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Eureka Math Geometry Module 3 Lesson 3 Answer Key
Engage ny eureka math geometry module 3 lesson 3 answer key, eureka math geometry module 3 lesson 3 exercise answer key.
THE SCALING PRINCIPLE FOR AREA: Answer: If similar figures A and B are related by a scale factor of r, then their respective areas are related by a factor of r 2 .
Eureka Math Geometry Module 3 Lesson 3 Problem Set Answer Key
a. List five unique sets of dimensions of your choice for a rectangle with an area of 18, and enter them in column 1. Answer:
b. If the given rectangle is dilated from a vertex with a scale factor of \(\frac{1}{2}\) what are the dimensions of the images of each of your rectangles? Enter the scaled dimensions in column 3. Answer:
c. What are the areas of the images of your rectangles? Enter the areas in column 4. Answer:
d. How do the areas of the images of your rectangles compare to the area of the original rectangle? Write the value of each ratio in simplest form in column 5. Answer:
e. Write the values of the ratios of area entered in column 5 in terms of the scale factor \(\frac{1}{2}\). Enter these values in column 6. Answer:
f. If the areas of two unique rectangles are the same, x, and both figures are dilated by the same scale factor r, what can we conclude about the areas of the dilated Images? Answer: The areas of the dilated images would both be r 2 x and thus equal.
Question 5. A piece of carpet has an area of 50 yd 2 . How many square inches will this be on a scale drawing that has 1 in. represent 1 yd.? Answer: One square yard will be represented by one square inch. So, 50 square yards will be represented by 50 square inches.
Question 8. A square region S is scaled parallel to one side by a scale factor of r, r ≠ 0, and is scaled in a perpendicular direction by a scale factor of one-third of r to yield its image S’. What is the ratio of the area of S to the area of S’? Answer: Let the sides of square S be s. Therefore, the resulting scaled image would have lengths rs and \(\frac{1}{3}\)rs. Then the area of square S would be s 2 , and the area of S’ would be \(\frac{1}{3}\) rs(rs) or \(\frac{1}{3}\)(rs) 2 or \(\frac{1}{3}\)r 2 s 2 . The ratio of areas of S to S’ is then s 2 : \(\frac{1}{3}\)r 2 s 2 or 1: \(\frac{1}{3}\) r 2 or 3: r 2 .
Question 9. Figure T’ is the image of figure T that has been scaled horizontally by a scale factor of \(\frac{1}{3}\). and vertically by a scale factor of \(\frac{1}{3}\). If the area of T’ is 24 square units, what is the area of figure T? Answer: Area(T’) = \(\frac{1}{3}\) · 4 · Area(T) 24 = \(\frac{4}{3}\) Area(T) \(\frac{3}{4}\) · 24 = Area(T) 18 = Area(T) The area of T is 18 square units.
Question 10. What is the effect on the area of a rectangle if … a. Its height is doubled and base left unchanged? Answer: The area would double.
b. Its base and height are both doubled? Answer: The area would quadruple.
c. Its base is doubled and height cut in half? Answer: The area would remain unchanged.
Eureka Math Geometry Module 3 Lesson 3 Exit Ticket Answer Key
b. What is the scale factor of the similarity transformation that takes ∆ ABC to ∆ CDE? Answer: r = \(\frac{4}{11}\)
c. What is the value of the ratio of the area of ∆ ABC to the area of ∆ CDE? Explain how you know. Answer: r 2 = \(\left(\frac{4}{11}\right)^{2}\) , or \(\frac{16}{121}\) by the scaling principle for triangles.
d. If the area of ∆ ABC is 30 cm 2 , what is the approximate area of ∆ CDE? Area(∆ CDE) = \(\frac{16}{121}\) × 30 cm 2 4 cm 2
Eureka Math Geometry Module 3 Lesson 3 Exploratory Challenge Answer Key
Complete parts (i)-(iii) of the table for each of the figures in questions (a)-(d): (i) Determine the area of the figure (pre image), (ii) determine the scaled dimensions of the figure based on the provided scale factor, and (iii) determine the area of the dilated figure. Then, answer the question that follows.
e. Make a conjecture about the relationship between the areas of the original figure and the similar figure with respect to the scale factor between the figures. Answer: It seems as though the value of the ratio of the area of the similar figure to the area of the original figure is the square of the scale factor of dilation.
THE SCALING PRINCIPLE FOR TRIANGLES: Answer: If similar triangles S and T are related by a scale factor of r, then the respective areas are related by a factor of r 2 .
THE SCALING PRINCIPLE FOR POLYGONS: Answer: If similar polygons P and Q are related by a scale factor of r, then their respective areas are related by a factor of r 2 .
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Lesson 3 Skills Practice Angles of Triangles Find the value of x in each triangle with the given angle measures. 1. 81 x 84 2. 24 36 x 3. 49 x 4. 38 38 x 5. 65 x 6. 71 x 45 7. 57˚, 51˚, x˚ 8. x˚, 126˚, 22˚ 9. 90˚, x˚, 50˚ Find the value of x in each triangle. 10. 37° 72° x° 11. 66° 33° x° 12. 50° 40° x° 13. 60° 60° x° 14 ...
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To get the answer, divide the total (180°) by 3. 6 8. 1 1. 1 1. 1 1. 1 4. 6 1. 10. 2013 ath ls otion, ® Date: Name: Triangles athantiso TRI 5 Instructions: For each triangle, find the unknown angle (X). Remember that for each triangle, the three interior angles must add up to 180 degrees.
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Eureka Math Geometry Module 3 Lesson 3 Problem Set Answer Key. Question 1. A rectangle has an area of 18. Fill in the table below by answering the questions that follow. Part of the first row has been completed for you. Answer: a. List five unique sets of dimensions of your choice for a rectangle with an area of 18, and enter them in column 1 ...
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