March 23-March 27

Math Tutorial Questions For the week of March 23 – March 27

Applying the Perpendicular Bisector Theorems Questions (5.1.1) – March 23, 2009 ,[object Object],[object Object],[object Object]

Applying the Perpendicular Bisector Theorems Solutions (5.1.1) – March 23, 2009 ,[object Object],[object Object],[object Object],MN = LN (Perpendicular Bisector Theorem) MN = 2.6 (Substitute 2.6 in for LN) BD = DC (Converse of Perpendicular Bisector Theorem) BD = 12 (Substitute 12 in for DC) BC = BD + BC (Segment Addition Postulate) BC = 12 + 12 (Substitute 12 in for BD & DC) MN = 24 (Add 12 & 12) UV = UT (Perpendicular Bisector Theorem) 7x - 17 = 3x + 9 (Substitute equations in for UT & UV) 4x - 17 = 9 (Subtract 3x from both sides of equation) 4x = 26 (Add 17 to both sides of equation) x = 6.5 (Divide both sides of equation by 4) UT = 3x + 9 UT = 3(6.5) + 9 (Substitute 6.5 in for x) UT = 19.5 + 9 (Multiply 3 & 6.5) UT = 28.5 (Add 19.5 & 9)

Applying the Angle Bisector Theorems Questions (5.1.2) – March 24, 2009 ,[object Object],[object Object]

Applying the Angle Bisector Theorems Solutions (5.1.2) – March 24, 2009 ,[object Object],[object Object],BC = CD (Angle Bisector Theorem) BC = 7.2 (Substitute 7.2 in for CD) Angle EFH = Angle HFG (Converse of Angle Bisector Theorem) Angle EFH + Angle HFG = Angle EFG (Angle Addition Postulate) Angle EFH + Angle HFG = 50 (Substitute in 50 for Angle EFG) Angle EFH + Angle EFH = 50 (Substitute in Angle EFH for Angle HFG) 2(Angle EFH) = 50 (Combine Angles EFH) Angle EFH = 25º (Divide both sides of equation by 2)

Using the Pythagorean Theorem Questions (5.7.1) – March 25, 2009 Pythagorean Theorem: a 2 + b 2 = c 2 ,[object Object],[object Object],a b c ,[object Object],[object Object],Pythagorean Theorem: a 2 + b 2 = c 2 ,[object Object],[object Object],Pythagorean Theorem: a 2 + b 2 = c 2 a b c a b

Using the Pythagorean Theorem Solutions (5.7.1) – March 25, 2009 ,[object Object],[object Object],a 2 + b 2 = c 2 (Pythagorean Theorem) 2 2 + 6 2 = x 2 (Substitute values in for a, b, & c) 4 + 36 = x 2 (Square 2 and 6) 40 = x 2 (Add 4 and 36) x = √40 ≈ 6.32 (Take the square root of 40 to solve for x) a 2 + b 2 = c 2 (Pythagorean Theorem) (x – 2) 2 + 4 2 = x 2 (Substitute values in for a, b, & c) x 2 – 4x + 4 + 16 = x 2 (Square (x – 2) & 4) x 2 – 4x + 20 = x 2 (Add 4 & 16) -4x + 20 = 0 (Subtract x 2 from both sides of equation) -4x = -20 (Subtract 20 from both sides of equation) x = 5 (Divide both sides of equation by -4)

Pythagorean Theorem Application Questions (5.7.2) – March 26, 2009 ,[object Object],[object Object],Pythagorean Theorem: a 2 + b 2 = c 2 ,[object Object],[object Object],c a b

Pythagorean Theorem Application Solutions (5.7.2) – March 26, 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],[object Object],[object Object],L W 12 52 inches 40 inches 60 inches ?

Classifying Triangles Questions (5.7.4) – March 27, 2009 ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Classifying Triangles Solutions (5.7.4) – March 27, 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],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

March 23-March 27

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March 23-March 27