Discovering Geometry Chapter 4 Test Review HGSH. 1 and 2 are Vertical Angles. 1 2 3 4 3 and 4...

GEOMETRYCHAPTER 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 writtenA and X are corresponding anglesC 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 ConjectureB = 180° – (60° + 45°) B = 75°

120° is the exterior angle of the triangle.By the Triangle Exterior Angle Conjecture,120° = 90° + a120° – 90° = a30° = a

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

120°


50°

65°

By the Triangle Exterior Angle Conjecturex= 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 supplementaryx= 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 increaseso that the sum of the two is greater than 52 cm

65°55°

Side-Angle Inequality Conjecture180° – ( 65° + 55°) = 60°60°

From least to greatest:j, k, l


11 cm 11 cm

zA B

C

D E

F

13°

25°

13°

25°

A D180° = A + 25° + 13°180° – 38° = A142° = A (CAB) so, by CPCTC142° = D

Remember that CPCTC

By CPCTC,AB DEBC EFAC DF


110°

55° 45° 88° 180°

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

B

C D

E T

U V

W


AD CD ADB CDBand 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 triangleare 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 legsthat include vertex P are equal, therefore, PQ is 28 mm.


20 ft.

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

58°62°

mC = 180- (62+58) = 60Segment BC


Segment BD, the Angle Bisector of Isosceles Triangle ABC,is also the perpendicular bisector of segment AC makingpoint D the midpoint. Since segment BD starts at vertex Band goes through midpoint D, then by definition, segmentBD 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 HGSH. 1 and 2 are Vertical Angles. 1 2 3 4 3 and 4...

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