discovering geometry chapter 4 test review...

Discovering Geometry Chapter 4 Test Review HGSH


1 and 2 are Vertical Angles.

1 2

3

4

3 and 4 are Alternate Interior Angles and are equal by the Parallel Line Conjecture.

S

A

A


Because of the way the triangle congruence is written A and X are corresponding angles

C and Z are corresponding angles

BC and YZ are corresponding, So: BC = YZ

8x + 3 = 7x + 5 , subtract 7x

x+3 = 5 , subtract 3 x=2

BC = (8X +3) YZ = (7X +5)


45°

Triangle Sum Conjecture B = 180° – (60° + 45°) B = 75°

120° is the exterior angle of the triangle. By the Triangle Exterior Angle Conjecture, 120° = 90° + a 120° – 90° = a 30° = a

OR, 120° and b are supplementary So, b= 180° - 120° b= 60° And, a = 180 – (90 +b) = 180 – (90-60) a = 30°

120°


50°

65°

By the Triangle Exterior Angle Conjecture x= 50° + 65° x= 115°

Another way, by the Triangle Sum Conjecture, the sum of the interior angles of a triangle equals 180°. So the third angle is: 180° – ( 65° +50°) = 65° and x + 65° = 180° since they are supplementary x= 180° – 65° x= 115°


22

29

52

By the Triangle Inequality Conjecture, the sum of two of the legs has to be greater than the third. 22cm + 29cm = 51cm and it’s not greater than 52cm. One of the two, 22cm or 29cm legs need to increase so that the sum of the two is greater than 52 cm

65° 55°

Side-Angle Inequality Conjecture 180° – ( 65° + 55°) = 60°

60°

From least to greatest: j, k, l


11 cm 11 cm

z A B

C

D E

F

13°

25°

13°

25°

A D 180° = A + 25° + 13° 180° – 38° = A 142° = A (CAB) so, by CPCTC 142° = D

Remember that CPCTC

By CPCTC, AB DE BC EF AC DF


110°

55° 45° 88° 180°

By the Triangle Exterior Angle Conjecture x + x= 110° 2x= 110° x=55°

B

C D

E T

U V

W


AD CD ADB CDB and BD is common to both triangles. Therefore we have SAS.


4 cm, 15 cm, 20 cm

5 cm, 5 cm, 10 cm 14 cm, 5 cm, 20 cm

12 cm, 11 cm, 20 cm

x=79°; y=101°

101° x=79°; y=68° x=22°; y=79°

x=22°; y=101°

Because the Triangle is an Isosceles Triangle, the base angles are equal. So, then y = 180° – 101° (Supplementary Angles) y=79° Therefore, 2y+x= 180° 2(79°) + x = 180° x = 180° – 158° = 22°


100°

By the Isosceles Triangle Conjecture, the base angles of this Isosceles triangle are equal. So 2x + 100° = 180° 2x = 80° x=40°

28

By the Converse of the Isosceles Triangle Conjecture, The Triangle is an Isosceles Triangle so the legs that include vertex P are equal, therefore, PQ is 28 mm.


20 ft.

By the Converse of the Isosceles Triangle Conjecture The Triangle is an Isosceles Triangle so the legs that include vertex G are equal, therefore, The perimeter = 20 + 6 + 6 = 32 ft.

58° 62°

Segment BC


Segment BD, the Angle Bisector of Isosceles Triangle ABC, is also the perpendicular bisector of segment AC making point D the midpoint. Since segment BD starts at vertex B and goes through midpoint D, then by definition, segment BD it is the median of triangle ABC.


By the Converse of the Isosceles Triangle Conjecture, the Triangle is an Isosceles, but since the base angles are 60°, then by the Triangle Sum Conjecture, 60 + 60 + x = 180, So x = 60, therefore making the triangle an equiangular, equilateral triangle. Therefore, 3y = 33 yards. y = 11 yds.

15 in 16 in

17 in

discovering geometry chapter 4 test review...

Documents