Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales.

Geometry Review

• Lines and Angles

• Triangles

• Polygons

1

• Circles

• 3-D Geometry

• Similarity & Scales

Lines and Angles

2

• Intersecting Lines:

Vertical angles: congruent

Linear angles: supplementary

Lines and Angles

3

• Parallel Lines:

Corresponding angles: congruent

Alternate angles: congruent

Example

Points A, B, C, D, E and F lie, in that order, on AF, dividing it into five segments, each of length 1. Point G is not on line . Point H lies on GD, and point J lies on GF. The line segments HC, JE and AG are parallel. Find HC/JE.

4

A B C D E F

G

JH

5

A B C D E F

G

JH

EJ // AG, we get: FEJ = FAG, FJE = FGAThus FEJ FAG

CH // AG, we get: DCH = DAG, DHC = DGA

Thus DCH DAG

Example

6

A B C D E F

G

JH

FEJ FAG, we get: EJ/AG = EF/AF = 1/5Hence EJ = 1/5 AG -------------------- (1)

DCH DAG, we get: CH/AG = CD/AD = 1/3

Hence CH = 1/3 AG ------------------- (2)

From (2), (1), we get: CH / EJ = 5/3

Example

Triangles

7

a + b > ca + c > bb + c > a

b ca

a - b < ca - c < bb - c < a

Triangle Inequality:

Area of Triangle:

area(ABC) = ½ * h * c

A

BCh

ExampleIn a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?

8

By triangle inequality, we have: 3 X – X < 15 Hence 2X < 15 X < 15/2 = 7.5

X 153X

X must be an integer, so the largest X = 7And the largest perimeter = 7 + 3*7 + 15 = 43

ExampleIn quadrilateral ABCD, AB=5, BC=17, CD=5, DA=9, and BD is an integer. What is BD?

9

By triangle Inequality: BD < AD + AB = 9 + 5 = 14

By triangle Inequality: BD > BC – CD = 17 – 5 = 12

Hence we have: 12 < BD < 14, and BD is an integer

Answer: BD = 13

Angles of TriangleSum of the interior angles:

10

a

b

c

a + b + c = 180

Exterior angle = sum of the non-adjacent angles

a

b

c

c = a + b

Special Triangles

11

|AB | = |AC|

Isosceles triangle

b cB

A

C b

a

c

A

B C

Equilateral triangle

|AB| = |BC| = |CA|

b + c a + b + c = 180a = b = c = 60

Height h = 3/2 R

Area S = 3/4 R2

h

Special Right Triangles

12

|AB | = ½ * |AC|

30 -60 Right Triangle 45 - 45 Right Triangle

|AB| = |BC| = 2/2 *|AC|

|BC| = 3 * |AB| = 3/2 * |AC|

Area S = ½ |AB| * |BC| = ¼ |AC|2

30

B

A

C

60

A

B C45

45

What’s the area of the stop sign whose sides are 1 foot long?

13

Note that ABC is a 45-45 right triangleArea of ABC = ¼ |BC| = ¼ * 1 = 1/4And |AB| = |AC| = 2/2 |BC| = 2/2 * 1 = 2/2Area of the green square = (1 + 2/2 + 2/2)2 = 3 + 22

STOP

A

B

C

Area of the stop-sign = 3 + 22 – 4 ( ¼) = 2 + 22

Example

Regular Hexagon ABCDEF has sides measuring 6 cm. What’s the area of triangle ACE?

14

Draw AD intersecting EC at PBy symmetry, we have AD ECEPD is a 30-60 right triangle!ED = 6cm |EP| = 3/2 * |EC| = 3/2 * 6 = 33ACE is equilateral. |CE| = 2 * |EP| = 2 * 33 = 63

A B

C

D

F

EP

Area of ACE = 3/4 |CE|2 = 3/4 *(63) 2 = 273

Example

Bi-Sector Theorem

15

|AB | |BD|--------- = ------------ |AC| |DC|

For any ABC. AD bisects BAC, we have:

B

A

CD

ExampleNon-degenerate ABC has integer side lengths, BD is an angle bisector, AD = 3, and DC = 8. What is the smallest possible value of the perimeter?

16

From Bi-Sector theorem, we have AB / BC = 3/ 8Hence AB = 3/8 * BCThe smallest BC to make AB an integer will be 8This gives AB = 3/8 * 8 = 3, and AB + BC = 11

A

B

CD3

8

and we get: AB + BC = AC --- invalid triangle!

ExampleNon-degenerate ABC has integer side lengths, BD is an angle bisector, AD = 3, and DC = 8. What is the smallest possible value of the perimeter?

17

So BC = 8 is not a valid answer!!!The next smallest BC to make AB = 3/8 * BC an integer is 16.This gives AB = 3/8 * BC = 3/8 * 16 = 6And we get the smallest perimeter: 6 + 16 + 11 = 33

A

B

CD3

8

18

|AB|2 + |BC|2 = |AC|2

For any right ABC, where ABC forms the right angle. We have:

B

A

C

Pythagorean Theorem

Right triangles whose side lengths are integers:3-4-5 triangles: right triangle with side length 3, 4, 5.5-12-13 triangles: right triangle with side length 5, 12, 13.7-24-25 triangles: right triangle with side length 7, 24, 25.8-15-17 triangles: right triangle with side length 8, 15, 17.

A rectangle with diagonal length 1, height is twice as long as it is wide. What is the area of the rectangle?

19

Assume h and w are height & width of the rectangleWe have: h = 2 * wBy Pythagorean theorem, we have: w2 + (2w)2 = 1 5 w2 = 1 w = 5/5

A B

CD

Thus the area of the rectangle: A = w * h = w * (2w) = 5/5 * 2 * 5/5 = 2/5

1

Example

In ABC, we have AB=AC=7 and BC=2. Suppose that D is a point on line BC such that C lies between B and D and AD=8. What is CD?

20

ABC is isosceles. Draw AE BC, we have: BE = EC = 1Also, ABE a right triangle, and by AE2 = AB2 – BE2: |AE| = (72 – 12 ) = (48)

B

A

C

7 7

2 D

8

E

Also, ADE is a right triangle, and by ED2 = AD2 – AE2: |ED| = (82 – 48 ) = (64 - 48) = = (16) = 4Hence CD = ED – EC = 4 – 1 = 3

Example

Let XOY be a right triangle with XOY=90. Let M and N be the midpoints of legs OX and OY, respectively. Given that XN=19 and YM=22, find XY.

21

Let OM=X, ON=Y. By Pythagorean Theorem on XON, MOY: (2X)2 + Y2 = 192 ------- (1) X2 + (2Y)2 = 222 ------- (2)

Sum up (1) & (2), we get: 5X2 + 5Y2 = 845 X2 + Y2 = 169

O

X

Y

M

N

19

22

?

By Pythagorean Theorem on OXY: XY2 = (2X)2 + (2Y)2 = 4*(X2 + Y2) = 4*169 XY = 2*13 = 26

Example

Polygon

22

Sum of interior angles of a N-polygon 180 * (N – 2)

A N-polygon can be divided into N – 2 triangles.

Sum of exterior angles = 360

A regular polygon has interior angles measuring 179, and a side length of 2cm. What is the perimeter of the polygon?

23

Since the interior angle is 179, the exterior angle is 1 Thus there are 360 / 1 = 360 exterior angles.And there are 360 sides, each of length 2 cm.Hence the perimeter = 2 * 360 = 720

Example

Equiangular hexagon ABCDEF has side lengths AB=CD=EF=1 and BC=DE=FA=r. The area of ACE is 70% of the area of the hexagon. What is the sum of all possible values of r ?

24

A B

C

DE

F

1

11

rr

r

Example

Equiangular hexagon ABCDEF has side lengths AB=CD=EF=1 and BC=DE=FA=r. The area of ACE is 70% of the area of the hexagon. What is the sum of all possible values of r ?

25

A B

C

DE

F

1

11

rr

r

Extend AB to form right BMC

M

We have CBM = 360 / 6 = 60, BCM=90. MC = 3/2 * r ; BM = ½ rArea of ABC = ½ * AB * MC = ½ * 1 * 3/2 * r = ¾* r Hence AC = (MC2 + (BM + 1)2) = (r2 + r + 1)Hence AC = (MC2 + (BM + 1)2) = (r2 + r + 1) = CD = EAArea ACE = ¾ AC2 = ¾(r2 + r + 1) = 7/10 (area of ACE + 3 * area of ABC)

Example

Equiangular hexagon ABCDEF has side lengths AB=CD=EF=1 and BC=DE=FA=r. The area of ACE is 70% of the area of the hexagon. What is the sum of all possible values of r ?

26

A B

C

DE

F

1

11

rr

r

M

Area ACE = ¾ AC2 = ¾(r2 + r + 1) = 7/10 (area of ACE + 3 * area of ABC) Hence ¾(r2 + r + 1) = 7/10 (¾(r2 + r + 1) + 3 * ¾ * r) 10r2 + 10r + 10 = 7r2 + 7r + 7 + 21r3r2 - 18r + 3 = 0 r2 - 6r + 1 = 0 From the sum of the roots theorem, we get: The sum of the possible values of r = 6

Example

Special Polygons

27

rhombus – parallelograms that’s equilateral, parallelogram – two pairs of parallel sidestrapezoids – quadrilateral with two parallel bases

rectangle – parallelograms that’s equal angular, with diagonals of same lengthsquare – regular rectangle

with diagonals perpendicular to each other

28

Let BO = OD = X, we have AO = OC = 3 * X

The long diagonal of a rhombus is 3 times the length of the short diagonal. The perimeter of the rhombus is 40cm. What is its area?

A

B

D

CO

X2 + (3X)2 = (40/4)2 = 10010 X2 = 100 X = 10Hence OB = 10, AO = 310, area(ABO) = ½*10*310 = 15

Area of rhombus = 4 * 15 = 60 (cm2)

Example

29

Trapezoid ABCD has bases AB and CD and diagonals intersecting at K. Suppose that AB=9, DC=12, and the area of AKD is 24. What is the area of trapezoid ABCD?

ABK CDK BK/DK = AB/CD = 9/12 = ¾ AK/CK = AB/CD = 9/12 = ¾ ABK & ADK share same height, hence: area(ABK) / area(ADK) = BK/DC = 3/4Area(ADK) = 24 area(ADK) = ¾ * 24 = 18

A B

D C

K

Example

30

Trapezoid ABCD has bases AB and CD and diagonals intersecting at K. Suppose that AB=9, DC=12, and the area of AKD is 24. What is the area of trapezoid ABCD?

Similarly, ADK & CDK share same height, hence: area(ADK) / area(CDK) = AK/DK = 3/4Area(ADK) = 24 area(CDK) = 4/3 * 24 = 32

A B

D C

K

Note that area(ADC) = area(BDC), we get:area(BCK) = area(BDC) - area(KDC) = area(ADC) - area(KDC) = area(ADK) = 24

Example

31

Trapezoid ABCD has bases AB and CD and diagonals intersecting at K. Suppose that AB=9, DC=12, and the area of AKD is 24. What is the area of trapezoid ABCD?

area(ABCD) = area(ABK) + area(CDK) + area(ADK) + area(BCK)

A B

D C

K

area(BCK) = 18 + 32 + 24 + 24 = 98

Example

Circle

32

Definitions: chord secant tangent

o

radius

Area of the circle = R2 *

Perimeter of the circle = 2 * R *

Circle

33

Inscribed α = ½ x where x is the opposite arc.

x yxα

scant angle = ½ (y - x )

34

Hence Y = 360 – 150 – 100 – 80 = 30

The measure of the circle = 360

Find the angle measure X in the diagram below

Hence X = ½ (100 – Y) = ½ (100 – 30) = ½ * 70 = 35

150

100

80

xy

Example

35

Observe that X + 360 = 125 * 2 + 80 * 2

The measure of the circle = 360

Find the measure of arc X in the diagram below

We have: X = 250 + 160 – 360 = 50

80

125

x

Example

36

Let P be center of small circle. By AOB = 60,

Points A and B lie on a circle centered at O, AOB= 60. A second circle is internally tangent to the first and tangent to both OA and OB. What is the ratio of the area of the smaller circle to that of the larger circle?

OMP is 30-60 right triangle OP = 2 PM

O

A

B

P

M

C

OC = OP + PC = OP + PM = 3 PM PM/OC = 1/3 Area(P)/Area(O) = (PM2* )/(OC2* ) = 1/9

Example

Inscribed Triangle

37

For any inscribed triangle that has one edge through the center, it must be a right triangle.

O

Inscribed Quadrilateral

38

For any inscribed quadrilateral, the sum of the two

opposite angles = 180

x

180 - X

39

In the given circle, the diameter EB is parallel to DC, and AB is parallel to ED. The angles AEB and ABE are in the ratio 4:5. What is the degree measure of angle BCD?

Example

40

In the given circle, the diameter EB is parallel to DC, and AB is parallel to ED. The angles AEB and ABE are in the ratio 4:5. What is the degree measure of angle BCD?

Hence ABE +AEB = 90Since AEB : ABE = 4:5 AEB = 40, ABE = 50Now AB//ED BED = ABE = 50

Since BAE over the center BAE = 90

BCDE is an inscribed quadrilateral BED + BCD = 180BCD = 180 - BED = 180 - 50 = 130

Example

Power of a Pointer

41

AE * EB = CE * ED

A

B

C

D

E

A

B

C

DE

AB * AC = AD * AE

42

In the figure below, AB and CD are diameters of the circle with center O, ABCD, and chord DF intersects AB at E. If DE=6 and EF=2, what’s the area of the circle?

Example

43

In the figure below, AB and CD are diameters of the circle with center O, ABCD, and chord DF intersects AB at E. If DE=6 and EF=2, what’s the area of the circle?

By Power of a Pointer, DE * EF = AE * EBLet X = OE, and AO = OB = R we have: AE = X+R, EB= X-R Hence: 6 * 2 = (R + X) (R – X) R2 – X2 = 12 ----- (1) DOE is right triangle, and OD = OE = R R2 + X2 = DE2 = 62 = 36 -------- (2)By (1) + (2), 2 R2 = 12 + 36 R2 = 24 Area = 24

Example

44

Let R1 and R2 be the radii of the two circles A & B.

A 45 arc of circle A is equal in length to a 30 arc of circle B. What is the ratio of circle A's area and circle B's area?

The arc length opposite to 45 in R1 = 45/360 * 2 R1 = 1/4 R1 The arc length opposite to 30 in R2 = 30/360 * 2 R2 = 1/6 R2 Hence we get: 1/4 R1 = 1/6 R2 R1 / R2 = 3/2

area(A) / area(B) = (R12 )/ (R2

2 ) = 9/4

Example

3-D Geometry

45

Volume of prism = height * area-of-base

Volume of pyramid = ⅓ * height * area-of-base

Volume of cylinder = height * R2 *

Volume of cone = ⅓ * H * R2 *

Hh

Surface are of cone = S * R * + R2 *

S

R

3-D Geometry

46

Volume of sphere V = 4/3 * R3 *

Surface are of sphere A = 4 * R2 *

47

The height of a cone is 4m and its radius is 3m. What’s the surface area and the volume of the cone?

Volume = ⅓ * H * R2 * = ⅓ * 4 * 9 * = 12 (m3)Surface Area = r * s * + r2 * = 3 * 5 * + 9 * = 24 (m2)

Example

48

A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?

Example

49

A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?

Volume of whole cube = 3 * 3 * 3 = 27Volume of the prism = 3 * 2 * 2 = 12There are 3 prisms, total volume = 3 * 12 = 36Note that the intersection of 3 prisms is a 2x2x2 cubeAnd we need to add two of them back.Hence the answer = 27 – 36 + 2 * 8 = 7

Example

Scale & Similarity

50

When the side lengths of any polygon (or radius of circle) are increased by a factor S, then the area is increased by a factor S2

The ratio between any corresponding edges is the same between any similar polygons

Rules for similar triangles: SAS, AAA, SSS

When the side lengths of any 3-D solid are increased by a scale factor of S, then the surface area is increased by a scale factor of S2, and the volume is increased by a scale factor of S3

Correspond angles have the same measure between any similar polygons

51

Rectangle ABCD has AB=4 and BC=3. Segment EF is constructed through B so that EF is perpendicular to DB. A and C lie on DE and EF, respectively. What is EF?

D C

AB

34

E

F

Example

52

Rectangle ABCD has AB=4 and BC=3. Segment EF is constructed through B so that EF is perpendicular to DB. A and C lie on DE and EF, respectively. What is EF?

D C

AB

34

E

F

By Pythagorean theorem: DB = 5Observe that ABE ABD EB/AB=DB/ADEB = AB * DB / AD = 4 * 5 / 3 = 20/3Observe that BCF ABD FB/CB=DB/ABFB = CB * DB / AB = 3 * 5 / 4 = 15/4EF = EB + BF = 20/3 + 15/4 = 125/12

Example

53

Points E and F are located on unit square ABCD so that BEF is equilateral. What is the ratio of the area of DEF to that of ABE?

Example

54

Points E and F are located on unit square ABCD so that BEF is equilateral. What is the ratio of the area of DEF to that of ABE?

Observe that BEF is equilateral, and EA = FCWe have ABE BCF DE = DE & DEF = 45

We get: EF = EB = FB = 2XLet DE = DE = X

From ABE, we get: (2X)2 = (1 –X)2 + 12

2X2 = 1 – 2X + X2 + 1 X2 + 2X -2 = 0 X = 3 - 1area(DEF)/area(ABE) = (½ X2)/(½(1- X)) = X2/(1-X) =2

Example