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Eureka Math Geometry Module 3 Lesson 11

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Eureka Math Geometry Module 3 Lesson 13 Exit Ticket Answer Key

Eureka Math, Grade 2, Module 3, Lesson 11

Lamar is using a 3D printer to construct a circular cone that has a base with radius 6 in.

a. If his 3D printer prints in layers that are 0. 004 in. thick , what should be the change in radius for each layer in order to construct a cone with height 4 in.?Answer:\x = 5.994The change in radius between consecutive layers is 0.006 inch.

b. What is the area of the base of the 27th layer?Answer:The 27th layer of the cone will have a radius reduced by 0.006 inch 26 times.26 = 0.156The radius of the 27th layer is 5.844 in.A = 2A 107.3The area of the base of the disk in the 27th layer is approximately 107.3 in2.

c. Approximately how much printing material is required to produce the cone?Answer:The volume of printing material is approximately equal to the volume of a true cone with the same dimensions.V = \2 V = 48

Eureka Math Grade 3 Module 2 Lesson 11 Problem Set Answer Key

Question 1.The total weight in grams of a can of tomatoes and a jar of baby food is shown to the right.a. The jar of baby food weighs 113 grams. How much does the can of tomatoes weigh?

Answer:

Explanation:The total weight in grams of a can of tomatoes and a jar of baby food is 671g and 113g.To find the weight of the can of tomatoes, subtract 113 from 671.671-113=558g.The weight of the can of tomatoes is 558g.

b. How much more does the can of tomatoes weigh than the jar of baby food?

Answer:

The weight of the can of tomatoes is 558g.A jar of baby food is 113gSo, subtract 113 from 558558-113=445Therefore, 445g more does can of tomatoes weigh than the jar of baby food.

Question 2.The weight of a pen in grams is shown to the right.a. What is the total weight of 10 pens?

Answer:As shown in the picture the weight of pen is 6g.If 1 pen is 6 grams then the weight of 10 pens is 6 x 10=60gTherefore the total weight of 10 pens is 60grams.

b. An empty box weighs 82 grams. What is the total weight of a box of 10 pens?

Answer:

Explanation:As we know from the above sum the weight of 10 pens is 60g and an empty box weighs 82 grams then the total weight will be 60+82=142grams.

Question 3.The total weight of an apple, lemon, and banana in grams is shown to the right.a. If the apple and lemon together weigh 317 grams, what is the weight of the banana?

Answer:

b. If we know the lemon weighs 68 grams less than the banana, how much does the lemon weigh?

Answer:

Answer:

Answer:

Answer:

Answer:

Eureka Math Geometry Module 2 Lesson 16 Example Answer Key

Example 1.Given ABC ~ ABC, find the missing side lengths.Answer: How many people used the ratio of lengths 4: 12 or 12: 4 and \) to determine the measurements of \ and \?

Are there any other length relationships we could use to set up an equation and solve for the missing side lengths?Provide students time to think about and try different equations that would lead to the same answer.

In addition to the ratio of corresponding lengths AC: AC, we can use a ratio of lengths within one of the triangles. For example, we could use the ratio of AC: AB to find the length of \. Equating the values of the ratios, we get the following:\AB = 15.

To find the length of \, we can use the ratios A C: B C and AC: BC. Equating the values of the ratios, we get the following:

Why is it possible to use this length relationship, AC: AB, to solve for a missing side length?Take student responses, and then offer the following explanation:

There are three methods that we can use to determine the missing side lengths.

Method 1 : Find the scale factor, and use it to compute for the desired side lengths.Since we know AC = 4 and the corresponding side AC = 12, the scale factor of r satisfies 4r = 12. So,r = 3, AB = rAB = 3 · 5 = 15, and BC \ = 2.

Method 2 : Equate the values of the ratios of the corresponding sides.

Method 3 : Equate the values of the ratios within each triangle.

Provide students time to explain why Method 3 works, and then offer the following explanation:

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Eureka Math Geometry Module 3 Lesson 11 Problem Set Answer Key

Question 1.What is the volume formula for a right circular cone with radius r and height h?Answer:Area = Bh = \h

Question 2.Identify the solid shown, and find its volume.Answer:The solid is a triangular pyramid.Volume = \BhVolume = \ · 3 · 6) · 4Volume = 12The volume of the triangular pyramid is 12 unit3.

Question 3.Find the volume of the right rectangular pyramid shown.Answer:Volume = \BhVolume = \ · 6Volume = 768The volume of the right rectangular pyramid is 768 unit3.

Question 4.Find the volume of the circular cone in the diagram. as an approximation of pi.)Answer:Volume = \BhVolume = \hVolume = \ · 142) · 27Volume = \ · 27Volume = 5544The volume of the circular cone is, 544 unit3.

Question 5.Find the volume of a pyramid whose base is a square with edge length 3 and whose height is also 3.Answer:Volume = \BhVolume = \ · 3 = 9The volume of pyramid is 9 units3.

Question 6.Suppose you fill a conical paper cup with a height of 6 with water. If all the water is then poured into a cylindrical cup with the same radius and same height as the conical paper cup, to what height will the water reach in the cylindrical cup?Answer:A height of 2

Question 8.A pyramid has volume 24 and height 6. Find the area of its base.Answer:Let A be the area of the base.Volume = \Bh24 = \A · 6A = 12The area of the base is 12 units2.

The volume of the frustum is the difference of the volumes of the total cone and the smaller cone between the frustum and the vertex.

Eureka Math Grade 3 Module 2 Lesson 11 Homework Answer Key

Eureka Math Geometry Module 1 Lesson 21 Answer Key  CCSS ...

Question 1.Karina goes on a hike. She brings a notebook, a pencil, and a camera. The weight of each item is shown in the chart. What is the total weight of all three items?

Item

Together a horse and its rider weigh 729 kilograms. The horse weighs 625 kilograms.Subtract the weight of horse from the total to find the weight of the rider729-625=104Therefore, The rider weighs 104 kilograms

Question 3.Theresas soccer team fills up 6 water coolers before the game. Each water cooler holds 9 liters of water. How many liters of water do they fill?

Answer:

Explanation:Theresas soccer team fills up 6 water coolers before the game. Each water cooler holds 9 liters of waterIf each cooler holds 9L and soccer team fills 6 colers, to find the total number of liters of water the filled,multiply 9 by 6=9 x 6=54Therefore, liters of water they filled is 54.

Question 4.Dwight purchased 48 kilograms of fertilizer for his vegetable garden. He needs 6 kilograms of fertilizer for each bed of vegetables. How many beds of vegetables can he fertilize?

Answer:

Dwight purchased 48 kilograms of fertilizer for his vegetable garden.If he needs 6 kilograms of fertilizer for each bed of vegetables,To find the number beds of vegetables he can fertilize divide 48 by 648/6=8Therefore, 8 beds of vegetables he can fertilize.

Question 5.Nancy bakes 7 cakes for the school bake sale. Each cake requires 5 milliliters of oil. How many milliliters of oil does she use?

Answer:

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Eureka Math Geometry Module 1 Lesson 11 Exercise Answer Key

Opening ExerciseA proof of a mathematical statement is a detailed explanation of how that statement follows logically from other statements already accepted as true.A theorem is a mathematical statement with a proof.

Consider taking a moment to mention that theorems can be stated without reference to any specific, labeled diagram. However, we cannot take steps to prove a statement without a way of referring to parts. Students observe situations where the labels are provided and situations where they must draw diagrams and label parts.

DiscussionOnce a theorem has been proved, it can be added to our list of known facts and used in proofs of other theorems.For example, in Lesson 9, we proved that vertical angles are of equal measure, and we know that if a transversal intersects two parallel lines, alternate interior angles are of equal measure. How do these facts help us prove that corresponding angles are equal in measure?Answer:Answers may vary.

In the diagram to the right, if you are given that \ || \, how can you use your knowledge of how vertical angles and alternate interior angles are equal in measure to prove that x=w?Answer:w=z z=x x=w

You now have available the following facts: Vertical angles are equal in measure. Alternate interior angles are equal in measure. Corresponding angles are equal in measure.

What is the absolute shortest list of facts from which all other facts can be derived?

Eureka Math Geometry Module 2 Lesson 16 Problem Set Answer Key

Question 1. DEF ~ ABC All side length measurements are in centimeters. Use between-figure ratios and/or within-figure ratios to determine the unknown side lengths.Answer:Using the given similarity statement, D corresponds with A, andC corresponds with F, so it follows that \ corresponds with \, \ with \, and \ with \.

Question 2.Given ABC XYZ, answer the following questions:a. Write and find the value of the ratio that compares the height \ to the hypotenuse of ABC.Answer:\

b. Write and find the value of the ratio that compares the base \ to the hypotenuse of ABC.Answer:\

c. Write and find the value of the ratio that compares the height AC to the base AB of ABC.Answer:

d. Use within-figure ratios to find the corresponding height of XYZ.Answer:\XZ = 1\

e. Use within-figure ratios to find the hypotenuse of XYZ.Answer:\YZ = 4\

Question 3.Right triangles A, B, C, and D are similar. Determine the unknown side lengths of each triangle by using ratios of side lengths within triangle A.a. Write and find the value of the ratio that compares the height to the hypotenuse of triangle A.Answer:\ 0.866

b. Write and find the value of the ratio that compares the base to the hypotenuse of triangle A.Answer:\ = 0.5

c. Write and find the value of the ratio that compares the height to the base of triangle A.Answer:\ = 3 1.73

f. Find the unknown lengths of triangle C.Answer:The base of triangle C is 3.The height of triangle C is 3.

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Eureka Math Geometry Module 3 Lesson 9 Exit Ticket Answer Key

Question 1.Two circular cylinders are similar. The ratio of the areas of their bases is 9:4. Find the ratio of the volumes of the similar solids.Answer:If the solids are similar, then their bases are similar as well. The areas of similar plane figures are related by the square of the scale factor relating the figures, and if the ratios of the areas of the bases is 9:4, then the scale factor of the two solids must be \, or \. The ratio of lengths in the two solids is, therefore, 3:2.By the scaling principle for volumes, the ratio of the volumes of the solids is the ratio 33 : 23, or 27 : 8.

Question 2.The volume of a rectangular pyramid is 60. The width of the base is then scaled by a factor of 3, the length of the base is scaled by a factor of \, and the height of the pyramid is scaled such that the resulting image has the same volume as the original pyramid. Find the scale factor used for the height of the pyramid.Answer:The scaling principle for volumes says that the volume of a solid scaled in three perpendicular directions is equal to the area of the original solid times the product of the scale factors used in each direction. Since the volumes of the two solids are the same, it follows that the product of the scale factors in three perpendicular directions must be 1.Lets represent the scale factor used for the height of the image:Volume = · s) )60 = \ · s · s = \

Eureka Math Geometry Module 3 Lesson 9 Opening Exercise Answer Key

Eureka Math Grade 1 Module 3 Lesson 11

a. For each pair of similar figures, write the ratio of side lengths a: b or C: d that compares one pair of corresponding sides. Then, complete the third column by writing the ratio that compares the areas of the similar figures. Simplify ratios when possible.Answer:

b.i. State the relationship between the ratio of sides a: b and the ratio of the areas Area: Area.Answer:When the ratio of side lengths is a: b, then the ratio of the areas is a2 : b2.

ii. Make a conjecture as to how the ratio of sides a: b will be related to the ratio of volumes Volume: Volume. Explain.Answer:When the ratio of side lengths is s: t, then the ratio of the volumes will probably be s3 : t3. Area is twodimensionaI, and the comparison of areas was raised to the second power. Since volume is three-dimensional, I think the comparison of volumes will be raised to the third power.

c. What does it mean for two solids In three-dimensional space to be similar?Answer:It means that a sequence of basic rigid motions and dilations maps one figure onto the other.

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Eureka Math Geometry Module 3 Lesson 11 Exercise Answer Key

Exercise 1.A cone fits inside a cylinder so that their bases are the same and their heights are the same, as shown in the diagram below. Calculate the volume that Is inside the cylinder but outside of the cone. Give an exact answer.Answer:The volume of the cylinder is V = 52 = 300.The volume of the cone is V = \ 52 = 100.The volume of the space that is inside the cylinder but outside the cone is 200.

Alternative solution:The space between the cylinder and cone is equal to \ the volume of the cylinder. Then, the volume of the space isV = 52 = 200.

Exercise 2.A square pyramid has a volume of 245 in3. The height of the pyramid is 15 in. What is the area of the base of the pyramid? What is the length of one side of the base?Answer:V = \ 245 = \ 245 = 549 = area of baseThe area of the base is 49 in2, and the length of one side of the base is 7 in.

Exercise 3.Use the diagram below to answer the questions that follow.a. Determine the volume of the cone shown below. Give an exact answer.Answer:Let the length of the radius be r.112 + r2 = 2r = 4The volume of the cone is V = \42 = \.

b. Find the dimensions of a cone that is similar to the one given above. Explain how you found your answers.Answer:Student answers vary. A possible student answer:A cone with height 33 and radius 12 is similar to the one given because their corresponding sides are equal in ratio, that is, 11:33 and 4:12. Therefore, there exists a similarity transformation that would map one cone onto the other.

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Eureka Math Geometry Module 5 Lesson 11 Exercise Answer Key

ExercisesExercise 1.\ and \ are tangent to circle A at points D and E, respectively. Use a two-column proof to prove a = b.Answer:Draw radii \ and \and segment \.CD = a, CE = b GivenADC and AEC are right angles. Tangent lines are perpendicular to the radius at the point of tangency.ADC and AEC are right triangles. Definition of a right triangleAD = AE Radii of the same circle are equal in measure.AC = AC Reflexive propertyCD = CE Corresponding sides of congruent triangles are equal in length.a = b Substitution

In circle A, the radius is 9 mm and BC = 12 mm.a. Find AC.

\ is tangent to the circle at point C, and CD = DE.a. Find x ).Answer:CDE is an inscribed angle, so \ is two times the measure of the intercepted arc \ = 152° and \ = 208°. Since CD = DE, then \ = m \. Therefore, 2x = 208° and x = 104°.

b. Find y .Answer:\ = 208°, so mCFE must be one half this value since it is an inscribed angle that intercepts the arc. Therefore, y = 104°.

Question 6.Construct two lines tangent to circle A through point B.Answer:

Question 7.Find x, the length of the common tangent line between the two circles .Answer:x = 12.17

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