Areas of Triangles

Objective:

This topic gives an overview of Areas of Triangles.

Triangles on the same Base and between the same Parallels

Let us look at the figure. In it, we have two triangles ABC and PBC on the same base BC and between the same parallels BC and AP. What can you say about the areas of such triangles? To answer this question, you may perform the activity of drawing several pairs of triangles on the same base and between the same parallels on the graph sheet and find their areas by the method of counting the squares. Each time, you will find that the areas of the two triangles are (approximately) equal. This activity can be performed using a geoboard also. You will again find that the two areas are (approximately) equal.

To obtain a logical answer to the above question, you may proceed as follows: In the figure above, draw CD || BA and CR || BP such that D and R lie on line AP.

From this, you obtain two parallelograms PBCR and ABCD on the same base BC and between the same parallels BC and AR. 

Therefore, ar (ABCD) = ar (PBCR)

Now Δ ABC ≅ Δ CDA and Δ PBC ≅ Δ CRP

So, ar (ABC) =«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/math»ar (ABCD)and ar (PBC) =«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/math» ar (PBCR).
Therefore, ar (ABC) = ar (PBC)

In this way, you have arrived at the following theorem:

Theorem 1: Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Now, suppose ABCD is a parallelogram whose one of the diagonals is AC. Let AN ⊥ DC. Note that

Δ ADC ≅  Δ CBA

So, ar (ADC) = ar (CBA)

Therefore, ar (ADC) =«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/math» ar (ABDC)
=«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/math»(DCxAN)
So,area of Δ ADC =«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/math»× base DC × corresponding altitude AN

In other words, area of a triangle is half the product of its base (or any side) and the corresponding altitude (or height). From this formula, you can see that two triangles with same base (or equal bases) and equal areas will have equal corresponding altitudes.

For having equal corresponding altitudes, the triangles must lie between the same parallels. From this, you arrive at the following converse of Theorem 1

Theorem 2 : Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.

Let us now take some examples to illustrate the use of the above results.

Example 1 : Show that a median of a triangle divides it into two triangles of equal areas.

Solution : Let ABC be a triangle and let AD be one of its medians.

You wish to show that

ar (ABD) = ar (ACD).

Since the formula for area involves altitude, let us draw AN ⊥ BC.Now ar(ABD) =«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/math»× base × altitude (of Δ ABD)
=«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/math» BD x AN
=«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/math»×CD ×AN (As BD = CD)
=«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/math»× base × altitude (of Δ ACD) = ar(ACD)

Example 2

In Figure, ABCD is a quadrilateral and BE || AC and also BE meets DC produced at E. Show that area of Δ ADE is equal to the area of the quadrilateral ABCD.

Solution : Observe the figure carefully .

Δ BAC and Δ EAC lie on the same base AC and between the same parallels AC and BE.

Therefore, ar(BAC) = ar(EAC)(By Theorem 1)

So, ar(BAC) + ar(ADC) = ar(EAC) + ar(ADC) (Adding same areas on both sides)

or ar(ABCD) = ar(ADE)

Summary

  • Triangles on the same base (or equal bases) and between the same parallels are equal in area.
  • Area of a triangle is half the product of its base and the corresponding altitude.
  • Triangles on the same base (or equal bases) and having equal areas lie between the same parallels.
  • A median of a triangle divides it into two triangles of equal areas.

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