Triangle on Same Base and between Same Parallels

Triangle on same base and between same parallels is equal in area.

In the adjoining figure, ∆ABD and ∆DEF are having equal base ‘a cm’ and are between the same parallels BF and AD.

Triangle on Same Base and between Same Parallels






Therefore, area of ∆ABD = Area of ∆DEF

Prove that the triangles on same base and between same parallels are equal in area.

Let ∆ABC and ∆ABD be on the same base AB and between the same parallel AB and CD. It is require to prove that ∆ABC = ∆ABD.

Construction: A parallelogram ABPQ is constructed with AB as base and lying between the same parallels AB and CD.

Triangles on Same Base and between Same Parallels






Proof: Since ∆ABC and parallelogram ABPQ are on the same base AB and between the same parallels AB and Q,

Therefore, ∆ABC = ½(Parallelogram ABPQ)

Similarly, ∆ABD = ½(Parallelogram ABPQ)

Therefore, ∆ABC = ∆ABD.

Note: Since the relationship between the areas of a triangle and a parallelogram on the same base and between the same parallels in known to us, so that parallelogram ABPQ is constructed]


Solved examples for the triangle on same base and between same parallels:

1. Shaw that the medians of the triangle divide it into triangles of equal area.

Solution:  

Triangle on Same Base






AD is the median of the ∆ABC and AE is the altitude of ∆ABC and also ∆ADC.

(AE ┴ BC)

AD is the median of ABC              

Therefore, BD = DC

Multiply both sides by AE,           

Then BD × AE = DC × AE                

1/2 BD × AE = 1/2 DC × AE              

Area of ∆ABD = Area of ∆ADC   


2. AD is the median of ∆ABC and ∆ADC. E is any point on AD. Show that area of ∆ABE = area of ∆ACE.

Solution:

Solved Examples for the Triangle on Same Base






Since, AD is the median of ∆ABC, therefore BD = DC

Since, ∆ABD and ∆ADC have equal bases BD = DC and are between the same parallels BC and l,

Therefore Area of ∆ABD = Area of ∆ADC

Since, E lies on AD,

Therefore, ED is the median of the BEC

Now, BED and CED have equal bases BD = DC and between the same parallels BC and m.

Therefore, area of ∆BED = Area of ∆CED

On subtracting (1) and (2), we get

Area of ∆ABD - Area of ∆BED = Area of ∆ACD - Area of ∆CED

Area of ∆ABE = Area of ∆ACE

Figure on Same Base and between Same Parallels

Parallelograms on Same Base and between Same Parallels

Parallelograms and Rectangles on Same Base and between Same Parallels

Triangle and Parallelogram on Same Base and between Same Parallels

Triangle on Same Base and between Same Parallels






8th Grade Math Practice

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