CHALLENGE QUESTION 7 and 8

Let D,E,F be points on the sides BC,CA, AB respectively of a triangle ABC. The lines AD,BE, CF are concurrent at the point P. If AP = PD= 2, BP = 3, PE=1 and CP = 5, then

Length of PF is

(a) $\cfrac {3}{2}$ (b) 2 (c) $\cfrac{4}{3}$
(d) $\cfrac{5}{3}$

The area of the triangle ABC is

(a) 8 (b) 9 (c) 10 (d) 12

Question

CHALLENGE QUESTION 7 and 8

Let D,E,F be points on the sides BC,CA, AB respectively of a triangle ABC. The lines AD,BE, CF are concurrent at the point P. If AP = PD= 2, BP = 3, PE=1 and CP = 5, then

Length of PF is

(a) $\cfrac {3}{2}$ (b) 2 (c) $\cfrac{4}{3}$
(d) $\cfrac{5}{3}$

The area of the triangle ABC is

(a) 8 (b) 9 (c) 10 (d) 12

vishweshshrimali5 · Answer

@experimentX 
@satellite73 
@sami-21 
@mukushla 
@waterineyes

vishweshshrimali5 · Answer

@myininaya  please help

experimentx · Answer

i think the best to start this problem is to draw figure.

vishweshshrimali5 · Answer

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