A) \[~15\,c{{m}^{2}}\]
B) \[~20\,c{{m}^{2}}\]
C) \[~35\,c{{m}^{2}}\]
D) \[~30\,c{{m}^{2}}\]
Correct Answer: C
Solution :
Given,\[ar(\Delta DPA)=15\,c{{m}^{2}}\] \[ar(\Delta \Alpha PC)=20\,c{{m}^{2}}\] Now, \[ar(\Delta ADC)=ar(\Delta DPA)+ar(\Delta APC)\] \[ar(\Delta ADC)=(15+20)c{{m}^{2}}=35\,c{{m}^{2}}\] Diagonals of a parallelogram divide it into tow triangles of equal area. \[\therefore \]\[ar(\Delta \Alpha DC)=ar(\Delta ABC)\] Also, triangles on the same base and between the same parallels are equal in area. \[\therefore \]\[ar(\Delta \Alpha PB)=ar(ABC)=35c{{m}^{2}}\]You need to login to perform this action.
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