# In A Triangle ABC Let N Be A Point On AB Such That CN Divides AB In the ratio 2:3 And M Be A Point On BC That AM Divides BC In The Ratio 2:5.Let P Be Intersection Point Of AM And CN  If Area Of Triangle ABC = 105 Then Find The Value Of(Area Of Trianglle APC-Area Of Quadrilateral PNBM)a) 33b) 35c) 28d) 21

Arun Kumar IIT Delhi
8 years ago
First draw the diagram yourself

we know CN and AM will divide the triangle areas equal to 2:3 and 2:5
So,
$\Delta_{ABC}=105 \\\Delta_{ACN}=(2/5)105 \\\Delta_{ANB}=(3/5)105 \\\Delta_{ACM}=(5/7)105 \\\Delta_{ABM}=(2/7)105 \\\Delta_{APC}=\Delta_{1} \\\Delta_{CPM}=\Delta_{2} \\\Delta_{APN}==\Delta_{3} \\\Delta_{NPBM}=\Delta_{4} \\\Delta_{1}+\Delta_{3}=\Delta_{ACN} \\\Delta_{3}+\Delta_{4}=\Delta_{AMB} \\\Delta_{1}+\Delta_{2}=\Delta_{AMC} \\\Delta_{2}+\Delta_{4}=\Delta_{CNB}$
Four unknown four variable

Solve it

Arun Kumar
IIT Delhi