1 degree 1 minute = 60

1 minute = 60 seconds

1 minute 1 second = 60

Symbolically:

1° = 60' 1” = 60”

1 degree = 60 × 60 seconds = 3600 seconds

1 second =

1 3600

1 1" = 3600

1° = 3600 "

degrees °

Example 1:

a) 36.5°= 36° + 0.5 x 60’

b) 68°45’18” = 68° +

= 36°30’

45 60

°

+

18 3600

°

= 68° + 0.75° + 0.005° = 68.755°

Special Angles Acute Angle: An angle is acute if it measures between 0° and 90°

Obtuse Angle: An angle is obtuse if it measures between 90° and 180°.

Right Angle: A right angle measures exactly 90°.

.

In a diagram, the square marking at an angle indicates a right angle. Straight Angle: an angle which measures exactly 180°. ∠ABC = 180°

Complementary Angles: Two angles are complementary if they sum to 90°. ∠ x + ∠ y = 90°

Supplementary Angles: Two angles are supplementary if they sum to 180°. ∠ α + ∠ β = 180°

C B y A

Chapter 1: Geometry Review

x

α

β

Page 1

Example 2:

a)

If ∠A = 23° and ∠B = 67° then ∠A and ∠B are complementary since ∠A + ∠B = 90°

b)

If ∠X = 15°, ∠Y = 65° and ∠Z = 10°, even though ∠X + ∠Y + ∠Z = 90° they are NOT complementary angles. A complementary angle refers only to pairs of angles.

c)

What is the complement of the supplement of the angle 145°? Supplement of 145° = (180 – 145)°

Solution:

= 35° Complement of 35° = (90 – 35)° = 55° The complement of the supplement of the angle 145° is 55° d)

Twice the supplement of a certain angle is 5 times its complement. Find the angle. Let the angle be x

Solution:

Then:

supplement of x = (180 – x) complement of x = (90 – x)

and: 2 (supplement of x) = 5 (complement of x) Hence:

2(180 − x ) = 5(90 − x ) 360 − 2 x = 450 − 5 x − 2 x + 5 x = 450 − 360 3 x = 90 x = 30

Supplement of x (180 – x)

x (90 – x) Complement of x

Hence the angle is 30°.

Chapter 1: Geometry Review

Page 2

Exercises 18.1.1: 1. Convert to degrees and minutes:

(a) 57.5°

(b) 114.2°

(c) 34.75°

2. Convert to decimal degrees:

(a) 74°15’

(b) 17°48’

(c) 56°60’

3. State whether each angle given is acute, obtuse, right, or straight. (a) 97°55’

(b) 48.6°

(c) 180°

(d) 88.6°

(e) 90°

(f) the supplement of 45°

(g) the complement of 45°

(h) the complement of any acute angle

(i) the supplement of any acute angle

(j) the supplement of a right angle

4. What angle has the same measure as its supplement? 5. What angle has the same measure as its complement? 6. Find the complement of each angle:

(a)74° (b)56° (c)34.5°

(d)88.6°

(e)17°25’

(f)6°34’

7. Find the supplement of each angle:

(a)104° (b)8°

(d)88.6°

(e)97°55’

(f)16°14’

(c)132.2°

8. What angle has a supplement that measures three times its complement? 9. What angle has a supplement that measures ten times its complement? 10. Find the supplement of the complement of the angle measuring 40°. 11. Find the complement of the supplement of the angle measuring 120°. 12. Find the complement of a right angle.

Chapter 1: Geometry Review

Page 3

18.1.2: TRIANGLES General A triangle has three angles, or vertices, and three sides. Each vertex of the triangle is named using capital letters: A, B, C, etc. The angle-sum of a triangle is 180°. That is, the sum of the 3 angles of the triangle is 180°. The longest side of the triangle is always opposite the largest angle of the triangle; the shortest side is opposite the smallest angle. Example 1:

a)

b) 65°

67°

90°

x 78°

35°

Since x ° + 65° + 90° = 180° x ° + 155° = 180° x ° + 155° = 180° x ° = 25° The third angle of the triangle is 25°

67° + 35° + 78° = 180°

c)

β

d) 63°43’ 46°39’

44.8°

β x Since 44.8° + β + β = 180° 44.8° + 2β = 180° 2β = 180° − 44.8° 2β = 135.2° 135.2° 2 β = 67.6° The third angle of the triangle is 67.6° β=

46°39’ + 63°43’ + x = 180° 110°22’ + x = 180° ⎛ Since 46°39' + 63°43' = 109°82' ⎞ ⎟ ⎜⎜ = 110°22' ⎟⎠ ⎝ x = 180° - 110°22’ ⎛ Since 180° − 110°22' = 179°60' −110°22' ⎞ ⎜⎜ ⎟⎟ = 69°38' ⎝ ⎠

x = 69°38'

Chapter 1: Geometry Review

Page 4

The Components of a Triangle Median: A line from the vertex of an angle to the midpoint of the opposite side.

AD is a median of ∆ABC.

Altitude: A line from the vertex of an angle perpendicular to the opposite side.

Angle Bisector: A line from the vertex of an angle to the opposite side which bisects that angle.

XW is an altitude of ∆XYZ.

PT is an angle bisector of ∆PQR. P

Y

A

W B

Q

Z

D

T

X

C

R

Naming the Sides and Angles of a Triangle Angles: There are three common systems of naming the angles of a triangle:

A

1. Using the vertices, with the vertex of the angle in the center.

α

2. Using a number or letter (often a Greek letter) inside the angle. 3. According to the vertex of the triangle at that angle. In the diagram the angle at the vertex A can be called ∠BAC, ∠ α, or ∠ A.

C B

Sides: There are two common ways to name the sides of a triangle: 1. According to two vertices it connects. The sides of ∆ABC (diagram above) are AB, BC and AC. 2. According to the vertex of the angle opposite the side. Each side of the triangle is named using the lower case letter corresponding to the letter of the vertex of the triangle opposite that side. B c

a

In the triangle ∆ABC the vertices of the triangle are A, B, and C and the sides of this triangle must be a, b, and c, such that: The side opposite the vertex A is called a. The side opposite the vertex B is called b. The side opposite the vertex C is called c.

A b

C We often use this second method in trigonometry because the name of each side of a triangle indicates its specific relationship to the angles of the triangle.

Chapter 1: Geometry Review

Page 5

Classifying Triangles 1. Types of Triangles Defined by Angle Size Acute Angled Triangle: All angles of the triangle are acute. Obtuse Angled Triangle: One angle of the triangle is obtuse.

Acute Triangle

Obtuse Triangle

Right Triangle: One angle of the triangle a right angle.

The other two angles of the triangle are acute and complementary. The side of the triangle opposite the right angle is called the hypotenuse. The other sides are called the legs of the triangle. The hypotenuse is always the longest side of the triangle. Pythagoras’ Theorem states the important relationship between the lengths of the sides of a right triangle.

A

b

C

hypotenuse

c

B

a

2. Types of Triangles Defined by Side Length Scalene Triangle: A triangle with three sides of different lengths. apex

Isosceles Triangle: A triangle with two sides of equal length.

Properties: ; The angles opposite to the equal sides are congruent. ; The third (unequal) angle of the triangle is called the apex. A line from the apex to the midpoint of the opposite side of the triangle bisects the apex angle from which it passes and is perpendicular to the opposite side. Any altitude of the triangle from the apex to the base is also a median and an angle bisector.

base

Equilateral Triangle: A triangle with three sides of equal length.

Properties: All properties of the isosceles triangle, and: 60°

; All angles measure 60°. ; Any altitude of the triangle is also a median and an angle bisector.

60°

; Any median or altitude of the triangle divides the triangle into 2 congruent right triangles, with angles of 30°, 60°, and.90°. We will use this special property in the study of Trigonometry.

60°

30°

60° Chapter 1: Geometry Review

30°

60° Page 6

Relationships between Triangles Congruent Triangles:

Two triangles are congruent if their three sides and the three corresponding angles are congruent. Example 2: C

The two triangles ∆ABC and ∆RPQ are congruent, written ∆ABC ≅ ∆RPQ, since

x

three corresponding angles are equal:

b

and the corresponding sides are equal:

R

a

∠ A = ∠ R, ∠ B = ∠ P, ∠ C = ∠ Q,

p A

a = r, b = p, and c = q.

q

c

x

B

(Note: the order in which the vertices are written when we name the triangles reflect the corresponding angles.)

Q

r P

Similar Triangles

Two triangles are similar if their angles are equal. The three corresponding pairs of sides are then in fixed proportion. Example 3:

The triangles ∆DEF and ∆ZXY are similar, written ∆ABC ≈ ∆ZXY, since ∠ D = ∠ Z, ∠ F = ∠ Y, then (using the angle sum of the triangle) we must have ∠ E = 50° = ∠ X. So three corresponding angles are equal.

E

1.7 cm

and: ED 1.7 1 = = XZ 5.1 3 DF 1.8 1 = = YZ 5.4 3 EF 2. 0 1 = = XY 6.0 3

2.0 cm

75° D

55° 1.8 cm

F

Y 55°

so the sides are in fixed proportion of 1:3. 5.4 cm

6.0 cm

75° X

Chapter 1: Geometry Review

5.1 cm

Z

Page 7

Exercise 18.1.2: 1.

Find x.

2. 57°

Find x in degrees, correct to one decimal place. 90°

x

45°

3.

Find α in degrees and minutes.

4.

Find y. 56°

α

65°40

5.

x

36.4°

y°

55°40

Label the sides.

6.

y°

Label the vertices. ?

R

d

?

e

?

P

7.

?

Q

?

f

?

Draw and label an isosceles right triangle ∆ABC in which ∠A = 90° and b = 3 cm. Find the size of the remaining angles and label these on the diagram.

8.

Draw and label a right triangle ∆UVW in which: ∠U = 90° ∠W = 59° u = 8.5 mm v = 6.3 mm

Find the size of the remaining angles and label these on the diagram. Chapter 1: Geometry Review

Page 8

18.1.3: RIGHT TRIANGLES Some Properties Of Right Triangles: THEOREM: Proof:

In any right triangle, the altitude from the right angle to the hypotenuse forms three similar triangles. C In ∆ABC, the line CD is an altitude from the right angle ∠ACB to the hypotenuse AB. b

a

All triangles are right triangles, so: ∠ACB = ∠ADC = ∠BDC = 90°

x

90 - x

A

Let ∠CAB = x°

D

c

B

Since ∆ABC is a right triangle, ∠CAB and ∠ABC are complementary, so: ∠ABC = 90 - x° Since ∆BCD is a right triangle, ∠ABC and ∠BCD are complementary, so: ∠ BCD = 90 - (90 - x°) = x° Since ∠ BCD = x°, and ∠BCD and ∠ACD are complementary, ∠ ACD = 90 - x° So ∠CAB = ∠CAD = ∠ BCD = x° ∠ ABC = ∠ACD = ∠ CBD = 90 - x°

and

Three angles of each triangle are equal to three angles of each of the others, hence the three triangles ∆ABC, ∆ADC, and ∆BCD are similar. The three triangles can be broken apart: C

C

C

(90 – x)° b

A

x°

b

a

a

(90 – x)°

x° c

x°

B

D

D

A

(90 – x)°

B

and then rotated, in order to see the corresponding sides more readily. A x°

A x°

C x°

C

Chapter 1: Geometry Review

(90 – x)°

B

D

(90 – x)°

C

D

(90 – x)°

B

Page 9

In similar triangles, the ratio of corresponding sides is fixed. Specifically, since ∆ABC and ∆ADC are similar, then: CD AD AC CD AC AD = = = BC AC AB BC AB AC Rearranging these ratios (see below) we get: AC AD = BC CD

BC CD = AB AC

AC AD = AB AC

Since ∆ABC and ∆BCD are similar, and ∆ADC and ∆BCD are similar, we can apply the same process and arrive at the final result that: AC AD CD = = BC CD DB

AB AC BC = = BC CD BD

AB AC BC = = AC AD CD

This result will be important to us when we derive the trigonometric ratios. ASIDE: Rearranging the Ratios: If Cross − multiply Divide both sides by CD ⋅ BC Canceling

Example 1: Solution:

CD AD = BC AC CD ⋅ AC = AD ⋅ BC CD ⋅ AC AD ⋅ BC = CD ⋅ BC CD ⋅ BC AC AD = BC CD

If CD is a altitude of ∆ABC (see diagram) find x. By the above theorem, the three right triangles formed by the altitude CD are similar. So BD CD = CD AD

C

Hence 4 x = x 16 x 2 = 4 ⋅ 16 x = 64 2

x B

4

D

16

A

x =8

Chapter 1: Geometry Review

Page 10

Pythagoras’ Theorem Originally Pythagoras’ Theorem dealt with area of squares.

Area = c

B

2

2

Area = a

Historically the theorem stated: “In any right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.”

c

a

c a

Since the area of a square on one of the sides of the triangle is equal to the square of the length of that side, it is logical to restate the theorem in its more useful form:

C

A

b 2

Area = b

b

A

PYTHAGORAS’ THEOREM

In any right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

c

b

That is, in any right triangle ∆ABC, where ∠C is the right angle, a2 + b 2 = c 2

C

B

a

There are a number of different proofs of Pythagoras’ Theorem. The proof given below depends upon the theory of similar right triangles. Proof of Pythagoras’ Theorem:

C

In a right triangle ∆ABC the line CD is an altitude from the right angle ∠ACB to the hypotenuse AB. By the theorem above, the three right triangles formed, ∆ABC, ∆ADC, and ∆BCD are similar and the sides are in fixed proportion.

b

Hence and

c a = a x a 2 = cx

c-x

A

c b = b c−x b 2 = c (c − x )

a

x

c

D

B

b 2 = c 2 − cx cx + b 2 = c 2

Substituting for cx, cx + b 2 = c 2 a2 + b 2 = c 2

Using Pythagoras’ Theorem to find the length of the hypotenuse, c: Example 2:

Solution:

Find the length of the hypotenuse of a right triangle ∆ABC if the two legs of the triangle are 6 and 8 units in length. Applying Pythagoras’ Theorem: a2 + b2 = c 2

B

62 + 82 = c 2 36 + 64 = c 2 100 = c 2 c = 100 c = 10 Chapter 1: Geometry Review

c

6

C

8

A

Page 11

Example 3:

Solution:

Find the length of the hypotenuse of an isosceles right triangle ∆ABC if the two equal sides of the triangle are 2 units in length. Applying Pythagoras’ Theorem:

B

a2 + b 2 = c 2 c

2 2 + 22 = c 2 4+4=c

2

2

8 = c2

c= 8 c=2 2

A

C

2

Using Pythagoras’ Theorem to Find the Length of a Leg of a Right Triangle: Example 4: Solution:

∆PQR has a right angle at ∠P. If q = 0.5 units and p = 1.3 units, find r. Applying Pythagoras’ Theorem: 2

2

q +r = p

2

0.52 + r 2 = 1.32 r 2 = 1.32 − 0.52

Q

1.3

r

r 2 = 1.69 − 0.25 r 2 = 1.44

P

R

0.5

r = 1.44 r = 1.2

Using Pythagoras’ Theorem in Solving Applications with Right Triangles: Example 5: Solution:

The diagonal of a cube is 2 3 units. Find its side length. Let the sides of the cube be x units in length. First consider the right triangle ∆BCD: x 2 + x 2 = BC 2

A

BC 2 = 2 x 2 BC = 2 x

x B

In the right triangle ∆ABC: AC 2 + BC 2 = AB2

x C

x

D

( )2

x 2 + 2x 2 = 2 3 3 x 2 = 12

x2 = 4 x=2

Therefore the side length of the cube is 2 unit.

Chapter 1: Geometry Review

Page 12

Example 6: Solution:

Find the area of the equilateral triangle which has sides of 2 units in length. Let AB be the base of the equilateral triangle ∆ABC, which we know to be 2 units in length. Draw an altitude CD, the height of ∆ABC of length h, creating two right triangles.

C

By the properties of equilateral triangle, AD = BD = 1 unit Applying Pythagoras’ Theorem to the right triangle ∆BCD: 2

2

h 2 + 12 = 22

h

h2 + 1 = 4 h2 = 3 h= 3

So the height of the triangle is

A

1

B

1

D

3 units and the area is:

A = 1 bh 2

= 1 ⋅2⋅ 3 2

= 3

Therefore the area of the equilateral triangle is Example 7:

Solution:

3 square units.

A rectangular field is three times as long as it is wide and has a diagonal measure of 40 meters. Find the length of the field to the nearest tenth of a meter. 3x

Let the field have a width of x meters. Then its length is 3x meters. Using the upper right triangle of the rectangle, and applying Pythagoras’ Theorem,

40 m

x

x 2 + (3 x )2 = 402 x 2 + 9 x 2 = 1600 10 x 2 = 1600 x 2 = 160 x = 160 x = 12.649 3 x = 37.947

Hence the field has an approximate width of 12.6 m and length of 37.9 m. Example 8:

Solution:

Chapter 1: Geometry Review

M

At noon two airplanes are 100 miles from the same airport. One plane is due north of the airport and traveling south at 450 miles per hour; the other airplane is due west and traveling east at 600 miles per hour. At what time, to the nearest minute, will these airplanes be 20 miles apart? After t hours:

the plane traveling south has traveled 450t miles, and is (100 – 450t) miles from the airport;

the plane traveling east has traveled 600t miles, and is (100 – 600t) miles from the airport.

450t 100 miles 20 miles 100 - 450t

M

600t

100 - 600t

AIRPORT

100 miles

Page 13

The two airplanes and the airport are the vertices of a right triangle. Applying Pythagoras’ Theorem to the right triangle,

(100 − 450t )2 + (100 − 600t )2 = 202 10000 − 90000t + 202500t 2 + 10000 − 120000t + 360000t 2 = 400 20000 − 210000t + 562500t 2 = 400 562500t 2 − 210000t + 19600 = 0 562500t 2 210000t 19600 0 − + = 100 100 100 100 5625t 2 − 2100t + 196 = 0

Using the quadratic formula t=

2100 −

(− 2100)2 − 4 ⋅ (5625) ⋅ (196)

2 ⋅ 5625 ≈ 0.186667 hours = 0.186667 × 60 min utes ≈ 11.2

Therefore, after approximately 11 minutes the planes will be 20 miles apart. A ladder 12 ft long is placed along a wall, with the base of the ladder 2 ft from the base of the wall. How far up the wall, to the nearest tenth of a foot, will the ladder reach?

Example 9:

Let the height that the ladder reaches up the wall be x ft.

Solution:

Applying Pythagoras’ Theorem to the right triangle formed by the ladder, the ground and the wall: x 2 + 22 = 12 2 x 2 + 4 = 144

12 ft x

x 2 = 140 x = 140 x ≈ 11.8

2 ft

The ladder reaches 11.8 ft up the wall.

Special Triangles. The Isosceles Right Triangle: 45°- 45°- 90°

Consider a right isosceles triangle, with two equal sides of length 1 unit . Since the equal angles sum to 90°, they are both 45°.

45°

Using Pythagoras’ Theorem to find the length of the third side of the triangle, x:

x

1

12 + 12 = x 2 x2 = 2 x= 2

45° 1

Hence the hypotenuse has a length of

Chapter 1: Geometry Review

2 units.

Page 14

Since any isosceles right triangle must have angles of 45°- 45°- 90° it will be similar to this triangle and will therefore have sides in the same fixed ratio. This leads us to the following theorem: The sides of any isosceles right triangle are in the fixed ratio of 1 : 1 : 2 . That is, in any isosceles right triangle the hypotenuse is of the legs of the triangle.

45°

2 times the length 2

1

45° 1

If the sides of a square are

Example 10:

2 units long, find the length of the diagonal.

Let the diagonal of the square be x units in length.

Solution:

2

The diagonal of a square divides the square into two isosceles right triangles. Therefore the length of the diagonal is as long as the length of the sides. So

2

2 times

x = 2⋅ 2 =2

The square has a diagonal of 2 units length. The 30° - 60° - 90° Triangle

Consider an equilateral triangle, with sides of length 2 units. [Diagram 1]

60° 2

2

60°

60° 2

30° 30°

An altitude will bisect both the angle and the opposite side. [Diagram 2] We now have a triangle divided into two right triangles with angles of 30°, 60° and 90°.

2

2

60°

60° 1

Diagram 1

1

Diagram 2

Using Pythagoras’ Theorem to find the length of the third side of the triangle, x [Diagram 3] 30° x

30°

x 2 + 12 = 22 x2 + 1 = 4

2

x2 = 3 x= 3

60°

Diagram 3

2

3

60° 1

Therefore, the sides of the 30o - 60o - 90o triangle (given in the order of the sides opposite the angles) are 1,

Diagram 4

3 , 2.

Furthermore, any right triangle with one angle of either 30° or 60° will be similar to the above and will therefore have sides in the same fixed ratio. This leads us to the following theorem: In any 30o - 60o - 90o triangle the ratio of the leg opposite the 30° angle to the leg opposite the 60° angle to the hypotenuse is 1 : 3 : 2 Chapter 1: Geometry Review

Page 15

Example 11: Solution:

In ∆ABC, ∠C = 90o , ∠A = 30o and b = 6 units. Find a and c. Compare ∆ABC to a 30o - 60o - 90o triangle ∆PQR with sides of 1 unit, units. Since

BC AC

=

QR PR

3 units and 2

then A

a 1 = 6 3 6 a= 3 6 3 = ⋅ 3 3 6 3 = 3 a=2 3

P

30°

30°

6

c 2

3

60°

60° C

a

B

Q

1

R

Since the hypotenuse of the triangle is twice the length of the side opposite the 30° angle, then c 2 = a 1 c

=2 2 3 c=4 3

Example 12:

Solution:

A cell tower has a guy wire attached to the top of it which makes a 60° angle with the ground. If the wire is anchored 30 meters from the base of the tower, how tall is the tower to the nearest tenth of a meter? Let the cell tower be x meters in height. Since the tower, ground and wire form a 30o - 60o - 90o triangle, the ratio of the sides is such that: the side opposite the 60o angle o

the side opposite the 30 angle

=

3 1

then: x 3 = 30 1 x = 30 3 ≈ 51.96

Therefore the tower is approximately 52.0 meters tall.

Chapter 1: Geometry Review

x

60° 30 m

Page 16

Using Pythagoras’ Theorem to Prove a Triangle has a Right Angle:

The converse of Pythagoras’ Theorem can also be proved, and is useful in establishing the existence of a right angle or perpendicular relationship between lines. PYTHAGORAS’ THEOREM - CONVERSE

A

If a triangle has sides such that the square of the longest side is equal to the sum of the squares of the other two sides, then it is a right triangle.

Example 13: Solution:

c

b

That is, if ∆ABC has sides a, b, c, such that a 2 + b 2 = c 2 then ∆ABC is a right triangle. Furthermore we know ∠C is the right angle, and c is the hypotenuse.

C

a

B

Is the triangle with sides measuring 4 cm, 8 cm and 10 cm a right triangle? Since the longest side of the triangle is 10 cm, let a = 4, b = 8 and c = 10. a 2 + b 2 = 4 2 + 82 = 16 + 64

8 4

= 80 c 2 = 102 = 100

10

Therefore a 2 + b 2 ≠ c 2 and the triangle is not a right triangle.

Example 14:

Pythagorean Triads

A set of numbers Triad.

{ a, b , c} which obeys the rule

a 2 + b 2 = c 2 is called a Pythagorean

{ 3, 4, 5} is a Pythagorean Triad since 32 + 42 = 9 + 16 = 25 = 52

{ 1, 2, 3} is not a Pythagorean Triad since 12 + 22 = 1 + 4 =5 ≠ 32

Chapter 1: Geometry Review

Page 17

Exercises 18.1.3: 1.

A right triangle ∆ABC has a right angle at ∠C. Find the length of the hypotenuse c if: (i) a = 10, b = 24

2.

(iii) a = b = 3

A right triangle ∆ABC has a right angle at ∠C. Find the length of the side a if: (i) b = 24, c = 25

3.

(ii) a = 1, b = 2

(ii) b = 1.5, c = 2.5

1 2

(iii) b =

,c=

3 2

Find x in each of the following right triangles: (i)

(ii) x

30

n

(iii)

0.05

x x

0.13

2n

40

4.

C

If BD = 9 and AD = 25 find the area of ∆ABC

B

9

D

25

A

5.

A runner and a bicyclist leave home at the same time. The runner is traveling due south at 6 miles per hour, and the cyclist is traveling due west at 16 miles per hour. How far apart (to the nearest tenth of a mile) are they after 15 minutes?

6.

Which of the following are Pythagorean Triads? (i)

{ 9, 40, 41}

(ii)

{ 16, 36, 49} (iii) { 21, 72, 75}

7.

∆ABC has a right angle at ∠C, ∠A = 30° and a =

8.

A square has a diagonal of length 16 cm. Find its side length correct to the nearest tenth of a cm.

Chapter 1: Geometry Review

2 . Find the length of b and c.

Page 18