Question

d and e are the mid points of bc and ad respectively. if area of ∆ABC =12cm then, area of ∆BDE is?​

Answer 1

Answer:

D &E are the mid points of the sides AC & BC respectively of ΔABC.

$\sfΔBED=12cm {}^{2}$

F, the mid point of the side AB, is joined to D.

D is the mid point of AC  and E  is the mid point of BC.

So, by mid point theorem, 

$\bf \: DE||AB..$

Again,  F is the mid point of AB.

$\bf \: So \:  DF||BC.$

i.e BEDF is a parallelogram and BD is its diagonal.

So BD divides BEDF  into two equal areas.

$\sf∴ar.ΔBED=ar.ΔBDF=12cm {}^{2}$

Again, in  ΔADB 

DF is the median since F is the mid point of AB.

$\sf \therefore \: ar.ΔBDF=ar.ΔAFD=12cm {}^{2}$

(median divides a triangle into two equal areas).

$\sf \: Now ar.ABED=ar.ΔAFD+ar.ΔBDF+ar.ΔBED$

$\sf = 12 {cm}^{2} + 12 {cm}^{2} + 12 {cm}^{2}$

$\sf = 36 {cm}^{2}$