question_answer

ABCD is a parallelogram. P is any point on CD. If \[ar(\Delta DPA)=15\,c{{m}^{2}}\]and \[ar(\Delta APC)=20\,c{{m}^{2}},\]then \[ar(\Delta APB)=\]

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. \[\therefore \]\[ar(\Delta \Alpha PB)=ar(ABC)=35c{{m}^{2}}\]