Question

Given: ΔABC ≅ ΔEFD

What is the length of Segment line FE rounded to the nearest tenth?

Answer 1

The answer is A. 2.8. Explained: I used the distance formula (or the top part of Pythagorean theorem) of d=square root of (x2-x1)+(y2-y1). Using points A (0,2) as (x1,y1) and point B (2,4) as (x2,y2) to put inside the distance formula. d=square root of (2-0)^2+(4-2)^2= square root of (2)^2+(2)^2= square root of (4+4)= square root of 8 which equals 2.8. Since triangle ABC is congruent to triangle EFD then since side AB equals 2.8 then side FE equals 2.8.

Answer 2

Line segment FE is equals to Line segment BA or FE=BA.

B = (2,4) A = (0,2)

First we have to subtract the x and y coordinates.

x = |2 - 0|
x = 2

y = |4-2|
y = 2

Then, we are going to use the Pythagorean Theorem a^2 + b^2 = c^2.

2^2 + 2^2 = 8

Get the square root of 8. The answer will be 2.83. Rounded off to the nearest tenth, the answer will be 2.80

I hope this helped you.