Question

If $\triangle ABC and \triangle DEF$ are two triangles such that $\frac{AB}{DE} = \frac{BC}{EF}= \frac{CA}{FD} = \frac{3}{4}$, then write Area ($\triangle ABC$) : Area ($\triangle DEF$)

Answer 1

Answer:

The ratio of ar(ΔABC): ar(∆DEF) is 9 : 16 .

Step-by-step explanation:

Given:

ΔABC and  ΔDEF are two triangles.  

AB/DE = BC/EF = CA/FD = ¾

It is given that corresponding sides of ΔABC and  ΔDEF are proportional, Then  

ΔABC ~  ΔDEF  

[Two triangles are similar if their corresponding sides are proportional]

ar(ΔABC)/ar(∆DEF) = (AB/DE)²

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

ar(ΔABC)/ar(∆DEF) = (¾)²

ar(ΔABC)/ar(∆DEF) = 9/16

ar(ΔABC): ar(∆DEF) = 9 : 16

Hence , the ratio of ar(ΔABC): ar(∆DEF) is 9 : 16 .

HOPE THIS ANSWER WILL HELP YOU ..

 

Answer 2

Answer

ΔABC congruent to ΔDEF

AB = 3 cm, BC = 2 cm

CA = 2.5 cm , EF = 4cm

AB/DE = BC/EF = CA/FD

corresponding sides of both triangles are in proportional

AB/DE = BC/EF

3/DE = 2/4

2 DE = 3 × 4

2 DE = 12

DE = 12/2

DE = 6 cm

also, here

BC/EF = CA/FD

2/4 = 2.5/FD

2 FD = 4 × 2.5

2 FD = 10

FD = 10/2

FD = 5 cm

per= ∆ DEF = DE + EF + FD

∆ DEF = DE + EF + FD= 6 + 4 + 5

∆ DEF = 15 cm this is the perimeter