How is the formula for a triangle derived from a parallelogram? Please give a good explanation.

At first, we divide the parallelogram into two triangles by joining any two opposite vertices. These two triangles are exactly the same (congruent) and thus have equal areas. The area of the parallelogram is the summation of the individual areas of the two triangles. We drop a perpendicular from a vertex to its opposite side to get an expression for the height of the triangles. The area of the individual triangle is 12×base×height12×base×height .The area of the parallelogram being twice the area of the triangle, thus becomes after evaluation base×heightbase×height .

Complete step by step

The parallelogram can be divided into two triangles by constructing a diagonal by joining any two opposite vertices.





In the above figure, ΔABDΔABD and ΔBCDΔBCDare the two such triangles. These two triangles have:

AB=CDAB=CD (as opposite sides of a parallelogram are equal)

AD=BCAD=BC (opposite sides of a parallelogram are equal)

BDBD is common

Thus, the two triangles are congruent to each other by SSS axiom of congruence. Since, the areas of two congruent triangles are equal,

⇒area(ΔABD)=area(ΔBCD)⇒area(ΔABD)=area(ΔBCD)

Now, we need to find the area of ΔABDΔABD . We draw a perpendicular from DD to the side ABAB and name it as DEDE . Thus, ΔABDΔABD is now a triangle with base ABAB and height DEDE .

Then, the area of the ΔABD

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step-by-step explanation: