practice and homework lesson 10.3 answer key 3rd grade

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Lesson 10 3

Lesson 10 3 - Displaying top 8 worksheets found for this concept.

Some of the worksheets for this concept are Answer key, Lesson 10 work 1 counting 12345 12345, Student work and activity, Lesson 12 length area and volume, Lesson practice b for use with 698704, Homework practice and problem solving practice workbook, Grade 2 lesson 10, Language handbook work.

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1. Answer Key

2. lesson 10 worksheet 1 counting 12345 12345, 3. student worksheets and activity sheets, 4. lesson 12: length, area, and volume, 5. lesson practice b 10.3 for use with pages 698704, 6. homework practice and problem-solving practice workbook, 7. grade 2 lesson 10 -, 8. language handbook worksheets.

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Problem Solving - Organize Data - Lesson 2.1

Use Picture Graphs - Lesson 2.2

Make Picture Graphs - Lesson 2.3

Use Bar Graphs - Lesson 2.4

Make a Bar Graph - Lesson 2.5

Solve Problems Using Data - Lesson 2.6

Use and Make Line Plots - Lesson 2.7

Number Patterns - Lesson 1.1

Round to Nearest Ten or Hundred - Lesson 1.2

Estimate Sums - Lesson 1.3

Mental Math Strategies for Addition - Lesson 1.4

Use Properties to Add - Lesson 1.5

Use the Break Apart Strategy to Add - Lesson 1.6

Use Place Value to Add - Lesson 1.7

Estimate Differences - Lesson 1.8

Mental Math Strategies for Subtraction - Lesson 1.9

Use Place Value to Subtract - Lesson 1.10

Combine Place Values to Subtract - Lesson 1.11

Describe Plane Shapes - Lesson 12.1

Describe Angles in Plane Shapes - Lesson 12.2

Identify Polygons - Lesson 12.3 ​

Describe Sides of Polygons - Lesson 12.4

​ Classify Quadrilaterals - Lesson 12.5

​Draw Quadrilaterals - Lesson 12.6

Describe Triangles - Lesson 12.7

Chapter 12 Performance Task Review For Test

Problem Solving - Compare Fractions - Lesson 9.1

Compare Fractions with the Same Denominator - Lesson 9.2

Compare Fractions with the Same Numerator - Lesson 9.3

Compare Fractions - Lesson 9.4

Compare and Order Fractions - Lesson 9.5

Model Equivalent Fractions - Lesson 9.6

Equivalent Fractions - Lesson 9.7

Divide by 2 - Lesson 7.1

Divide by 10 - Lesson 7.2

Divide by 5 - Lesson 7.3

Divide by 3 - Lesson 7.4

Divide by 4 - Lesson 7.5

Divide by 5 - Lesson 7.6

Mid-Chapter 7 Checkpoint on Division Facts and Strategies

Divide by 7 - Lesson 7.7

Divide by 8 - Lesson 7.8

Divide by 9 - Lesson 7.9

Problem Solving - Two-Step Problems - Lesson 7.10

Order of Operations - Lesson 7.11

Problem Solving - Model Division - Lesson 6.1

Size of Equal Groups - Lesson 6.2

Number of Equal Groups - Lesson 6.3

Model (Division) with Bar Model - Lesson 6.4

Relate Subtraction and Division - Lesson 6.5

Mid-Chapter 6 Checkpoint

Model (division) with Arrays - Lesson 6.6

Relate Multiplication and Division - Lesson 6.7

Write Related Facts - Lesson 6.8

Division Rules for 1 and 0 - Lesson 6.9

Chapter 6 Review for Test - Understanding Division

Multiply with 2 and 4 - Lesson 4.1

Multiply with 5 and 10 - Lesson 4.2

Multiply with 3 and 6 - Lesson 4.3

Distributive Property - Lesson 4.4

Multiply with 7 - Lesson 4.5

Associative Property of Multiplication - Lesson 4.6

Patterns on the Multiplication Table - Lesson 4.7

Multiply with 8 - Lesson 4.8

Multiply with 9 - Lesson 4.9

Review For Test on Chapter 4

Describe Patterns - Lesson 5.1

Find Unknown Factors - Lesson 5.2

Problem Solving: Using the Distributive Property - Lesson 5.3

Multiplication Strategies with Multiples of 10 - Lesson 5.4

Multiply Multiples of 10 by 1-Digit Numbers - Lesson 5.5

Chapter 5 Review on Multiplication Facts

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Model Perimeter - Lesson 11.1

Find Perimeter - Lesson 11.2

Find Unknown Side Lengths - Lesson 11.3

Understanding Area - Lesson 11.4

Measure Area - Lesson 11.5

Use Area Models - Lesson 11.6

Problem Solving - Area of Rectangles - Lesson 11.7

Area of Combined Rectangles - Lesson 11.8

Same Perimeter - Different Area - Lesson 11.9

Same Area - Different Perimeter - Lesson 11.10

Chapter 11 Review for Test on Perimeter and Area

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Count Equal Groups - Lesson 3.1

Relate Addition and Multiplication - Lesson 3.2

Skip Count on a Number Line - Lesson 3.3

Problem Solving - Model Multiplication - Lesson 3.4

Model with Arrays - Lesson 3.5

Commutative Property of Multiplication - Lesson 3.6

Multiply with 1 and 0 - Lesson 3.7

Time to the Minute - Lesson 10.1

A.M. and P.M. - Lesson 10.2

Measure Time Intervals - Lesson 10.3

Use Time Intervals - Lesson 10.4

Problem Solving - Time Intervals - Lesson 10.5

Measure Length - Lesson 10.6

Estimate and Measure Liquid Volume - Lesson 10.7

Estimate and Measure Mass - Lesson 10.8

Equal Parts of a Whole - Lesson 8.1

Equal Shares - Lesson 8.2

Unit Fractions of a Whole - Lesson 8.3

Fractions of a Whole - Lesson 8.4

Fractions on a Number Line - Lesson 8.5

Relate Fractions and Whole Numbers - Lesson 8.6

Fractions of a Group - Lesson 8.7

Find Part of Group Using Unit Fractions - Lesson 8.8

Problem Solving: Find the Whole Using Unit Fractions - Lesson 8.9

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Into Math Grade 7 Module 10 Lesson 3 Answer Key Describe and Analyze Cross Sections of Circular Solids

We included  HMH Into Math Grade 7 Answer Key  PDF   Module 10 Lesson 3 Describe and Analyze Cross Sections of Circular Solids to make students experts in learning maths.

HMH Into Math Grade 7 Module 10 Lesson 3 Answer Key Describe and Analyze Cross Sections of Circular Solids

I Can identify the shapes of cross sections of circular solids and solve problems involving the areas of cross sections.

Spark Your Learning

Build Understanding

Question 1. An art class is cutting foam figures for a project.

Connect to Vocabulary A plane is a flat surface that has no thickness and extends forever. When a plane intersects a solid, the two-dimensional figure formed is called a cross-section. Some cross-sections of cones and cylinders are circles.

Step It Out

B. Find the area of the vertical cross-section of the cone. Explain how you arrived at your answer. Answer: Area = 7.875 sq in Explanation: A = (1/2)bh A = 0.5 x 2.625 x 6 A = 7.875 sq in

Check Understanding

Question 1. Ralph has a cylindrical container of parmesan cheese. The diameter of the base of the container is 2.75 inches, and the height is 6 inches. What is the area of a horizontal cross-section of the cylinder to the nearest tenth of a square inch? Use 3.14 for π. Answer: Area = 5.94 sq in Explanation: d = 2.75 in r = 2.75 / 2 = 1.375 in Area A = πr 2 A = 3.14 x 1.375 x 1.375 A = 5.94 sq in

On Your Own

Question 4. Use Repeated Reasoning Brock has a cylindrical metal tin where he keeps his coins. The radius of the base is 5.5 inches, and the height is 4 inches. A. What is the circumference of a horizontal cross-section of the cylinder? Use 3.14 for π. Answer: Circumference = 34.55 in Explanation: C = 2πr r = 5.5 in C = 2 x 3.14 x 5.5 C = 34.55 in

B. What is the area of a horizontal cross-section of the cylinder to the nearest hundredth? Use 3.14 for π. Answer: Area = 94.985 sq in Explanation: radius  = 5.5 in Area A = πr 2 A = 3.14 x 5.5 x 5.5 A = 94.985 sq in

Question 9. Reason Why can’t the dimensions of a horizontal cross-section of a cone be determined from just the dimensions of the cone? Answer: A horizontal plane passing through the vertex will cut the cone into two parts with a cross section of a triangle. Explanation: As, the vertical shape of the cone is of triangle. The horizontal cross section of a cone is triangle but varies the radius from base to top edge.

Question 10. A cone-shaped paperweight is 5 inches tall, and the base has a circumference of about 12.56 inches. What is the area of a vertical cross-section through the center of the base of the paperweight? Use 3.14 for π. Answer: Area = 10 sq in Explanation: Circumference = 12.56 in C = 2πr h = 5 in 12.56 = 2 x3.14 x r r = 12.56/6.28 r = 2 in d = 2r = 2 x 2 = 4 in Area of a triangle shape cone A = (1/2) base x height A = 0.5 x 4 x 5 A = 10sq in

Question 12. Find the area of a vertical cross-section through the center of the base of a cone with a height of 5 feet and a circumference of about 28.26 feet. Use 3.14 for π. Answer: Area = 22.5 sq ft Explanation: height = 5 ft Circumference = 28.26ft radius = 28.26/2 x 3.14 r = 28.26/6.28 = 4.5 ft A = (1/2) x 9 x 5 A = 0.5 x 45 A = 22.5 sq ft

Question 13. Find the area of a vertical cross-section through the center of a sphere with a diameter of 16 centimeters. Use 3.14 for π. Answer: Area = 200.96 sq cm Explanation: diameter = 16 cm radius = d/2 r = 16/2 cm r = 8 cm A = πr 2 A = 3.14 x 8 x 8 A = 3.14 x 64 A = 200.96 sq cm

Question 14. Find the area of a vertical cross-section through the centers of the bases of a cylinder with a height of 24 inches and a circumference of about 43.96 inches. Use 3.14 for π. Answer: Area = 336 sq in Explanation: Circumference = 2 π r 43.96 = 2 π r 43.96 / 2 x 3.14 = r radius = 43.96/6.28 r = 7 in d = 2r = 2 x 7 = 14 in A = length x width A = 14 x 24 A = 336 sq in

Question 15. Find the area of a horizontal cross-section of a cylinder with a height of 34 centimeters and a circumference of about 131.88 centimeters. Use 3.14 for π. Answer: Area = 882 sq cm Explanation: Circumference = 2 π r 131.88 = 2 π r 131.88 / 2 x 3.14 = r radius = 131.88/6.28 r = 21 cm d = 2r = 2 x 21 = 42 cm A = length x width A = 42 x 21 A = 882 sq cm

I’m in a Learning Mindset!

What strategies do I use to stay on task when working on my own? Answer: One should be through with the concept and formulas of circle and their cross sections.

Lesson 10.3 More Practice/Homework

Question 1. Clyde has a cone-shaped party hat. The height of the hat is 10 inches, and the radius of the base of the hat is 4 inches. What is the area of a vertical cross-section through the center of the base of the party hat? Answer: Area = 40 sq in Explanation: height = 10 in radius = 4 in diameter = 2r = 2 x 4 = 8 in Area = (1/2) base x width A = 0.5 x 10 x 8 A = 40 sq in

Question 2. A cylindrical swimming pool has a height of 4 feet and a circumference of about 75.36 feet. What is the area of a vertical cross-section through the center of the pool? Use 3.14 for π. Answer: Area = 96 sq ft Explanation: Circumference = 75.36 ft C = 2πr 75.36/2×3.14 = r radius = 12 ft diameter = 2r = 2 x 12 = 24 ft height = 4 ft Area  = length x width Area= 24 x 4 Area = 96 sq ft

Question 6. Find the area of a vertical cross-section through the centers of the bases of a cylinder with a height of 27 inches and a circumference of about 47.1 inches. Use 3.14 for π. Answer: Area = 702 sq in Explanation: height = 27 in Circumfrence = 47.1 in C = 2πr = 81.64 m 81.64 = 2πr 81.64/ 2 x 3.14 = r 81.64/6.28 = r radius = 13 in diameter = 2r = 2 x 13 = 26 in A = length x width A = 27 x 26 A = 702 sq in

Question 7. Find the area of a horizontal cross-section of a cylinder with a height of 11 meters and a circumference of about 81.64 meters. Use 3.14 for π. Answer: Area = 530.66 sq m Explanation: Circumference = 2πr = 81.64 m C = 2 x 3.14 x r 81.64 = 6.28 r r = 81.64 / 6.28 r = 13 Area of the circle A = π.r.r A = 3.14 x 13 x 13 A = 530.66 sq m

Question 8. Two chefs are working on cylindrical cakes. David wants to make a stripe of frosting in his cake, and he makes a cut that shows a rectangle. Terri wants to put a layer of frosting in the middle of her cake, so she makes a cut that shows a circle. How was Terri’s cut different from David’s? Answer: David’s cut It is rectangular in shape. It has 4 corners, 6 sides and 8 faces. Terri’s cut  It is circular in shape. It has no corners, no edges, no sides. Explanation: Rectangle is a simple four sided polygon with internal angles equal to 90 degrees. The two sides at each corner meet at right angles and parallel to each other. A circle is a round shaped figure that has no corners or edges.

Spiral Review

Question 11. A straight path from the edge of a circular garden to the center of the garden is 7 meters long. What is the area of the garden? Use 3.14 for π. Answer: Area = 153.86 sq m Explanation; radius = 7 m Area A = πr 2 A = 3.14 x 7 x 7 A = 3.14 x 49 A = 153.86 sq m

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Texas Go Math Grade 7 Lesson 10.3 Answer Key Lateral and Total Surface Area

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 10.3 Answer Key Lateral and Total Surface Area.

Step 1 Find the lateral area of the rectangular prism. There are two 2 in. by 5 in. rectangles, and two 2 in. by 6 in. rectangles.

2 . (2 . 5) = 20 in 2 2 . (2 . 6) = 24 in 2 The lateral area is 20 + 24 = 44 in 2 .

Step 2 Find the total surface area of the rectangular prism. Each base is 5 inches by 6 inches. 2 . (5 . 6) = 60 in 2 . The total surface area is 44 + 60 = 104 in 2 .

Math Talk Mathematical processes Explain how the total surface area of a prism differs from the lateral area.

Step 1 Find the lateral area of the rectangular pyramid. There are four triangles with base 16 in. and height 17 in. The lateral area is 4 × \(\frac{1}{2}\)(16)(17) = 544 in 2 .

Step 2 Find the total surface area of the rectangular pyramid. The area of the base is 16 × 16 = 256 in 2 . The total surface area is 544 + 256 = 800 in 2 .

Question 2. How many surfaces does a triangular pyramid have? What shape are they? Answer: A triangular pyramid has four surfaces and they have the shape of triangle.

Example 3. Shoshanna’s team plans to build stands to display sculptures. Each stand will be in the shape of a rectangular prism. The prism will have a square base with side lengths of 2\(\frac{1}{2}\) feet, and it will be 3\(\frac{1}{2}\) feet high. The team plans to cover the stands with metallic foil that costs $0.22 per square foot. How much money will the team save on each stand if they cover only the lateral area instead of the total surface area?

Step 2 Find the lateral and total surface areas of the prism. The four lateral faces are 3\(\frac{1}{2}\) feet by 2\(\frac{1}{2}\) feet rectangles. The two bases are 2\(\frac{1}{2}\) feet by 2\(\frac{1}{2}\) feet squares. Lateral area: 4\(\frac{1}{2}\) . (3\(\frac{1}{2}\) . 2\(\frac{1}{2}\)) = 35 square feet Total surface area: 2 (2\(\frac{1}{2}\) . 2\(\frac{1}{2}\)) + 35 = 47\(\frac{1}{2}\) square feet

Add the area of the bases to the lateral area.

Step 3 Find and compare the prices. Cost for total surface area: $0.22(47\(\frac{1}{2}\)) = $10.45 Cost for lateral area: $0.22(35) = $7.70 The team will save $10.45 – $7.70, or $2.75, for each stand by covering only the lateral area.

Texas Go Math Grade 7 Lesson 10.3 Guided Practice Answer Key 

A three-dimensional figure is shown sitting on a base. (Example 1)

Question 1. The figure has a total of _____ rectangular faces. Answer: The figure has a total of six rectangular faces.

Question 2. Of the total number of faces, ___ are lateral faces. Answer: Of the total number of faces, four are lateral faces

Question 3. The figure is a ____. Answer: The figure is a rectangular prism

Question 5. The lateral area of the prism is ___. Answer: For lateral area: Two 6 by 3 rectangtes Two 7 by 3 rectangles. 2 . (6 . 3) = 36 2 . (7 . 3) = 42 Lateral area: 36 + 42 = 78

The lateral area of the prism is 36 + 42 = 78

Question 6. The total surface area is ___. Answer: The base is in the shape of 6 by 7 rectangle. Base area: 2 (6 . 7) = 84 The total surface area is 78 + 84 = 162

A triangular prism is shown. (Example 2)

Question 7. Identify the number and type of faces of the prism. Answer: A triangular prism has five face: Two 6 cm. by 4.3 cm. rectangles; One 6 cm. by 3 cm. rectangle; Two triangles with base of 3 cm. and height of 4 cm.

Question 8. Find the lateral area of the prism. ____ Answer: For lateral area: Two 6 cm. by 4.3 cm. rectangles; One 6 cm. by 3 cm. rectangle; 2 . (6 . (4.3)) = 51.6 1 . (6 . 3) = 18 Lateral area: 51.6 + 18 = 69.6 cm 2 .

The lateral area of the prism is 69.6 cm 2 .

Question 9. Find the total surface area of the prism. Answer: The base is in the shape of triangle with length of 3 cm. and height of 4 cm. Base area 2 . (\(\frac{1}{2}\) . 3 . 4) = 12 ft 2 The total surface area is 69.6 + 12 = 81.6 cm 2 .

Go Math Lesson 10.3 7th Grade Areas of Similar Shapes Answer Key Question 10. Use a net to find the total surface area of the pyramid. Then find the cost of wrapping the pyramid completely in gold foil which costs $0.05 per square centimeter. (Examples 2 and 3) Answer: Find the lateral area of the rectangular pyramid. we have four triangles with base of 10 cm. ain height of 12 cm. b = 10 cm Length of base h B = 12 cm Height of base Lateral area: 4 . (\(\frac{1}{2}\) . b . h b ) = 4 . \(\frac{1}{2}\) . 10 . 12 = 240 cm 2 Base area: 10 . 10 = 1oo cm 2 The total surface area is 240 + 100 = 340 cm 2 . Wrapping the pyramid completely in gold foil will cost: 340 . (0.05) = $17

Essential Question Check-In

Question 11. How do you find the lateral and total surface area of a triangular pyramid? Answer: Lateral surface area of any three-dimensional figure means the sum of area of all sides except the area of base. And the total surface area means the area of all the sides of figure including the area of base.

Now in case of triangular pyramid it will have total four faces including the base triangle. So the lateral surface area will. be the sum of areas of all the three triangle which are faces of triangular pyramid except the base triangle. And the total surface area will be the sum of areas of all the four triangle including the base triangle

Texas Go Math Grade 7 Lesson 10.3 Independent Practice Answer Key 

The lateral area of the carton is 168 in 2 . The total surface area of the carton is 264 in 2 .

Question 15. Victor wrapped this gift box with adhesive paper (with no overlaps). How much paper did he use? Answer: The gift box is in the shape of a rectangular prism. Find the lateral area of the rectangular prism. We have: Two 6 in. by 5 in. rectangle; Two 8 in. by 5 in. rectangle. 2 . (6 . 5) = 60 2 . (8 . 5) = 80 The lateral area is 60 + 80 = 140 in 2 . Each base is 8 in. by 6 in.\ 2 . (8 . 6) = 2 . 48 = 96 in 2 The total surfance area is 140 + 96 = 236 in 2 .

He used 236 in 2 of paper.

Question 16. Vocabulary Name a three-dimensional shape that has four triangular faces and one rectangular face. Answer: Three-dimensional The shape that has four triangular faces and one rectangular face is a rectangular pyramid

Question 18. A glass paperweight has the shape of a triangular prism. The bases are equilateral triangles with side lengths of 4 inches and heights of 3.5 inches. The height of the prism is 5 inches. Find the lateral area and the total surface area of the paperweight. Answer: For lateral area: There are three 4 in. by 5 in. rectangles. 3 . (4 . 5) = 3 . 20 = 60 in 2 Lateral area: 60 in 2 Base area: The bases are equilateral triangles with length of 4 in. and height of 3.5 in. 2 . (\(\frac{1}{2}\) . 4 . (3.5)) = 14 in 2 The total surface area is 60 + 14 = 74 in 2 .

Lateral area: 60 in 2 . The total surface area is 74 in 2 .

The total surface area of the doghouse is 90 ft 2 .

Question 20. Describe the simplest way to find the total surface area of a cube. Answer: The surface area of a cube is the area of the six squares that cover it. The area of one of them is b. b. Since these are alt the same, multiply one of them by six, so the surface area of a cube is 6 times one of the sides squared.

Lesson 10.3 Answer Key 7th Grade Go Math Question 21. Communicate Mathematical Ideas Describe how you approach a problem involving lateral area and total surface area. What do you do first? In what ways can you use the figure that is given with a problem? What are some shortcuts that you might use when you are calculating these areas? Answer: When solving a problem involving lateral area and total surface area, determine first the lateral area of the given figure. This is because the lateral area is needed to determine the total surface area of the given figure. It is easier to determine the lateral area of the given figure if you know what figure are you dealing with, is it a pyramid or a prism, and what is the base of the given figure. The total surface area is the sum of the lateral area and the area of the base. The lateral area should be determined first before determining the total surface area of the given figure.

Solve for the lateral area first.

Texas Go Math Grade 7 Lesson 10.3 H.O.T. Focus on Higher Order Thinking Answer Key 

Question 23. Communicate Mathematical Ideas The base of Prism A has an area of 80 ft 2 , and the base of Prism B has an area of 80 ft 2 . The height of Prism A is the same as the height of Prism B. Is the base of Prism A congruent to the base of Prism B? Explain. Answer: The base of Prism A has the same area as the base of Prism B. They could be congruent however, it depends on the shape of the base. If the bases of Prism A and Prism B are both squares, then they are congruent. But if the bases of Prism A and Prism B are triangles or rectangles, there could be a difference in the dimensions of the triangle base and the rectangle base, which makes them not congruent It is because congruent figures have the same shape and size.

It depends on the shape of the base.

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