1}{2}PQ \times AM = \frac{1}{2}\left( {\frac{1}{3}BC} \right) \times AM\)
\( = \frac{1}{6}BC \times AM = \frac{1}{6}{\rm{area}}\,{\rm{of}}\parallel {\rm{gm}}\,ABCD \ldots (1)\)
And, the area of \(\Delta DPQ = \frac{1}{2}PQ \times DN = \frac{1}{2}\left( {\frac{1}{3}BC} \right) \times AM\)
\( = \frac{1}{6}BC \times AM = \frac{1}{6}{\rm{area}}\,{\rm{of}}\parallel {\rm{gm}}\,ABCD \ldots (2)\)
From the equation \((1)\) and \((2),\) we get,
Area of \(\Delta APQ = \) area of \(\Delta DPQ = \frac{1}{6}\) area of \(\parallel {\rm{gm}}\,ABCD.\)

Summary

This article taught us about the figures on the same base and between the same parallel lines. We learned that if two parallelograms, or a parallelogram and a rectangle, shares the same base and lies between the same parallel lines, their area is equal.

In addition to this, we also learned that if a parallelogram and a triangle lie on the same base and between the same parallels, the triangle area is half the area of the parallelogram.

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Frequently Asked Questions (FAQs)

Frequently asked questions related to figures sharing the same base and are within same parallels are listed as follows:

Q.1. When two figures are said to be on the same base and between the same parallels?
Ans: Any two geometric figures are said to be on the same base and between the same parallels if they both have a common side as the base, and the vertices opposite to the common base lie on the line parallel to this base.

Q.2. What happens if parallelograms are on the same base and between the same parallels?
Ans: If the parallelograms lie on the same base and between the same parallel lines, then their areas are equal.

Q.3. What is the relationship in triangles if they are in the same base and between the same parallels? Write the theorem statement?
Ans: If two triangles are on the same base and between the same parallels, their areas will be equal.

Q.4. What does the word parallel mean in math?
Ans: When two straight lines on the same plane do not intersect at any point of time, they are parallel lines.

Q.5. What happens if a parallelogram and a rectangle lie on the same base and between the same parallels?
Ans: If a parallelogram and a rectangle lie on the same base and between the same parallel lines, they are equal in area.

We hope this detailed article on the figures between the same parallels and related theorems helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!