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8.1 Ratio and Proportion. Day 1 Part 1 CA Standards 5.0. Warmup. Find the slope of the line that passes through the points. A(0,0), B(4,2) C(-1,2), D(6,5) E(0,3), F(-4, -8). Computing Ratios. Ratio of a to b : Ex: Simplify the ratio of 6 to 8. Simplify the ratio of 12 to 4.
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8.1 Ratio and Proportion Day 1 Part 1 CA Standards 5.0
Warmup • Find the slope of the line that passes through the points. • A(0,0), B(4,2) • C(-1,2), D(6,5) • E(0,3), F(-4, -8)
Computing Ratios • Ratio of a to b : • Ex: Simplify the ratio of 6 to 8. Simplify the ratio of 12 to 4.
Simplifying Ratios • Simplify the ratios.
The perimeter of rectangle ABCD is 60 cm. The ratio of AB:BC is 3:2. Find the length and width of the rectangle. • Hint: draw a rectangle ABCD.
Using Extended Ratios • The measure of the angles in ∆JKL are in the extended ratio of 1:2:3. Find the measures of the angles. 2x 3x x
Using Ratios • The ratios of the side lengths of ∆DEF to the corresponding side lengths of ∆ABC are 2:1. Find the unknown lengths. A D 3 L B C L F E 8
Properties of proportions • Cross product property then ad = bc. • Reciprocal property
Given that the track team won 8 meets and lost 2, find the ratios. • What is the ratio of wins to loses? • What is the ratio of losses to wins? • What is the ratio of wins to the total number of track meets?
Simplify the ratio. • 3 ft to 12 in • 60 cm to 1 m • 350g to 1 kg • 6 meters to 9 meters
8.2 Problem Solving in Geometry with Proportions Day 1 Part 2 CA Standards 5.0
Additional Properties of Proportions • If , then • If , then
Using Properties of Proportions • Tell whether the statement is true.
In the diagram . Find the length of BD. A 30 16 B C x 10 D E
In the diagram • Solve for DE. A 5 2 D B 3 E C
Geometric mean • The geometric mean of two positive numbers a and b is the positive number x such that • Find the geometric mean of 8 and 18.
Geometric Mean • Find the geometric mean of 5 and 20. • The geometric mean of x and 5 is 15. Find the value of x.
Different perspective of Geometric mean • The geometric mean of ‘a’ and ‘b’ is √ab • Therefore geometric mean of 4 and 9 is 6, since √(4)(9) = √36 = 6.
Geometric mean • Find the geometric mean of the two numbers. • 3 and 27 √(3)(27) = √81 = 9 • 4 and 16 √(4)(16) = √64 = 8 • 5 and 15 √(5)(15) = √75 = 5√3
A scale model of the Titanic is 107.5 inches long and 11.25 inches wide. The titanic itself was 882.75ft long. How wide was it? • A model truck is 13.5 inches long and 7.5 inches wide. The original truck was 12 feet long. How wide was it?
Handout 8.1 • Handout 8.2
8.3 Similar Polygons Day 2 Part 1 CA Standard 11.0
Identifying Similar Polygons • When there is a correspondence between two polygons such that their corresponding angles are congruent and the lengths of corresponding sides are proportional the two polygons are called similar polygons.
Similar polygons • If ABCD ~ EFGH, then G H C D F B A E
Similar polygons • Given ABCD ~ EFGH, solve for x. G H C D 6 x B F 2 4 A E 2x = 24 x = 12
Is ABC ~ DEF? Explain. B D 6 E 13 12 5 7 A C F 10 ABC is not similar to DEF since corresponding sides are not proportional. ? ? yes no
Similar polygons Given ABCD ~ EFGH, solve for the variables. G H C D 5 x B F 10 2 y 6 A E
If two polygons are similar, then the ratio of the lengths of two corresponding sides is called the scale factor. Ex: Scale factor of this triangle is 1:2 9 4.5 3 6
Quadrilateral JKLM is similar to PQRS. Find the value of z. R K L S 15 Q z 6 P J M 10 15z = 60 z = 4
Theorem • If two polygons are similar, then the ratio of their perimeters is equal to the ratios of their corresponding side lengths. • If KLMN ~ PQRS, then
Given ABC ~ DEF, find the scale factor of ABC to DEF and find the perimeter of each polygon. E P = 8 + 12 + 20 = 40 B P = 4 + 6 + 10 = 20 12 20 10 6 A C D F 8 4 CORRESPONDING SIDES 4 : 8 1 : 2
8.4 Similar Triangles Day 2 Part 2 CA Standards 4.0, 5.0, 17.0
In the diagram, ∆BTW ~ ∆ETC. • Write the statement of proportionality. • Find m<TEC. • Find ET and BE. T 34° E C 3 20 79° B W 12
Postulate 25 • Angle-Angle (AA) Similarity Postulate If two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar.
Similar Triangles • Given the triangles are similar. Find the value of the variable. )) m )) ) 6 8 11m = 48 ) 11
Similar Triangles • Given the triangles are similar. Find the value of the variable. Left side of sm Δ Base of sm Δ Left side of lg Δ Base of lg Δ = 6 5 > 2 6h = 40 > h
∆ABC ≈ ∆DBE. A 5 D y 9 x B C E 8 4
Determine whether the triangles are similar. 6 32° 33° 9 18 No, because two angles of one triangle are not congruent to two angles of another triangle.
Determine whether the triangles are similar. 60° 60° 60° 60° Yes, because two angles of one triangle are congruent to two angles of another triangle.
Given two triangles are similar, solve for the variables. 2b - 8 a + 3 14 15 16 ) ) 10 15(a+3) = 10(16) 15a + 45 = 160 15a = 115
Decide whether two triangles are similar, not similar, or cannot be determined. A 92° 31° S 47° 41° 92° 57° S + 92 + 41 = 180 S + 133 = 180 S = 47 A + 92 + 57 = 180 A + 149 = 180 A = 31 Not similar
Handout 8.3 • Pg. 483 # 6 – 48 even
8.5 Proving Triangles are Similar Day 3 Part 1 CA Standards 5.0, 16.0
Warmup • Solve.
Side-Side-Side (SSS) Similarity Theorem • If the lengths of the corresponding sides of two triangles are proportional, then the triangles are similar. If AB= BC= CA PQ QR RP, then, ΔABC ~ΔPQR P A Q R C B