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Aleks04 [339]
3 years ago
14

Use the Pythagorean theorem to find the length of the line

Mathematics
1 answer:
Elanso [62]3 years ago
6 0

Answer:

This is one of my specialties so sit back, relax while I work on this for 5 mins :)

Step-by-step explanation:

The formula is a^2 + B^2 = c^2

Sooo

7^2 + 5^2 = c^2

49 + 25 = 74

√74 = √c^2

8.60232526 = c

8.60232526 rounded to the nearest tenth is 8.6

So there you go. Hope it helps!!

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Given AG bisects CD, IJ bisects CE, and BH bisects ED. Prove KE = FD.
OverLord2011 [107]

When a line is bisected, the line is divided into equal halves.

See below for the proof of \mathbf{KE \cong FD}

The given parameters are:

  • <em>AC bisects CD</em>
  • <em>IJ bisects CE</em>
  • <em>BH bisects ED</em>

<em />

By definition of segment bisection, we have:

  • \mathbf{CK \cong KE}
  • \mathbf{EF \cong FD}
  • \mathbf{CE \cong ED}

By definition of congruent segments, the above congruence equations become:

  • \mathbf{CK = KE}
  • \mathbf{EF = FD}
  • \mathbf{CE = ED}

By segment addition postulate, we have:

  • \mathbf{CE = CK + KE}
  • \mathbf{ED = EF + FD}

Substitute \mathbf{ED = EF + FD} in \mathbf{CE = ED}

\mathbf{CK + KE = EF + FD}

Substitute \mathbf{CK = KE} and \mathbf{EF = FD}

\mathbf{KE + KE = FD + FD}

Simplify

\mathbf{2KE = 2FD}

Apply division property of equality

\mathbf{KE = FD}

By definition of congruent segments

\mathbf{KE \cong FD}

Read more about proofs of congruent segments at:

brainly.com/question/11494126

6 0
3 years ago
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