Question

Given ∆ABC~ ∆DEF, what is the ratio of

ar(∆ABC)/ar(∆DEF)​

Answer 1

Answer:

9.39

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Answer 2

Given:

∆ABC ~ ∆DEF

AB = 3 cm

DF = 12 cm

EF = 13 cm

To find:

What is the ratio of AREA(∆ABC)/ AREA(∆DEF)​

Solution:

To find the area we need to find the length of the sides of both the triangles

In ∆DEF as it is a right-angled triangle

(Hypotenuse)² = (base)² +  (height)²

(EF)² = (DF)² +  (DE)²

Substituting the values

(13)² = (12)² +  (DE)²

(13)² - (12)²  = (DE)²

169- 144 = (DE)²

25 =(DE)²

5 = DE

DE = 5

As we know,

∆ABC~ ∆DEF

AB/ DE = BC / EF = AC/ DF

Putting the values

AB = 3 cm

DF = 12 cm

EF = 13 cm

DE = 5

3/ 5 = BC / 13 = AC/ 12

3/ 5 = BC / 13

39/ 5 = BC

3/ 5 = AC/ 12

36/ 5 = AC

AREA (∆ABC)

$\frac{1}{2}$ × HEIGHT × BASE

= $\frac{1}{2}$ × AB ×  AC

=  $\frac{1}{2}$ × 3 ×  36/ 5

AREA(∆DEF)​

$\frac{1}{2}$ × HEIGHT × BASE

= $\frac{1}{2}$ × DE ×  DF

= $\frac{1}{2}$ ×5 ×  12

AREA (∆ABC) / AREA(∆DEF)​

= $\frac{1}{2}$ × 3 ×  36/ 5 ÷ = $\frac{1}{2}$ ×5 ×  12

= 9 / 25

Therefore, the ratio of ar(∆ABC)/ar(∆DEF)​ is equal to 9 / 25.