Triangles on the Same Base and Between the Same Parallels: Meaning, Theorem, Problems

Triangles on the Same Base and between the Same Parallels: A triangle is a three-sided polygon. Parallel lines are two or more lines at an equal distance from each other and never meet. Triangles on the same base mean the base of both triangles will be the same or equal. Triangles made by the same parallels show that the base and the line connecting the vertices on the opposite side of the base are parallel.

When a parallelogram and a triangle share the same base and are connected by parallel lines, the triangle’s area is half that of the parallelogram. In this article, we are going to learn about the triangles on the same base and between the same parallels. 

Triangles

A three-sided polygon is known as a triangle. It is defined as a figue bordered or encircled by three-line segments. By definition, a triangle has three sides and three vertices.

Parallel Lines

Parallel lines are defined as two lines at an equal distance from each other and never meet. Two or more lines that are at the same distance apart and never meet, intersect or join are called parallel lines.

Parallel lines are those that, no matter how far they are stretched, never meet. Parallel lines are denoted by the symbol \(\parallel \).

In the figure, lines \(m\) and \(n\) are parallel lines.

Parallel lines \(m\) and \(n\) are symbolically written as \(m\parallel n\).

The symbol ∦ denotes non-parallel lines.

Triangles on the Same Base and Between the Same Parallels

Consider the following two triangles, which have the same base and lying between the same parallel lines.

We can see that both the triangles have the same base. The line joining their vertices opposite to the base and the base are parallel to each other. We know that if a triangle and a parallelogram have the same base and are parallel to each other, the triangle’s area is half that of the parallelogram.

So, in the above figure, each triangle’s area is half that of any parallelogram with the same base and parallels. As a result, the two triangles’ areas are equal. 

Theorem: The area of two triangles with the same base and parallels is equal.

Construct \(EA || FH\) and \(F B || E G\)

So, \(E H\) is the diagonal of the parallelogram \(A H F E\). And \(G F\) is the diagonal of the parallelogram \(GBFE\).

We know, diagonal of the parallelogram divides it into two congruent triangles.

So, \(\triangle A E H=\triangle E H F\) and \(\triangle E G F=\triangle F B G\)

We have already seen how to calculate the area of any triangle.

Area of any triangle \(=\frac{1}{2} \times \text {base} \times \text {height}\)

Consider a triangle. Take the side \(E F\) to be the base of this triangle. Now, measure the height of this triangle, which will be the distance between \(E F\) and the line parallel to \(E F\) through \(G\).

In other words, the height of this triangle will be the length of the perpendicular \(G D\).

Thus, the area of this triangle \(\triangle E G F\) can be written as:

\(A=\frac{1}{2} \times \text {Base} \times \text {Height}\)

\(A=\frac{1}{2} \times E F \times G D……(1)\)

Now, consider the triangle \(\triangle E H F\). The altitude of this triangle will also be the distance between the parallel lines.

Then, the area of this triangle can be written as:

\(A=\frac{1}{2} \times \text {Base} \times \mathrm{Height}\)

\(A=\frac{1}{2} \times E F \times H P……(2)\)

We know, the distance between two parallel lines is always the same.

So, \(G D=H P\) (Altitudes of both the triangles are same)

By comparing the equation \((1)\) and \((2)\)

\(\text {Area} (\triangle E G F)= \text {Area} (\triangle F E H)\)

Hence, it is proved.

Examples of Triangles on the Same Base and between the Same Parallels

In this figure, two triangles are \(\triangle A B E\) and \(\triangle C B E\) on the base \(A C\) between the common parallels.

In the above figure, two triangles are \(\triangle A B D\) and \(\triangle A C D\) on the common base \(B C\) between the common parallels.

In the above figure, two triangles are \(\triangle A C B\) and \(\triangle A D B\) on the same base \(A B\) between the same parallels.

In the above figure, two triangles are \(\triangle A B C\) and \(\triangle D E F\) on the common base \(B F\) between the common parallels.

Solved Examples – Triangles on the Same Base and between the Same Parallels

Q.1. Consider the following figure, in which \(EFGH\) is a parallelogram. \(B C\) has been produced too \(Q\), such that \(A D=C Q \cdot E Q\) intersects \(H G\) at \(P:\) Show that \(\text {area} (\triangle F P H)=\operatorname{area}(\Delta G P Q)\).

Ans: Join \(E H\), as shown below:

\(EHQG\) is a parallelogram so that \(P\) is the midpoint of \(E Q\) and \(H G\). Thus,
\(\operatorname{area}(\Delta G P Q)=\operatorname{area}(\Delta E P H) \ldots \ldots.(1)\)
We also note that \(\triangle E P H\) and \(\triangle F P H\) are on the same base \(P H\) and between the same parallels \(E F\) and \(H G\), so that:
\(\operatorname{area}(\Delta E P H)=\operatorname{area}(\Delta F P H) \ldots \ldots(2)\)
From the above two relations \((1)\) and \((2)\), we have:
\(\operatorname{area}(\Delta G P Q)=\operatorname{area}(\Delta F P H)\)
Hence, the given area \((\triangle F P H)=\operatorname{area}(\triangle G P Q)\) is proved.

Q.2. Prove that two triangles on the same base (or equal bases) and between the same parallels are equal in area.
Ans: Now, suppose \(A B C D\) is a parallelogram whose one of the diagonals is \(A C\). Let \(A P \perp D C\)
Note that:
\(\triangle A D C \cong \triangle C B A\)
So, \(\text {area} (\triangle A D C)= \text {area} (\triangle C B A)\)
Therefore, \(\text {area} (A D C)=\frac{1}{2} \text {area} (A B C D)\)
\(\Rightarrow \text {area} (\triangle A D C)=\frac{1}{2}(D C \times A P)\)

So, the area of \(\triangle A D C=\frac{1}{2} \times \text {base} D C \times \text {corresponding altitude} \,A P\)
If we put it another way, the area of a triangle equals half the product of its base (or any other side) and the altitude corresponding to it (or height).
Two triangles having the same base (or equal bases) and equal areas will have the same corresponding altitudes, according to this formula.
For having equal corresponding altitudes, the triangles must lie between the same parallels

Q.3. In what ratio (of areas) does any median divide a triangle?
Ans: Consider the following figure, in which \(E D\) is the median through \(E\), and we have drawn a line through \(E\) parallel to \(B C:\)

\(\triangle E B D\) and \(\triangle E D C\) are on equal bases \((B D=D C)\) and between the same parallels. This means that their areas must be equal.
Therefore, any median will of a triangle will divide into two triangles of equal areas.

Q.4. \(O\) is any point in the interior of a parallelogram \(A B C D\). Show that:
\(\operatorname{area}(\Delta A O B)+ \text {area} (\Delta C O D)= \text {area} (\Delta A O D)+ \text {area} (\Delta B O C)=1 / 2 \operatorname{area}(A B C D)\)
Ans: Consider the following figure, which shows \(O\) to be any point inside the parallelogram \(A B C D\). We have drawn a line through \(O\), which is parallel to \(AB:\)

We note that \(A B Q P\) and \(C D P Q\) are themselves parallelograms. Now, the area of \(\triangle A O B\) will be half of the area of parallelogram \(A B Q P\), because they are on the same base \(A B\) and between the same parallels \(A B\) and \(P Q\):
\(\operatorname{area}(\triangle A O B)=\frac{1}{2} \operatorname{area}(A B F E)\)
Similarly, the area of \(\triangle C O D\) will be half of the area of parallelogram \(C D P Q\) because they are on the same base \(C D\) and between the same parallels \(C D\) and \(Q P\):
\(\operatorname{area}(\triangle C O D)=\frac{1}{2} \operatorname{area}(C D P Q)\)
Thus,
\(\operatorname{area}(\triangle A O B)+\operatorname{area}(\triangle C O D)\)
\( = \frac{1}{2}\{ \operatorname{area} (ABQP) + \operatorname{area} (CDPQ)\} \)
\(=\frac{1}{2} \text {area} (A B C D)\)
We have shown that the areas of \(\triangle A O B\) and \(\triangle C O D\) add up to half of the area of the parallelogram \(A B C D\). The remaining area is the sum of the areas of \(\triangle A O D\) and \(\triangle B O C\).
So, the sum will also be equal to half of the area \((A B C D)\).
Hence, the given condition is proved.

Q.5. Suppose that \(E\) is the midpoint of the median \(P D\) in \(\triangle A B C\). What is the ratio of the areas \(\triangle B E D\) and \(\triangle A B C\)?
Ans: Consider the following figure:

We note that \(\text {Area} (\triangle B E D)= \text {Area} (\triangle C E D)\) because these triangles have equal bases \((B D=D C)\), and they are between the same parallels.
Note also that \(E\) is the midpoint of \(P D\), so that \(P E=E D\).
This means that \(\text {area} (\triangle A B E)= \text {area} (\triangle D B E)\) since these triangles are on equal bases \((P E=E D)\) and between the same parallels (they have the same third vertex, point \(B)\).

Similarly, \(\operatorname{area}(\triangle A C E)=\operatorname{area}(\triangle D C E)\)
Thus, we have:
\(\operatorname{area}(\triangle B E D)=\operatorname{area}(\triangle C E D)=\operatorname{area}(\Delta P E B)=\operatorname{area}(\triangle P E C)\)
In other words, the areas of the four smaller triangles are equal. Also, the areas of these four triangles must add up to the total area of \(\text {Area} (\triangle P B C)\). Therefore,
\(\operatorname{area}(\triangle B E D)=\operatorname{area}(\triangle C E D)=\operatorname{area}(\Delta P E B)=\operatorname{area}(\triangle P E C)=\frac{\operatorname{area}\left(\triangle P B C\right)}{4}\)
The required ratio is \(1:4\).

Summary

Triangles with the same base (or equal bases) and that are parallel to each other have the same area. A triangle’s area is equal to half of the sum of its base and altitude. Between the same parallels are triangles with the same (or equal) base and area. This article includes the definition of triangles, parallel lines, theorem statements, and proof about triangles on the same base and between the same parallels.

Learn About Figures Between the Same Parallels

Frequently Asked Questions (FAQs)

Q.1. Do triangles with the same base have the same area?
Ans: Triangles with the same base (or equal bases) connected by the same parallels have the same area. Between the same parallels are triangles with the same base (or equal bases) and equal areas.

Q.2. What happens if parallelograms are on the same base and between the same parallels?
Ans: Parallelograms on the same base and between the same parallels are equal in area.

Q.3. Is a triangle half of a parallelogram?
Ans: When a triangle and a parallelogram have the same base and altitude, the triangle’s area is half that of the parallelogram. They will lie between the same parallels if they are at the same altitude. As a result, the triangle’s area will be half that of the parallelogram.

Q.4. How do you prove that the two triangles have the same area?
Ans: Two triangles with the same base do not need to have the same area. Their base and attitudes must be the same for the triangle to have the same area. The area of two triangles on the same base and connected by the same parallel lines is the same.

Q.5. Do similar triangles have the same size and shape?
Ans: Similar figures would be those that have the same shape as each other. However, they do not have to be the same size. The circles with radii \(3 \,\text {cm} \,\) & \(4 \,\text {cm}\) are similar, but they are not of the same size.

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