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# March09 March13

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### March09 March13

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