geometry review lines and angles triangles polygons 1 circles 3-d geometry similarity & scales
TRANSCRIPT
![Page 1: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/1.jpg)
Geometry Review
• Lines and Angles
• Triangles
• Polygons
1
• Circles
• 3-D Geometry
• Similarity & Scales
![Page 2: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/2.jpg)
Lines and Angles
2
• Intersecting Lines:
Vertical angles: congruent
Linear angles: supplementary
![Page 3: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/3.jpg)
Lines and Angles
3
• Parallel Lines:
Corresponding angles: congruent
Alternate angles: congruent
![Page 4: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/4.jpg)
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
![Page 5: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/5.jpg)
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
![Page 6: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/6.jpg)
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
![Page 7: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/7.jpg)
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
![Page 8: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/8.jpg)
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
![Page 9: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/9.jpg)
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
![Page 10: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/10.jpg)
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
![Page 11: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/11.jpg)
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
![Page 12: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/12.jpg)
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
![Page 13: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/13.jpg)
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
![Page 14: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/14.jpg)
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
![Page 15: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/15.jpg)
Bi-Sector Theorem
15
|AB | |BD|--------- = ------------ |AC| |DC|
For any ABC. AD bisects BAC, we have:
B
A
CD
![Page 16: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/16.jpg)
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!
![Page 17: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/17.jpg)
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
![Page 18: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/18.jpg)
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.
![Page 19: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/19.jpg)
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
![Page 20: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/20.jpg)
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
![Page 21: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/21.jpg)
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
![Page 22: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/22.jpg)
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
![Page 23: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/23.jpg)
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
![Page 24: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/24.jpg)
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
![Page 25: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/25.jpg)
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
![Page 26: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/26.jpg)
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
![Page 27: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/27.jpg)
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
![Page 28: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/28.jpg)
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
![Page 29: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/29.jpg)
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
![Page 30: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/30.jpg)
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
![Page 31: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/31.jpg)
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
![Page 32: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/32.jpg)
Circle
32
Definitions: chord secant tangent
o
radius
Area of the circle = R2 *
Perimeter of the circle = 2 * R *
![Page 33: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/33.jpg)
Circle
33
Inscribed α = ½ x where x is the opposite arc.
x yxα
scant angle = ½ (y - x )
![Page 34: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/34.jpg)
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
![Page 35: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/35.jpg)
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
![Page 36: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/36.jpg)
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
![Page 37: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/37.jpg)
Inscribed Triangle
37
For any inscribed triangle that has one edge through the center, it must be a right triangle.
O
![Page 38: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/38.jpg)
Inscribed Quadrilateral
38
For any inscribed quadrilateral, the sum of the two
opposite angles = 180
x
180 - X
![Page 39: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/39.jpg)
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
![Page 40: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/40.jpg)
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
![Page 41: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/41.jpg)
Power of a Pointer
41
AE * EB = CE * ED
A
B
C
D
E
A
B
C
DE
AB * AC = AD * AE
![Page 42: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/42.jpg)
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
![Page 43: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/43.jpg)
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
![Page 44: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/44.jpg)
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
![Page 45: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/45.jpg)
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
![Page 46: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/46.jpg)
3-D Geometry
46
Volume of sphere V = 4/3 * R3 *
Surface are of sphere A = 4 * R2 *
![Page 47: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/47.jpg)
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
![Page 48: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/48.jpg)
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
![Page 49: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/49.jpg)
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
![Page 50: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/50.jpg)
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
![Page 51: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/51.jpg)
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
![Page 52: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/52.jpg)
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
![Page 53: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/53.jpg)
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
![Page 54: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/54.jpg)
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
![Page 55: Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales](https://reader033.vdocument.in/reader033/viewer/2022061611/551c53005503469d6a8b4cc9/html5/thumbnails/55.jpg)