Unit 5 relationships in triangles homework 1 triangle midsegments

Students were asked to answer a question at education and to express what is most important for them to succeed. The one that response stood out from the rest was practice. Persons who are definitely successful do not become successful by being born. They work hard and commitment their lives to succeeding. If you want to get your goals, keep this in mind! shown below some question and answer examples that you would certainly make use of to expand your knowledge and gain insight that will assist you to maintain your school studies.

Question:

Unit 5 relationships in triangles homework 1 triangle midsegments

Answer:

Based on the definition of a parallel line and the Midsegment Theorem the following are the right answers:

1. a.) BD║AE

b.) BF ║CE

c.) DF║CA

2. a.) YZ║RT

b.) RS ║XZ

c.) XY║TS

3. a.) FH = 24

b.) JL = 74

c.) KJ = 60

d.) FJ = 30

4. a.) AE = 26

b.) AN = 58

c.) CT = 21.5

d.) Perimeter of ΔAEN = 127

5. x = 15

6. x = 6

What are Parallel lines?

Parallel lines coplanar straight lines that do not meet each other and are equal distance from each other.

The Triangle Midsegment Theorem

  • A midsegment is a line that connects the midpoints of the two sides of a triangle together.
  • Every triangle three midsegments.
  • Based on the Midsegment Theorem of a triangle, the third side of a triangle is always parallel to the midsegment, and thus, the third side is twice the size of the midsegment. In order words, length of midsegment = ½(length of third side).
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Applying the definition of a parallel line and the Midsegment Theorem the following can be solved as shown below:

1. The pairs of parallel lines in ΔAEC (i.e. the midsegment is parallel to the third side) are:

a.) BD║AE

2. The segment parallel to the given segments are:

a.) YZ║RT

3. Given:

FG = 37; KL = 48; GH = 30

a.) FH = ½(KL)

  • Substitute

FH = ½(48)

FH = 24

b.) JL = 2(FG)

  • Substitute

JL = 2(37)

JL = 74

c.) KJ = 2(GH)

  • Substitute

KJ = 2(30)

KJ = 60

d.) FJ = ½(KJ)

  • Substitute

FJ = ½(60)

FJ = 30

4. Given:

PT = 13

EN = 43

CP = 29

a.) AE = 2(PT)

  • Substitute

AE = 2(13)

AE = 26

b.) AN = 2(CP)

  • Substitute

AN = 2(29)

AN = 58

c.) CT = ½(EN)

  • Substitute

CT = ½(43)

CT = 21.5

d.) Perimeter of ΔAEN = EN + AN + AE

  • Substitute

Perimeter of ΔAEN = 43 + 58 + 26

Perimeter of ΔAEN = 127

5. 10x + 44 = 2(8x – 23) (midsegment theorem)

10x + 44 = 16x – 46

  • Add like terms together

10x – 16x = -44 – 46

-6x = -90

Divide both sides by -6

x = 15

6. 19x – 28 = 2(6x + 7) (midsegment theorem)

19x – 28 = 12x + 14

  • Add like terms together

19x – 12x = 28 + 14

7x = 42

  • Divide both sides by 7

x = 6

Learn more about midsegment theorem on:

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