Geometry 201

10.0 Compute areas of polygons, including rectangles, scalene triangles, equilateral triangles, rhombi, parallelograms, and trapezoids. (cont)

Example 2 Find the area of Trapezoid ABCD

Solution: The area of a trapezoid is • height • (base 1 + base 2)

= ( 6) (8 + 18)

= ( 6) ( 26)

= 78

Example 3 Find the area of the triangle shown.

Solution: The area of a triangle is • b • h

The base is 8, but we do not have the height. Draw an altitude. The altitude cuts the base into two equal parts. Now find the height using Pythagoras 42 + (height) 2 = 82 16 + h2 = 64 h2 = 48 h = =

So, Area of triangle = • 8 •

= Homework 1. Find the area of the trapezoid shown.

2. Find the area of the trapezoid shown.

(cont)

8 8

8

8 8

4 4

x

y

(4,0)

(3,2) (0,2)

13

18

13

8

8

6

18 B

C D

A

Geometry 202

2 2

2

2 2

2 2

2

10.0 Compute areas of polygons, including rectangles, scalene triangles, equilateral triangles, rhombi, parallelograms, and trapezoids. (cont)

3. The square shown has side lengths of 10. Four triangles are removed from the corners as shown. What is the area of the shaded region? (Hint: Find the area of the square and subtract the areas of the 4 triangles.) 4. Rhombus ABCD has side lengths of 10. BE = 6, CE = 8 Find the area of ABCD.

B C

D A

6 8 E

Geometry 203

11.0 Determine how changes in dimensions affect the perimeter, area, and volume of common geometric figures and solids. Rule If two polygons are similar, and the sides are in the ratio of a: b, then

Perimeter Ratio = side ratio a : b

Area Ratio = (side ratio)2 a2 : b2

Volume Ratio = (side ratio)3 a3 : b3 Example 1 The circumferences of the bases of two similar cylinders are in the ratio of 3: 5. Find the ratio of their lateral areas.

Solution: Areas are in the ratio of a2 : b2 So, 32 : 52 = 9 : 25 Example 2 If the sides of a regular square pyramid are tripled, what happens to the volume? Solution: Assume the sides of the original pyramid have a length of 1 If the sides are tripled, then the new side length would be (3 • 1) or 3. The side ratio would then be 1: 3 The formula says that the Volumes would be in the ratio of a3 : b3 So, 13 : 33 = 1 : 27. The volume would be multiplied by 27 Example 3 Two cartons and their dimensions are shown. How many more cubic centimeters can the larger one hold? Solution: Find the volume of each and then subtract.

Larger Carton Area Smaller Carton Area

(Base Area) • (Height) (Base Area) • (Height) L • W • H L • W • H 4 • 4 • 8 3 • 3 • 5 64 45

64 – 45 = 19 more cubic centimeters in the larger carton.

(cont)

4

8

4

3

5

3

Geometry 204

A

B

D

E

C

11.0 Determine how changes in dimensions affect the perimeter, area, and volume of common geometric figures and solids. (cont)

Homework

1. If the sides of two similar triangles are in the ratio 4:9, what is the ratio of their perimeters? 2. If the sides of a cube are multiplied by 4, the volume of the cube is multiplied by how much?

3. Find the ratio of the lateral area of a cylinder with radius ‘r’ to its volume.

Hint: Use the cylinder formulas found in Worksheet 9.0.

Substitute the formulas into the ratio and then cancel common units.

4. The dimensions of Cone A are shown.

Cone B’s dimensions are double those of Cone A.

Cone C’s dimensions are double those of Cone B.

How many times greater is the volume of Cone C when compared to the volume of Cone A? Cone A Cone B Cone C

5. In the diagram AE = 5 and ED = 3. What is the ratio of the areas of ∆ABE to ∆ACD?

5

4

3

Geometry 205

12.0 Find and use measures of sides and of interior and exterior angles of triangles and polygons to classify figures and solve problems. Rule Triangles can be classified in two ways. By their sides or by their angles.

BY THEIR SIDES Scalene – no congruent sides.

Isosceles – two congruent sides Equilateral – three congruent sides

BY THEIR ANGLES Acute- all angles measure less than 90º

Right – one angle measures exactly 90º

Obtuse – one angle measures more than 90º

Equiangular – all angles measure the same (60º)

Rule An exterior angle of a triangle is formed by a side of the triangle and an extension of another side. The formula to find the measure of an exterior angle is

Exterior Angle Measure

Exterior = sum of the two non-adjacent interior angles of the triangle In the triangle to the right, 4 is an exterior angle. It’s measure is equal to the sum of the 2 interior angles to which it is not adjacent (not next to). m 4 = m 2 + m 3 Example 1 In ∆DEF, find the measure of the exterior angle at F.

Solution: Exterior = sum of two non-adjacent interior

(8x + 15)° = (3x + 20)° + (4x + 5)°

(8x + 15)° = (7x + 25)° x = 10° m F = (8x + 15)° = 8 (10) + 15 = 95º

4 2

1 3

D E

F (8x+15)°

(3x+20)° (4x+5)°

Geometry 206

12.0 Find and use measures of sides and of interior and exterior angles of triangles and polygons to classify figures and solve problems. (cont)

Rule A Polygon is a many sided figure. It has the same number of angles as sides. Listed below are some of the more common polygons whose names you should know.

Triangle 3 sides Quadrilateral 4 sides Pentagon 5 sides Hexagon 6 sides Heptagon 7 sides Octagon 8 sides Nonagon 9 sides Decagon 10 sides Dodecagon 12 sides Rule The formula we use to find the sum of all the interior angles of a polygon with ‘n’ number of sides is Sum of All Interior Angles = 180(n – 2)

Rule An exterior angle of a polygon is formed by extending one side of the polygon. In the diagram, angle 1 is an exterior angle. It was formed by extending side ED. Rule No matter what type of polygon, the sum of the exterior angles (one angle at each vertex) is ALWAYS equal to 360º. Sum of Exterior Angles = 360º Note: While it is possible to draw two exterior angles at each vertex, the sum of the exterior angles uses only ONE exterior angle at each vertex. 1 + 2 + 3 + 4 + 5 + 6 = 360º

Rule A polygon is called a regular polygon when all of its sides are of the same length and all of its angles are of the same measure. . Rule The formula we use to find the measure of each interior angle of a regular polygon with ‘n’ sides is

Each interior measure of a regular polygon =

B

D

C

E

F

A

1

4

5

3 2

6 1

Geometry 207

12.0 Find and use measures of sides and of interior and exterior angles of triangles and polygons to classify figures and solve problems. (cont) Rule

The formula we use to find the measure of each exterior angle of a regular polygon with ‘n’ sides is

Each exterior measure of a regular polygon =

Example 2 Find the number of degrees in each interior angle of a regular pentagon.

Solution: We know that the number of sides for a pentagon is 5. We use the formula and use 5 for ‘n’

= = = 108º

Example 3 Each interior angle of a regular polygon measures 135º. How many sides does the polygon have?

Solution: Use the formula: interior angle measure =

Substitute 135 =

Cross multiply 135n = 180(n – 2) 135n = 180n – 360 – 45n = – 360 n = 8

The polygon has 8 sides.

Homework 1. Find the measure of each interior angle of a regular hexagon.

2. The measure of each exterior angle of a regular polygon is 45º. How many sides does the polygon have? 3. What type of polygon has the sum of the interior angles equal to 360º and the sum of the exterior angles equal to 360º?

4. In ∆ABC, A = 48º, C = 24º, what type of triangle is ABC? (Multiple choice) a. acute b. right c. obtuse

5. The vertex angle of an isosceles triangle measure 8 times the measure of a base angle. Find the measure of a base angle.

6. In ∆DEF, D = 37º and F = 56º. Find the measure of an exterior angle at E.

7. Find the sum of the measures of all the interior angles in a 9-sided polygon.

8. In hexagon ABCDEF, ,

B + C + D + E + F = ____

C

A

B

D

E

F

Geometry 208

13.0 Prove relationships between angles in polygons by using properties of complementary, supplementary, vertical, and exterior angles.

Rule Complimentary angles are two angles with a sum of 90º Supplementary angles are two angles with a sum of 180º

Vertical angles are congruent

Example 1

Find the measure of x in the figure.

Solution: Vertical angles are congruent so the missing angle in the pentagon has measure x. In a pentagon the total number of degrees is (5-2) (180º) or 540º. So, 100º + 150º + 60º + 120º + x = 540º 430º + x = 540º x = 110º Example 2

In the figure 1 and 3 are compliments. m 1 = 63º

Find the measure of 2

Solution: Since 1 and 3 are compliments, they sum to 90º.

1 + 3 = 90 63º + 3 = 90º 3 = 27º

By vertical angles, in the top triangle 27º + 10º + 2 = 180º

37º + 2 = 180º

2 = 143º

(cont)

1 2

10º

3

1 2

100° 150° 60°

120° x°

100° 150° 60°

120° x°

x°

10º 27º

63º 2 2

Geometry 209

13.0 Prove relationships between angles in polygons by using properties of complementary, supplementary, vertical, and exterior angles. (cont)

Homework 1. bisects EFB. EFB and DFC are vertical angles with

DFC = (2x – 10)° and EFB = (x + 36)° Find AFB

2. In the figure, AD and BD intersect at G. BGC = 65º and 5 = 55º, find AGF 3. In the figure, if , then x = _____ 4. In the figure, find the value of x. 5. In the figure 1 and 2 are complements. Find m 2. 6. ABCED is a regular pentagon. Find the measure of x.

42°

81°

3x°

1 2

40º

72º

D

E

C

A

B x

D

B

A

C

E

F

D

B

A C

E 25°

x° 65°

D

C

B

A

F

E G

2

1

3

4 5

Geometry 210

14.0 Prove the Pythagorean theorem.

Over 2,500 years ago, a Greek mathematician named Pythagoras developed the relationship between the hypotenuse and the legs in a right triangle. This relationship can be stated as a, b are legs c is the hypotenuse There are several different ways of proving the Pythagorean theorem. Here’s one way. Using the given a, b, c triangle draw 4 of them and put them together as shown. Now, let’s find the area of the new larger figure. There are two ways you could find it.

Area Method 1 The large figure is a square with sides lengths a+b The formula for area of a square is side • side So, (a+b) • (a+b)

After foiling Area = a2 + 2ab + b2

Area Method 2 The area of the large figure can be found by summing the areas of all the interior regions.

Four of the regions are triangles. One triangle area is • base • height

= • a • b

Four triangle areas = 4 • ( • a • b )

= 2 a b The 5th region is a square, its area is side • side

= c • c = c2

Summing all 5 regions together, we get Area = 2ab + c2 --------------------------------------------------------------------------------------------------------------------- Now, we know that Area = Area So Method 1’s answer and Method 2’s answer must be

equal. So, a2 + 2ab + b2 = 2ab + c2

Subtracting the common term from each side a2 + b2 = c2 This is what the Pythagorean Theorem states.

a c

b

c c

a

b

b

a b

a a

b c c a c

b

a

a

b a

a

b

b

b

a2 + b2 = c2

1

3 4

2

5 c

c

b a

Geometry 211

14.0 Prove the Pythagorean theorem. (cont)

Homework In proving the Pythagorean Theorem, many properties or concepts were used. Answer ‘yes’ or ‘no’in each of the following as to whether the item was used. 1. The area of a square equals side2

2. The inner square and the large outer square are congruent

3. The four triangles are congruent.

4. The area of a triangle equals • a • b

5. The triangles are 30º, 60º, 90º triangles

6. The area of a large region is equal to the sum of the areas of its internal regions.

7. The total of the areas of the four triangles is equal to the area of the inner square.

Geometry 212

9

7

x

15.0 Use the Pythagorean theorem to determine distance and find missing lengths of sides of right triangles. Rule In any right triangle, the lengths of the sides are in the relationship where

leg2 + leg2 = hypotenuse2 This is normally stated as:

a2 + b2 = c2 Example 1 Find the measure of AB in ∆ABC. Solution: 6 2 + 8 2 = (AB)2 36 + 64 = (AB)2 100 = (AB)2 10 = AB

Example 2 A triangle has sides 6, 7, and 10. Is it a right triangle?

Solution: If the triangle is a right triangle, the longest side must be the hypotenuse.

Now check to see if the Pythagorean Theorem is true.

6 2 + 7 2 ? 10 2

36 + 49 ? 100

85 100 Since the Pythagorean Theorem is NOT true, this triangle is NOT a right triangle.

Example 3 A ramp was constructed to load a truck. If the ramp is 9 feet long and the horizontal distance from the bottom of the ramp to the truck is 7 feet, what is the vertical height of the ramp?

Solution: Since the ramp is described as having horizontal and vertical measurements, a right angle is implied. Solve using the Pythagorean Theorem.

a 2 + b 2 = c 2

7 2 + x 2 = 9 2

49 + x 2 = 81

x 2 = 32

x = or

(cont)

leg

hypotenuse leg

b

c a

8

6

C B

A

Geometry 213

15.0 Use the Pythagorean theorem to determine distance and find missing lengths of sides of right triangles. (cont) Example 4 To get from point A to point B you must avoid walking through a pond. To avoid the pond, you must walk 30 meters south and 40 meters east. How many meters would be saved if it were possible to walk through the pond?

Solution: When walking south 30 meters and east 40 meters you walk a total of 70 meters. If you walked through the pond the distance could be found using Pythagoras. 30 2 + 40 2 = (through distance)2 900 + 1600 = (through distance)2 2500 = (through distance)2 50 = through distance

The number of meters saved would be 70

Homework 1. A right triangle has one leg with length 6 and the other leg with a length of 4. What is the length of the hypotenuse? 2. A baseball diamond is a square with sides of 90 feet. What is the shortest distance between first base and third base?

3. Two joggers run 8 miles north and then 5 miles west. What is the shortest distance, to the nearest tenth of a mile, they must travel to return to their starting point? (Multiple Choice)

a. 5.7 b. 8.5 c. 9.4 d. 13.1

4. In a computer catalog, a computer monitor is listed as being 26 inches. This distance is the diagonal distance across the screen. If the screen measures 10 inches in height, what is the actual width of the screen? 5. A tent has sides that are both 5 feet long and the bottom of the tent is 6 feet across. What is the height of the tent, at its tallest point? Hint: Draw the triangle. The height cuts the base in half.

1st Base

2nd Base

3rd Base

Home

5 feet 5

feet

6 feet

5

3

x

40

30 X

Geometry 214

16.0 Perform basic constructions with a straightedge and compass, such as angle bisectors, perpendicular bisectors and the line parallel to a given line through a point off the line. For constructions, we will construct geometric figures using only a straightedge and a compass

Copy A Line Segment 1. To draw a line segment congruent to , begin by drawing a reference line, l , with endpoint K

2. Place the point of the compass on point A. Stretch the compass so that the pencil is exactly on B. Make a mark. 3. Without changing the span of the compass, place the compass point on K and swing the pencil so that it crosses the reference line. Label this crossing point Y.

Your copy, KY, and the original, AB, are congruent line segments.

(cont)

A B

K l �

A B

K l �

Y

Geometry 215

For constructions, we will construct geometric figures using only a straightedge and a compass.

Copy An Angle

1. To draw an angle congruent to A, begin by drawing a ray with endpoint D.

2. Place the compass on point A and draw an arc across both sides of the angle. Without changing the compass radius, place the compass on point D and draw a long arc crossing the ray. Label the three intersection points as shown.

3. Set the compass so that its radius is BC. Place the compass on point E and draw an arc intersecting the one drawn in the previous step. Label the intersection point F.

4. Use the straightedge to draw ray DF.

EDF BAC

Geometry 216

For constructions, we will construct geometric figures using only a straightedge and a compass.

Construct the Perpendicular Bisector Of A Line Segment

1. Begin with line segment XY.

2. Place the compass at point X. Adjust the compass radius so that it is more than (1/2)XY. Draw two arcs as shown here.

3. Without changing the compass radius, place the compass on point Y. Draw two arcs intersecting the previously drawn arcs. Label the intersection points A and B.

4. Using the straightedge, draw line AB. Label the intersection point M. Point M is the midpoint of line segment XY .

Segment XF has now been bisected and XM YM

M

Geometry 217

For constructions, we will construct geometric figures using only a straightedge and a compass

Bisect An Angle 1. Let point P be the vertex of the angle. Place the compass on point P and draw an arc across both sides of the angle. Label the intersection points Q and R.

2. Place the compass on point Q and draw an arc across the interior of the angle.

3. Without changing the radius of the compass, place it on point R and draw an arc intersecting the one drawn in the previous step. Label the intersection point W.

4. Using the straightedge, draw ray PW. This is the bisector of QPR.

has now been bisected.

Geometry 218

For constructions, we will construct geometric figures using only a straightedge and a compass

Construct a line parallel to a given line through a point off the line

(We are going to construct a corresponding angle at point P that is congruent to angle Q)

1. Begin with point P and line k.

2. Draw an arbitrary line through point P, intersecting line k. Call the intersection point Q. Now the task is to construct an angle with vertex P, congruent to the angle of intersection.

3. Center the compass at point Q and draw an arc intersecting both lines. Without changing the radius of the compass, center it at point P and draw another arc.

4. Set the compass radius to the distance between the two intersection points of the first arc. Now center the compass at the point where the second arc intersects line PQ. Mark the arc intersection point R.

5. Line PR is parallel to line k.

PR is the line through P parallel to line k

Geometry 219

Homework: 1. Construct a segment whose measure is twice the measure of the segment below. 2. Segment and are given. Use them to construct + 3. Construct the perpendicular bisector of LM. 4. Construct a line through point X perpendicular to line m 5. Construct an angle congruent to 6. Construct a 90° angle. 7. Describe the procedure that could be used to construct a 45° angle. 8. This drawing shows how to ____ A. Bisect segment AB. B. Find the perpendicular bisector to C. Copy segment AB D. Construct an angle 9. This drawing shows how to ______ A. Copy an angle B. Copy a segment C. Bisect an angle D. Find the perpendicular to a given line

(cont)

W X Y Z

L M

m

X

E F

D

A B

A B

C

D E

F

Geometry 220

10. This drawing shows how to ______ A. Construct an angle bisector B. Construct a perpendicular bisector of a segment C. Copy a segment D. Copy an angle. 11. This drawing shows how to ____ A. Bisect a segment B. Bisect an angle C. Copy an angle D. Construct an altitude. 12. This drawing shows how to ____ A. Bisect an angle B. Construct the perpendicular bisector of an angle C. Copy a segment D. Construct a line parallel to a given line 13. To construct the angle bisector of , point Y would be constructed by opening your compass _______. A. more than half of B. less than half of C. equal to D. equal to 14. Which procedure should be followed to construct a 90° angle? A. Construct 3 angles of 30° B. Construct 2 parallel lines. C. Construct the perpendicular bisector of a line D. Find the midpoint of a line segment.

A

C B

Y

Geometry 221

17.0 Prove theorems by using coordinate geometry, including the midpoint of a line segment, the distance formula, and various forms of equations of lines and circles.

Midpoint Formula Distance Formula Slope Formula

(x,y) = d = m =

Parallel Lines Perpendicular Lines

same slope negative reciprocal slopes (to get a negative reciprocal of a number, flip the number over and change the sign) Slope Intercept Form of a Line Equation of a Circle with Center at (h,k)

and radius r y = mx + b (x – h)2 + (y – k)2 = r2

Example 1 What is the equation of a circle whose diameter is 24 and whose center is at the origin?

Solution: If the diameter is 24, the radius is 12. The center is at (0,0) Plugging into the formula

for a circle (x – 0)2 + (y – 0)2 = 12 2

x 2 + y 2 = 144

Example 2 The point (5,4) lies on a circle. What is the length of the radius of this circle if the center is located at (3,2) ? Solution: Using the distance formula: = = =

(cont)

Geometry 222

17.0 Prove theorems by using coordinate geometry, including the midpoint of a line segment, the distance formula, and various forms of equations of lines and circles. (cont) Example 3 ∆ABC has coordinates A(0, 0), B (2,3) , C (4, 0). Prove ∆ABC is isosceles.

Solution: If a triangle is isosceles it has 2 sides congruent. Find the length of all 3 sides of ∆ABC and if it is isosceles, 2 sides will be the same length.

Using A(0, 0), B (2,3) find AB = =

Using A(0, 0), C (4, 0) find AC = = 4

Using B(2,3), C (4, 0) find BC = =

Since two sides have lengths , the triangle is isosceles.

Example 4 ABCD is a square with the given coordinates. Find the coordinates of point D.

Solution: The x-coordinate of point D gives the right/left movement. The y-coordinate of point D gives the up/down movement. Point D is ‘n’units right and ‘n’units up. The coordinates are (n,n) Homework 1. The coordinates of rectangle ABCD are A(0,2), B (4,8), C(7, 6) and D (3, 0). Show that the diagonals are equal in length. 2. The coordinates of a quadrilateral are A (1,1), B (2,5) , C (5,7), and D (7,5). Which statement would prove that ABCD is a trapezoid? (multiple choice) a. Find the distance of all 4 sides and show 2 sides are congruent b. Find the slopes of all 4 sides and show 2 sides are parallel and the other 2 sides are not parallel. c. Find the lengths of the 2 diagonals and show they are congruent d. Show the diagonals bisect each other. 3. State the coordinates of the center of this circle. (multiple choice) (x + 3) 2 + (y – 5) 2 = 16

a. (–3, 5) b. (–3, – 5) c. (3, – 5) d. (3, 5)

(cont)

Geometry 223

17.0 Prove theorems by using coordinate geometry, including the midpoint of a line segment, the distance formula, and various forms of equations of lines and circles. (cont) 4. ∆ABC is shown. Which statement would prove that ∆ABC is a right triangle? (multiple choice) a. Show distance from A to B = distance from B to C b. Show distance from A to B = distance from A to C c. Show (slope of AB) = (slope of AC) d. Show (slope of AB) • (slope of AC) = – 1 5. Give the coordinates of point P in parallelogram LMNP.

6. Give the coordinates of the point of intersection of the diagonals.

B

C

A

N (c,0)

L (b,d)

M (a,0)

P

(n,n)

Geometry 224

18.0 Know the definitions of the basic trigonometric functions defined by the angles of a right triangle. Know and use elementary relationships between them.

Rule In a right triangle there are three relationships called the sine (sin), cosine (cos), and tangent (tan).

SOH CAH TOA

sin = cos = tan =

Example 1

In ∆ABC, find cos A

Solution: By the formula, cos A = =

The hypotenuse is not given in the figure. We must find it using Pythagoras 82 + 152 = hyp2 64 + 225 = hyp2 289 = hyp2

17 = hyp

So, cos A = =

Example 2

If cos x = , find sin x

Solution: Since cos = , we know that = ,

so the side adjacent to x is 3 and hypotenuse is 5 Find the third missing side, 32 + leg2 = 52 9 + leg2 = 25 leg2 = 16 leg = 4

sinx = =

(cont)

x°

C B

A

8

15

x° 3

5

4

x° 3

5

Geometry 225

18.0 Know the definitions of the basic trigonometric functions defined by the angles of a right triangle. Know and use elementary relationships between them. (cont)

Example 3 Determine if the following statement is true or false.

sin A =

Solution: Draw a right triangle and label one of the angles A. (angle A cannot be the right angle)

Since no numbers were given let’s use the common 3,4,5. (Remember the largest is the hypotenuse) Now, using SOH CAH TOA, get the fraction values.

sin A = cos A = tan A =

Now plug the fractions into the original problem and see if the two sides are the

same.

sin A =

=

=

Change divide to multiply by the reciprocal =

= which is

FALSE

(cont)

A 3

5 4

Geometry 226

18.0 Know the definitions of the basic trigonometric functions defined by the angles of a right

triangle. Know and use elementary relationships between them. (cont) Example 4

M is a right angle in ∆LMN, where m L = 15º. Select the one true statement below.

a. (MN) sin 15º = NL b. (ML) sin 15º = MN

c. (NL) sin 15º = MN d. (LN) sin 15º = ML

Solution: First sketch the triangle.

Since all the choices involve sin 15º, we know sin =

From the triangle, put the letters of the side opposite and the letters of the hypotenuse

sin L =

Now, take each multiple choice and get sin 15º by itself and find the one that is

Choice ‘a’ (MN) sin 15º = NL sin 15º = not a match

Choice ‘b’ (ML) sin 15º = MN sin 15º = not a match

Choice ‘c’ (NL) sin 15º = MN sin 15º = is a match. So choice c

(cont)

N

M L 15°

Geometry 227

18.0 Know the definitions of the basic trigonometric functions defined by the angles of a right triangle. Know and use elementary relationships between them. (cont)

Example 5 In ∆ABC, cos A = 0.9, find 3 possible side lengths for side AC.

Solution: cos A = We need to write 0.9 as a fraction, so 0.9 =

cos A = = Putting the numbers on the triangle

One possible side length for AC is 10

A second possibility we could write as . Then the triangle looks like

Another possible side length for AC is 20

A third possibility we could write as . Then the triangle looks like

A third possible side length for AC is 30

Homework Using ∆JKL find the requested values in problems 1 – 3 1. sin J 2. tan2L 3. sin L • cos L

Using ∆CMK, decide if each statement is true or false in problems 4-6.

4. tan C =

5. sin2 M + cos2 M = 1 6. (CM) cos C = CK 7. Using ∆CMK find the numerical fraction value of sin M

if tan M =

8. In ∆UVW, tanW = 0.4 What is the length of ? (multiple choice) a. 10 b. 8 c. 20 d. 1

A

B C

9

A

B C

10

18

A

B C

20

27

A

B C

30

6

L

J K

10

M

C K

M

C K

V 20

U

W

Geometry 228

19.0 Use trigonometric functions to solve for an unknown length of a side of a right triangle, given an angle and a length of a side. RULE In a right triangle there are three relationships called the sine (sin), cosine (cos),

and tangent (tan).

SOH CAH TOA sin = cos = tan =

Example 1 Find the length of side x in right triangle ABC. Solution: In relation to the given angle of 40º the two sides are opposite the angle and adjacent to the angle

The trig relationship for opposite and adjacent is tan. tan =

Substituting the numbers we write tan 40º =

Using a calculator we can find the tan 40º to be the decimal approximately 0.84

We then substitute the decimal value in for tan 40º and get. 0.84 =

Now cross multiply and solve for x 0.84 =

x = 16.8

(cont)

C B

A

x

20 40°

Geometry 229

19.0 Use trigonometric functions to solve for an unknown length of a side of a right triangle, given an angle and a length of a side. (cont) Example 2 SBC puts up a 72 foot pole. To help support the pole, a wire is attached on the side. The wire forms a 65º angle with the ground as shown. How long is the wire? sin 65º 0.91 cos 65º 0.42 tan 65º 2.14 Solution: In relation to the angle the two sides are opposite and the hypotenuse. The trig

function with these two is the sin. sin =

Substituting the numbers we write sin 65º =

You don’t need a calculator because the sin 65º is given. (The cos 65º and the tan 65º are also given but they are not needed.)

Substituting in the decimal value for sin 65º you get 0.91 =

Now, cross multiply. 0.91x = 72

Solve for x x =

x = 79.1 The wire is 79.1 feet long.

Example 3 If you were finding the length of side x in ∆LMN, which equation would you use? (multiple choice) a. x = 17 sin 37º b. x = 17 cos 37º

c. x = 17 tan 37º d. x =

Solution: In relation to the given angle of 37º the two sides are adjacent and hypotenuse.

The trig relationship for adjacent and hypotenuse is cos. cos =

Substituting the numbers we write cos 37º =

Now solve for x by cross multiplying cos 37º =

17 cos 37º = x This is choice ‘b’.

72

65°

37°

17

M

N

L x

Geometry 230

19.0 Use trigonometric functions to solve for an unknown length of a side of a right triangle, given an angle and a length of a side. (cont) Homework 1. Find the length of side x in right triangle PQR. 2. Tell which equation you would use to find the length of side x in right triangle a. x = 7 sin 25º b. x = 7 cos 25º

c. x = 7 tan 25º d. x =

3. A jet climbs at a steady 20º. When it has traveled 2 km through the air, what would its altitude be? sin 20º = 0.34 cos 20º = 0.94 tan 20º = 0.37 4. A 16 foot ladder is leaning against a building. The angle the ladder makes with the ground is 70º. How far up the building is the ladder?

81° 20 R

Q P x

25°

7 U

V

W

x

20°

2 x

16

70°

x

Geometry 231

20.0 Use angle and side relationships in problems with special right triangles, such as 30º, 60º, and 90º triangles and 45º, 45º, and 90º triangles.

Rule There are 2 special right triangles. One is a 45º, 45º, 90º triangle. The other is a 30º, 60º, 90º triangle. The lengths of the sides can be obtained using special formulas. 45º,45º,90º Triangle 30º, 60º, 90º Triangle

Note: Always put ‘a’ on the sides that are opposite the 45º angles. Note: Always put ‘a’ on the side opposite the 30º Put ‘2a’ always opposite the right angle Put ‘ a ’ always opposite the 60º angle Example 1 Find the measures of sides ‘k’ and ‘m’ in the given triangle. Solution: You’ll notice that one side is given. Find that side in the formula picture for the 30º, 60º, 90º triangle. So the value of ‘a’ is 7. In the formula triangle the hypotenuse is 2a or 2(7) = 14. This would be the value of k. In the formula triangle the long leg is a• or This is the value of m

(cont)

45°

45° a

a

30°

60° a

2a

30°

60° 7

k

m

30°

60° a

2a

30°

60° 7

2 7

Geometry 232

20.0 Use angle and side relationships in problems with special right triangles, such as 30º, 60º, and 90º triangles and 45º, 45º, and 90º triangles. (cont) Example 2 Find the length of side x in the given triangle. Solution: The one side that is given is the hypotenuse. By the formula we see that the hypotenuse is normally 2a. So 2a = 28 a = 14 The side we want is opposite the 60º angle. By the formula that side is normally ‘ a ‘ and since ‘a’ is 14 we get an answer of 14 Example 3 In the given figure ABCD is a square with side length 3. Find the perimeter of ∆DEF. Solution: Fill in the remaining angles and congruent sides. In the two small triangles, the hypotenuse would be ‘a ’ and the ‘a’ value is 3, so we get Now, add up all exterior segments

EF + DF + ED

( + ) + (3 + 3) + (3 + 3)

= + 12

(cont)

30°

28 x

B

A C

D

E F 45° 45°

3 3

B

A C

D

E F 45° 45°

3 3

45° 45° 3 3 3 3

3 3

Geometry 233

20.0 Use angle and side relationships in problems with special right triangles, such as 30º, 60º, and 90º triangles and 45º, 45º, and 90º triangles. (cont)

Homework

1. Find the values of x and y in the triangle shown.

2. ∆ABC is an equilateral triangle with side lengths 8. Altitude BD is shown. Find the length of BD.

3. The diagonal of a square has length 6, find the length of a side of the square. 4. The lower base angles in this isosceles trapezoid each measure 45º. The length of the shorter base is 10 inches and the altitude is 8 inches. Find the length of the longer base. 5. The lengths of the two adjacent sides of parallelogram ABCD are 8 and 14 inches. If the degree measure of the included angle is 60º, what is the length of the height of the parallelogram? 6. A 40 foot telephone pole has wires on each side supporting the pole. Find the length of each wire, represented in the diagram by m and k.

30°

60° y

x

60° 60° A C

B

D

8 8

6

45° 45°

10

8

x

60° A

C

B

D

8

14

45° 60°

9 m

k

Geometry 234

21.0 Prove and solve problems regarding relationships among chords, secants, tangents, inscribed angles, and inscribed and circumscribed polygons of circles. Formulas

Central Angle A central angle of a circle is an angle formed by two intersecting radii with the vertex at the center of the circle . Central Angle = Intercepted Arc AOB is a central angle

Its intercepted arc is the minor arc from A to B

AOB =

Inscribed Angle An inscribed angle is an angle with its vertex “on” the circle and whose sides contain chords of the circle

Inscribed Angle = Intercepted Arc

ABC is an inscribed angle

Its intercepted arc is the minor arc from A to C.

ABC =

Tangent Chord Angle An angle formed by an intersecting tangent and chord has its vertex “on” the circle.

Tangent Chord Angle = Intercepted Arc

ABC is a tangent chord angle

Its intercepted arc is the minor arc from A to B.

ABC =

B A

O

x

x

x

B

A C

B

A

x

x C

Geometry 235

Angle Formed Inside of a Circle by Two Intersecting Chords

When two chords intersect “inside” a circle, the measure of each angle is related to one-half the sum of the measures of the intercepted arcs.

Angle Formed Inside by Two Chords = Sum of Intercepted Arcs

x = ( c + d )

Angle Formed Outside of a Circle When chords and/or tangents intersect to form an angle ‘outside’ a circle, the measure of the angle is related to one-half the difference of the intercepted arcs.

Angle Formed Outside = Difference of Intercepted Arcs Two Tangents Two Secants

x = (d – c) x = (d – c)

a Tangent and a Secant

x = (d – c)

x d c

c d

x

c d x

x d c

Geometry 236

Line Segments Created by Intersecting Chords When two chords intersect inside a circle, the product of the segment of one chord equals the product of the segments of the other chord.

r • s = u • w

Line Segments Created by Two Secants Drawn to a Circle From an External Point

When two secants are drawn to a circle from an external point, the product of one secant segment and its external segment equals the product of the other secant segment and its external segment.

Whole segment • Outside piece = Whole segment • Outside piece

a • b = x • y

w

u r

s

y

a

b

x

Geometry 237

Line Segments Created by a Secant and a Tangent Drawn to a Circle From an External Point

When a secant and a tangent are drawn to a circle from an external point, the product of the secant segment and its external segment equals the square of the tangent segment. (a) (b) = x 2

Example 1 Find the length of segment x in the diagram. Solution: the formula states that the product of the pieces of one segment equals the product of the pieces of the other segment. So, 4 • x = 3 • 8 4x = 24 x = 6 Example 2 Find the measure of angle x given the arc lengths shown.

Solution: The formula says m x = ( 168 – y)

To find the measure of the arc ‘y’ we add all arcs to 360º. y + 118 + 168 = 360

y + 286 = 360

y = 74

Now, substitute into the original equation: m x = (168 – 74)

m x = (168 – 74)

m x = ( 94)

mx = 47

b

a

x

x

4 3

8

118°

168°

x

y

118°

168°

x

Geometry 238

Example 3 Find the measure of the arc length labeled ‘x’.

Solution:

STEP 1 STEP 2 STEP 3 First, find the angles in the The third angle in the triangle The arc length ‘x’ is triangle. The 80º arc is cut to make 180º would be 68º double the68º angle, so in half and yields an angle 136º. of 40º Example 4 Find the measure of ABC if mOBC = 31º and BD is a diameter of circle O.

Solution:

STEP 1 STEP 2 Step 3

would be twice the 31º angle. Since is a diameter mABC is half the So, = 62° measures 180º and measure of

since measures 62º So, mABC =

= 180 – 62 = 118 (118) = 59º

80°

68°

x

40°

72°

80°

x

72°

40°

80°

x

72°

80°

68°

136°

40°

72°

D

B

31°

A O

C

62°

D

B

31°

A O

C

62°

D

B

31°

A O

C 118°

62°

D

B

31°

A O

C 118°

59°

Geometry 239

Example 5 Find the area of the shaded region in circle F.

= 5, = 6

Solution:

STEP 1 STEP 2

AF is a radius of the circle and its measure is 5. The third side of the ∆ would Therefore AC the diameter measures 10. measure 8 by Pythagoras mABC = 90º since it intercepts the diameter AC (AB) 2 + 6 2 = 10 2 and m = 180º and the angle is half the arc.

STEP 3

To find the shaded area, find the area of the circle and subtract the area of the triangle.

–

= • 5 2 – • 6 • 8

= – 24

A

5

B C

F

6

A

10

8

B C

F

6

180° A

10

B C

F

6

180°

F 5

10 8

6

Geometry 240

Example 6 Find the measure of Solution:

STEP 1 STEP 2

In a quadrilateral, opposite angles The formula says the arc is twice and angle. are supplementary, So, = 2 • 95 = 190º so mF = 180 – 85 = 95° Homework 1. Find the length of the segment labeled ‘x’. 2. Find the measure of angle ‘x’ 3. Find the measure of angle ‘x’ 4. A square with side length 12 is circumscribed about a circle.

Find the area of the shaded region.

5. ∆BCD is an isosceles ∆ with BC BD 6. Find the length of segment x Find m CBD

F

85°

E

G

D

F

85°

E

G

D

95°

190°

F

85°

E

G

D

95°

O

x

3 5

4

70°

x

170°

x 260°

B

146°

C

D

x

9

3

Geometry 241

22.0 Know the effect of rigid motions on figures in the coordinate plane and space, including rotations, translations, and reflections. Rule Translations are SLIDES

The original object and its translation have the same shape and size, and they face the same direction. The word ‘translate’ means ‘carried across’. Example 1 Translate ∆ABC 5 units to the right and one unit down. Solution In math, the translation of an object is called its image. If the original object was labeled with letters, such as ∆ ABC, the image may be labeled with the letters ∆A’B’C’ Take point A and slide it 5 units right and 1 down and label it A’ Take point B and slide it 5 units right and 1 down and label it B’ Take point C and slide it 5 units right and 1 down and label it C’ Connect points A’, B’, and C’ and you have ∆A’B’C’ (A translation moves an object without changing its size or shape and without turning it or flipping it.) Example 2 Line segment AB has endpoints A (5,1) and B (3, – 4). Give the coordinates of A’B’ if AB is translated (x,y) (x – 3, y + 2)

Solution: Take the original points and subtract 3 to the x coordinate and add 2 to the y coordinate. A’ becomes (5 – 3, 1 + 2) and B’ becomes ( 3 – 3, –4 + 2)

A’ (2,3) and B’ (0,–2) .

A

B

C A’

B’

C’

A

B

C

Geometry 242

Rule Reflections are FLIPS An object and its reflection have the same shape and size, But the figures face in opposite directions. In a mirror, for example, right and left are switched. Example 3 Reflect polygon ABCD about line m.

Solution:

The line (where a mirror may be placed) is called the line of reflection. A reflection can be thought of as a ‘flipping’ of an object over the line of reflection.

Example 4 Line segment AB has endpoints A (1,3 ) and B (5, 2). Give the coordinates of A’B’ if AB is reflected over the x-axis. Solution: You ‘flip’ line AB over the x axis as shown.

A

B

A’

B’

A

B

m

A

B

C

D

m

A

B

C

D

C’

D’

B’

A’

Geometry 243

Rule Rotations are TURNS. An object and its rotation are the same shape and size, but the figures may be turned in different directions. An example of a rotation is when a person who is standing up lies down or stands on their head. A rotation of 90º turns the object a quarter turn right or left. A rotation of 90º clockwise turns the

object a quarter turn right.

A rotation of 90º clockwise turns the object a quarter turn left.

A rotation of 180º turns the object a

half turn.

Example 5 Rotate the triangle 90º clockwise.

Solution: First put a line at the top that we can use

as a reference and can rotate or turn. Now, turn the entire figure

turn right.

90º rotation counterclockwise

90º rotation clockwise

180º rotation

L I N E

L I N E

Geometry 244

Example 6 Rotate the triangle 180º Solution: First put a line at the top that we can use as a reference and can rotate or turn. Now, turn the entire figure ½ turn.

Example 7 Rotate the triangle 90º counterclockwise.

Solution: First put a line at the top that we can use as a reference and can rotate or turn. Now, turn the entire figure ¼ turn left

L I N E

L I N E

L I N E

LI N E

Geometry 245

Homework In problems 1-3 identify whether each figure pair represents a translation, reflection, or rotation.

1. 2. 3.

4. If line AB is rotated 180º about the origin, what are the coordinates of A’ ? 5. Find the coordinates of M’ if point M (7, –5) is translated 2 units in the negative x direction and 6 units in the positive y direction. 6. a. Give the coordinates of point A’ if trapezoid ABCD is reflected over the y-axis. b. Give the coordinates of point B’ if trapezoid ABCD is reflected over the x-axis. c. Give the coordinates of point C’ if trapezoid ABCD is rotated 90º to the left.

A

B

A B

C D