question_answer

We know that parallelogram is also quadrilateral. Let us also split such a quadrilateral into two triangles, find their areas and hence that of the parallelogram. Does this agree with the formula that you know already?

Answer:

AD = BC = b

CO = AO? = h

ABCD is parallelogram divided into 2 triangles i.e., \[\Delta ABC+\Delta ACD\]

The area of parallelogram ABCD also divided into

two parts = Area of \[\Delta ABC\text{ }+\text{ }Area\text{ }of\text{ }\Delta ACD\]

Area of \[\Delta ABC\text{ }=\text{ }\frac{1}{2}\times BC\times AO'~\]

{Base BC = b, height = h}

\[=\frac{1}{2}\times b\times h=\frac{1}{2}bh\]    ? (i)

Area of \[\Delta ACD\text{ }=\text{ }\frac{1}{2}\times AD\times OC~~\]

{Base AD = b, height = h}

\[=\frac{1}{2}\times b\times h\]                                       ...(ii)

On adding eq. (i) & (ii) we get,

Area of \[\Delta \]ABC + Area of \[\Delta \]ACD \[=\frac{1}{2}\]bh + \[\frac{1}{2}\]bh

Area of AABC + Area of \[\Delta \]ACD = bh      ?(iii)

We know that,

Area of parallelogram ABCD \[=\text{ }b\times h\]       ?(iv)

Area of parallelogram ABCD = Area of \[\Delta \]ABC + Area of \[\Delta \]ACD

bh = bh

L.H.S. = R.H.S.