March02 March06

Math Tutorial Questions For the week of March 02 – March 06 ***No Lesson for March 3 (ELA TAKS Testing)

Classifying Triangles—Application Questions (4.1.4) – March 02, 2009 ,[object Object],[object Object]

Classifying Triangles—Application Solutions (4.1.4) – March 02, 2009 ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],18 ft 18 ft 18 ft 2 cm 2 cm 2 cm

Angle Relationships in Triangles—Application Questions (4.2.1) – March 04, 2009 ,[object Object],[object Object],[object Object],[object Object]

Angle Relationships in Triangles—Application Solutions (4.2.1) – March 04, 2009 ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],78º ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Finding Angle Measures in Right Triangles Questions (4.2.2) – March 05, 2009 ,[object Object],[object Object],[object Object],[object Object]

Finding Angle Measures in Right Triangles Solutions (4.2.2) – March 05, 2009 ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Applying the Exterior Angle Theorem Questions (4.2.3) – March 06, 2009 ,[object Object],[object Object]

Applying the Exterior Angle Theorem Solutions (4.2.3) – March 06, 2009 ,[object Object],[object Object],15º + (2x + 3)º = (5x – 60)º (Exterior Angle Theorem) 2x + 18 = 5x – 60 (Add 15 & 3) 18 = 3x – 60 (Subtract 2x from both sides of equation) 78 = 3x (Add 60 to both sides of equation) x = 26 (Divide both sides of equation by 3) Angle B = 2x + 3 Angle B = 2(26) + 3 (Substitute 26 in for x) Angle B = 52 + 3 (Multiply 2 & 26) Angle B = 58º (Add 52 & 3) 58º + (x + 12)º = (2x + 16)º (Exterior Angle Theorem) x + 70 = 2x + 16 (Add 58 & 12) 70 = x + 16 (Subtract x from both sides of equation) 54 = x (Subtract 16 from both sides of equation) Angle ABD = 2x + 16 Angle ABD = 2(54) + 16 (Substitute 54 in for x) Angle ABD = 108 + 16 (Multiply 2 & 54) Angle ABD = 124º (Add 108 & 16)

March02 March06

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