A) \[36\text{ }c{{m}^{2}}\]
B) \[~48\,c{{m}^{2}}\]
C) \[24\text{ }c{{m}^{2}}\]
D) None of these
Correct Answer: C
Solution :
Since, diagonals of a parallelogram divides it into two triangles of equal areas. \[\therefore \]\[ar(\Delta \Alpha BC)=\frac{1}{2}ar(ABCD)\] \[\Rightarrow \]\[2ar(\Delta \Alpha \Beta C)=ar(ABCD)\] \[\Rightarrow \]\[2ar(\Delta \Alpha \Beta C)=ar(APCD)+ar(\Delta PBC)\] \[\Rightarrow \]\[2ar(\Delta \Alpha \Beta C)=36+ar(\Delta PBC)\] ?(i) CP is the median of \[\Delta \Alpha CB,\]and median divides a triangle into two triangles of equal area. \[\therefore \]\[ar(\Delta PBC)=\frac{1}{2}ar(\Delta \Alpha \Beta C)\] ?(ii) Now, from (i) and (ii), we get \[2ar(\Delta \Alpha \Beta C)=36+\frac{1}{2}ar(\Delta \Alpha \Beta C)\] \[\Rightarrow \]\[2\,ar\,(\Delta \Alpha \Beta C)-\frac{1}{2}ar(\Delta \Alpha \Beta C)=36\] \[\Rightarrow \]\[\frac{3}{2}ar(\Delta \Alpha \Beta C)=36\] \[\Rightarrow \]\[ar(\Delta \Alpha \Beta C)=\frac{36\times 2}{3}=24\,c{{m}^{2}}\]
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