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:

answer is 3

Step-by-step explanation:

they are correct

Answer 2

Answer:

3

Step-by-step explanation:

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

$= > \frac{de}{xy} = \frac{ef}{yz} = \frac{df}{xz}$

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

Putting all these values gives ,

$\frac{4}{4(n + 1)} = \frac{5}{7n - 1}$

$= > \frac{1}{n + 1} = \frac{5}{7n - 1}$

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

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

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

$= > 2n = 6$

$= > n = \frac{6}{2} = 3$