Areas of Triangles

Two triangles on the same base or equal bases and between the same parallels are equal in area.




Given  : Triangles ΔABC and ΔABD are on a common base AB and between same parallels AB and CD.

To Prove : ar(ΔABC) = ar(ΔABD)

Construction : Draw BL||AC , BM||AD

Proof : ABLC is a parallelogram
           (AC||BL and AB||CL)
          ABMD is a parallelogram
           (AD||BM and AB||DM)

   ar(||gm ABLC) = ar(||gm ABMD) ..........(1)
In (||gm ABLC) , ar(ΔABC) = ar (ΔCLB)

ar(||gm ABLC) = 2ar(ΔABC) ..........(2)

In (||gm ABMD) , ar(ΔABD) = ar (ΔBDM)
ar(||gm ABMD) = 2ar(ΔABD) ..........(3)
     2ar(ΔABC) = 2 ar(ΔABD)..........(4)
⇒ ar(ΔABC) = ar(ΔABD)

Area of a triangle is half the product of its base and the corresponding altitude.

Median of a triangle divides it into two triangles of equal area.

Two triangles with same base (or equal bases) and equal areas will have equal corresponding altitudes.

Triangles on the same base and having equal areas lie between the same parallels.

If triangles ABC and ABD are on a common base AB and ar( ABC) = ar( ABD), then AB || CD.



Diagonals of a parallelogram divide it into four triangles of equal area.

Summary

Two triangles on the same base or equal bases and between the same parallels are equal in area.




Given  : Triangles ΔABC and ΔABD are on a common base AB and between same parallels AB and CD.

To Prove : ar(ΔABC) = ar(ΔABD)

Construction : Draw BL||AC , BM||AD

Proof : ABLC is a parallelogram
           (AC||BL and AB||CL)
          ABMD is a parallelogram
           (AD||BM and AB||DM)

   ar(||gm ABLC) = ar(||gm ABMD) ..........(1)
In (||gm ABLC) , ar(ΔABC) = ar (ΔCLB)

ar(||gm ABLC) = 2ar(ΔABC) ..........(2)

In (||gm ABMD) , ar(ΔABD) = ar (ΔBDM)
ar(||gm ABMD) = 2ar(ΔABD) ..........(3)
     2ar(ΔABC) = 2 ar(ΔABD)..........(4)
⇒ ar(ΔABC) = ar(ΔABD)

Area of a triangle is half the product of its base and the corresponding altitude.

Median of a triangle divides it into two triangles of equal area.

Two triangles with same base (or equal bases) and equal areas will have equal corresponding altitudes.

Triangles on the same base and having equal areas lie between the same parallels.

If triangles ABC and ABD are on a common base AB and ar( ABC) = ar( ABD), then AB || CD.



Diagonals of a parallelogram divide it into four triangles of equal area.

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