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

What is the length of Segment line FE rounded to the nearest tenth?-example-1
User Mtfk
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8.9k points

2 Answers

1 vote

In order to find the length of line FE, determine the corresponding points and plug them into the distance formula.

The corresponding point for F is C, F = (3, 2). The corresponding point for E is A, E = (0, 1).

Plug into the distance formula: square root of (3-0) squared plus square root of (1-2) squared. The outcome is the square root of 10.

Square root of 10 = 3.16227766017. Rounded to the nearest tenth is 3.2.

The answer is 3.2

User Shekit
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8.4k points
4 votes
Well, Aml says they're congruent. I don't think they CAN be ... There's no way
you can slide one around and make it fit exactly on top of the other one. They
appear to be mirror-images of each other ... maybe.

But AML says they're congruent. So let's assume that the longest sides of both
triangles have the same length. Then the length of FE in the standing-up triangle
is the same as the distance between points 'A' and 'C' in the lying-down triangle.

Point A . . . . . (0, 1)
Point C . . . . . (3, 2) .

The distance between them is

square root of (square of difference in 'x' + square of difference in 'y').

Difference in 'x' . . . . . 3
Difference in 'y' . . . . . 1

Distance = √(3² + 1²)

= √(9 + 1)

= √10 = 3.1622776...

Rounded to the nearest tenth: 3.2

User Bastaspast
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8.2k points