Two triangles ABC and PBC on the same base BC and between the same parallels BC and AP. in following fig.
Draw CD || BA and CR || BP such that D and R lie on line AP in following fig.
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 ∆ ABC ≅ ∆ CDA and ∆ PBC ≅ ∆ CRP
So, ar (ABC) = `1/2` ar (ABCD) and ar (PBC) = `1/2` ar (PBCR)
Therefore, ar (ABC) = ar (PBC)
Theorem : Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
Proof: Now, suppose ABCD is a parallelogram whose one of the diagonals is AC (see Fig.).
Let AN ⊥ DC. Note that
∆ ADC ≅ ∆ CBA
So, ar (ADC) = ar (CBA)
Therefore, ar (ADC) = `1/2` ar (ABCD)
=`1/2` (DC × AN)
So, area of ∆ ADC = `1/2` × 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).
The two triangles with same base (or equal bases) and equal areas will have equal corresponding altitudes.
Shaalaa.com | Theorem : Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that
(i) BDEF is a parallelogram.
(ii) ar (DEF) = 1/4ar (ABC)
(iii) ar (BDEF) = 1/2ar (ABC)
In a ΔABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point
of AP. Prove that :
(1) ar (Δ PBQ) = ar (Δ ARC)
(2) ar (Δ PRQ) =`1/2`ar (Δ ARC)
(3) ar (Δ RQC) =`3/8` ar (Δ ABC) .
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
Can you answer the question that you have left in the ’Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?
[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into n triangles of equal areas.]