Eureka Math Geometry Module 2 Lesson 19 Answer Key

Engage NY Eureka Math Geometry Module 2 Lesson 19 Answer Key

Eureka Math Geometry Module 2 Lesson 19 Opening Exercise Answer Key

Show x: y = x’: y’ is equivalent to x: x’ = y: y’.  Eureka Math Geometry Module 2 Lesson 19 Exercise Answer Key

Exercise 1. x = 4

Exercise 2. x = 1.5

Eureka Math Geometry Module 2 Lesson 19 Problem Set Answer Key

Question 1.
Given the diagram shown, $$\overline{A D}\|\overline{G J}\| \overline{L O} \| \overline{Q T}$$ and $$\overline{A Q}\|\overline{B R}\| \overline{C S} \| \overline{D T}$$ Use the additional information given in each part below to answer the questions: a. If GL = 4, what is HM?
GHML forms a parallelogram since opposite sides are parallel, and opposite sides of a parallelogram are equal in length; therefore, HM = GL = 4.

b. If GL = 4, LQ = 9, and XV = 5, what is YZ?
Parallel lines cut transversals proportionally; therefore, it is true that = $$\frac{G L}{L Q}=\frac{X Y}{Y Z}$$, and likewise YZ = $$\frac{L Q}{G L}$$ (XV).
YZ = $$\frac{9}{4}$$(5)
YZ= 11$$\frac{1}{4}$$

c. Using information from part (b), If CI = 18, what is WX?
By the same argument used in part (a), IN = GL = 4. Parallel lines cut transversals proportionally; therefore, it is true that $$\frac{C I}{I N}=\frac{W X}{X Y}$$, and likewise WX = $$\frac{C I}{I N}$$(XY).
WX = $$\frac{18}{4}$$(5)
WX = 22$$\frac{1}{2}$$

Question 2.
Use your knowledge about families of parallel lines to find the coordinates of point P on the coordinate plane below. The given lines on the coordinate plane are parallel because they have the same slope m = 3. First, draw a horizontal transversal through points (8, -2) and (10, -2) and a second transversal through points (10,4) and (10, -4). The transversals intersect at (10, -2). Parallel lines cut transversals proportionally, so using horizontal and vertical distances, $$\frac{6}{2}=\frac{2}{x}$$, where x represents the distance from point (10, -2) to P.
x = $$\frac{2}{6}$$(2)
x = $$\frac{4}{6}=\frac{2}{3}$$
Point P is $$\frac{2}{3}$$ unit more than 10, or 10$$\frac{2}{3}$$, so the coordinates of point Pare (1o$$\frac{2}{3}$$, -2).

Question 3.
ACDB and FCDE are both trapezoids with bases $$\overline{A B}, \overline{F E}$$, and $$\overline{C D}$$. The perimeter of trapezoid ACDB is 24$$\frac{1}{2}$$. If the ratio of AF: FC is 1: 3, AB = 7, and ED = 5$$\frac{5}{8}$$, find AF, FC, and BE.  Question 4.
Given the diagram and the ratio of a: b is 3: 2, answer each question below: a. Write an equation for an in terms of bn.
an = $$\frac{3}{2}$$bn

b. Write an equation for bn in terms of an.
bn = $$\frac{2}{3}$$an

c. Use one of your equations to find b1 in terms of a if a1 = 1.2(a).
bn = $$\frac{2}{3}$$an
b1 = $$\frac{2}{3}\left(\frac{12}{10} a\right)$$
b1 = $$\frac{4}{5}$$a

d. What is the relationship between b1 and b?
Using the equation from parts (a) and (c), a = $$\frac{3}{2}$$b, so b1 = $$\frac{4}{5}\left(\frac{3}{2} b\right)$$; thus, b1 = $$\frac{6}{5}$$b.

e. What constant, c, relates b1 and b? Is this surprising? Why or why not?
The constant relating b1 and b is the same constant relating a1 to a, c = $$\frac{12}{10}$$ = 1.2.

f. Using the formula an = c · an – 1, find a3 in terms of a. g. Using the formula bn = c · bn-1, find b3 in terms of b.
b3 = $$\frac{216}{125}$$b

h. Use your answers from parts (f) and (g) to calculate the value of the ratio of a3: b3. Question 5.
julius wants to try to estimate the circumference of the earth based on measurements made near his home. He cannot find a location near his home where the sun Is straight overhead. Will he be able to calculate the circumference of the earth? If so, explain and draw a diagram to support your claim.
Note to the teacher: This problem is very open-ended, requires critical thinking, and may not be suitable for all students. The problem may be scaffolded by providing a diagram with possible measurements that Julius made based on the description in the student solution below. Possible solution: If Julius can find two locations such that those locations and their shadows lie in the same straight path, then the difference of the shadows’ angles can be used as part of the 360° in the earth’s circumference. The distance between those two locations corresponds with that difference of angles.

Eureka Math Geometry Module 2 Lesson 19 Exit Ticket Answer Key

Question 1.
Given the diagram shown, $$\overline{A G}$$ || $$\overline{B H}$$ || $$\overline{C I}$$, AB = 6.5 cm, GH = 7.5 cm, and HI = 18 cm. Find BC.  