Chapter 7 Similar Polygons (page 240)

Chapter 7 Similar Polygons (page 240)

Lesson 7-1: Ratio and Proportion (page 241)
Essential Question How can similar polygons be used to solve real life problems?

two ; ; ; ; Not just 4. It’s a ratio!
RATIO: the quotient of _________ numbers, ab, usually written _____ or a:b, where b  0. Examples: ; ; ; ; Not just 4. It’s a ratio!

fraction simplest compare units units comparison compare
A ratio is simply ________________. A ratio is usually expressed in ________________ form. Ratios can be used to ________________ two numbers, ie. a : b . The quantities being compared must be in the same _________. Note: ratios have no _________, it is a ___________________ of numbers. Ratios can be used to ________________ more than two numbers. To compare more than two numbers, use the notation a : b : c : d . fraction simplest compare units units comparison compare

PROPORTION: an _______________ stating that two ratios are __________.
equation equal and a : b = c : d The 1st term is ____, the 2nd term is ____, the 3rd term is ____ , & the 4th term is ____. a b c d When 3 or more ratios are equal, you can write an ________________ proportion. extended and a : b = c : d = e : f

Example #1. What is the ratio of 250 mL to 0.5 L?
1000 mL = 1 L ∴ 0.5 L = 500 mL Example #1. What is the ratio of 250 mL to 0.5 L? 1 2 CAUTION! The units must be the same! 1 : 2 ______

Do you know another way to solve this problem?
Example #2. Two complementary angles have measures in the ratio 2 : 7. Find the measure of each angle. Let m∠1 = 2 x then m∠2 = 7 x = 2 (10) = 7 (10) = 20º = 70º = 90º √ 2 x + 7x = 90 WHY? 9x = 90 x = 10 Do you know another way to solve this problem?

9x + 8x + 6x + 4x + 3x = 540 30x = 540 x = 18 Let m∠1 = 9 x
Example #3. Five angles of a pentagon have their measures in the ratio 9 : 8 : 6 : 4 : 3. Find the measure of each angle. Let m∠1 = 9 x m∠2 = 8 x m∠3 = 6 x m∠4 = 4 x m∠5 = 3 x = 9 (18) = 162º = 8 (18) = 144º = 6 (18) = 108º = 4 (18) = 72º = 3 (18) = 54º = 540º √ WHY? 9x + 8x + 6x + 4x + 3x = 540 30x = 540 x = 18

How can similar polygons be used to solve real life problems?
Assignment Written Exercises on pages 243 & 244 1 to 31 odd numbers, 33 to 36 How can similar polygons be used to solve real life problems?

Lesson 7-2 Properties of Proportions (page 245)

a d extremes b c means In the proportion, a : b = c : d
the 1st & last terms, ____ & ____, are called the ______________, the middle terms, ____ & ____, are called the ____________. a d extremes b c means

Properties of Proportions - #1
The product of the means is equal to the product of the __________________. This is called the ___________________ property of proportions. extremes In the middle school, you called this Cross-Multiplying! NOT in the high school! means-extremes b c = a d

If you interchange the means, you get an equivalent _____________________. proportion This also works when interchanging the extremes. c b = a d b c = a d

If you take the reciprocals of the _________, the result is an equivalent proportion. ratios a d = b c b c = a d

If you add _________ to both sides, you still have an equivalent proportion. one “So what Mr. Kline! Who didn’t know that?”

like adding the denominator
“That’s cool! It is just like adding the denominator to the numerator.”

In an extended proportion, the ratio of the sum of all the numerators to the ______ of all the denominators is equivalent to each of the ______________ ratios. sum original

Example #1. Use the proportion to complete the statement.
3 b 5 a = ____ 5 a = 3 b

=

9 3 9 2 6 If CE = 2, EB = 6, and AD = 3, then DB = ______.
Example #5. In the figure, If CE = 2, EB = 6, and AD = 3, then DB = ______. 9 A 3 D 9 C 2 E 6 B

Example #6. In the figure, If AB = 10, DB = 8, and CB = 7.5, then EB = ______. 6 A 2 D 10 8 1 x C 7.5-x E B 7.5 4 Let EB = x, then CE = x

Assignment Written Exercises on pages 247 & 248 1 to 27 odd numbers, 30, 33, 35, 37 BE PREPARED FOR SURPRISES! How can similar polygons be used to solve real life problems?

Lesson 7-2 HW, page 247, #30 - What can you conclude?

Quiz Problem from Lesson 7-2 , page 248, #38
(x - 1)(3 x - 2) = 10 (x + 2) 3 x2 - 5 x + 2 = 10 x + 20 3 x x - 18 = 0 (divide both sides by 3) x2 - 5 x - 6 = 0 (x - 6)(x + 1) = 0 x - 6 = 0 OR x + 1 = 0 x = x = -1

Lesson 7-3 Similar Polygons (page 248)

Similar Polygons: congruent proportion
Two polygons are similar if their vertices can be paired so that … corresponding angles are ___________________ and corresponding sides are in ___________________. congruent proportion

shape size Similar figures have the same ___________,
but not necessarily the same __________. shape size

If polygon ABCDE ~ polygon FGHIJ, then …

∠A ≅ ∠F A B G F E J H I D C

∠A ≅ ∠F ; ∠B ≅ ∠G A B G F E J H I D C

∠A ≅ ∠F ; ∠B ≅ ∠G ; ∠C ≅ ∠H A B G F E J H I D C

∠A ≅ ∠F ; ∠B ≅ ∠G ; ∠C ≅ ∠H ; ∠D ≅ ∠I A B G F E J H I D C

∠A ≅ ∠F ; ∠B ≅ ∠G ; ∠C ≅ ∠H ; ∠D ≅ ∠I ; ∠E ≅ ∠J A B G F E J H I D C

∠A ≅ ∠F ; ∠B ≅ ∠G ; ∠C ≅ ∠H ; ∠D ≅ ∠I ; ∠E ≅ ∠J and A B G F E J H I D C

corresponding A’ is read “A prime” Take notice to the order to get
SCALE FACTOR: the ratio of the lengths of two _____________________ sides. corresponding A’ is read “A prime” Quad. ABCD ~ Quad. A’B’C’D’ 24 A B 18 A’ B’ z x 28 32 D’ 27 C’ C D y Take notice to the order to get the SF (L to R). Scale Factor = SF =

Quad. ABCD ~ Quad. A’B’C’D’
SCALE FACTOR: the ratio of the lengths of two corresponding sides. Quad. ABCD ~ Quad. A’B’C’D’ 24 A B 18 A’ B’ z x 28 32 D’ 27 C’ C D y Scale Factor = SF =

SCALE FACTOR: the ratio of the lengths of two corresponding sides. Quad. ABCD ~ Quad. A’B’C’D’ 24 A B 18 A’ B’ x = 24 z 28 32 D’ 27 C’ C D y Scale Factor = SF =

SCALE FACTOR: the ratio of the lengths of two corresponding sides. Quad. ABCD ~ Quad. A’B’C’D’ 24 A B 18 A’ B’ x = 24 z 28 32 D’ 27 C’ C D y = 36 Scale Factor = SF =

SCALE FACTOR: the ratio of the lengths of two corresponding sides. Quad. ABCD ~ Quad. A’B’C’D’ 24 A B 18 A’ B’ z = 21 x = 24 28 32 D’ 27 C’ C D y = 36 Scale Factor = SF =

m∠B =____ ; m∠D =____ ; m∠Y =____ ; m∠Z =____
Quad. ABCD ~ Quad. WXYZ , find the unknown angle measures. m∠D + 90º + 60º + 130º = 360º D 80º Z C 130º 80º Y 130º 360º 60º 60º B W X A m∠B =____ ; m∠D =____ ; m∠Y =____ ; m∠Z =____ 60º 80º 130º 80º

Quad. GEOM ~ Quad. G’E’O’M’. Find the scale factor,
the perimeter of GEOM, and the perimeter of G’E’O’M’. O’ O M’ 24 M 24 18 21 G 27 E G’ E’ scale factor = SF =

the perimeter of GEOM, and the perimeter of G’E’O’M’. O’ O M’ 24 M 24 18 21 G 27 E G’ E’ Perimeter of GEOM = = 90

the perimeter of GEOM, and the perimeter of G’E’O’M’. Perimeter of GEOM = 90 O’ O M’ 24 M 24 18 21 G 27 E G’ 36 E’ Perimeter of G’E’O’M’ = ? Find the lengths of the other sides.

the perimeter of GEOM, and the perimeter of G’E’O’M’. Perimeter of GEOM = 90 O’ O M’ 24 M 28 24 18 21 G 27 E G’ 36 E’ Perimeter of G’E’O’M’ = ? Find the lengths of the other sides.

the perimeter of GEOM, and the perimeter of G’E’O’M’. Perimeter of GEOM = 90 O’ 32 O M’ 24 M 28 24 18 21 G 27 E G’ E’ 36 Perimeter of G’E’O’M’ = ? Find the lengths of the other sides.

the perimeter of GEOM, and the perimeter of G’E’O’M’. Perimeter of GEOM = 90 O’ 32 O M’ 24 M 28 24 18 21 G 27 E G’ E’ 36 Perimeter of G’E’O’M’ = = 120

the perimeter of GEOM, and the perimeter of G’E’O’M’. O’ 32 O M’ 24 Perimeter of G’E’O’M’= 120 M 28 Perimeter of GEOM = 90 24 18 21 G 27 E G’ E’ 36 The ratio of perimeters = Scale Factor.

the perimeter of GEOM, and the perimeter of G’E’O’M’. O’ 32 O M’ 24 Perimeter of G’E’O’M’= 120 M 28 Perimeter of GEOM = 90 24 18 21 G 27 E G’ E’ 36

Classroom Exercises on page 250 1 to 9 Assignment Written Exercises on pages 250 to 252 1 to 21 odd numbers, 25, 27, 34, 35 Prepare for Quiz on Lessons 7-1 to 7-3 How can similar polygons be used to solve real life problems?

Golden Rectangle Read page 253 about the Golden Rectangle
and Golden Ratio.

Golden Rectangle ~ Golden Ratio ~

Parthenon in Athens, Greece

Floor Plan for the Parthenon

Lincoln Memorial Washington DC Nashville Parthenon

What rectangles interest you?

Lesson 7-4 A Postulate for Similar Triangles (page 254)

six prove similar corresponding
To prove two triangles congruent, you didn’t need to compare all _______ pairs of corresponding parts. Postulates and theorems gave easier ways to ____________ congruence. There are also postulates and theorems for proving triangles _________ without comparing all of the _________________ parts. six prove similar corresponding

AA Similarity Postulate
If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. AA ~ Postulate

Example #1. Tell whether the triangles are similar or not similar.
If you can’t reach a conclusion, write no conclusion is possible. 54º 62º 60º 58º Not Similar 60º 66º

If you can’t reach a conclusion, write no conclusion is possible. Similar

No Conclusion is Possible
Example #3. Tell whether the triangles are similar or not similar. If you can’t reach a conclusion, write no conclusion is possible. 90º 45º 45º No Conclusion is Possible

If you can’t reach a conclusion, write no conclusion is possible. 75º 75º 40º 30º 70º 70º Not Similar

Similar ∆’s by the AA ~ Postulate
Example #5 Find the values of “x” and “y”. 8 10 Similar ∆’s by the AA ~ Postulate 12 10 + y x ➤ S.F. = 4 y ➤ 15 10 x = _____ y = _____ 3 x = 2 • 15 2 (10 + y) = 3 • 10 5 3 x = 30 20 + 2y = 30 x = 10 2y = 10 y = 5

(Corresponding Sides in Proportion in Similar ∆’s)
(6) Given: ∠1 ≅ ∠2 Prove: BD • CA = BC • DE Proof: D 1 2 A E B Statements Reasons ____________________________________ _____________________________________________ ∠1 ≅ ∠2 Given ∠B ≅ ∠B Reflexive Property ∆ ABC ∆ EBD AA ~ Postulate CSPST (Corresponding Sides in Proportion in Similar ∆’s) BD • CA = BC • DE Means - Extremes Property

Assignment Written Exercises on pages 257 to 259 1 to 15 ALL #’s, 20, 31, 32 How can similar polygons be used to solve real life problems?

Lesson 7-5 Theorems for Similar Triangles (page 263)

Way to Prove ANY Two Triangles Similar
AA ~ Postulate

Theorem 7-1 SAS ~ Theorem If an angle of one triangle is congruent to an angle of another triangle and the sides including those angles are in proportion , then the triangles are similar. D Given: ∠A ≅∠D Prove: ∆ ABC ∆ DEF A E F B C

Theorem 7-2 SSS ~ Theorem If the sides of two triangles are in proportion , then the triangles are similar. D Given: Prove: ∆ ABC ∆ DEF A E F B C

Ways to Prove ANY Two Triangles Similar
AA ~ Postulate SAS ~ Theorem SSS ~ Theorem

To prove two polygons similar, you may need to compare corresponding sides. A useful technique is to compare the longest sides, the shortest sides, and so on.

Perimeters of similar polygons are in the same ratio as the corresponding sides. By using similar triangles, you can prove that corresponding segments, such as diagonals, also altitudes, medians, etc. of similar polygons, also have this ratio .

∴ ∆PQR ~ ∆TSP by the SSS ~ Theorem Compare the longest sides:
Example #1 Can the information given be used to prove ∆PQR ~ ∆TSP? If so, how? SP = 8, TS = 6, PT = 12, QR = 12, PQ = 9, RP= 18. Q ∴ ∆PQR ~ ∆TSP by the SSS ~ Theorem 12 9 18 P R 12 T 8 6 S Compare the longest sides: Compare the shortest sides: Compare the other sides:

SP = 7, TS = 6, PQ = 9, QR = 10.5, m∠S = 80º, m∠QPR + m∠QRP = 100º.
Example #2 Can the information given be used to prove ∆PQR ~ ∆TSP? If so, how? SP = 7, TS = 6, PQ = 9, QR = 10.5, m∠S = 80º, m∠QPR + m∠QRP = 100º. Q ∴ ∆PQR ~ ∆TSP by the SAS ~ Theorem 80º 10.5 9 If m∠QPR + m∠QRP = 100º, then m ∠Q = 80º. m∠QPR + m∠QRP = 100º P R T 7 6 80º S Compare the longer sides: Compare the shorter sides:

Classroom Exercises on pages 264 & 265 1 to 6 Assignment Written Exercises on pages 266 & 267 1 to 7 ALL numbers, 9, 14, 20 Prepare for Quiz on Lessons 7-4 & 7-5 How can similar polygons be used to solve real life problems?

Lesson 7-6 Proportional Lengths (page 254)

If AB : BC = XY : YZ, then AC and XZ
are said to be divided proportionally . A B C X Y Z

Theorem 7-3 Triangle Proportionality Theorem ➤ ➤
If a line parallel to one side of a triangle intersects the other two sides, then it divides those sides proportionally . T Given: ∆ RST; PQ || RS Prove: P ➤ Q ➤ S R

Triangle Proportionality Theorem
Given: ∆ RST; PQ || RS Prove: PT QT RP SQ T Top Left Top Right P ➤ Q Bottom Left Bottom Right ➤ S R

many equivalent proportions can be justified …
NOTE: Since ∆RST ~ ∆PQT, many equivalent proportions can be justified … T P ➤ Q ➤ S R by the AA ~ Postulate

➤ ➤ Here are many equivalent proportions along with
informal statements describing them. P ➤ Q ➤ S R

T P ➤ Q ➤ S R

This is the only way to get the parallel sides.
➤ Q ➤ S R DO NOT FORGET THESE RATIOS! This is the only way to get the parallel sides.

Example #1. Find the value of “x”.
20 15 ➤ 8 x ➤

Theorem REVIEW!!! If three parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal. Given: Prove: A ➤ X B ➤ Y C ➤ Z

Corollary If three parallel lines intersect two transversals, then they divide the transversals proportionally . R ➤ X Given: Prove: S ➤ Y T ➤ Z

➤ ➤ ➤ Example #2. Find the value of “x”. 25 - x = 20 25 - x 16 x = 5 x
4 ➤

Theorem 7-4 Triangle Angle-Bisector Theorem DG bisects ∠FDE
If a ray bisects an angle of a triangle, then it divides the opposite side into segments proportional to the other two sides . F Given: ∆ DEF ; DG bisects ∠FDE Prove: G E D

27 45 x = 15 40 - x = 25 40

12 18 x 24

Assignment Written Exercises on pages 272 & 273 1 to 11 ALL numbers, 13, 15, 17, 20, 21, 25 Prepare for Quiz on Lessons 7-4 to 7-6 How can similar polygons be used to solve real life problems?

Lesson 7-6 Proportional Lengths
Page 273 #25

Triangle Angle-Bisector Theorem
If a ray bisects an angle of a triangle, then it divides the opposite side into segments proportional to the other two sides. F Given: ∆ DEF ; DG bisects ∠FDE Prove: G E D

Find MP. B 12 14 x 13 - x C A P 13

• Since M is the midpoint CM = MA = 6.5.
Find MP. Since M is the midpoint CM = MA = 6.5. B 12 14 x 13 - x • C A P M 13 Since x = 6, CP = 6, then AP = = 7. ∴ MP = CM - CP = = 0.5 or MP = PA - MA =

Prepare for Quiz on Lessons 7-4 to 7-6 Prepare for Test on Chapter 7: Similar Polygons How can similar polygons be used to solve real life problems?

Chapter 7 Similar Polygons (page 240)

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