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Transcript
Geometry Semester 1 Final Review packet continued Geometry Review Semester 1 1. A proof is given (3) I have split the statements into a statement and then a reason in a separate column, as we have been doing. Notice we have switched to congruent triangles in the following problems. 1. ABCD is a parallelogram Prove: AB = DC and AD =Reason BC Statement: 1. AB||DC and AD||BC 1. Definition of a parallelogram 2. Angle b is congruent to angle d and angle a is congruent to angles c 2. For parallel lines cut by a transversal, alternate interior angles are congruent 3. DB is congruent to DB 3. Reflexive property 4. Triangle DAB is congruent to Triangle BCD 4. ASA congruence postulate 5. AB congruent to DC and AD congruent to BC 5. Corresponding parts of congruent triangles are congruent The reason for step 5 of the proof was That corresponding parts of congruent triangles are congruent. 2. What is the flaw in the proof? Angle Angle Angle is a triangle similarity postulate - not a triangle congruence postulate. Remember we noticed that for similar triangles corresponding angles were congruent, but corresponding sides were proportional, but could be different size triangles. For congruent triangles, corresponding sides have to be congruent. Knowing all angles are congruent, is not enough to prove triangles are congruent. 3. Draw out the information given. N E D F M O 3. We can see that if the angles at E and N are congruent then we will have SAS congruence N E D F M O Answer D) Angle E is congruent to angle N. 4. Which triangle congruence theorem should be used... Label the diagram with the given information. A C B D 4. Which triangle congruence theorem should be used... BD is congruent to itself (reflexive property) A C B D SSS congruence postulate applies 5. Complete the 4th line of the proof (Diagram needs an F where angle 2 is.) 1. Given 2. Perpendicular lines meet to form right angles. 3. Right angles are congruent. 4. 5. Corresponding parts of congruent triangles are congruent. Statement: 4. Triangle ABC is congruent to triangle EFD Reason: 4. Triangle ACB is congruent to triangle CFD ASA Congruence postulate Geometry: Semester 1 Review (4) 1) If triangle BDF is similar to triangle CEG, what is the ratio from triangle BDF to triangle CEG? BD = 8 CE = 10 corresponding sides DF = 4 EG = 5 corresponding sides Ratio 4 : 5 Geometry: Semester 1 Review (4) 1) If triangle BDF is similar to triangle CEG, what is the ratio from triangle BDF to triangle CEG? BD = 8 CE = 10 corresponding sides DF = 4 EG = 5 corresponding sides Ratio 4 : 5 (Scale factor 5/4, check 6 x 5/4 = 7.5, yes). Rectangle - 90 degree vertices, opposite sides are congruent 2) 100° 40° 40° AAS congruence 3. Mark the information on the diagram. E D * C W X* Y 3. If we know XY congruent to DC, then we have SAS congruence E D * C W X* Or if we know that the angle at E is congruent to the angle at W, then we have ASA congruence Y 4. Diagonals of a quadrilateral are congruent and bisect each other. Review page 62 in your Unit 2 workbook 4. Diagonals of a quadrilateral are congruent and bisect each other. P 62 Unit 2 workbook 4. Diagonals of a quadrilateral are congruent and bisect each other. P 62 Unit 2 workbook 4. Diagonals of a quadrilateral are congruent and bisect each other. Best description of the quadrilateral is a rectangle. Question 5 is the same as qu. 3 on sheet (3) Geometry Review Semester 1 (5) 1. Find geometric mean. 1a 5=x x 15 x2 = 75 = 3(25) x = √3√25 = 5√3 Geometry Review Semester 1 (5) geometric mean 1b 5=x x 45 x2 = 225 x = √225 = 15 Geometry Review Semester 1 (5) 1c 9=x x 4 x2 = 36 x = √36 = 6 Geometry Review Semester 1 (5) 1d 8=x x 10 x2 = 80 = 16(5) x = √16√5= 4√5 2. Determine x, y and z. Exact answers. y is the geometric mean of 4 and 9 (altitude of the large right triangle) 4cm = y y 9cm y2 = 36cm2 y = √36cm2 = 6cm 2. Determine x, y and z. Exact answers. Using Pythagorean theorem x2 = y2 + (4cm)2 x2 = (6cm)2 + (4cm)2 x2 = 36cm2 +16cm2 = 52cm2 x = √52cm2 = √4√13cm2 = 2√13cm 2. Determine x, y and z. Exact answers. Using Pythagorean theorem 132 = 2√132 + z2 169 = 52 + z2 z2 = 169 - 52 =117= 9(13) z = √9(13) = 3√13 3. Determine x, y and z. Exact answers. y is the geometric mean of 3 and 8 (altitude of the large right triangle) 3cm = y y 8cm y2 = 32cm2 y = √2(16)cm2 = 4√2cm 3. Determine x, y and z. Exact answers. Using Pythagorean theorem x2 = (4√2)2 + (3cm)2 x2 = (32 + 9)cm2 x = √41cm 3. Determine x, y and z. Exact answers. Using Pythagorean theorem z2 = (4√2)2 + (8cm)2 z2 = (32 + 64)cm2 z = √98cm2 = √2(49)cm2 = 7√2cm 4. Determine x, y and z. Exact answers. Using Pythagorean theorem 152 = 92 + x2 225 - 81 = x2 x = √144 = 12 4. Determine x, y and z. Exact answers. 12 is the geometric mean of 9 and y (altitude of the large right triangle) 9 = 12 12 y y = 12(12) = 144 = 16 9 9 4. Determine x, y and z. Exact answers. Using Pythagorean theorem z2 = (12)2 + (16)2 z2 = (144 + 256)cm2 z = √400 = 20 Extra work on Dilations Unit 3A Workbook, page 17 Dilations (pg 17) 1. On Graph A plot triangle ABC with A(2,1), B(3, 2) and C (1,3) 2. On the SAME graph, Plot D-3 as triangle A’B’C’ A(2,1)(-3) maps to A’(-6, -3) B(3, 2)(-3) maps to B’ C’(1, 3)(-3) maps to C’ Dilations (pg 17) 1. On Graph A plot triangle ABC with A(2,1), B(3, 2) and C (1,3) 2. On the SAME graph, Plot D(-3, O)as triangle A’B’C’ A(2,1)(-3) maps to A’(-6, -3) B(3, 2)(-3) maps to B’ (-9, -6) C’(1, 3)(-3) maps to C’ (-3, -9) Dilations cont. 3.Looking at Graph A and your two triangles we can see that angle ABC is congruent to angle A’B’C’ angle BCA is congruent to angle B’C’A’ angle CAB is congruent to angle C’A’B’ Dilations cont. 4.Find the length of line segment AB 2 2 2 AB =(1 +1 ) = 2 So length of AB = √2 Count squares - it’s easier! 5.Find points A’B’. Find the length of line segment A’B’ 2 2 2 A’B’ = {(-9 - -6) +(-6 - -3) } = (32 + 32) = (9 + 9) = 18 A’B’ = √ (2)(9) = 3√ 2 A’B’= √ (2)(9) = 3√ 2 6. How do the lengths of AB and A’B’ relate to our original dilation? The original dilation was scale factor -3, and we see the length of AB multiplied by 3 gives us the length of A’B’ ; 3AB = A’B’