REVIEW 4

Congruence, Parallelograms, and Polygons

1.   At least one pair of congruent sides There is a great deal of information about congruent angles. Because , alternate interior angles are congruent, so ∠A ≅ ∠C and ∠B ≅ ∠D. In addition, ∠AOB ≅ ∠COD because vertical angles are congruent. But in order to prove the triangles congruent, it is necessary to have at least one pair of congruent sides.

2.   Given that bisects ∠WZY, it is possible to conclude that ∠WZX ≅ ∠YZX. Given bisects ∠WXY, it is possible to conclude that ∠WXZ ≅ ∠YXZ. and ΔWZX ≅ ΔYZX by ASA.

3.   If , as given, the ∠A ≅ ∠C by Alternate Interior Angle theorem, and in addition, ∠AOB ≅ ∠COD because vertical angles are congruent. Given that O is the midpoint of , , and ΔAOB ≅ ΔCOD by ASA.

4.    bisects ∠PQR is given and implies ∠PRT ≅ ∠QRT by the definition of bisector. ∠PRQ ≅ ∠QTR is also given and . ΔPTR ≅ ΔQTR by ASA, and by CPCTC.

5.   Given that and ∠ABC ≅ ∠DCB, it is necessary only to add (reflexive) and ΔABC ≅ ΔDCB by SAS. By CPCTC, and ∠A ≅ ∠D. Because vertical angles are congruent, ∠BEA ≅ ∠CED, and then ΔBEA ≅ ΔCED by AAS. Finally, by CPCTC.

6.   

ΔABC is isosceles with and is the perpendicular bisector of . ∠ADB and ∠CDB are right angles formed by the perpendiculars, and because all right angles are congruent, ∠ADB ≅ ∠CDB. by the definition of bisector. ΔABD ≅ ΔCDB by SAS, and ∠ABD ≅ ∠CDB by CPCTC. By the definition of angle bisector, bisects ∠ABC.

7.   Rectangle and square

8.   Rhombus and square

9.   Parallelogram, rectangle, rhombus, square

10.   Rhombus and square

11.   Trapezoid, isosceles trapezoid, parallelogram, rectangle, rhombus, square

12.   If M is the midpoint of and the midpoint of , then diagonals and bisect each other, and ABCD is a parallelogram. In a parallelogram, opposite sides are congruent, so . Alternately, and by the definition of midpoint and ∠AMD ≅ ∠BMC because vertical angles are congruent. ΔAMD ≅ ΔBMC by SAS and by CPCTC.

13.   10   In a parallelogram, opposite angles are congruent and consecutive angles are supplementary. Use those facts to write two equations, each involving x and y. Solve the system to find the values of x and y.

14.   In a parallelogram, opposite angles are congruent. If ABCD is a parallelogram, ∠A ≅ ∠C. If AEFG is a parallelogram, ∠A ≅ ∠F. By transitivity, ∠C ≅ ∠F. ∠C and ∠D are consecutive angles of a parallelogram, and so they are supplementary, which means m∠C + m∠D = 180°. ∠C ≅ ∠F, so m∠C = m∠F and by substituting, m∠F + m∠D = 180°. ∠F and ∠D are supplementary.

15.   

First, verify that PQRS is a parallelogram by checking that opposite sides are parallel. Find the slopes and show that opposite sides have the same slope. For For , For , For ,

Opposite sides have the same slope, and so are parallel, therefore PQRS is a parallelogram. To show that PQRS is a rectangle, show that adjacent sides are perpendicular. It is only necessary to show that there is one right angle.

For For ,

Because the slopes of and multiply to so ∠PQR is a right angle. Because it is a parallelogram with a right angle, PQRS is a rectangle.

16.   10   sides Solve to find 144n = 180(n − 2) and n = 10. The polygon has 10 sides.

17.   128.6°   In a regular heptagon, each interior angle measures

18.   2,340°   In a polygon with 15 sides, the total of the interior angles is 180°(n − 2) = 13(180) = 2,340°.

19.   

20.   374.1 square centimeters   If the radius of the circle is 12 cm, the radius of the hexagon is 12 cm, the side of the hexagon is 12 cm, divided into 2 segments of 6 cm by the apothem.

Because the right triangle is a 30°-60°-90° right triangle (or by using the Pythagorean theorem) the apothem is 6cm. square centimeters.

CHAPTER 9

Similarity

EXERCISE 9.1

1.   32.2 cm 6 cm × 0.7 = 4.2 cm. 17 cm × 0.7 = 11.9 cm. P = 2(4.2) + 2(11.9) = 8.4 + 23.8 = 32.2 cm.

2.   R(3, −4)→R′(6, −8). S(5, −1)→S′(10, −20. T(2, 4)→T′(4, 8).

3.   A(−2, 2)→A′(−3, 3). B(4, 2)→B′(6, 3). C(4, −4)→C′(6, −6). D(−2, −4)→D′(−3, −6).

4.    To carry A′B′C′D′ back onto ABCD use a scale factor of k, where k(−3) = −2, so .

5.   

6.   

7.   A negative scale factor would represent a dilation combined with a rotation of 180° about the origin.

EXERCISE 9.2

1.   If ΔXYZ ~ ΔRTS, then ∠X ≅ ∠R, ∠Y ≅ ∠T and ∠Z ≅ ∠S. corresponds to , to , and to .

2.   The two statements involve the same triangles, but establish different correspondences. In the first, X corresponds to R, Y to T and Z to S. In the second X again corresponds to R, but Y corresponds to S, and Z to T.

3.   If m∠A = 43° and m∠B = 112°, then m∠C = 180° − (43° + 112°) = 25°. If m∠X = 25° and m∠Y = 43°, then m∠Z = 180° − (25° + 43°) = 112°. ΔABC ~ ΔYZX

4.   If a correspondence exists, it will map shortest side to shortest side (12→6), middle to middle (22→12), and longest to longest (30→18). But these do not show the same scale factor so the triangles are not similar.

5.   Check scale factors: , , and . ΔARM ~ ΔGEL with a scale factor of .

6.   Locate the largest and smallest angle of each polygon and the longest and shortest side of each. Rotate and/or reflect the polygons until these coincide, largest with largest, smallest with smallest. CLAMP ~ NDTRE

7.   Perimeter: 2.4, Area: 5.76 The original is 6 by 10, so it has P = 2(6) + 2(10) = 32, and A = 6(10) = 60. The dilated rectangle is 14.4 by 24, so it has P = 2(14.4) + 2(24) = 76.8, and A = 14.4(24) = 345.6. The perimeter was enlarged by a factor of and the area was enlarged by

EXERCISE 9.3

1.   By the Triangle Sum theorem, m∠A + m∠B + m∠C =180° and m∠X + m∠Y + m∠Z =180°, so m∠A + m∠B + m∠C = m∠X + m∠Y + m∠Z. Given that m∠A = m∠X and m∠B = m∠Y, subtract m∠A + m∠B from one side and m∠X + m∠Y from the other, leaving m∠C = m∠Z.

2.   ΔABC ~ ΔDEC by AA Given , cut by a transversal, alternate interior angles ∠A ≅ ∠D and alternate interior angles ∠B ≅ ∠E. ΔABC ~ ΔDEC.

3.   ΔABC ~ ΔDEC by AA Given that , when these parallel lines are cut by a transversal, corresponding angles are congruent, so ∠A ≅ ∠EDC. ∠BCA and ∠ECD are in fact the same angle, so ∠BCA ≅ ∠ECD. ΔABC ~ ΔDEC by AA.

4.   ΔCDE ~ ΔBDF by AA Given , cut by a transversal, corresponding angles are congruent, so ∠DCE ≅ ∠DBF and ∠DEC ≅ ∠DFB. ΔCDE ~ ΔBDF by AA.

5.    Given that , ∠QTP and ∠RTS are right angles because perpendiculars form right angles, and because all right angles are congruent, ∠QTP ≅ ∠RTS. That ∠PQT ≅ ∠SRT is given and so ΔPQT ~ ΔSRT. Then because corresponding sides of similar triangles are in proportion.

6.    Given ∠X ≅ ∠Y, and ∠XYW ≅ ∠VYZ because vertical angles are congruent, ΔXYW ~ ΔVYZ by AA. Then because corresponding sides of similar triangles are in proportion.

7.   ΔABC ~ ΔJKL by AA Given that ΔACD ~ ΔJLM, corresponding angles will be congruent, so ∠C ≅ ∠L. and are altitudes, so by the definition of altitude, and . Perpendiculars form right angles ∠ABC and ∠JKL, and all right angles are congruent so ∠ABC ≅ ∠JKL. ΔABC ~ ΔJKL by AA.

8.    and are given. By the Base Angle theorem, ∠QPR ≅ ∠QRP and ∠SRT ≅ ∠STR. If (given), then ∠QPR ≅ ∠SRT because alternate interior angles are congruent, and by transitivity, ∠QRP ≅ ∠SRT ≅ ∠STR ≅ ∠QPR. ΔPQR ~ ΔRST by AA. Then because corresponding sides of similar triangles are in proportion. Because T is the midpoint of , TR = , or , and by transitivity or .

EXERCISE 9.4

1.   x = 4 Cross-multiply to get 21x = 84 and x =4.

2.    Cross-multiply to get 11x + 33 = 90, and 11x = 57, so .

3.   The proportion cross-multiplies to ad = bc. Variant becomes ad = bc directly, becomes ad + bd = bc + bd. Subtract bd from both sides and ad = bc. Similarly, becomes ad – bd = bc – bd. Add bd to both sides and ad = bc.

4.    becomes . Cross-multiply to get 49x = 798 and .

5.   4 so A′B′ = 4.

6.    becomes . Cross-multiply and solve to get . becomes . Cross-multiply and solve to get

7.   5:12 Cross-multiply to get . Solve by factoring. factors to , and when each factor is set equal to zero, the solutions are x = −12, which is rejected as unreasonable, and x = 5. The proportion becomes . The scale factor is 5:12.

EXERCISE 9.5

1.   If , then . CA = CX + XA, so CA – CX = XA. Likewise, CB = CY + YB, so CB – CY = YB. The proportion becomes .

2.   In ΔCWV, and or . In ΔCAB, and or . Rewrite XA and YB to get or . Because from above, subtract: and or . Finally, in ΔCAB, and . Rewrite: or . Subtract: or .

3.    is so is correct.

4.    Let x = WZ and 5 – x = ZY. Because bisects ∠X, . Substitute to get and cross-multiply to get 3x = 20 – 4x. Solve to get 7x = 20 and .

5.   BC = 25 cm and AB = 40 cm Let x = BC and x + 15 = AB. Then . Cross-multiply to get 32x = 20x + 300. Solve to get 12x = 300 and x = 25 cm. BC = 25 cm and AB = 40 cm.

6.   Focus on ΔABX. In this triangle, bisects ∠B and divides so that the segments and are proportional to the sides of the triangle and . Therefore, .

EXERCISE 9.6

1.   x = 4 becomes . Cross-multiply and solve. 49x = 196 and x = 4.

2.   x = 9 becomes . Cross-multiply and solve. 9 + 3x = 36 and x = 9.

3.   ΔC = 18 and AC = 26 becomes . Cross-multiply and solve. 8x = 144 and x = 18. DC = 18 and AC = AD + DC = 8 + 18 = 26.

4.   x = 9 becomes . Cross-multiply and solve. 25x = 225 and x = 9.

5.    becomes 8DC = 256 and DC = 32. AC = AD + DC = 8 + 32 = 40. becomes Cross-multiply and solve. and . Finally, becomes so and

6.    becomes and becomes .

7.   The Pythagorean theorem.

CHAPTER 10

Right Triangle Trigonometry

EXERCISE 10.1

1.   ΔPQR ~ ΔZXY

2.   

3.   25.5 cm. becomes .

Cross-multiply and solve to get x = 25.5 cm.

4.   5:1 becomes . Cross-multiply and solve to get y = 4.2 in. The scale factor is 5:1.

5.   6.5

6.   

7.   

EXERCISE 10.2

1.   

2.   

3.   

4.   

5.   

6.   

7.   By definition, but if ∠X is the right angle, the hypotenuse is the also the opposite side. but if ∠X is the right angle both legs are adjacent to ∠X. has the same ambiguity. The current definitions are not workable for right angles.

8.   a) sin 30° = 0.5

b) cos 45° ≈ 0.707

c) tan 82° ≈ 7.115

d) cos 11° ≈ 0.982

e) sin 75° ≈ 0.966

EXERCISE 10.3

1.   a) sin 78° ≈ 0.978

b) cos 15° ≈ 0.966

c) cos 43° ≈ 0.731

d) tan 62° ≈ 1.881

e) sin 21° ≈ 0.358

2.   19 cm so XY = 38 cos(30°) ≈ 32.9 cm. Then so YZ = 38 sin(30°) = 19 cm.

3.   35.7 cm so AC sin(79°) = 35 and

4.   16.97 inches In an isosceles right triangle with a leg of 12 inches, both legs measure 12 inches, and the length of the hypotenuse could be found, using the Pythagorean theorem, as inches. Working instead with trig ratios, can be solved to find inches.

5.   13.4 inches so YZ = 16 tan(40°) ≈ 13.4 inches.

6.   11.6 inches so AD = 12 cos(15°) ≈ 11.6 inches.

7.   7.3 meters so and meters.

EXERCISE 10.4

1.   a)

b.  

c.  

d.  

e.  

2.   57.8°

3.   37.5°

4.   41.2°

5.   30° so

6.   45°

7.   53.1°

EXERCISE 10.5

1.   

2.   

3.   

4.    and

Simplify:

5.   

6.    implies

7.    and cos(40°) = (1.192)(0.643) ≈ 0.766

8.   cot(15°) = 3.732

9.   sin(30°) = 0.5

10.   0.267

11.   a) 0.954

b)  0.314

c)  3.18

d)  1.048

e)  3.333…

f)  0.954

g)  0.3

h)  3.18

i)  0.314

j)  1.048

c)  cot(∠A)

d)  sec(∠A)

e)  csc(∠A)

f)  sin(∠C) = cos(∠A) =0.954

g)  cos(∠C) = sin(∠A) =0.3

h)  tan(∠C) = cot(∠A) = 3.18

i)  cot(∠C) = tan (∠A) = 0.314

j)  sec(∠C) = csc(∠A) =

k)  csc(∠C) = sec(∠A) = 1.048

EXERCISE 10.6

1.   

2.   

3.   1.221 (0.700)2 + 1 = 1.49 so

4.   1.11 so cot²(42°) = and

5.   0.515 To find cos(59°), first find sec(59°). and . Then

6.   1.803 If cos(61°) = 0.485, then sec(61°) = Then tan²(61°) = sec²(61°) − 1 = so tan(61°) =

7.   0.966 Given tan(75°), to find sin(75°), first find cot(75°), then find csc(75°), and finally sin(75°). If tan(75°) = 3.732, then cot(75°) = and Finally,

EXERCISE 10.7

1.   525.3 feet so x = 250 tan(35°) ≈ 175.1 yards or 525.3 feet.

2.   27.8 feet so x = 25 tan(48°) ≈ 27.8 feet.

3.   37.9° so

4.   1,343.8 feet so 1,343.8 feet or about a quarter of a mile.

5.   5.2 feet so x = 25sin(12°) ≈ 5.2 feet.

6.   513 feet 513 feet.

7.   83.9 feet so x = 100tan(40°) ≈ 83.9 feet.

8.   8.1 feet so y = 100tan(42.6°) ≈ 92.0 feet. 92 feet – 83.9 feet = 8.1 feet is the height of the flagpole.

CHAPTER 11

Trigonometry in Other Triangles

EXERCISE 11.1

1.   9.6

2.   32.3

3.   17.6

4.   33.5

5.   82°

6.   59°

7.   77°

EXERCISE 11.2

1.   10.5

2.   12.6

3.   108.5

4.   60.2

5.   No such triangle

No solution is possible because sine can only take values between −1 and 1. No such triangle exists.

6.   Two triangles are possible:

The triangle may have m∠Z = 98°, m∠X = 62°, and m∠Y = 20°. Also consider ∠Z = 98°, m∠X = 180° − 62° = 118°, but that already exceeds 180°, so a second triangle is not possible.

7.   Two triangles are possible:

Consider both 21° and 180° − 21° = 159°. Triangle # 1: 14°, 21°, 145°. Triangle # 2: 14°, 159°, 7°. Both triangles are possible. There are two solutions.

EXERCISE 11.3

1.   m∠A = 24°, m∠B = 49°, m∠C = 103°

m∠A = 24°, m∠B = 49°, m∠C = 103°

2.   

m∠A = 62° and m∠B = 88° (given) so m∠C = 180° − (62° + 88°) = 30°.

3.   XZ = 68.7,

With side XZ = 68.7, find m∠Z.

4.   

Triangle 1: m∠R = 64°, m∠S = 180° – (64° + 43°) = 73°, m∠T = 43°

Triangle 2: m∠R = 180° − 64° = 116°, m∠S = 180° – (116° + 43°) = 21°, m∠T = 43°

Find side :

5.    Let x = length of side opposite 65° angle and y = length of side opposite 80°angle.

6.   m∠A = 54°, m∠B = 61°, m∠C = 65°

m∠A = 54°, m∠B = 61°, m∠C=180° − (54° + 61°) = 65°

7.   m∠A = 55°, m∠B = 33°, m∠C = 92°

m∠A = 55°, m∠B = 180° − (55° + 92°) = 33°, m∠C = 92°

EXERCISE 11.4

1.   101.9

2.   87.0

3.   1,051.9

4.   6,990.4

5.   65°

6.   65

7.   24

REVIEW 5

Similarity, Right Triangle Trigonometry, and Trigonometry in Other Triangles

1.   ΔABC ~ ΔQRP

2.    ΔWXY ~ ΔVXZ, so

3.      If (given), then ∠A ≅ ∠E and ∠B ≅ ∠D because alternate interior angles are congruent. ∆ABC ~ ∆EDC by AA, and corresponding sides are in proportion, so .

4.   ΔXWZ ~ ΔWYZ by AA   ∠X ≅ ∠ZWY (given) and ∠Z ≅ ∠Z (reflexive), so ΔXWZ ~ ΔWYZ by AA.

5.      Given that ∠DEA ≅ ∠EAC, it is possible to conclude that by Alternate Interior Angles converse. Then by the Corresponding Angle theorem, ∠BDE ≅ ∠BAC and ∠BED ≅ ∠BCA, so ∆BDE ~ ∆BAC by AA. Corresponding sides are in proportion, so .

6.      If (given), then ∠Q ≅ ∠SRQ. Also given is ∠PRS ≅ ∠Q, and by transitivity, ∠SRQ ≅ ∠PRS. By the definition of angle bisector, bisects ∠PRQ. Because the bisector of an angle of a triangle divides the opposite side in a way that is proportional to the adjacent sides, .

7.   20.4 cm    becomes .

Cross-multiply to get 25x = 510, and solve to get x = 20.4 cm.

8.   63 cm    becomes . Cross-multiply to get 21x − 63 = 6x + 12, and solve to get 15x = 75 and x = 5. AC = 5 − 1 = 4 and XY = 2(5) + 4 = 14. becomes . Solve to find 3y = 84 and y = 28. The perimeter of ΔXYZ = XY + YZ + XZ = 14 + 21 + 28 = 63 cm.

9.   7.2 feet   Let x = the distance from the point where the man stands to the tip of his shadow.

Solve to get x = 10.8 feet. To have the tip of the man’s shadow match the tip of the tree’s shadow, he must stand 18 - 10.8 = 7.2 feet from the base of the tree.

10.   29.75 cm   Convert all measurements to centimeters to avoid confusion if you wish, but you can work with numerators in meters and denominators in centimeters, if you work carefully. becomes 15,300x = 38.25(11,900) and x ≈ 29.75 cm.

11.   a) sin∠R =

b)  tan∠T =

c)  cos∠R =

d)  cos∠T =

e)  csc∠R =

12.   21.5 centimeters    becomes

centimeters.

13.   36.9°    so

14.   5   

15.   5   

16.   0.98   

17.   19.5 cm m∠J = 63°, m∠K = 180° − (63 + 51) = 66°, and m∠L = 51°. JL = 20 cm. The triangle is not a right triangle, so use the Law of Sines.

becomes . Cross-multiply to get xsin(66°) = 20sin(63°) and solve for cm.

18.   61.7°

19.   195.8 square inches

square inches.

20.   25°

CHAPTER 12

Circles

EXERCISE 12.1

1.   

2.   

3.   

4.   

5.   Radius

6.   Common internal tangent

7.   Diameter (or chord)

EXERCISE 12.2

1.   14.1 feet Diameter = 54 in = 4.5 feet, so C = 4.5π ≈14.1 feet.

2.   r = 42 cm C = 2πr = 84π, so 2r = 84 and r = 42 cm.

3.    square inches.

4.   r = 17 meters so and r = 17 meters.

5.   796 square meters C = 100 = 2πr, so . Then square meters.

6.   10-inch pizza A pizza with a diameter of 12 inches has a radius of 6 inches and an area of 36π square inches. Divide the cost of $12 by 36π to get $0.1061 per square inch. A pizza with a diameter of 10 inches has a radius of 5 inches and an area of 25π square inches, so costs $8 ÷ 25π = $0.1019 per square inch. The 10-inch pizza is a slightly better buy.

7.    Solve C = 2πr for r: . Substitute in the area formula .

EXERCISE 12.3

1.   m∠DBC = 19° The measure of inscribed angle ∠DBC is half the measure of the central angle with the same arc. m∠DBC = m∠DOC = 19°.

2.   m∠PAB= (50°) = 25°.

3.    ∠ABC is an inscribed angle, so its measure is half the measure of its intercepted arc; therefore .

4.   m∠QAB = 155° ∠QAB intercepts . m= 360° − 50° = 310°. m∠QAB = (310°) = 155°.

5.   m= 230° m + m + m=50° + 42° + 38° = 130°. m= 360° − 130° = 230°.

6.   115° m∠BAD = (230°) = 115°.

7.   12.5° m∠BPD = (m − m) = (40° – 15°) = (25°) = 12.5°.

8.   90° m∠RQT = (m − m) = (270° − 90°) = (180°) = 90°.

9.   135° m∠SRT = (m) = (30° + 40° + 110° + 15° + 75°) = (270°) = 135°.

10.   22.5° m∠DPQ = (m − m) = (120° - 75°) = (45°) = 22.5°.

11.   135° m∠RTP = (m) = (75° + 15° + 110° + 40° + 30°) = (270°) = 135°.

EXERCISE 12.4

1.   EC = 10 AE ⋅ EB = DE ⋅ EC becomes 5 ⋅ 8 = 4 ⋅ EC, so EC = 10.

2.   OJ = 8 bisects so FJ = JG = 6. OK =10 so OF = 10 (all radii are congruent).

becomes , so OJ = 8.

3.   2 JK = OK – OJ = 10 – 8 = 2.

4.    becomes so

5.    and are two tangents drawn to the circle from the same point, and so are congruent. Therefore, ΔPDK is isosceles and by the Base Angle theorem ∠PDK ≅ ∠PKD.

6.    PA ⋅ PB = PC ⋅ PD becomes 9(9 + 16) = x ⋅ 27 or 27x = 225, so

7.   PT = 15 becomes and and PT = 15.

8.   QS = 6 becomes 144 = QS ⋅ 24 and QS = 6.

9.   18 The diameter = QV – QS = 24 − 6 = 18.

10.   17 If x(x + 1) = 56, solve by factoring. Setting each factor in (x + 8) (x – 7) = 0 equal to zero will produce solutions of x = −8, which is rejected, and x = 7. Then AB = x + (x + 1) = 7 + 8 = 15, and CD = (x + 1) + (x + 2) = 8 + 9 = 17.

EXERCISE 12.5

1.   Draw a radius from C to A and extend. Scribe a segment which has A as its midpoint. Construct the perpendicular bisector.

2.   Draw and bisect it. Scribe an arc centered at the midpoint of with a radius that is . Draw tangents to the two points where the arc intersects the circle.

3.   Draw a radius. From a point on the circle, with the compass set to the radius of the circle, scribe arcs that divide the circle into six congruent arcs. Connect the points.

4.   Draw and bisect it. With the midpoint as center and as radius, scribe a circle that intersects the original circle in two points. Draw from P to each point of intersection.

5.   

6.   

7.   Draw a diameter then construct a perpendicular to it, through the center, to divide the circle into four congruent segments. Construct a perpendicular at a point where the diameter crosses the circle. Repeat at each point.

EXERCISE 12.6

1.   67 meters Length of meters, and length of 67 meters.

2.   48π square meters square meters.

3.   49.1 meters The perimeter of the sector is the combined lengths of two radii and the arc. 2 ⋅ OA + length of = 2(12) + (24π) = 24 + 8π ≈ 49.1 meters

4.   29.8 cm cm

5.   13.2 square inches If d = 12, r = 6.

square inches

6.   The first arc is longer. The first arc has a length of cm, and the second arc has a length of cm. The first arc is longer.

7.   The first sector has an area twice as large as the second. The first sector has an area of square centimeters, and the second sector has an area of square centimeters. The first sector has an area twice as large as the second.

8.   22.3 square centimeters The sector with the larger radius has an area of

With the smaller radius, the area is . The shaded area is square centimeters.

EXERCISE 12.7

1.    radians becomes 360r = 60π and radians. (approximately 0.524 radians)

2.    radians becomes 360r = 270π and radians. (approximately 2.356 radians)

3.    radians becomes 360r = 160π and radians. (approximately 1.396 radians)

4.   60° becomes 2πd =120π and d = 60°.

5.   180° becomes 2πd = 360π and d =180°.

6.   143° becomes 2πd = 900 and d ≈ 143°.

7.   34° becomes 2πd = 216 and d ≈ 34°.

8.   62.8 inches s = θr becomes inches, approximately 62.8 inches.

9.   1.5 radians s = θr becomes 108 = θ ⋅ 72, so radians.

10.   19.6 square meters becomes square meters, or approximately19.6 square meters.

CHAPTER 13

Conic Sections

EXERCISE 13.1

1.   C(3, 2), r = 2

2.   C(−3, −5), r = 11

3.   

4.   

5.   

6.   

7.   

EXERCISE 13.2

1.   10 units   The major axis is 2a and , so the major axis is 10 units long.

2.   14 units   The minor axis is 2b and b = 7, so the minor axis is 14 units.

3.      C(0, 0), minor axis (horizontal) = 12 so b = 6, major axis (vertical) = 18, so a = 9.

4.      C(−3, −2), major axis (horizontal) = 10 so a = 5, minor axis (vertical) = 6 so b = 3.

5.   A rotation of 90° about the origin will carry one onto the other. Both are ellipses centered at the origin. One has a major axis of length 8 that is horizontal and a minor axis of length 6 that is vertical. The other has a major axis of length 8 that is vertical and a minor axis of length 6 that is horizontal. A rotation of 90° about the origin will carry one onto the other.

6.   

7.   

8.   

9.   6π, 10π The area of the ellipse is A = πab = π ⋅ 3 ⋅ 5 = 15π. The area of the smaller circle is . The difference is 6π. The area of the larger circle is , so the difference between the area of the larger circle and the area of the ellipse is 10π.

EXERCISE 13.3

1.   a) down

b)  right

c)  up

2.   a) (−2, 6)

b)  (−4, 3)

c)  (−5, 4)

3.    The parabola opens down, has a vertex of (0, 5), and passes through the point (1, 3). Begin with and replace h with 0 and k with 5. or Then replace x with 1 and y with 3 to solve for a. says a = −2. The equation is or .

4.    The parabola opens to the right, has a vertex of (−6, −2), and passes through the point (−4, 0). Begin with and substitute −6 for h and −2 for k. With substitute −4 for x and 0 for y, and solve for a. becomes 2 = 4a, so a = . The equation is or .

5.   

6.   

7.   

EXERCISE 13.4

1.   a) (0, 3)

b)  (−2, 3)

c)  (1, −1)

2.   a) left/right

b)  up/down

c)  left/right

3.      The hyperbola is centered at the origin, opens to the left and right, and touches the edges of the rectangle at (±2, 0). The rectangle is 4 units wide and 6 units tall, so b = 2 and a = 3.

4.      The hyperbola is centered at the origin, opens up and down, and touches the edges of the rectangle at (0, ±1). The rectangle is 20 units wide and 2 units tall, so a = 10 and b = 1.

5.      The hyperbola is centered at (2, −1), and opens to the left and right. The rectangle is 10 units wide and 12 units tall, so b = 5 and a = 6.

6.   

7.   

REVIEW 6

Circles and Conics

1.   

2.   ∠RST

3.   m∠RST =

4.   r = 12

5.    of the large circle is shaded Let D = diameter of the larger circle, R = radius of the larger circle, d = diameter of the smaller circle, and r = the radius of the smaller circle. . Area of the larger circle: Area of the smaller circle: . The shaded area is . This is what part of ? of the large circle is shaded.

6.   2π   centimeters ∠OAE is an inscribed angle in circle C that measures 30° and so arc measures 60°. ∠OAE is also an inscribed angle in circle O, and so arc also measures 60°. The central angles that intercept these arcs will measure 60°. The length of arc is π cm, so set up and solve for r. , so r = 3 cm and R = 6 cm. The length of arc is centimeters.

7.   m∠PQR = m∠PRQ = 80°   Two tangents drawn from the same point will be congruent, so ΔQPR is isosceles and ∠PQR ≅ ∠PRQ. m∠P = 20°, so there is 180° − 20° = 160° for the two base angles. m∠PQR = m∠PRQ = 80°.

8.   CD = 14   AE ⋅ EB = CE ⋅ ED becomes x(3x) = 2x(x + 2) or . Simplify to and solve by factoring to get x = 0, which is rejected, and x = 4. CD = 2x + (x + 2) = 8 + 6 = 14.

9.   m∠AED = 112°    m∠AEC = 180° − m∠AED = 180° − 68° = 112°

10.   

11.   x = 7 PB ⋅ AP = PC ⋅ PD becomes 3(17 + 3) = 5(5 + x) or 60 = 25 + 5x. Solve to find x = 7.

12.   Draw and bisect it. From the midpoint of , scribe a circle that passes through O and P and intersects the circle in two points. Those two points are the points of tangency. Draw from P to each point of tangency.

13.    To inscribe a regular hexagon in a circle, the circle is divided into 6 congruent arcs of 60° each.

14.   

15.   The length of ≈ 10 inches. The area of the sector is 34 square inches. so The radius inches. The length of is inches.

16.    radians cross-multiplies to 360r = 240π and radians.

17.   d = 135°    cross-multiplies to 2πd = 270π and d = 135°.

18.   

19.    The center is C(4, 3) and the radius is r = 5. The equation of the circle is .

20.   a) hyperbola

b)  circle

c)  ellipse

d.  parabola

CHAPTER 14

Three-Dimensional Figures

EXERCISE 14.1

1.   Triangular pyramid

2.   Pentahedron

3.   Heptahedron

4.   E = 18 F + V = E + 2 becomes 8 + 12 = E + 2, so E = 18.

5.   F = 4 F + V = E + 2 becomes F + 4 = 6 + 2, so F = 4.

6.   V =8 F + V = E + 2 becomes 6 + V = 12 + 2, so V =8.

7.   Hexagonal prisms have 8 faces, 18 edges, and 12 vertices.

EXERCISE 14.2

1.   Choice a) is a net for a triangular pyramid, which has 4 triangular faces.

2.   A possible net for a hexagonal prism:

3.   555 square centimeters S = 2(aP) + Ph = 2.75(20) + 20(25) = 55 + 500 = 555 square centimeters.

4.   720π square centimeters square inches.

5.   h ≈ 16 cm S = 2B + Ph becomes 1,100.55 = 2(166.28) + 48h. Solve to find 48h = 1,100.55 − 332.56 = 767.99 and h ≈ 16 cm.

6.   r = 4 inches becomes , which simplifies to . Solve by factoring. has solutions of r = −13, which is rejected, and r = 4 inches.

7.   The cylinder has the larger surface area If the radius, from center to vertex, of the square, is r, then the side of the square is and the area of each square base is The surface area of the square prism is The surface area of the cylinder is Comparing with means comparing 4 with 2π and with 2π. Since 4 < 2π and <2π, the cylinder has the larger surface area.

EXERCISE 14.3

1.   2,494.2 cubic inches If the side of the hexagonal base is 8, then the apothem is and the area of the base is . Then the volume is V = Bh = cubic inches (approximately 2,494.2 cubic inches).

2.   3,769.9 cubic centimeters If the diameter is 20, then r = 10 and cubic centimeters.

3.   h = 24 V = Bh, so solve 25,807.2 = 1,075.3h to find h = 24.

4.   h ≈ 32 inches becomes 787,320 =. Solve to find h ≈ 32 inches.

5.   4,608π cubic inches with a diameter of 24 and therefore r =12 inches, becomes Simplify to and h = 32 inches. Then cubic inches.

6.   The ratio is 5. The 3-4-5 base has an area of 6, and the 5-12-13 base has an area of 30. If the volumes are to be equal and

7.   The height must be divided by 4, If the radius is doubled, the volume is multiplied by. If the volume is going to remain the same, the height must be divided by 4.

EXERCISE 14.4

1.   1,036.75 square centimeters square centimeters.

2.   933.1 square inches square inches (or approximately 933.1 square inches.)

3.   1,433.1 square inches The side of the hexagon is 96 ÷ 6 = 16 inches, and the apothem of the hexagon is . The area of the hexagonal base is square inches. The slant height is 16 inches. The surface area is square inches.

4.   18.5 inches becomes 1,024.5 = 377 + (70)l. Solve to find 35l = 647.5 and l = 18.5 inches.

5.   The second cone has the larger surface area. A cone with a radius of 3 and a height of 4 has a surface area of square centimeters. A cone with a radius of 4 and a height of 3 has a surface area of square centimeters. The second cone has the larger surface area.

6.   5.4 square feet larger The shorter pyramid has a slant height of and the taller pyramid has a slant height of . The surface area of the shorter pyramid is square feet. The surface area of the taller pyramid is square feet. The taller pyramid has a surface area 79.3 − 73.9 = 5.4 square feet larger.

7.   The cone has the larger surface area. The square pyramid has a perimeter of 40, a side of 10, an apothem of 5 and a slant height of Its surface area is square centimeters. The cone, with a slant height of , has a surface area of square centimeters. The cone has the larger surface area.

EXERCISE 14.5

1.   36 square centimeters square centimeters.

2.   804.2 square inches square inches.

3.   r = 6 cm becomes . Solve to find that and r = 6 cm.

4.   h = 11 cm becomes . Solve to find h = 11 cm.

5.   37.7 cubic centimeters If then slant height is 5 cm, and the cone has a surface area of 24π square centimeters then and . Solve by factoring or quadratic formula, and r = −8 (reject) or r = 3. With a radius of 3 and a slant height of 5, use the Pythagorean theorem to find h = 4. A cone with a radius of 3 and a height of 4 has a volume of cubic centimeters, or approximately 37.7 cubic centimeters.

6.   531.7 square centimeters becomes . Solve to find B = 144 square centimeters. The square base has a side of 12 and a perimeter of 48. The surface area of the pyramid is square centimeters.

7.   r = 5 inches The volume of the cone is the volume of the cylinder, so the empty space is of the volume of the cylinder. If then . If h = 18, becomes and r = 5 inches.

EXERCISE 14.6

1.   2,463 square centimeters square centimeters.

2.   268.1 cubic inches cubic inches.

3.   r = 6 cm becomes . Solve to find and r = 6 cm.

4.   18 inches . Solve to find and r = 9. The diameter is 18 inches.

5.   1,809.6 square meters Solve to find and r = 12 meters.

Then square meters.

6.   r = 3 cm If , then so and . The equation has two solutions, r = 0, which is impossible, and r = 3 cm.

7.   5,575.3 cubic centimeters If the diameter is 22 cm, the radius is 11 cm. The volume is cubic centimeters.

EXERCISE 14.7

1.   6 centimeters If , then and r = 3, so the diameter is 6 cm.

2.   h = 12 cm becomes . Solve for h = 12 cm.

3.   12 feet 10,000 gallons ÷ 7.48 gallons per cubic foot ≈ 1,336.9 cubic feet. A cylindrical tank with a height of 12 feet needs to have a volume of 1,336.9 cubic feet. becomes . Solve to find so r ≈ 5.95. The radius of the tank should be about 6 feet and the diameter approximately 12 feet.

4.   The radius of 3 cm gives a smaller surface area for the same volume. If the volume of the cylinder is 250 cubic centimeters and the radius is 3 cm, solving says the height must be approximately 8.84 cm. Those dimensions would result in a surface area of square centimeters. If the same volume is desired with a radius of 4 cm, gives h ≈5.0 cm, and yields a surface area of The radius of 3 cm gives a smaller surface area for the same volume.

5.   180 billion square miles If the radius of the Earth is 3,959 miles, the surface area will be square miles. If 71% of that surface is water, that equals 0.71() ≈ square miles. (180 billion square miles)

6.   The pyramid in Mexico has the larger volume. The pyramid in Egypt has a volume of cubic meters. The pyramid in Mexico has a volume of cubic meters. The pyramid in Mexico has the larger volume.

7.   The height of the cone must be twice the radius. The cylindrical portion of the volume can be ignored because it is the same in both structures, but the volume of the hemisphere must equal the volume of the cone. . The radius is determined by the radius of the cylinder and is the same in both structures. simplifies to or . A solution of r = 0 must be rejected, but 2r − h = 0 means h = 2r. The height of the cone must be twice the radius, or put another way, the height of the cone is equal to its diameter.

EXERCISE 14.8

1.   3,720.6 grams The volume of the cylinder is cubic centimeters. If cast from aluminum, it will have a mass of 2.7(1,378) = 3,720.6 grams.

2.   An increase of 7,372.3 grams If the cylinder in exercise 50 is instead cast from steel, its mass will be 8.05(1,378) = 11,092.9 grams. This is an increase of 7,372.3 grams.

3.   1.11 grams per cubic centimeter If 355 cubic centimeters of a cola have a mass of 394 grams, the density of the cola is grams per cubic centimeter.

4.   1.0003 grams per cubic centimeter If 355 cubic centimeters of diet cola has a mass of 355.1 grams, the density of diet cola is grams per cubic centimeter.

5.   The key difference between cola and diet cola is the sweetening agent used. The artificial sweeteners used in diet cola are less dense than the sugar used to sweeten cola.

6.   The difference is 4.556 grams. The volume of the bead is the volume of a sphere of diameter = 1 cm minus the volume of a tiny cylinder with diameter = 0.1 cm and height equal to the diameter of the sphere. The radius of the sphere is 0.5 cm, and the radius of the cylinder is 0.05 cm, and the height of the cylinder is 1. cubic centimeters. If cast in gold, the bead will have a mass of 19.32(0.516) = 9.969 grams, but if cast in silver, the mass is 10.49(0.516) = 5.413 grams. The difference is 4.556 grams.

REVIEW 7

Three-Dimensional Figures

1.   F = 8 F + V = E + 2 becomes F + 12 = 18 + 2. Solve to find F = 8.

2.   A possible map for a triangular pyramid:

3.   56.8 square inches   Perimeter = 12 means each side of the hexagon is 2 and the apothem is , and S = 2B + Ph square inches.

4.   1,105.8 square centimeters    ≈ 1,105.8 square centimeters.

5.   1,320 cubic feet   square feet. V = Bh = (110)(12) = 1,320 cubic feet.

6.   18,472.6 cubic inches    cubic inches.

7.   85 inches    inches.

8.   2,358 square centimeters   The area of the base octagon is square centimeters. The slant height is cm, and the surface area is square centimeters.

9.   678.6 square inches   If diameter = 18, then radius = 9, height = 12 and slant height = 15 inches. ≈ 678.6 square inches.

10.   32 cubic feet    cubic feet.

11.   7,938.8 cubic inches   If the circumference is 38π, then the diameter is 38 and the radius is 19 inches. cubic inches.

12.   332 square feet   S = 2B + Ph = 2(2)(8) + 20(15) = 32 + 300 = 332 square feet. (Alternate method: S = 2lw + 2wh + 2lh = 2(2)(8) + 2(8)(15) + 2(2)(15)=32 + 240 + 60 = 332 square feet.)

13.   1,017.9 cubic centimeters    , and , so . Simplify to or . Solve by factoring. (r + 12)(r − 9) = 0 gives a solution of r = −12, which is rejected, and r = 9. If r = 9 then l = 15 and h = 12. cubic centimeters.

14.   14.7 inches   The area of the equilateral triangle base is inches. becomes . Simplify and solve to find

inches.

15.   7 meters    becomes , which simplifies to . Factor to get (r + 12)(r − 7) = 0, and solve, rejecting the negative solution and concluding that r = 7 meters.

16.   400 cubic centimeters    becomes or and the slant height l = 13. Then cubic centimeters.

17.   The cone has the larger surface area. For the cylinder, d = 20 inches, so r = 10 inches and h = 20 inches. The surface area of the cylinder square inches. For the cone, the radius r = 15 inches and height h = 36 inches, so the slant height is = 39 inches. The surface area The cone has the larger surface area.

18.   5,652,833 cubic meters The volume of the TransAmerica pyramid is approximately 5,652,833 cubic meters.

19.   39,219,187.2 cubic feet V = Bh = (79,515π)(157) = 12,483,855π ≈ 39,219,187.2 cubic feet.

20.   11.3 grams per cubic centimeter The diameter of the cylinder is 6 cm, the radius is 3 cm and the height is 2 cm. The volume of the cylinder is ≈ 56.5 cubic centimeters. If cast from iron, the cylinder will have a mass of 18π(7.9) ≈ 446.7 grams. Casting it in lead would increase its mass by 192 grams, making the mass of the lead cylinder approximately 638.7 grams. The density of lead is approximately 638.7 grams per 56.5 cubic centimeters, or approximately 11.3 grams per cubic centimeter.

Posttest

1.   C   Translation will not change the orientation.

2.   B    = 30°

3.   B   Perimeter of ΔADE = AD + DE + AE = 4 + 8 +7 = 19

4.   A   PQRT is a trapezoid but because and , PQRS is a parallelogram, so m∠PSR = m∠Q = 108°. Because linear pairs are supplementary, m∠RST = 72°. If ΔSRT is isosceles with vertex ∠T, then TS = TR and m∠RST = m∠SRT = 72°.

5.   C   Because their bases are squares, the cube and the square pyramid could have square cross-sections. A rectangular prism could possibly have a square cross section if one of its bases is square. The triangular prism would have only triangular cross sections.

6.   C   cos(∠A) = sin(∠C) =

7.   D   The exterior angle is 115° and is equal to m∠ABC + m∠BAC, so A cannot be true. Choice B might be true but there is no information to support it. Choices C and D are similar but only D has the right pair of angles.

8.   C   To find the length of the legs, use a proportion with the leg as the geometric mean between the hypotenuse (4 + 12 =16) and the segment of the hypotenuse nearest the side. solves to give AB = 8, and solves to give BC = ≈ 13.9. Perimeter ≈ 16 + 8 + 13.9 = 37.9.

9.   B   Find the length of each side using the distance formula. Perimeter = 2.8 + 7.2 + 7.2 =17.2.

10.   B   Choice A says the diagonals of the parallelogram are congruent, which is sufficient. Choice C says the parallelogram has a right angle and choice D says that consecutive angles, which are supplementary in any parallelogram, are also congruent, so both right angles. Choice C says the diagonals are perpendicular, which would prove the quadrilateral is a rhombus, but would not assure that it is a rectangle.

11.   C   The sine of an angle is equal to the cosine of its complement, so sin(2x + 7) = cos(4x − 1) can become cos(90 − (2x+7)) = cos(4x − 1) and that means that (90 − (2x+7)) = 4x − 1. Simplify to get 83 − 2x = 4x − 1 and solve to get 6x = 84 and x = 14.

12.   B    Solve to find x = 33°.

13.   D    divides sides proportionally, so if, then . To find DE, use the Pythagorean theorem. becomes. DE = .

14.   B   The lighthouse, distance out to sea, and line of sight form a right triangle. so which rounds to 2°.

15.   C   To divide the segment in ratio 2:3, find the points that divide it into 5 equal lengths and then group into two-fifths and three-fifths. Horizontal change is and vertical change is . Starting from (1, 9) take 2 steps to (1 + 2(0.6), 9 − 2(2)) = (2.2, 5).

16.   C   The reflecting line will be the perpendicular bisector of the line segments connecting a vertex to its image. Those would be (2, 7), (4.5, 4.5), and (5, 4). The slope of the line that passes through those points is , so the equation of the line is y − 7 = −1(x − 2) or y = 9 − x.

17.   D   No conclusion can be drawn about congruent triangles because there is no information about congruent sides; however the parallel lines guarantee that the triangles are similar by AA.

18.   A    square centimeters.

19.   D    becomes Solve to find that SV ≈ 20.7 cm.

20.   B   The central angles of a regular octagon measure 45°, so rotations of multiples of 45° will carry the octagon onto itself.

21.   B   Rotating a right triangle about its hypotenuse will not produce a cone, so eliminate choice C. For the remaining choices, determine the radius and height of the resulting cone. In choice A, the radius is 24 and the height is 18. In choice B, r = 18 and h = 24, and . Choice D requires first finding the length of the altitude (14.4) and the lengths of the two segments of the hypotenuse created by the altitude (10.8 and 19.2). Luckily, choice B is correct.

22.   D   The endpoints of the segment are (0, −3) and (8, 3). The slope of this segment is so the slope of the perpendicular bisector is . The midpoint of the given segment is

The equation of the perpendicular bisector is . Choice A is the midpoint so that lies on the bisector. and , so B and C are points on the perpendicular bisector. Choice D is the point that does not lie on the perpendicular bisector.

23.   D   To find the population (number of people). multiply . Results are: New Hampshire 1,325,044; Indiana 6,591,984; Louisiana 4,666,032; Georgia 10,179,801.

24.   A   To prove the triangles congruent, a third piece of information is necessary, specifically, a pair of congruent angles, to complete AAS. That eliminates choices B and C. To choose between A and D, look at the congruence statement ΔABC ≅ ΔXZY. ∠A corresponds to ∠X. Choice A is correct.

25.   C   The diagonals of a rhombus divide the rhombus into 4 right triangles, each with legs of 12 and 4.5. Use the Pythagorean theorem to find the hypotenuse of the triangle, which is a side of the rhombus. so The perimeter of the rhombus is 4(12.8) = 51.2 centimeters.

26.   D   The equation of a circle centered at (h, k) with radius r is Eliminate B and C, which equal r, not . The center is (3, −4) so becomes or . Choice D is correct.

27.   B   Find the third angle of each triangle, assure that there is a correspondence, and make sure that the similarity statement spells out the correct correspondence.

28.   B   If the diameter is 30 cm, then the radius is 15 cm. The measure of ∠AOB is equal to the measure of The area of the sector is calculated by and it is given that . Simplify to and solve to find The length of is

29.   B   The intersection points C and D define the perpendicular bisector because they are equidistant from A and B. All that remains is to connect them by drawing .

30.   C   Choices B and D are not true. Choice A is true, but not enough to ensure that the line is the perpendicular bisector.

31.   

32.   

33.   If ABCD is an isosceles trapezoid (given), then by the definition of isosceles trapezoid. If F is the midpoint of , then by the definition of midpoint. ∠B ≅ ∠C because base angles of a trapezoid are congruent. ΔABF ≅ ΔDCF by SAS, and by CPCTC. Given that ∠EAD ≅ ∠EDA, by the Base Angle Converse theorem, . (reflexive) and ΔAFE ≅ ΔDFE by SSS.

34.   First, find the midpoint of the diagonals. In a parallelogram, the diagonals bisect each other, so expect them to have the same midpoint, which we’ll label X.

.

Next, find the midpoints of opposite sides of the parallelogram, which we’ll label M and N.

and

Now find the equation of the line that connects M and N, starting with the slope.

The slope of the line is zero, so it is a horizontal line and has the form . Point X has a y-coordinate of , so it is a point on passes through the midpoint of both diagonals, so it bisects both diagonals.

35.   Smallest to largest: square pyramid, hexagonal pyramid, cone. Each of the figures has a volume equal to one-third the area of the base times the height. All three calculations have the and the same height, so the difference will be in the area of the base. For the cone, . In the hexagonal pyramid, each of the small triangles created by central angles is equilateral, because the central angle is 60°. If the radius of the hexagon is r, the side is r, and the apothem is . The area of the hexagon is For the square, if the radius is r, the side is and the area of the square base is

Comparing , , and means ordering π, , and 2. Because π ≈ 3.14 and , we can conclude the square pyramid has the smallest volume

the hexagonal pyramid ranks second, and the cone has the largest volume.

36.   Circle with a radius of 9, centered at (−6, 3). The equation does not immediately look like the equation of a circle, but if you compare it to the form you expect, it can be transformed.

A little factoring will convert the given equation to . The equation represents a circle with a radius of 9, centered at (−6, 3).

37.   The best location for a pool would be at the rear of the property, in the space behind the garage and deck. The largest pool would be a rectangle with vertices at (50, 20), (65, 20), (65, 75), and (50, 75). It would have an area of 825 square feet. The best circular option would be centered at (50, 55) with a radius of ≈ 15.8 feet. The area would be slightly over 784 square feet. A circle centered at (55, 55) with a radius of 15 would give an area of almost 707 square feet, and a rectangle with vertices (35, 45), (65, 45), (65, 65) and (35, 65) would have an area of 600 square feet.

38.   The building covers an area of approximately 115,849 square meters. The side of the Pentagon is 281 meters, so its perimeter is 1,405 meters. To find the apothem, focus on the isosceles triangle created by one central angle. The central angle will measure 360° ÷ 5 = 72°, so the base angles of the triangle will be Half of a side will be 140.5 meters. so a = 140.5tan(54°) ≈ 193.38 meters. Then square meters. This is the area enclosed by the outer wall of the building, including the courtyard. Subtract the 20,000 square meters of courtyard, and the building covers an area of approximately 115,849 square meters.

The side of the courtyard is about 108 meters long. To find the side of the courtyard, express all the quantities in the area formula in terms of the side length. The apothem can be found from or . Perimeter is easier: P = 5s. Then and we know that the area of the courtyard is 20,000 square meters. Solve: