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.
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