Fundations Level 1 Unit 6
Fundations Level 1 Unit 6

1
Mrs. Rivas 𝟐(𝟐𝒙 + 𝟏) = 𝟏𝟖 𝟒𝒙 + 𝟐 = 𝟏𝟖 𝟒𝒙 = 𝟏𝟔 𝒙 = 𝟒
(5-1) Algebra Find the value of x. 1. 𝟐(𝟐𝒙 + 𝟏) = 𝟏𝟖 𝟒𝒙 + 𝟐 = 𝟏𝟖 𝟒𝒙 = 𝟏𝟔 𝒙 = 𝟒

2
Mrs. Rivas 𝟐(𝟑𝒙) = 𝟑𝟎 𝟔𝒙 = 𝟑𝟎 𝒙 = 𝟓 (5-1) Algebra Find the value of x.
2. 𝟐(𝟑𝒙) = 𝟑𝟎 𝟔𝒙 = 𝟑𝟎 𝒙 = 𝟓

3
Mrs. Rivas (5-1) Algebra Find the value of x. 3. 𝟐(𝟑𝒙) = 𝟐𝟏 𝟔𝒙 = 𝟐𝟏 𝒙 = 𝟑.𝟓

4
Mrs. Rivas 𝟗 X is the midpoint of 𝑴𝑵 . Y is the midpoint of 𝑶𝑵 .
4. Find XZ. 𝟗

5
Mrs. Rivas 𝟐𝟎 X is the midpoint of 𝑴𝑵 . Y is the midpoint of 𝑶𝑵 .
5. If XY = 10, find MO. 𝟐𝟎

6
Mrs. Rivas 𝟔𝟒 𝟔𝟒 X is the midpoint of 𝑴𝑵 . Y is the midpoint of 𝑶𝑵 .
6. If 𝑚𝑀 is 64, find 𝑚𝑌. 𝟔𝟒 𝟔𝟒

7
Mrs. Rivas 𝟓.𝟓 Use the diagram at the right for Exercises 7 and 8.
7. What is the distance across the lake? 𝟓.𝟓

8
Mrs. Rivas 𝟒 𝟓.𝟓 BC is shorter. BC is half od 8 and AB is half od 11.
Use the diagram at the right for Exercises 7 and 8. 8. Is it a shorter distance from A to B or from B to C? Explain. 𝟒 BC is shorter. BC is half od 8 and AB is half od 11. 𝟓.𝟓

9
Mrs. Rivas 𝟓𝒙 + 𝟑 = 𝟕𝒙 − 𝟏 𝟓𝒙 𝟓𝒙 𝟑 = 𝟐𝒙 − 𝟏 + 𝟏 + 𝟏 𝟒 = 𝟐𝒙 𝟐 = 𝒙
(5-2) Algebra Find the indicated variables and measures. 10. x, EH, EF 𝟓𝒙 + 𝟑 = 𝟕𝒙 − 𝟏 𝟓𝒙 𝟓𝒙 𝟑 = 𝟐𝒙 − 𝟏 + 𝟏 + 𝟏 𝟒 = 𝟐𝒙 𝟐 = 𝒙 𝑬𝑯=𝟓𝒙+𝟑=𝟓 𝟐 +𝟑=𝟏𝟑 𝑬𝑭=𝟕𝒙−𝟏=𝟕 𝟐 −𝟏=𝟏𝟑

10
Mrs. Rivas 𝟐𝒙 – 𝟔 = 𝟑𝒙 – 𝟐𝟓 𝟐𝒙 𝟐𝒙 – 𝟔 = 𝒙 – 𝟐𝟓 + 𝟐𝟓 + 𝟐𝟓 𝟏𝟗 = 𝒙
(5-2) Algebra Find the indicated variables and measures. 11. x, mTPS, mRPS 𝟐𝒙 – 𝟔 = 𝟑𝒙 – 𝟐𝟓 𝟐𝒙 𝟐𝒙 – 𝟔 = 𝒙 – 𝟐𝟓 + 𝟐𝟓 + 𝟐𝟓 𝟏𝟗 = 𝒙 𝒎∠𝑻𝑷𝑺=𝟐𝒙−𝟔=𝟐 𝟏𝟗 −𝟔=𝟑𝟐 𝒎∠𝑹𝑷𝑺=𝟑𝒙−𝟐𝟓=𝟑 𝟏𝟗 −𝟐𝟓=𝟑𝟐

11
Mrs. Rivas 𝟑𝒂 – 𝟐 = 𝒂 + 𝟏𝟎 𝟐𝒂 – 𝟐 = 𝟏𝟎 𝟐𝒂 = 𝟏𝟐 𝒂 = 𝟔 𝟑𝒃 – 𝟏𝟓 = 𝟐𝒃 + 𝟓
(5-2) Algebra Find the indicated variables and measures. 12. a, b 𝟑𝒂 – 𝟐 = 𝒂 + 𝟏𝟎 𝟐𝒂 – 𝟐 = 𝟏𝟎 𝟐𝒂 = 𝟏𝟐 𝒂 = 𝟔 𝟑𝒃 – 𝟏𝟓 = 𝟐𝒃 + 𝟓 𝒃 – 𝟏𝟓 = 𝟓 𝒃 =𝟐𝟎

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Mrs. Rivas 1.

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Mrs. Rivas 2.

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Mrs. Rivas 3.

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Mrs. Rivas 4.

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Mrs. Rivas 5.

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Mrs. Rivas 6.

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Mrs. Rivas 7.

19
Mrs. Rivas 𝒙 + 𝟓 = 𝟑𝒙 + 𝟕 8. −𝟐𝒙 + 𝟓 = 𝟕 −𝟐𝒙 =𝟐 𝒙=−𝟏

20
Mrs. Rivas 𝒙 + 𝟐 = 𝟐𝒙 – 𝟑 9. −𝒙 + 𝟐 = – 𝟑 −𝒙 = –𝟓 𝒙 = 𝟓

21
Mrs. Rivas 10. 𝟓𝒙 + 𝟕 = 𝒙 + 𝟖 𝟒𝒙 + 𝟕 = 𝟖 𝟒𝒙 =𝟏 𝒙 = 𝟏 𝟒

22
Mrs. Rivas 𝑪𝑿= 𝟐 𝟑 𝑪𝑾 𝑪𝑿= 𝟐 𝟑 (𝟏𝟓) 𝑿𝑾=𝑪𝑾−𝑪𝑿 𝑪𝑿= 𝟑𝟎 𝟑 𝑿𝑾=𝟏𝟓−𝟏𝟎 𝑿𝑾=𝟓
(5-4) In ∆ABC, X is the centroid. 11. If CW = 15, find CX and XW. 𝑪𝑿= 𝟐 𝟑 𝑪𝑾 𝟏𝟓 𝑪𝑿= 𝟐 𝟑 (𝟏𝟓) 𝑿𝑾=𝑪𝑾−𝑪𝑿 𝑪𝑿= 𝟑𝟎 𝟑 𝑿𝑾=𝟏𝟓−𝟏𝟎 𝑿𝑾=𝟓 𝑪𝑿=𝟏𝟎

23
Mrs. Rivas 𝑩𝑿= 𝟐 𝟑 𝑩𝒀 𝟖= 𝟐 𝟑 𝑩𝒀 𝑿𝒀=𝑩𝒀−𝑩𝑿 𝟑 𝟐 𝟖= 𝟐 𝟑 𝑩𝒀 𝟑 𝟐 𝑿𝒀=𝟏𝟐−𝟖
(5-4) In ∆ABC, X is the centroid. 12. If BX = 8, find BY and XY. 𝑩𝑿= 𝟐 𝟑 𝑩𝒀 𝟖 𝟏𝟐 𝟖= 𝟐 𝟑 𝑩𝒀 𝟑 𝟐 𝟖= 𝟐 𝟑 𝑩𝒀 𝟑 𝟐 𝑿𝒀=𝑩𝒀−𝑩𝑿 𝑿𝒀=𝟏𝟐−𝟖 𝟐𝟒 𝟐 =𝑩𝒀 𝑿𝑾=𝟒 𝟏𝟐=𝑩𝒀

24
Mrs. Rivas 𝑿𝒁= 𝟏 𝟑 𝑨𝒁 𝟑= 𝟏 𝟑 𝑨𝒁 𝟑 𝟏 𝟑= 𝟏 𝟑 𝑨𝒁 𝟑 𝟏 𝑨𝑿=𝑨𝒁−𝑿𝒁 𝑨𝑿=𝟗−𝟑 𝟗=𝑨𝒁
(5-4) In ∆ABC, X is the centroid. 13. If XZ = 3, find AX and AZ. 𝑿𝒁= 𝟏 𝟑 𝑨𝒁 𝟗 𝟑 𝟑= 𝟏 𝟑 𝑨𝒁 𝟑 𝟏 𝟑= 𝟏 𝟑 𝑨𝒁 𝟑 𝟏 𝑨𝑿=𝑨𝒁−𝑿𝒁 𝑨𝑿=𝟗−𝟑 𝟗=𝑨𝒁 𝑨𝑿=𝟔

25
Mrs. Rivas Is 𝑨𝑩 a median, an altitude, or neither? Explain. 15. 14.
Altitude; 𝑨𝑩 is perpendicular to the opposite side. Median; 𝑨𝑩 bisects the opposite side.

26
Mrs. Rivas Is 𝑨𝑩 a median, an altitude, or neither? Explain. 17. 16.
Altitude; 𝑨𝑩 is perpendicular to the opposite side. Neither; 𝑨𝑩 is not perpendicular to nor does it bisect the opposite side.

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Mrs. Rivas 𝑪𝑱 In Exercises 18–22, name each segment.
18. a median in ∆ABC 𝑪𝑱

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Mrs. Rivas 𝑨𝑯 In Exercises 18–22, name each segment.
19. an altitude for ∆ABC 𝑨𝑯

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Mrs. Rivas 𝑰𝑯 In Exercises 18–22, name each segment.
20. a median in ∆AHC 𝑰𝑯

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Mrs. Rivas 𝑨𝑯 In Exercises 18–22, name each segment.
21. an altitude for ∆AHB 𝑨𝑯

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Mrs. Rivas 𝑨𝑯 In Exercises 18–22, name each segment.
22. an altitude for ∆AHG. 𝑨𝑯

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Mrs. Rivas 23. A(0, 0), B(0, 2), C(3, 0). Find the orthocenter of ∆ABC.

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Mrs. Rivas 24. In which kind of triangle is the centroid at the same point as the orthocenter? equilateral

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Mrs. Rivas 𝑴𝒆𝒅𝒊𝒂𝒏 𝑷𝒐𝒊𝒏𝒕 𝒐𝒇 𝒄𝒐𝒏𝒄𝒖𝒓𝒓𝒆𝒏𝒄𝒚 𝑪𝒆𝒏𝒕𝒓𝒐𝒊𝒅 𝑰𝒏𝒄𝒆𝒏𝒕𝒆𝒓
𝑪𝒐𝒏𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒍𝒊𝒏𝒆𝒔 𝑪𝒊𝒓𝒄𝒖𝒎𝒄𝒆𝒏𝒕𝒆𝒓 𝑨𝒍𝒕𝒊𝒕𝒖𝒕𝒆 𝑶𝒓𝒕𝒉𝒐𝒄𝒆𝒏𝒕𝒆𝒓 𝑽𝒆𝒓𝒕𝒆𝒙

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