Question

Find the value of n that makes ΔDEF ∼ΔXYZ when DE = 4, EF = 5, XY = 4(n+1), YZ = 7n - 1, and ∠E ≅∠Y. n =

Answer 1

answer is 3

Explanation:

they are correct

Answer 2

3

Explanation:

It's given that ΔDEF ∼ΔXYZ . So the corresponding sides of both triangles will be proportional to each other.

`= > (de)/(xy) = (ef)/(yz) = (df)/(xz)`

DE = 4 ; XY = 4(n + 1) ; EF = 5 ; YZ = 7n - 1

Putting all these values gives ,

`(4)/(4(n + 1)) = (5)/(7n - 1)`

`= > (1)/(n + 1) = (5)/(7n - 1)`

`= > 7n - 1 = 5(n + 1)`

`= > 7n - 1 = 5n + 5`

`= > 7n - 5n = 5 + 1`

`= > 2n = 6`

`= > n = (6)/(2) = 3`