Question

In D and E are the points on AC and AB respectively. BD and CE intersect at F. If the area FBE, BFC
FDC are 10cm 2 , 20 cm 2 and 16 cm 2 respectively, then find the area of
quadrilateral AEFD is

Answer 1

Answer:

Let:

AF meet triangle at X. Now in <ABX we have AE/EB + AD/DC = AF/FX

Now, AE/EB = ar (ACE)/ar (ECB) = (ar (ADFE) + ar (DCF)) / (ar (EFB) + ar (BFC))

= (ar (ADFE) + 16) / (10 + 20) = (ar (ADFE) + 16)/30 ---(1)

=  AD/DC = (ar (ADFE) + ar (EFB))/(ar (BFC) + ar (DFC))

= (ar (ADFE) + 10)/ (20+16) = (a r(ADFE) + 10)/(36) ---(2)

=  AF/FX = (ar (ABF) + ar (ACF))/(ar (BFX) + ar(XFC)))

= (ar (EFB) + ar (ADFE) + ar (DFC))/(ar (BFC)

= (10+ar (ADFE) + 16)/20 = (ar (ADFE)+26)/20 ---(3)

Substituting the values in (1), (2) and (3)  we get ,

=> ar (ADFE) = 36.