Question

If D, E, F are the mid-points of sides BC, CA and AB respectively of $\triangle ABC$, then the ratio of the areas of triangles DEF and ABC is
(a) 1 : 4
(b) 1 : 2
(c) 2 : 3
(d) 4 : 5

Answer 1

Answer:

The ratio of area(ΔDEF): area(ΔABC) is 1 : 4

Among the given options option (a) is 1 :4  is the correct answer.

Step-by-step explanation:

Given:  

D, E and F are the mid-points of the sides AB, BC and CA of the ΔABC.

Construction : Join the points D,E,F

 

In ΔABC, we have

By the mid-point theorem, we have

DF = 1/2BC  

DE = ½ CA …………..(1)

EF = ½ AB

In ΔDEF and ΔCAB

DF/BC = DE/CA = EF/AB = ½  

[From eq 1]

ΔDEF ~ ΔCAB

[By SSS similarity criterion]

ar(ΔDEF) / ar(ΔCAB) = DE²/CA²

[The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides]

ar(ΔDEF) / ar(ΔCAB) = (½ CA)² / CA²

[From eq 1]

ar(ΔDEF) / ar(ΔCAB) = ¼

ar(ΔDEF) / ar(ΔABC) = ¼

[ar(ΔCAB) = ar(ΔABC)]

ar(ΔDEF) :  ar(ΔABC) = 1 : 4

Hence, the ratio of the area(ΔDEF) : area(ΔABC) is 1:4

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Answer=1 : 4

Δ ABC, is the mid-point theorem

DF = 1/2BC

DE = ½ CA

EF = ½ AB

ΔDEF and ΔCAB

DF/BC = DE/CA = EF/AB = ½

ΔDEF congruent to ΔCAB

SSS criterion

ar(ΔDEF) / ar(ΔCAB) = DE²/CA²

[corresponding sides]

ar(ΔDEF) / ar(ΔCAB) = (½ CA)² / CA²

ar(ΔDEF) / ar(ΔCAB) = ¼

ar(ΔDEF) / ar(ΔABC) = ¼ar(ΔCAB) = ar(ΔABC)

ar(ΔDEF) : ar(ΔABC)

Hence,, Answer= 1 : 4