1. Rectangle ABCD is shown below. Find the midpoint of diagonal 𝐴𝐶 .
(𝟒,𝟒)
(−𝟑,𝟏)
(𝟎.𝟓,𝟐.𝟓)
− ,
𝑚= 𝑥 1 + 𝑥 2 2 , 𝑦 1 + 𝑦 2 2
1 2 =0.5, =2.5

2. Use the distance formula… Use the midpoint formula…
𝑑= (7−1) 2 + (4−2) 2
Use the distance formula…
𝑑= (6) 2 + (2) 2
𝑑= ( 𝑥 1 − 𝑥 2 ) 2 + ( 𝑦 1 − 𝑦 2 ) 2
𝑑= 36+4
𝑑= 40
Use the midpoint formula…
𝑑=2 10 ≈6.3
𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡= 𝑥 1 + 𝑥 2 2 , 𝑦 1 + 𝑦 2 2
𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡= ,
8 2 , 6 2
=(4, 3)

3. Use the midpoint formula… Use the distance formula…
𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡= 𝑥 1 + 𝑥 2 2 , 𝑦 1 + 𝑦 2 2
𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡= ,
Use the distance formula…
16 2 , 16 2
=(8, 8)
𝑑= ( 𝑥 1 − 𝑥 2 ) 2 + ( 𝑦 1 − 𝑦 2 ) 2
𝑑= (9−7) 2 + (10−6) 2
𝑑= (2) 2 + (4) 2
𝑑= 4+16
𝑑= 20
𝑑=2 5 ≈4.47

4. 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡= 𝑥 1 + 𝑥 2 2 , 𝑦 1 + 𝑦 2 2 7= 8+𝑥 2 , 3= 4+𝑦 2
Use the midpoint formula…
𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡= 𝑥 1 + 𝑥 2 2 , 𝑦 1 + 𝑦 2 2
Use the distance formula…
7= 8+𝑥 2 , 3= 4+𝑦 2
𝑑= ( 𝑥 1 − 𝑥 2 ) 2 + ( 𝑦 1 − 𝑦 2 ) 2
14=8+𝑥;6=4+𝑦
𝑑= (8−6) 2 + (4−2) 2
6=𝑥;2=𝑦
𝑑= (2) 2 + (2) 2
(6, 2)
𝑑= 4+4
Clear the denominator by multiplying each by two…
𝑑= 8
𝑑=2 2 ≈2.828

5. Use the Pythagorean Theorem6
8+6=14 𝑏𝑙𝑜𝑐𝑘𝑠 ∟ 8
Use the Pythagorean Theorem
Direct path
10 𝑏𝑙𝑜𝑐𝑘𝑠
𝒂 𝟐 + 𝒃 𝟐 = 𝒄 𝟐
= 𝑐 2
36+64= 𝑐 2
100= 𝑐 2
10=𝑐
14 𝑏𝑙𝑜𝑐𝑘𝑠−10 𝑏𝑙𝑜𝑐𝑘𝑠=4 𝑏𝑙𝑜𝑐𝑘𝑠 𝑠ℎ𝑜𝑟𝑡𝑒𝑟

6. Use the distance formula to find the distance on a coordinate plane.
𝑑= ( 𝑥 2 − 𝑥 1 ) 2 +( 𝑦 2 − 𝑦 1 ) 2
𝑑= (12− 4) 2 +(6−0 ) 2
𝑑= (8 ) 2 +(6 ) 2 𝑑=
𝑑= 100
𝑑=10
10×0.75=7.5 miles

7. Find the distance of 𝑨𝑩 for both 𝒙 and 𝒚 coordinates…
16−1=15 𝑎𝑛𝑑 8 −3=5
Since the point is partitioned from point A, subtract from point A’s coordinates…
Find 𝟐 𝟓 of each coordinate distance…
2 5 ×15=6 𝑎𝑛𝑑 2 5 ×5=2
16−6=10 𝑎𝑛𝑑 8−2=6
(𝟏𝟎, 𝟔)

8. (15.5, 0) 8+7.5=15.5 3÷2 𝑝𝑎𝑟𝑡𝑠=1.5 5 𝑝𝑎𝑟𝑡𝑠×1.5=7.5 2 parts 5 parts 3

9. Find the distance of 𝑨𝑩 for both 𝒙 and 𝒚 coordinates…
|−9−12|=21 𝑎𝑛𝑑 |2 −8|=6
Since the point is partitioned from point A, add to point A’s coordinates…
Find 𝟏 𝟑 of each coordinate distance…
1 3 ×21=7 𝑎𝑛𝑑 1 3 ×6=2
−9+7=−2 𝑎𝑛𝑑 2+2=4
(−𝟐, 𝟒)

10. Use the Point-Slope Formula
Find the slope of the line with points (4, 9) and (-3, 6)…
Use the Point-Slope Formula
𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 )
𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1
𝑦−8= 3 7 (𝑥−14)
Simplify
𝑚= 6−9 −3−4
= −3 −7 𝑜𝑟 3 7
𝑦−8= 3 7 𝑥−6
Add 8 to each side of the equation… +8 +8
𝑦= 3 7 𝑥+2

11. Use the Point-Slope Formula
𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 )
Find the slope of the line with points (4, 9) and (-3, 6)…
𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1
𝑦−8=− 7 3 (𝑥−14)
Simplify
𝑚= 6−9 −3−4
= −3 −7 𝑜𝑟 3 7
=− 7 3
𝑦−8=− 7 3 𝑥+ 98 3
Add 8 to each side of the equation…
Perpendicular slopes are opposite sign reciprocals. +8
+ 24 3 +8
𝑦=− 7 3 𝑥
8 1 × 3 3 = 24 3
Simplify
𝑦=− 7 3 𝑥

12. Use the Point-Slope Formula
𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 )
𝑦=3𝑥−5
The slope of a parallel line would be +𝟑
𝑦−5=3(𝑥−8)
Standard Form
𝐴𝑥+𝐵𝑦=𝐶
Simplify
𝑦−5=3𝑥−24
Subtract 3x from each side of the equation…
−3𝑥
−3𝑥
−3𝑥+𝑦−5=−24 +5 +5
Add 5 to each side of the equation…
−3𝑥+𝑦=−19 OR
3𝑥−𝑦=19

13. Use the Point-Slope Formula
𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 )
𝑦=3𝑥−5
The slope of a perpendicular line would be − 𝟏 𝟑 .
𝑦−5=− 1 3 (𝑥−8)
Simplify
𝑦−5=− 1 3 𝑥+ 8 3
Add 5 to each side of the equation…
Standard Form
𝐴𝑥+𝐵𝑦=𝐶 +5
+ 15 3 +5
5 1 × 3 3 = 15 3
𝑦=− 1 3 𝑥+ 23 3
Add 𝟏 𝟑 𝒙 to each side of the equation…
+ 1 3 𝑥
+ 1 3 𝑥
Multiply each term by 3…
(3)
(3)
1 3 𝑥+𝑦= 23 3
(3)
𝑥+3𝑦=23

14. 𝑎. 𝑏. 𝒚=−𝟓𝒙−𝟑 𝒚=−𝟓𝒙−𝟑 y= 1 5 𝑥+6 y=−5𝑥+6
The slope of a perpendicular line would be + 𝟏 𝟓 .
The slope of a parallel line would be −𝟓.
y= 1 5 𝑥+6
y=−5𝑥+6

15. Translate 𝐴𝐵 using the translation rule 𝑥, 𝑦 →(𝑥−2, 𝑦+4) then translate 𝐴′𝐵′ using the translation rule 𝑥, 𝑦 → 𝑥−2, 𝑦−6 .. 𝑨′
Translate 2 units left…
Translate 4 units up…
𝑨" 𝑩′
Translate 2 units left…
Translate 6 units down.
𝑩"

16. 𝑩" 𝑨′
𝑨" 𝑩′

17. Rotate 𝐴𝐵 90° counterclockwise around the origin.
𝑅 90° →(−𝑦,𝑥) 𝑩′ 𝑨′
(𝟑, 𝟏)
𝑨 ′ =(−𝟏, 𝟑)
𝑩 ′ =(𝟒, 𝟒)
(𝟒, −𝟒)

18. Rotate 𝐴𝐵 90° clockwise around the origin.
𝑅 −90° →(𝑦,−𝑥)
(𝟑, 𝟏)
𝑨 ′ =(𝟏, −𝟑)
𝑩 ′ =(−𝟒, −𝟒) 𝑨′ 𝑩′
(𝟒, −𝟒)

19. The scale factor would be 2.
a. What is the scale factor of the dilation of line segment 𝐴𝐵 ? 6
The scale factor would be 2. 3
𝑨𝑩 𝐢𝐬 𝐭𝐡𝐞 𝐩𝐫𝐞𝐢𝐦𝐚𝐠𝐞 𝐚𝐧𝐝
𝐡𝐚𝐬 𝐚 𝐥𝐞𝐧𝐠𝐭𝐡 𝐨𝐟 𝟑 𝐮𝐧𝐢𝐭𝐬.

20. b. Draw a segment demonstrating a dilation of 𝐴𝐵 by a factor of 1 3 .𝑩′ 𝑨′

22. Use a compass and straight edge to copy angle A.

24. (𝟕𝒙−𝟐𝟒)° Since 𝑴𝑶 is an angle bisector, 𝒎∠𝑳𝑴𝑶 is also 𝟐𝒙+𝟑𝟏.
𝑚∠𝐿𝑀𝑁= °
(𝟐𝒙+𝟑𝟏)°
(𝟐𝒙+𝟑𝟏)° ° °
Use the Angle Addition Postulate
Combine Like Terms
2𝑥+31+2𝑥+31=7𝑥−24
Subtract 4x from each side of the equation…
4𝑥+62=7𝑥−24
−4𝑥
−4𝑥
Add 24 to each side of the equation…
62=3𝑥−24
+24
+24
Divide each side of the equation by 3…
86=3𝑥 3 3
= 265 3
=88 1 3
Substitute 𝟖𝟔 𝟑 for x to find the 𝒎∠𝑳𝑴𝑶 𝐀𝐍𝐃 𝒎∠𝑵𝑴𝑶 …
86 3 =𝑥

25. Vertical Angle Theorem
Substitute
8𝑥−9=7𝑥+7
−7𝑥
−7𝑥
𝑥−9=7 +9 +9
𝑥=16
=119°

26. Vertical Angle Theorem
6𝑥−9=2𝑥+7
−2𝑥
−2𝑥
4𝑥−9=7 +9 +9 4 4
4𝑥=16
𝑥=4
2 4 +7=15°

27. Substitute back into the expression…
Linear Pair
(two angles that share a common ray that form a straight line)
144°
144°
36°
(8𝑥−64)°+36°=180°
Combine Like Terms…
8 26 −64=144°
Add 28 to each side of the equation…
8𝑥−28=180
+28
+28
8𝑥=208 8 8
Divide each side of the equation by 8…
𝑥=26
Substitute back into the expression…

28. Substitute back into the expression…
Linear Pair
(two angles that share a common ray that form a straight line)
135°
135°
45°
(5𝑥−20)°+45°=180°
Combine Like Terms…
5 31 −20=135°
5𝑥+25=180
−25
−25
Subtract 25 from each side of the equation…
5𝑥=155 5 5
𝑥=31
Divide each side of the equation by 5…
Substitute back into the expression…

29. 𝑎 ∥𝑏 𝑨𝒍𝒕. 𝑬𝒙𝒕. ∠𝟏 & ∠𝟓, ∠𝟐 & ∠𝟔 𝑨𝒍𝒕. 𝑰𝒏𝒕. ∠𝟒 & ∠𝟖, ∠𝟑 & ∠𝟕
𝑺𝒂𝒎𝒆−𝑺𝒊𝒅𝒆 𝑰𝒏𝒕. ∠𝟖 & ∠𝟕, ∠𝟑 & ∠𝟒
𝑪𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 ∠𝟏 & ∠𝟕, ∠𝟐 &∠𝟒, ∠𝟑 & ∠𝟓, ∠𝟖 & ∠𝟔

30. 𝑎 ∥𝑏 𝑪𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝑨𝒏𝒈𝒍𝒆𝒔: ∠𝟏≅∠𝟕;∠𝟐≅∠𝟒;∠𝟑≅∠𝟓;∠𝟔≅∠𝟖
𝑪𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝑨𝒏𝒈𝒍𝒆𝒔: ∠𝟏≅∠𝟕;∠𝟐≅∠𝟒;∠𝟑≅∠𝟓;∠𝟔≅∠𝟖
𝑨𝒍𝒕. 𝑰𝒏𝒕𝒆𝒓𝒊𝒐𝒓 𝑨𝒏𝒈𝒍𝒆𝒔: ∠𝟑≅∠𝟕;∠𝟒≅∠𝟖
𝑨𝒍𝒕. 𝑬𝒙𝒕𝒆𝒓𝒊𝒐𝒓 𝑨𝒏𝒈𝒍𝒆𝒔: ∠𝟏≅∠𝟓;∠𝟐≅∠𝟔
𝑺𝒂𝒎𝒆−𝑺𝒊𝒅𝒆 𝑰𝒏𝒕𝒆𝒓𝒊𝒐𝒓 𝑨𝒏𝒈𝒍𝒆𝒔 𝒂𝒓𝒆 𝑺𝒖𝒑𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒂𝒓𝒚: ∠𝟑+∠𝟒;∠𝟕+∠𝟖

31. ∠𝟔 ≅∠𝟐 as they are corresponding angles.

33. ∠𝟏=𝟏𝟏𝟐° by alternate exterior angles
∠𝟒=𝟏𝟏𝟐° by vertical angles
112°+112°=𝟐𝟐𝟒°

34. 3 angles that form a straight line
Vertical Angles
𝟕𝟖°
180°−40°−62°=78°
Alt. Interior Angles
𝟒𝟎°
𝟕𝟖°
3 angles that form a straight line
Vertical Angles
𝟏𝟏𝟖°
Remote Exterior Angle
𝟒𝟎°
78°+40°=118°
𝟕𝟖°
𝟒𝟎°
𝟒𝟎°
𝟕𝟖°
𝟏𝟏𝟖°

35. 3 angles that form a straight line
𝟕𝟖°
Vertical Angles
𝟒𝟎°
Alt. Interior Angles
𝟕𝟖°
3 angles that form a straight line
Vertical Angles
𝟒𝟎°
180°−40°−62°=78°

36. 𝐨𝐟 𝐭𝐡𝐞 𝐩𝐨𝐥𝐲𝐠𝐨𝐧. 𝐨𝐟 𝐭𝐡𝐞 𝐩𝐨𝐥𝐲𝐠𝐨𝐧. 𝐨𝐟 𝐭𝐡𝐞 𝐩𝐨𝐥𝐲𝐠𝐨𝐧.
Complete the table below.
𝟏𝟖𝟎 𝒏−𝟐 𝐰𝐡𝐞𝐫𝐞 𝒏 𝐢𝐬
𝐭𝐡𝐞 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐬𝐢𝐝𝐞𝐬
𝐨𝐟 𝐭𝐡𝐞 𝐩𝐨𝐥𝐲𝐠𝐨𝐧.
PENTAGON
𝟏𝟖𝟎(𝟓−𝟐)
𝟏𝟖𝟎(𝟑)
𝟓𝟒𝟎°
𝟏𝟖𝟎(𝟓−𝟐) 𝟓
𝟏𝟖𝟎(𝟑) 𝟓
𝟏𝟎𝟖°
𝟑𝟔𝟎 𝟓
𝟕𝟐°
OCTAGON
𝟏𝟖𝟎(𝟖−𝟐) 𝟖
𝟏𝟖𝟎(𝟔) 𝟖
𝟑𝟔𝟎 𝟖
𝟏𝟖𝟎(𝟖−𝟐)
𝟏𝟖𝟎(𝟔)
𝟏𝟎𝟖𝟎°
𝟏𝟑𝟓°
𝟒𝟓°
𝟏𝟖𝟎 𝒏−𝟐 𝒏 𝐰𝐡𝐞𝐫𝐞 𝒏 𝐢𝐬
𝐭𝐡𝐞 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐬𝐢𝐝𝐞𝐬
𝐨𝐟 𝐭𝐡𝐞 𝐩𝐨𝐥𝐲𝐠𝐨𝐧.
HEPTAGON
𝟏𝟖𝟎(𝟕−𝟐)
𝟏𝟖𝟎(𝟓)
𝟏𝟖𝟎(𝟕−𝟐) 𝟕
𝟏𝟖𝟎(𝟓) 𝟕
𝟗𝟎𝟎°
𝟏𝟐𝟖.𝟓°
𝟓𝟏.𝟓°
𝟑𝟔𝟎 𝟕
𝟑𝟔𝟎 𝒏 𝐰𝐡𝐞𝐫𝐞 𝒏 𝐢𝐬
𝐭𝐡𝐞 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐬𝐢𝐝𝐞𝐬
𝐨𝐟 𝐭𝐡𝐞 𝐩𝐨𝐥𝐲𝐠𝐨𝐧.
HEXAGON
𝟏𝟖𝟎(𝟔−𝟐) 𝟔
𝟏𝟖𝟎(𝟒) 𝟔
𝟑𝟔𝟎 𝟔
𝟏𝟖𝟎(𝟔−𝟐)
𝟏𝟖𝟎(𝟒)
𝟕𝟐𝟎°
𝟏𝟐𝟎°
𝟔𝟎°
TRIANGLE
𝟏𝟖𝟎(𝟑−𝟐) 𝟑
𝟏𝟖𝟎(𝟏) 𝟑
𝟏𝟐𝟎°
𝟑𝟔𝟎 𝟑
𝟏𝟖𝟎(𝟑−𝟐)
𝟏𝟖𝟎(𝟏)
𝟏𝟖𝟎°
𝟔𝟎°
𝟑𝟔𝟎 𝟏𝟎
DECAGON
𝟏𝟖𝟎(𝟏𝟎−𝟐) 𝟏𝟎
𝟏𝟖𝟎(𝟖) 𝟏𝟎
𝟑𝟔°
𝟏𝟖𝟎(𝟏𝟎−𝟐)
𝟏𝟖𝟎(𝟖)
𝟏𝟒𝟒𝟎°
𝟏𝟒𝟒°

37. What is the measure of the missing angle in the polygons below?
𝐒𝐮𝐦 𝐨𝐟 𝐭𝐡𝐞 𝐈𝐧𝐭𝐞𝐫𝐢𝐨𝐫 𝐀𝐧𝐠𝐥𝐞𝐬
𝟏𝟖𝟎(𝒏−𝟐)
180(5−2)
180(8−2)
180(3)
=540°
180(6)
=1080°
540°−90−145−95−165=45°
1080°−150−120−112−108−170−165−150=105°

38. B’ A’ 2 C’ 3

39. Translate 5 units left…
Translate 2 units down. B’ A’ C’

40. A’ C’ 4 4 1 B’ 1 4 4

41. 𝒙=−𝟏 B’ A’ C’

42. Rotations without specification are always counterclockwise!𝑨′ 𝑩′ 𝑪′ ∟
The distance from the pre-image point to the origin is the same distance as the image point to the origin. Note the 90 degree angle formed. ∟ ∟

43. Reflect across the y-axis.
The distance from the pre-image point to the origin is the same distance as the image point to the origin. Note the 180 degree angle formed. 𝐵′ 𝐴′ 𝐶′
𝐶′′
𝐵′′
𝐴′′
𝟏𝟖𝟎°
𝟏𝟖𝟎°
Reflect across the y-axis.
𝟏𝟖𝟎°

44. 𝑨′ 𝑪′ 𝑩′
𝑨"
𝑩"
𝑪"
𝑅 180° →(−𝑥,−𝑦)

45. What is the scale factor in the dilation shown below?
Pre Image
Since the image is getting smaller, the scale factor will be fraction…
4 units
2 units
The scale factor will be ½.

46. What is the scale factor in the dilation shown below?
Pre Image
Since the image is getting larger, the scale factor will be greater than 1…
3 units
6 units
The image is twice as large as the pre image, so the scale factor would be 2.

47. Draw the translated image (𝑥,𝑦)→(𝑥+6, 𝑦−1) and then reflect the image over the x-axis.
Translate triangle ABC 6 units right and 1 unit down. 𝐵′ 𝐴′ 𝐶′
Reflect triangle A’B’C’ over the x-axis.
𝐴"
𝐶"
𝐵"

48. By the Linear Pair Postulate… By the Triangle Angle Sum Theorem…
𝟏𝟖𝟎°−𝟏𝟏𝟔°=𝟔𝟒°
𝟔𝟒°
By the Triangle Angle Sum Theorem…
𝟏𝟖𝟎°−𝟕𝟓°−𝟔𝟒°=𝟒𝟏°

49. By the Linear Pair Postulate… By the Triangle Angle Sum Theorem
𝟏𝟖𝟎°−𝟏𝟑𝟏°=𝟒𝟗°
𝟒𝟗°
By the Triangle Angle Sum Theorem
(the sum of the interior angles of a triangle is equal to 𝟏𝟖𝟎°)…
𝟏𝟖𝟎°−𝟔𝟏°−𝟒𝟗°=𝟕𝟎°

50. Divide the shape into a rectangle and a triangle…
A softball home plate has been designed on the coordinate plane shown below. If each unit is 2 inches, what is the area of home plate?
𝐴= 1 2 (𝑙×𝑤)
Divide the shape into a rectangle and a triangle…
𝐴= ×3 =9
𝟗 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔
𝐴=𝑙×𝑤
𝐴=6×4=24
𝟐𝟒 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔
𝟐𝟒+𝟗=𝟑𝟑 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔
𝟑𝟑 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔× 𝟐 𝟐 =𝟏𝟑𝟐 𝒔𝒒. 𝒊𝒏𝒄𝒉𝒆𝒔

51. Divide the shape into a rectangle and 2 triangles…
If each unit in the coordinate plane below is 3 cm, what is the area of the polygon ABCDEF?
𝐴= 1 2 (𝑙×𝑤)
Divide the shape into a rectangle and 2 triangles…
𝐴= ×4 =16
𝟏𝟔 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔
𝐴=𝑙×𝑤
𝟐𝟒 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔
𝐴=8×3=24
𝟏𝟐 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔
𝐴= ×3 =12
𝟐𝟒+𝟏𝟔+𝟏𝟐=𝟓𝟐 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔
𝟓𝟐 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔× 𝟑 𝟐 =𝟒𝟔𝟖 𝒔𝒒 𝒄𝒎

52. Reflexive Property

53. Symmetric Property

54. Transitive Property

55. 27° 𝟏𝟑 𝟏𝟐
𝟔𝟕°
𝟗𝟎° 𝟓
If the triangles are congruent, then their corresponding parts are congruent. (CPCTC)

56. What other information do you need in order to prove the triangles congruent using the SAS Congruence Postulate? By AAS Congruence Theorem?
Hidden is the fact that 𝑨𝑪 ≅ 𝑨𝑪 by the reflexive property. 𝑨 𝑨
The missing information is
𝑨𝑩 ≅ 𝑨𝑫 so that the triangle are congruent by SAS. 𝑺 𝑺 𝑺 𝑨
The missing information is
∠𝑩≅∠𝑫 so that the triangle are congruent by AAS.

57. Hidden is the fact that 𝑨𝑪 ≅ 𝑨𝑪 by the reflexive property.𝑺 𝑺 𝑨
∠𝐵𝐴𝐶≅∠𝐷𝐶𝐴 𝑓𝑜𝑟 𝐴𝑆𝐴
𝐴𝐷 ≅ 𝐵𝐶 𝑓𝑜𝑟 𝑆𝐴𝑆

58. AB is congruent to DC and AC is congruent to DB by given information
AB is congruent to DC and AC is congruent to DB by given information. By the reflexive property, BC is congruent to BC. Triangle ABC is congruent to triangle DCB by the side, side, side postulate.

59. 𝐴𝐶 𝑏𝑖𝑠𝑒𝑐𝑡𝑠 ∠𝐷𝐴𝐵, 𝐶𝐴 𝑏𝑖𝑠𝑒𝑐𝑡𝑠 ∠𝐷𝐶𝐵
𝐺𝑖𝑣𝑒𝑛
∠𝐷𝐴𝐶≅∠𝐵𝐴𝐶, ∠𝐷𝐶𝐴≅∠𝐵𝐶𝐴
𝐷𝑒𝑓 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝑏𝑖𝑠𝑒𝑐𝑡𝑜𝑟
𝐴𝐶 ≅ 𝐴𝐶
𝑅𝑒𝑓𝑙𝑒𝑥𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦
∆𝐷𝐴𝐶≅∆𝐵𝐴𝐶
𝐴𝑆𝐴

60. 𝑺 𝑺 𝑨 𝐴𝐵 ≅ 𝐴𝐶 𝑎𝑛𝑑 ∠𝐵𝐴𝐷 ≅ ∠𝐶𝐴𝐷 𝐺𝑖𝑣𝑒𝑛 𝐴𝐷 ≅ 𝐴𝐷 𝑅𝑒𝑓𝑙𝑒𝑥𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦
𝐴𝐵 ≅ 𝐴𝐶 𝑎𝑛𝑑 ∠𝐵𝐴𝐷 ≅ ∠𝐶𝐴𝐷
𝐺𝑖𝑣𝑒𝑛
𝐴𝐷 ≅ 𝐴𝐷
𝑅𝑒𝑓𝑙𝑒𝑥𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦
∆𝐴𝐶𝐷≅∆𝐴𝐵𝐷
𝑆𝐴𝑆
𝐶𝐷 ≅ 𝐵𝐷
𝐶𝑃𝐶𝑇𝐶
𝐴𝐷 𝑏𝑖𝑠𝑒𝑐𝑡𝑠 𝐵𝐶
𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 𝑏𝑖𝑠𝑒𝑐𝑡𝑜𝑟

61. 𝑨𝑨~ ∠𝑨𝑫𝑬≅ ∠𝑨𝑩𝑪 𝒃𝒚 𝑪𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒊𝒏𝒈 𝑨𝒏𝒈𝒍𝒆 𝑷𝒐𝒔𝒕𝒖𝒍𝒂𝒕𝒆
∠𝑫𝑨𝑬≅ ∠𝑩𝑨𝑪 𝒃𝒚 𝒕𝒉𝒆 𝑹𝒆𝒇𝒍𝒆𝒙𝒊𝒗𝒆 𝑷𝒓𝒐𝒑𝒆𝒓𝒕𝒚

62. By the Triangle Proportionality Theorem…
In the diagram below, what is the length of 𝑆𝑇 ?
By the Triangle Proportionality Theorem…
𝟓 𝟒 = 𝟏𝟎 𝒙
Cross Multiply
The question is asking for ST not the value of x!!!
Divide each side of the equation by 5…
5𝑥=40 5 5
𝑥=8
𝑆𝑇=10+8=18

63. By the Triangle Proportionality Theorem…
𝒙 𝒙+𝟓 = 𝒙−𝟐 𝒙+𝟏
Cross Multiply
𝑥 2 +𝑥= 𝑥 2 +3𝑥−10
𝑥=+3𝑥−10
Subtract 3x from each side of the equation…
Divide each side of the equation by -2…
−3𝑥
−3𝑥
−2𝑥=−10 −2 −2
𝑥=5

64. ( 𝟏 𝟐 the length of the parallel side)
∆𝐵𝐷𝐹= =81 𝟒𝟐 𝟐𝟓 𝟕𝟎
∆𝐴𝐶𝐸= =162 𝟑𝟓 𝟐𝟏
Triangle Midsegment
( 𝟏 𝟐 the length of the parallel side) 𝟓𝟎

65. By the triangle Midsegment Theorem…
Simplify…
2 2𝑥−4 =20
Add 8 to each side of the equation…
4𝑥−8=20 +8 +8
4𝑥=28 4 4
Divide each side of the equation by 4…
Triangle Midsegment
( 𝟏 𝟐 the length of the parallel side)
𝑥=7

66. The tic marks indicate that this is the midsegment of the triangle.
136 𝑓𝑡.
The tic marks indicate that this is the midsegment of the triangle.
By the Triangle Midsegment Theorem, the midsegment is ½ the length of its parallel side (or the parallel side is double the length of the midsegment)…
68×2=136 𝑓𝑡.

67. The largest side must be less than the sum of the two other sides…
The smallest side must be greater than the difference of the two other sides…
The largest side must be less than the sum of the two other sides…
14−7=7
14+7=21
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒: 𝐺𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 7 𝑎𝑛𝑑 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 21

68. The largest side must be less than the sum of the two other sides…
Two sides of a triangle have lengths 3 and 9. What must be true about the length of the third side?
The smallest side must be greater than the difference of the two other sides…
The largest side must be less than the sum of the two other sides…
9−3=6
9+3=12
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒: 𝐺𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 6 𝑎𝑛𝑑 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 12

69. By the Hinge Theorem, Since 98°>82°, 𝐾𝐿 > 𝐿𝑀 .
<
98°
180°−82°=98°
By the Hinge Theorem, Since 98°>82°, 𝐾𝐿 > 𝐿𝑀 .

70. By the Hinge Theorem, Since 23°>29°, 𝐶𝐷 > 𝐴𝐵 .
A pair of scissors is opened to two positions, A and B.
<
By the Hinge Theorem, Since 23°>29°, 𝐶𝐷 > 𝐴𝐵 .

71. The Centroid partitions the medians in a ratio of 2 to 1.
𝑨𝑬=𝟏𝟖+𝟗 𝒐𝒓 𝟐𝟕 5 9 15 18 10

72. The centroid is the point of concurrency of the medians of a triangle, therefor…
2 2x−3 =x+18
4x−6=x+18
3x−6=18
3x=24
x=8

73. Use the Pythagorean Theorem to find the length of the side.
Find the length of the base in the rectangle below.
Use the Pythagorean Theorem to find the length of the side.
𝒂 𝟐 + 𝒃 𝟐 = 𝒄 𝟐
6 2 + 𝑏 2 = 10 2 8
36+ 𝑏 2 =100
−36
−36
𝑏 2 =64
𝑏=8

74. Use the Pythagorean Theorem to find the length of the side.
If a 25 foot ladder leans against the side of a building 24 feet from its base, how high up the building will the ladder reach?
Use the Pythagorean Theorem to find the length of the side. 7
𝒂 𝟐 + 𝒃 𝟐 = 𝒄 𝟐
𝑏 2 = 25 2
576+ 𝑏 2 =625
−576
−576
𝑏 2 =49
𝑏=7

75. In the triangle below, write the sine, cosine, and tangent ratios for angle B and angle C.
𝑠𝑖𝑛= opposite leg hypotenuse
hypotenuse
leg
𝑐𝑜𝑠= adjacent leg hypotenuse
leg
𝑡𝑎𝑛= opposite leg adjacent leg
sin ∠𝐵= 𝐴𝐶 𝐵𝐶
sin ∠𝐶= 𝐴𝐵 𝐵𝐶
cos ∠𝐵= 𝐴𝐵 𝐵𝐶
cos ∠𝐶= 𝐴𝐶 𝐵𝐶
tan ∠𝐵= 𝐴𝐶 𝐴𝐵
tan ∠𝐶= 𝐴𝐵 𝐴𝐶

76. 𝑨𝑪 ≅ 𝑨𝑪 𝒃𝒚 𝒕𝒉𝒆 𝒓𝒆𝒇𝒍𝒆𝒙𝒊𝒗𝒆 𝒑𝒓𝒐𝒑𝒆𝒓𝒕𝒚
𝑨𝑪 ≅ 𝑨𝑪 𝒃𝒚 𝒕𝒉𝒆 𝒓𝒆𝒇𝒍𝒆𝒙𝒊𝒗𝒆 𝒑𝒓𝒐𝒑𝒆𝒓𝒕𝒚
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
∆𝑩𝑨𝑪≅∆𝑫𝑨𝑪 𝒃𝒚 𝑯𝑳 (𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 −𝒍𝒆𝒈)
𝒍𝒆𝒈

77. 𝑳 𝑯 𝑆𝐷 ⊥ 𝐻𝑇 𝑎𝑛𝑑 𝑆𝐻 ≅ 𝑆𝑇 𝐺𝑖𝑣𝑒𝑛 ∠𝑆𝐷𝐻 𝑎𝑛𝑑 ∠𝑆𝐷𝑇 𝑎𝑟𝑒 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒𝑠
𝑆𝐷 ⊥ 𝐻𝑇 𝑎𝑛𝑑 𝑆𝐻 ≅ 𝑆𝑇
𝐺𝑖𝑣𝑒𝑛
∠𝑆𝐷𝐻 𝑎𝑛𝑑 ∠𝑆𝐷𝑇 𝑎𝑟𝑒 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒𝑠
𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑙𝑖𝑛𝑒𝑠
∆𝑆𝐷𝐻 and ∆𝑆𝐷𝑇 𝑎𝑟𝑒 𝑟𝑖𝑔ℎ𝑡 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠
𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑟𝑖𝑔ℎ𝑡 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠
𝑆𝐷 ≅ 𝑆𝐷
𝑅𝑒𝑓𝑙𝑒𝑥𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦
∆𝑆𝐻𝐷≅∆𝑆𝑇𝐷 𝐻𝐿

78. Definition of right triangles
Converse of Isosceles Triangle Theorem HL
Reflexive Property

79. In a 𝟒𝟓°−𝟒𝟓°−𝟗𝟎° triangle, the legs are congruent.
What are the lengths of the missing sides in the triangle?
𝟕 𝟐
leg
leg 𝟕 𝟕
𝟕 𝟐
hypotenuse
In a 𝟒𝟓°−𝟒𝟓°−𝟗𝟎° triangle, the legs are congruent.
In a 𝟒𝟓°−𝟒𝟓°−𝟗𝟎° triangle, the hypotenuse is the length of a leg times the 𝟐 .

80. Quilt squares are cut on the diagonal to form triangular quilt pieces
Quilt squares are cut on the diagonal to form triangular quilt pieces. The hypotenuse of the resulting triangles is 10 inches long. What is the side length of each piece? Leave your answer in simplified radical form.
In a 𝟒𝟓°−𝟒𝟓°−𝟗𝟎° triangle, the hypotenuse is the length of a leg times the 𝟐 .
𝟒𝟓°
10 𝑖𝑛.
So, the length of a leg is the length of the hypotenuse divided by 𝟐 . ∟
𝟒𝟓° × =
10 2
=5 2
Rationalize the denominator

81. Use a scientific calculator and divide 24 by the sin of 60 degrees…
A conveyor belt carries supplies from the first floor to the second floor, which is 24 feet higher. The belt makes a 60° angle with the ground. How far do the supplies travel from one end of the conveyor belt to the other? Round your answer to the nearest foot.
𝑠𝑖𝑛= opposite leg hypotenuse
24 𝑓𝑡.
60° ∟
sin 60°= 24 𝑥
≈27.7 𝑓𝑡.
Use a scientific calculator and divide 24 by the sin of 60 degrees…

82. In the triangle below, find the values of 𝑎 and 𝑏.
The altitude (a) of the right triangle is the geometric mean of the two parts of the hypotenuse.
𝑎 32 = 2 𝑎
Cross Multiply
The leg (b) of the right triangle is the geometric mean of the entire hypotenuse and the adjacent part of the hypotenuse..
𝑎 2 =64
Take the square root of each side…
Because distance cannot be negative
𝑎=±8
𝑏 34 = 2 𝑏
𝑎=8
Take the square root of each side…
Cross Multiply
𝑏 2 =68
𝑏=± 68

83. Use the Pythagorean Theorem to find the length of the hypotenuse.
Jason wants to walk the shortest distance to get from the parking lot to the beach.
Use the Pythagorean Theorem to find the length of the hypotenuse.
The altitude (a) of the right triangle is the geometric mean of the two parts of the hypotenuse.
𝒂 𝟐 + 𝒃 𝟐 = 𝒄 𝟐
32 𝑚
50 𝑚
= 𝑐 2
= 𝑐 2
𝑎 18 = 32 𝑎
2500= 𝑐 2
50=𝑐
𝑎 2 =576
𝑎=24 𝑚
How far is the spot on the beach from the parking lot?
b. How far will his place on the beach be from the refreshment stand?
50−18=32 𝑚

84. Kristen lives directly east of the park
Kristen lives directly east of the park. The football field is directly south of the park. The library sits on the line formed between Kristen’s home and the football field at the exact point where an altitude to the right triangle formed by her home, the park, and the football field could be drawn. The library is 2 miles from her home. The football field is 5 miles from the library
The altitude (a) of the right triangle is the geometric mean of the two parts of the hypotenuse.
The leg (b) of the right triangle is the geometric mean of the entire hypotenuse and the adjacent part of the hypotenuse..
10 𝑚
35 𝑚
𝑎 5 = 2 𝑎
𝑏 5 = 7 𝑏
𝑎 2 =10
𝑎= 10 𝑚
𝑏 2 =35
𝑏= 35 𝑚 a.
How far is library from the park? b.
How far is the park from the football field?

85. Use a scientific calculator and multiply the Sine of 17 degrees by 14…
Find the length of 𝐴𝐵 .
𝐒𝐢𝐧𝐞= 𝐨𝐩𝐩𝐨𝐬𝐢𝐭𝐞 𝐥𝐞𝐠 𝐡𝐲𝐩𝐨𝐭𝐞𝐧𝐮𝐬𝐞
𝟒.𝟎𝟗 𝒙
Use a scientific calculator and multiply the Sine of 17 degrees by 14…
𝑺𝒊𝒏 𝟏𝟕= 𝒙 𝟏𝟒
𝑺𝒊𝒏 𝟏𝟕 𝟏𝟒=
𝟒.𝟎𝟗

86. Use a scientific calculator 𝒕𝒂 𝒏 −𝟏 (𝟏𝟎𝟎÷𝟐𝟒𝟗)
A large totem pole in the state of Washington is 100 feet tall. At a particular time of day, the totem pole casts a 249-foot-long shadow. Find the measure of ∠to the nearest degree.
𝑡𝑎𝑛= opposite leg adjacent leg
𝟐𝟐°
tan 𝑥°=
≈21.88°
Use a scientific calculator 𝒕𝒂 𝒏 −𝟏 (𝟏𝟎𝟎÷𝟐𝟒𝟗)

87. 𝑥° 𝑥° 56.3° 𝑡𝑎𝑛= opposite leg adjancent leg 𝑡𝑎𝑛 𝑥 −1 = 1200 800 ≈56.3°
By the alternate interior angle theorem… 𝑥°
56.3°
𝑡𝑎𝑛= opposite leg adjancent leg 𝑥°
𝑡𝑎𝑛 𝑥 −1 =
≈56.3°
Using a scientific calculator, select 𝒕𝒂𝒏 −𝟏 …

88. 𝑥 𝑡𝑎𝑛= opposite leg adjancent leg 30° 21,651 𝑓𝑡. tan 30° = 12,500 𝑥
By the alternate interior angle theorem…
𝑡𝑎𝑛= opposite leg adjancent leg
30°
21,651 𝑓𝑡. 𝑥
tan 30° = 12,500 𝑥
≈21,651 𝑓𝑡.
Use a scientific calculator and divide 12,500 by the tan 30…