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# March09 March13

## on Feb 25, 2009

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## March09 March13Presentation Transcript

• Math Tutorial Questions For the week of March 09 – March 13
• Naming Congruent Corresponding Parts Questions (4.3.1) – March 09, 2009
• Given that Δ PQR is congruent to Δ STW. Identify all pairs of congruent corresponding parts.
• Angles
• Sides
• Given that polygon MNOP is congruent to polygon QRST, identify the congruent corresponding part.
• Segment NO is congruent to segment _____.
• Angle T is congruent to angle _____.
• Naming Congruent Corresponding Parts Solutions (4.3.1) – March 09, 2009
• Given that Δ PQR is congruent to Δ STW. Identify all pairs of congruent corresponding parts.
• Angles:
• Sides:
• Given that polygon MNOP is congruent to polygon QRST, identify the congruent corresponding part.
• Segment NO is congruent to segment RS .
• Angle T is congruent to angle P .
• Using Corresponding Parts of Congruent Triangles Questions (4.3.2) – March 10, 2009
• Δ ABC is congruent to Δ JKL. AB = 2x + 12. JK = 4x – 50. Find the following values.
• x
• AB
• Δ ABC is congruent to Δ DBC. Use the diagram to find the following values.
• x
• Angle DBC
• Using Corresponding Parts of Congruent Triangles Solutions (4.3.2) – March 10, 2009
• Δ ABC is congruent to Δ JKL. AB = 2x + 12. JK = 4x – 50. Find the following values.
• Δ ABC is congruent to Δ DBC. Use the diagram to find the following values.
• x
• AB = JK (Property of Congruent Polygon)
• 2x + 12 = 4x – 50 (Substitute equations in for AB & JK)
• 2x = 4x – 62 (Subtract 12 from both sides of equation)
• -2x = - 62 (Subtract 4x from both sides of equation)
• x = 31 (Divide both sides of equation by -2)
B. AB AB = 2(31) + 12 (Substitute 31 in for x) AB = 62 + 12 (Multiply 2 & 31) AB = 74 (Add 62 & 12)
• x
• Angle BCA = Angle BCD (Property of Congruent Polygon)
• 2x – 16 = 90 (Substitute values in for angles above)
• 2x = 106 (Add 16 to both sides of equation)
• x = 53 (Divide both sides of equation by 2)
B. Angle DBC Angle BDC + Angle DCB + Angle DBC = 180 (Triangle Sum Theorem) 49.3 + 90 + Angle DBC = 180 (Substitute values in for angles) 139.3 + Angle DBC = 180 (Add up 49.3 & 90) Angle DBC = 40.7º (Subtract 139.3 from both sides of equation)
• Verifying Triangle Congruence Questions (4.4.3) – March 11, 2009
• Show that the triangles are congruent for the given value of the variable.
• Δ MNO is congruent to Δ PQR, when x = 5.
• Δ STU is congruent to Δ VWX, when y = 4.
• Verifying Triangle Congruence Solutions (4.4.3) – March 11, 2009
• Show that the triangles are congruent for the given value of the variable.
• Δ MNO is congruent to Δ PQR, when x = 5.
• QR = x = 5
• PQ = x + 2 = 5 + 2 = 7
• RP = 3x – 9 = 3(5) – 9 = 15 – 9 = 6
• The 3 sides of Δ MNO are congruent to the 3 sides of Δ PQR, therefore the triangles are congruent by Side-Side-Side Postulate.
• Δ STU is congruent to Δ VWX, when y = 4.
• TU = y + 3 = 4 + 3 = 7
• ST = 2y + 3 = 2(4) + 3 = 8 + 3 =11
• Angle T = 20y + 12 = 20(4) + 12 = 80 + 12 = 92º
• 2 sides and an included angle in Δ STU are congruent to 2 sides and an included angle in Δ VWX, therefore the triangles are congruent by Side-Angle-Side Postulate.
• Finding the Measure of an Angle Questions (4.8.2) – March 12, 2009
• Find the measure of angle F.
• Find the measure of angle G.
• Finding the Measure of an Angle Solutions (4.8.2) – March 12, 2009
• Find the measure of angle G.
• Angle J = Angle G (Isosceles Triangle Theorem)
• 3x = x + 44 (Substitute values in for angles J & G)
• 2x = 44 (Subtract x from both sides of equation)
• x = 22 (Divide both sides of equation by 2)
• Angle G = x + 44 = 22 + 44 = 66
• Angle G = 66º
• Find the measure of angle F.
• Angle D = Angle F (Isosceles Triangle Theorem)
• Angle D + Angle E + Angle F = 180º (Triangle Sum Theorem)
• F + 22 + F = 180 (Substitute angle F in for angle D)
• 2F + 22 = 180 (Combine the 2 F’s)
• 2F = 158 (Subtract 22 from both sides of equation)
• F = 79 (Divide both sides of equation by 2)
• Angle F = 79º
• Using Properties of Equilateral Triangles Questions (4.8.3) – March 13, 2009
• Find the value of x.
• Find the value of y and the measure of each side of the triangle.
• Using Properties of Equilateral Triangles Solutions (4.8.3) – March 13, 2009
• Find the value of x.
• Angle L = Angle M = Angle K (Equilateral Triangle Corollary)
• Angle L + Angle M + Angle K = 180º (Triangle Sum Theorem)
• Since all angles are equal each angle is equal to 60º
• 2x + 32 = 60 (Set angle L equal to 60)
• 2x = 28 (Subtract 32 from both sides of equation)
• x = 14 (Divide both sides of equation by 2)
• Find the value of y and the measure of each side of the triangle.
• NP = ON = OP (Equiangular Triangle Corollary)
• 5y – 6 = 4y + 12 (Substitute values in for ON & OP)
• y – 6 = 12 (Subtract 4y from both sides of equation)
• y = 18 (Add 6 to both sides of equation)
• NP = ON = OP = 4y + 12 = 4(18) + 12 = 72 + 12 = 84