The Area of Parallelogram and Triangles is measured by multiplying its base and height. A triangle is a closed polygon bounded by three line segments. And a parallelogram is a closed polygon bounded by four line segments such that the opposite sides are the same in length and are parallel to each other. The triangle and the parallelogram are \(2\)-dimensional figures.

The area is the space occupied by a \(2\)-dimensional figure in space. So, both a triangle and a parallelogram occupy some space, and hence they have areas. In this article, we shall discuss the various relationships between the areas of a triangle and a parallelogram in different conditions.

Area Proportion

Two figures are said to be congruent if they are of the same shape and the same size. Here, two figures \(A\) and \(B\), are congruent. 

And two congruent figures have equal areas. Thus, above given two congruent figures \(A\) and \(B\), have equal areas.

But, the reverse is not true. Two figures having equal areas may not always be congruent. 

For example, rectangle \(ABCD\) and the square \(EFGH\) have equal areas \(9 \times 4 = 36\,{\rm{c}}{{\rm{m}}^2}\) and \(\left.6 \times 6=36 \mathrm{~cm}^{2}\right)\) But clearly, they are not congruent.

Look at the figure given below:

We can observe that the planar region formed by figure \(T\) comprises two planar regions formed by figures \(P\) and \(Q.\) We can write that the area of figure \(T =\) area of figure \(P +\) area of figure \(Q.\) 

The general representation of the area of figure \(A\) is \(ar(A)\), area of figure \(B\) is \(ar(B)\), area of figure \(T\) is \(ar(T)\), and so on. 

Same Base and Between the Same Parallel Lines

From the above figure, trapezium \(ABCD\) and parallelogram \(EFCD\) are on the same base \(DC.\) And, also the vertices \(A\) and \(B\) (of trapezium \(ABCD\) opposite to base \(DC\) and the vertices \(E\) and \(F\) (of parallelogram \(EFCD\) opposite to base \(DC\), lie on a line \(AF\) that is parallel to \(DC.\) Thus trapezium \(ABCD\) and parallelogram \(EFCD \) lie on the same base \(DC\) and between the same parallels \(AF\) and \(DC.\) 

Similarly, parallelograms \(PQRS\) and \(MNRS\) are on the same base \(SR\) and between the same parallels \(PN\) and \(SR.\)

Area of Parallelogram and Triangles

We know that magnitude of measure of the space occupied by the planar region is called the area. The parallelogram is the quadrilateral with two pairs of opposite side parallel and equal. The area of a parallelogram is obtained by multiplying its base and height.

\({\rm{Area = base \times height}}\)

The area of a triangle is given by \(ar(\Delta ABD) = \frac{1}{2} \times {\rm{base \times height}}\)

Theorem 1

A Triangle and a Parallelogram lying on the same Base and Between the same Parallel lines

Diagonal divides the parallelogram into two congruent triangles. Thus, the area of the parallelogram equals the sum of the areas of two congruent triangles.

\(ar(ABCD) = ar(\Delta ABD) + ar(\Delta BCD) = 2ar(\Delta ABD)\)

\(\Rightarrow ar(\Delta ABD) = \frac{1}{2}ar(ABCD)\)

So, if a triangle and a parallelogram lie on the same base and lie between the same two parallel lines, then the triangle area is half the area of the parallelogram.

Theorem 1

Two Parallelograms lying on the same Base and Between the same Parallel Lines

If two parallelograms lie on the same base and lie between the same parallel lines, they have equal areas. Parallelograms \(ABCD\) and \(EFCD\) are drawn on the same base \(DC\) and between the same parallels \(AF\) and \(DC.\)

In \(ADE\) and \(BCF\), 

\(DAE=CBF\) (Corresponding angles)

\(AED=BFC\) (Corresponding angles) 

And, \(AD=BC\) (The opposite sides of a parallelogram are equal in length)

So, \(\Delta AED \cong \Delta BCF\)  [By \(ASA\) rule]

Therefore, \(ar (ADE) = ar (BCF)\) (Congruent figures have equal areas) —–(1)

Now, \(ar (ABCD) = ar (ADE) + ar (EDCB) = ar (BCF) + ar (EDCB)\) [From(1)] 

\(ar (ABCD)=ar (EFCD)\)

So, parallelograms \(ABCD\) and \(EFCD\) have equal areas. 

Theorem 1

Two Triangles lying on the Same Base and Between the Same Parallel Lines

Two triangles lying on the same base (or equal bases) and between the same parallel lines are equal in area. 

Here, \(ABD\) and \(ABC\) are on the same base \(AB\) and same parallel lines \(AB\) and \(DC.\) We shall prove that \(ar ADB=ar ABC.\)

Study Area of Triangle Here

Draw \(AE∥BC\) and \(AD∥BF.\)

In quadrilateral \(ABCE, AE∥BC\) and \(AB∥EC\).
So \(ABCE\) is a parallelogram.

Similarly, in quadrilateral \(ABFD, AD∥BF\) and \(AB∥DF\).
So \(ABFD\) is also a parallelogram.

We observe that the two parallelograms are lying on the same base \(AB\) and the same two parallel lines \(AB\) and \(EF.\)
Hence, their areas are equal.

So, \(arABCE=arABFD\)

\(⟹2×ar ABC=2×ar ABD\) (diagonal divides a parallelogram in two triangles having equal areas)

\(⟹ar ABC=ar ABD.\) (proved)

Solved Examples

Q.1. \(ABCD\) is a parallelogram, and \(EFCD\) is a rectangle. And also, \(AL \bot DC.\) Prove that 
(i) \(ar\left( {ABCD} \right) = ar\left( {EFCD} \right)\) 
(ii) \(ar\left( {ABCD} \right) = DC \times AL\)

Ans:
From the above figure, we observe that parallelogram \(ABCD\) and the rectangle \(EFCD\) have the same base \(CD\) and lie between the parallel lines \(BE\) and \(CD.\) (We know that rectangle is also a parallelogram.)
We know that two parallelograms with the same base and lying between the same parallel lines are equal in area. 
So, parallelogram \(ABCD\) and rectangle \(EFCD\) have equal area.
\(ar (ABCD) = ar (EFCD) —– (1)\)
We know that area of the rectangle is the product of length and breadth.
\(ar EFCD=ED×CD=CD×AL\) (As, \(ED\) and \(AL\) have the same value) —– (2)
From (1) and (2),
\(ar (ABCD) = DC × AL.\)

Q.2. \(ABCD\) is a quadrilateral, and \(BE\parallel AC\) and also \(BE\) meets \(DC\) produced at \(E.\) Show that area of \(\Delta ADE\) is equal to the area of the quadrilateral \(ABCD.\) 

Ans:
Observe the given figure as shown above.
\(BAC\) and \(EAC\) lies on the same base \(AC\) and lies between the same parallels \(AC\) and \(BE.\)
We know that two triangles have an equal area when they lie between two parallel lines and having the same base.
Therefore, \(ar(BAC) = ar(EAC)\) 
Adding \(ar(ADC)\) on both sides of the equation, we get,
\(ar(BAC) + ar(ADC) = ar(EAC) + ar(ADC) \)
From the figure, \(ar(ABCD) = ar(ADE)\)
Hence, the area of \(ADE\) is equal to the area of the quadrilateral \(ABCD.\)

Q.3. Given \(E\) is any point on the median \(AD\) of a triangle \(ABC.\) Prove that \({\rm{area}}\left( {ABE} \right) = {\rm{area}}\left( {ACE} \right).\)

Ans:
Given \(E\) is any point on the median \(AD.\)
Join points \(EB\) and point \(EC.\)
Given, \(AD\) is the median in triangle \(AB.\) 
We know that the median divides the triangle into two triangles having equal areas.
So,  \(area (∆ ABD) = area (∆ ADC) ….. (i)\)
Similarly, in triangle \(EBC, ED\) is the median that divides the triangle into two equal congruent triangles.
So,  \(area (∆ EBD) = area (∆ EDC) ….. (ii)\)
Subtracting (i) from (ii), we get, 
\(area (ABD) – area (EBD) = area (ADC) – area (EDC)\)
\(⟹area (ABE) = area (AEC).\)
Hence, proved.

Q.4. Show that a median of a triangle divides it into two triangles of equal areas. 
Ans:
Let \(ABC\) be a triangle and let \(AD\) be the median.

Let us consider \(AN  BC\)
Now, the area of the triangle is \(\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times base \times height}}\)
\(\Rightarrow \operatorname{ar}(A B D)=\frac{1}{2} \times B D \times A N\)
As median bisects the opposite side, thus \(BD=CD\)
\(\Longrightarrow a r(A B D)=\frac{1}{2} \times C D \times A N\)
\(\Rightarrow \operatorname{ar}(A B D)=\operatorname{ar}(\Delta A C D)\)
Thus, the median divides the given triangle into two congruent triangles of equal areas.

Q.5. Observe the given figure, find the relation between the areas of two triangles \(ABC\) and \(PBC.\)

Ans:
From the given figure,
Triangles \(ABC\) and \(BPC\) have the same base as \(BC.\)
Both triangles lie between two parallel lines \(BC\) and \(AP.\)
We know that triangles having the same base and lies between two parallels are equal in area.
So, \(ar∆ABC=ar(∆BPC)\)
Hence, proved.

Summary

In this article, we have discussed the formulas for calculating the area of the triangle and parallelogram.

This article also gives the relationship between the areas of two parallelograms lying on the same base and between the same two parallel lines, two triangles lying on the same base and between the same two parallel lines, one parallelogram and one triangle lying on the same base and between the same two parallel lines.

Here we have discussed the solved examples, which helps us solve the problems easily.

FAQs

Q.1. What is the formula for finding the area of the triangle and parallelogram?
Ans: The area of the triangle is half of the product of base and height. And, area of the parallelogram is the product of the base and height or altitude.

Q.2. What is the relation between the areas of two parallelograms having the same base and between the same parallel lines?
Ans: The two parallelograms having the same base and lie between the same parallel lines have equal areas.

Q.3. What is the relation between triangles having the same base and between two parallels?
Ans: Two triangles having the same base and lying in between the same parallel lines are equal in areas.

Q.4. Have two congruent triangles have equal area?
Ans: Yes, two congruent triangles have equal areas.

Q.5. Is diagonal divides the parallelogram into two equal areas?
Ans: Yes. A diagonal divides the parallelogram into two congruent triangles, and we know that two congruent triangles have equal areas. Therefore, a diagonal divides the parallelogram into two equal areas.

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