Geometry: Answer Key

Answer Key

This provides the answers and solutions for the Put Me in, Coach! exercise boxes, organized by sections.

Taking the Burden out of Proofs

  1. Yes
  2. Theorem 8.3: If two angles are complementary to the same angle, then these two angles are congruent.

A and B are complementary, and C and B are complementary.

Given: A and B are complementary, and C and B are complementary.

Prove: A ~= C.

 StatementsReasons
1.A and B are complementary, and C and B are complementary.Given
2.mA + mB = 90º , mC + mB = 90ºDefinition of complementary
3.mA = 90 º - mB, mC = 90º - mBSubtraction property of equality
4.mA = mCSubstitution (step 3)
5.A ~= CDefinition of ~=

Proving Segment and Angle Relationships

  1. If E is between D and F, then DE = DF EF.
E is between D and F.

E is between D and F.

Given: E is between D and F

Prove: DE = DF EF.

 StatementsReasons
1.E is between D and FGiven
2.D, E, and F are collinear points, and E is on ¯DFDefinition of between
3.DE + EF = DFSegment Addition Postulate
4.DE = DF EFSubtraction property of equality

2. If BD divides ABC into two angles, ABD and DBC, then mABC = mABC - mDBC.

BD divides ABC into two angles, ABD and DBC.

Given: BD divides ABC into two angles, ABD and DBC

Prove: mABD = mABC - mDBC.

 StatementsReasons
1.BD divides ABC into two angles, ABD and DBCGiven
2.mABD + mDBC = mABCAngle Addition Postulate
3.mABD = mABC - mDBCSubtraction property of equality

3. The angle bisector of an angle is unique.

ABC with two angle bisectors: BD and BE.

Given: ABC with two angle bisectors: BD and BE.

Prove: mDBC = 0.

 StatementsReasons
1.BD and BE bisect ABCGiven
2.ABC ~= DBC and ABE ~= EBCDefinition of angel bisector
3.mABD = mDBC and mABE ~= mEBCDefinition of ~=
4.mABD + mDBE + mEBC = mABCAngle Addition Postulate
5.mABD + mDBC = mABC and mABE + mEBC = mABCAngle Addition Postulate
6.2mABD = mABC and 2mEBC = mABCSubstitution (steps 3 and 5)
7.mABD = mABC/2 and mEBC = mABC/2Algebra
8.mABC/2 + mDBE + mABC/2 = mABCSubstitution (steps 4 and 7)
9.mABC + mDBE = mABCAlgebra
10.mDBE = 0Subtraction property of equality

4. The supplement of a right angle is a right angle.

A and B are supplementary angles, and A is a right angle.

Given: A and B are supplementary angles, and A is a right angle.

Prove: B is a right angle.

 SatementsReasons
1.A and B are supplementary angles, and A is a right angleGiven
2.mA + mB = 180ºDefinition of supplementary angles
3.mA = 90ºDefinition of right angle
4.90º + mB = 180ºSubstitution (steps 2 and 3)
5.mB = 90ºAlgebra
6.B is a right angleDefinition of right angle

Proving Relationships Between Lines

  1. m6 = 105º , m8 = 75º
  2. Theorem 10.3: If two parallel lines are cut by a transversal, then the alternate exterior angles are congruent.

l m cut by a transversal t.

Given: l m cut by a transversal t.

Prove: 1 ~= 3.

 StatementsReasons
1.l m cut by a transversal tGiven
2.1 and 2 are vertical anglesDefinition of vertical angles
3.2 and 3 are corresponding anglesDefinition of corresponding angles
4.2 ~= 3Postulate 10.1
5.1 ~= 2Theorem 8.1
6.1 ~= 3Transitive property of 3.

3. Theorem 10.5: If two parallel lines are cut by a transversal, then the exterior angles on the same side of the transversal are supplementary angles.

l m cut by a transversal t.

Given: l m cut by a transversal t.

Prove: 1 and 3 are supplementary.

 StatementReasons
1.l m cut by a transversal tGiven
2.1 and 2 are supplementary angles, and m1 + m2 = 180ºDefinition of supplementary angles
3.2 and 3 are corresponding anglesDefinition of corresponding angles
4.2 ~= 3Postulate 10.1
5.m2 ~= m3Definition of ~=
6.m1 + m3 = 180ºSubstitution (steps 2 and 5)
7.1 and 3 are supplementaryDefinition of supplementary

 

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4. Theorem 10.9: If two lines are cut by a transversal so that the alternate exterior angles are congruent, then these lines are parallel.

Lines l and m are cut by a transversal t.

Lines l and m are cut by a transversal t.

Given: Lines l and m are cut by a transversal t, with 1 ~= 3.

Prove: l m.

 StatementReasons
1.Lines l and m are cut by a transversal t, with 1 ~= 3Given
2.1 and 2 are vertical anglesDefinition of vertical angles
3.1 ~= 2Theorem 8.1
4.2 ~= 3Transitive property of ~=.
5.2 and 3 are corresponding anglesDefinition of corresponding angles
6.l mTheorem 10.7

5. Theorem 10.11: If two lines are cut by a transversal so that the exterior angles on the same side of the transversal are supplementary, then these lines are parallel.

Lines l and m are cut by a t transversal t.

Lines l and m are cut by a t transversal t.

Given: Lines l and m are cut by a transversal t, 1 and 3 are supplementary angles.

Prove: l m.

 StatementReasons
1.Lines l and m are cut by a transversal t, and 1 are 3 supplementary anglesGiven
2.2 and 1 are supplementary anglesDefinition of supplementary angles
3.3 ~= 2Example 2
4.3 and 2 are corresponding anglesDefinition of corresponding angles
5.l mTheorem 10.7

Two's Company. Three's a Triangle

  1. An isosceles obtuse triangle
  2. The acute angles of a right triangle are complementary.

ABC is a right triangle.

Given: ABC is a right triangle, and B is a right angle.

Prove: A and C are complementary angles.

 StatementReasons
1.ABC is a right triangle, and B is a right angleGiven
2.mB = 90ºDefinition of right angle
3.mA + mB + mC = 180ºTheorem 11.1
4.mA + 90º + mC = 180ºSubstitution (steps 2 and 3)
5.mA + mC = 90ºAlgebra
6.A and C are complementary anglesDefinition of complementary angles

3. Theorem 11.3: The measure of an exterior angle of a triangle equals the sum of the measures of the two nonadjacent interior angles.

ABC with exterior angle BCD.

 StatementReasons
1.ABC with exterior angle BCDGiven
2.DCA is a straight angle, and mDCA = 180ºDefinition of straight angle
3.mBCA + mBCD = mDCAAngle Addition Postulate
4.mBCA + mBCD = 180ºSubstitution (steps 2 and 3)
5.mBAC + mABC + mBCA = 180ºTheorem 11.1
6.mBAC + mABC + mBCA = mBCA mBCDSubstitution (steps 4 and 5)
7.mBAC + mABC = mBCDSubtraction property of equality

4. 12 units2

5. 30 units2

6. No, a triangle with these side lengths would violate the triangle inequality.

Congruent Triangles

1. Reflexive property: ABC ~= ABC.

Symmetric property: If ABC ~= DEF, then DEF ~= ABC.

Transitive property: If ABC ~= DEF and DEF ~= RST, then ABC ~= RST.

2. Proof: If ¯AC ~= ¯CD and ACB ~= DCB as shown in Figure 12.5, then ACB ~= DCB.

 StatementReasons
1.¯AC ~= ¯CD and ACB ~= DCBGiven
2.¯BC ~= ¯BCReflexive property of ~=
3.ACB ~= DCBSAS Postulate

3. If ¯CB ¯AD and ACB ~= DCB, as shown in Figure 12.8, then ACB ~= DCB.

 StatementReasons
1.¯CB ¯AD and ACB ~= DCBGiven
2.ABC and DBC are right anglesDefinition of
3.mABC = 90º and mDBC = 90ºDefinition of right angles
4.mABC = mDBCSubstitution (step 3)
5.ABC ~= DBCDefinition of ~=
6.¯BC ~= ¯BCReflexive property of ~=
7.ACB ~= DCBASA Postulate

4. If ¯CB ¯AD and CAB ~= CDB, as shown in Figure 12.10, then ACB~= DCB.

 StatementReasons
1.¯CB ¯AD and CAB ~= CDBGiven
2.ABC and DBC are right anglesDefinition of
3.mABC = 90º and mDBC = 90ºDefinition of right angles
4.mABC = mDBCSubstitution (step 3)
5.ABC ~= DBCDefinition of ~=
6.¯BC ~= ¯BCReflexive property of ~=
7.ACB ~= DCBAAS Theorem

5. If ¯CB ¯AD and ¯AC ~= ¯CD, as shown in Figure 12.12, then ACB ~= DCB.

 StatementReasons
1.¯CB ¯AD and ¯AC ~= ¯CDGiven
2.ABC and DBC are right trianglesDefinition of right triangle
3.¯BC ~= ¯BCReflexive property of ~=
4.ACB ~= DCBHL Theorem for right triangles

6. If P ~= R and M is the midpoint of ¯PR, as shown in Figure 12.17, then N ~= Q.

 StatementReasons
1.P ~= R and M is the midpoint of ¯PRGiven
2.¯PM ~= ¯MRDefinition of midpoint
3.NMP and RMQ are vertical anglesDefinition of vertical angles
4.NMP ~= RMQTheorem 8.1
5.PMN ~= RMQASA Postulate
6.N ~= QCPOCTAC

Smiliar Triangles

  1. x = 11
  2. x = 12
  3. 40º and 140º
  4. If A ~= D as shown in Figure 13.6, then BC/AB = CE/DE.
 StatementReasons
1.A ~= DGiven
2.BCA and DCE are vertical anglesDefinition of vertical angles
3.BCA ~= DCETheorem 8.1
4.ACB ~ DCEAA Similarity Theorem
5.BC/AB = CE/DECSSTAP

5. 150 feet.

Opening Doors with Similar Triangles

  1. If a line is parallel to one side of a triangle and passes through the midpoint of a second side, then it will pass through the midpoint of the third side.

¯DE ¯AC and D is the midpoint of ¯AB.

Given: ¯DE ¯AC and D is the midpoint of ¯AB.

Prove: E is the midpoint of ¯BC.

 StatementReasons
1.¯DE ¯AC and D is the midpoint of ¯AB.Given
2.¯DE ¯AC and is cut by transversal ABDefinition of transversal
3.BDE and BAC are corresponding anglesDefinition of corresponding angles
4.BDE ~= BACPostulate 10.1
5.B ~= BReflexive property of ~=
6.ABC ~ DBEAA Similarity Theorem
7.DB/AB = BE/BCCSSTAP
8.DB = AB/2Theorem 9.1
9.DB/AB = 1/2Algebra
10.1/2 = BE/BCSubstitution (steps 7 and 9)
11.BC = 2BEAlgebra
12.BE + EC = BCSegment Addition Postulate
13.BE + EC = 2BESubstitution (steps 11 and 12)
14.EC = BEAlgebra
15.E is the midpoint of ¯BCDefinition of midpoint

2. AC = 43 , AB = 8 , RS = 16, RT = 83

3. AC = 42 , BC = 42

Putting Quadrilaterals in the Forefront

  1. AD = 63, BC = 27, RS = 45
  2. ¯AX, ¯CZ, and ¯DY
Trapezoid ABCD with its XB CY four altitudes shown.

Trapezoid ABCD with its XB CY four altitudes shown.

3. Theorem 15.5: In a kite, one pair of opposite angles is congruent.

Kite ABCD.

Kite ABCD.

Given: Kite ABCD.

Prove: B ~= D.

 StatementReasons
1.ABCD is a kiteGiven
2.¯AB ~= ¯AD and ¯BC ~= ¯DCDefinition of a kite
3.¯AC ~= ¯ACReflexive property of ~=
4.ABC ~= ADCSSS Postulate
5.B ~= DCPOCTAC

4. Theorem 15.6: The diagonals of a kite are perpendicular, and the diagonal opposite the congruent angles bisects the other diagonal.

Kite ABCD.<>Kite ABCD.

Given: Kite ABCD.

Prove: ¯BD ¯AC and ¯BM ~= ¯MD.

 StatementReasons
1.ABCD is a kiteGiven
2.¯AB ~= ¯AD and ¯BC ~= ¯DCDefinition of a kite
3.¯AC ~= ¯ACReflexive property of ~=
4.ABC ~= ADCSSS Postulate
5.BAC ~= DACCPOCTAC
6.¯AM ~= ¯AMReflexive property of ~=
7.ABM ~= ADMSAS Postulate
8.¯BM ~= ¯MDCPOCTAC
9.BMA ~= DMACPOCTAC
10.mBMA = mDMADefinition of ~=
11.MBD is a straight angle, and mBMD = 180ºDefinition of straight angle
12.mBMA + mDMA = mBMDAngle Addition Postulate
13.mBMA + mDMA = 180ºSubstitution (steps 9 and 10)
14.2mBMA = 180ºSubstitution (steps 9 and 12)
15.mBMA = 90ºAlgebra
16.BMA is a right angleDefinition of right angle
17.¯BD ¯ACDefinition of

5. Theorem 15.9: Opposite angles of a parallelogram are congruent.

Parallelogram ABCD.

Given: Parallelogram ABCD.

Prove: ABC ~= ADC.

 StatementReasons
1.Parallelogram ABCD has diagonal ¯AC.Given
2.ABC ~= CDATheorem 15.7
3.ABC ~= ADCCPOCTAC

6. 144 units2

7. 180 units2

8. Kite ABCD has area 48 units2.

Parallelogram ABCD has area 150 units2.

Rectangle ABCD has area 104 units2.

Rhombus ABCD has area 35/2 units2.

Anatomy of a Circle

  1. Circumference: 20 feet, length of ˆRST = 155/18 feet
  2. 9 feet2
  3. 15 feet2
  4. 28º

The Unit Circle and Trigonometry

  1. 3/34 = 334/34
  2. 1/3 = 3/3
  3. tangent ratio = 40/3, sine ratio = 40/7
  4. tangent ratio = 5/56 = 556/56, cosine ratio = 56/9

Excerpted from The Complete Idiot's Guide to Geometry © 2004 by Denise Szecsei, Ph.D.. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc.

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