Mathematics 6 Quarter 3 Module 9: Solving Word Problems Involving Surface Area • DepEd Tambayan

This is the Mathematics Grade 6 Quarter 3 Module 9 self-learning module, “Solving Word Problems Involving Measurement of Surface Area.” It is for Grade 6 learners studying at home or in class. After working through it, learners should be able to solve word problems about the surface area of different solid objects, such as gift boxes, milk cans, balls, tents, and cones.

The module was developed by the Department of Education Region VI as part of its Alternative Delivery Mode (ADM) materials. It is a First Edition, 2021, written by Rhona S. Macado.

MATH6-Q1-MODULE9

On this page 7 sections

What learners will learn

The module states one main objective. After going through it, learners are expected to:

Solve word problems involving measurement of surface area of different solid objects.

The module builds on earlier lessons, where learners studied the parts of solid figures and how to find their surface area. It now focuses on using those skills to solve real-life problems.

Surface area formulas covered

Surface area (S.A.) is the total area of all the surfaces of a solid object. The module reviews these formulas:

Solid figure	Surface area formula
Rectangular prism	S.A. = 2(lw + lh + wh), or 2lw + 2lh + 2wh
Cube	S.A. = 6s²
Cylinder	S.A. = L.A. + 2B, or 2πrh + 2πr²
Pyramid	S.A. = L.A. + B
Cone	S.A. = L.A. + B, or πrs + πr²
Sphere	S.A. = 4πr²

In these formulas, L.A. is the lateral area (the area of the sides), B is the area of the base, r is the radius, h is the height, and s is the slant height for a cone or the edge for a cube. The lateral area of a cylinder is the circumference of the base times the height. The module uses π = 3.14 in all problems.

The four steps for solving word problems

The module teaches a four-step method:

Understand: What is asked? What are given?
Plan: Which formula will you use?
Solution: Show your computation.
Check: Solve again in a different way and see if you get the same answer.

Worked example 1: Wrapping a gift

Mary wants to wrap a rectangular gift that measures 15 cm by 6 cm by 2 cm. What is the least amount of wrapper she can use?

Understand: The question asks for the least amount of wrapper. The given dimensions are 15 cm, 6 cm, and 2 cm.
Plan: The wrapper covers the outside of the box, so find the surface area of a rectangular prism.
Solution: S.A. = 2(15 × 6 + 15 × 2 + 6 × 2) = 2(90 + 30 + 12) = 2(132) = 264 cm²
Check: Using 2lw + 2lh + 2wh gives 180 + 60 + 24 = 264 cm². Both methods agree, so Mary needs 264 cm² of wrapper.

Worked example 2: A can of milk

Find the surface area of a can of milk with a diameter of 12 inches and a height of 20 inches.

Understand: The question asks for the surface area of the can. The diameter is 12 inches, so the radius is 6 inches. The height is 20 inches.
Plan: A can of milk is a cylinder, so use the cylinder formula.
Solution: S.A. = 2(3.14)(6)(20) + 2(3.14)(6²) = 753.6 + 226.08 = 979.68 in²
Check: First find the lateral area, 2πrh = 753.6 in². Then add the two circular bases, 2πr² = 226.08 in². The total is again 979.68 in².

Parts and activities of the module

What I Know (pretest): Find the surface area of six labeled figures (sphere, square pyramid, rectangular prism, cone, cube, cylinder) and solve two word problems about a ball and a milk can.
What’s In: A true-or-false review of solid figures, and matching each solid figure with its surface area formula.
What’s New and What is It: The two worked examples above.
What’s More: Two practice problems about a basketball (a sphere) and an ice cream cone.
What I Have Learned: A summary of the four steps and all the formulas.
What I Can Do: Problems about repainting a wooden cabinet and finding the surface area of a sphere.
Assessment: Five word problems about a pyramid-shaped tent, a gift box, a water pitcher, a milk can, and a cube-shaped music box.
Additional Activities: Five more problems, including a shoe box, a bar of soap, and whether a can of varnish is enough to coat a jewelry box.

How to use this module

Learners: Start with the pretest in What I Know to see how much you already know. Then go through each part in order. Use the four steps for every problem, and write all answers on a separate sheet of paper, not on the module. Use π = 3.14.
Parents: Help your child read each problem slowly and name what is asked and what is given before solving. The module asks learners to be honest when checking their work with the answer key.
Teachers: An answer key for all activities and tests is on the last pages. The Notes to the Teacher offer strategies for guiding home-based learners, and the Assessment can serve as the post-test.

Frequently asked questions

What grade and quarter is this module for?

It is for Grade 6 Mathematics, Quarter 3, Module 9. It is a First Edition, 2021 module from DepEd Region VI.

Does the module have an answer key?

Yes. The answer key at the end covers What I Know, What’s In, What’s More, What I Can Do, the Assessment, and the Additional Activities.

What value of pi does the module use?

The module uses π = 3.14 in all of its problems.

What steps does the module teach for solving surface area word problems?

It teaches four steps: Understand (what is asked and what is given), Plan (which formula to use), Solution (show your computation), and Check (solve another way to confirm the answer).

Which solid figures are covered?

The module covers rectangular prisms, cubes, cylinders, pyramids, cones, and spheres, with a surface area formula for each.

Full text of the module

The text below is a transcription of the PDF above, as printed.

Mathematics Quarter 3 – Module 9: Solving Word Problems Involving Measurement of Surface Area

[illustration: students learning outdoors and at home, with a globe and a lantern]

Mathematics – Grade 6 Alternative Delivery Mode Quarter 3 – Module 9: Solving Word Problems Involving Measurement of Surface Area. First Edition, 2021

Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties.

Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this book are owned by their respective copyright holders. Every effort has been exerted to locate and seek permission to use these materials from their respective copyright owners. The publisher and authors do not represent nor claim ownership over them.

Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio

Development Team of the Module

Author: Rhona S. Macado

Editors: Mae Joy M. Tan, Susana L. Lutero, Nora B. Rebadomia, Ritchel T. Maratas, Mary Jane O. Amante, Susan T. Dela Torre, Jem Rymon S. Chien

Reviewers: Ivy Joy A. Torres, Ma. Theresa L. Tabotabo, Mae Joy M. Tan, Jem Rymon S. Chien

Illustrators: Eldiardo E. Dela Peña

Layout Artists: Eldiardo E. Dela Peña

Management Team: Ramir B. Uytico, Pedro T. Escobarte, Jr., Elena P. Gonzaga, Donald T. Genine, Adonis A. Mosquera, Clarissa G. Zamora, Fevi S. Fanco, Ivy Joy A. Torres, Jason R. Alpay, Mae Joy M. Tan, Jem Rymon S. Chien, Ethel S. Gali

Printed in the Philippines by ________________________

Department of Education – Region VI

Office Address: Duran Street, Iloilo City Telefax: (033)336-2816, (033)509-7653 E-mail Address: region6@deped.gov.ph

Mathematics

Quarter 3 – Module 9: Solving Word Problems Involving Measurement of Surface Area

Introductory Message

This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.

Each SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.

Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these.

In addition to the material in the main text, Notes to the Teacher are also provided to our facilitators and parents for strategies and reminders on how they can best help you on your home-based learning.

Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.

If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.

Thank you.

What I Need to Know

This module was designed and written with you in mind. It is here to help you master the skills in solving word problems involving surface area. this module allows you to use it in various learning situations. The language used recognizes your vocabulary level. The lessons are arranged to follow the standard sequence of your course. But the order in which you read them can be changed to match with the textbook you are now using.

After going through this module, you are expected to be able to solve word problems involving measurement of surface area of different solid objects.

What I Know

A. On a separate sheet of paper, find the surface area of the following figures:

[illustration: sphere with radius labeled 7cm]
[illustration: square pyramid with slant height labeled 8cm, base edge 10cm, and side labeled 4cm]
[illustration: rectangular prism with dimensions 8 ft, 6 ft, and 4 ft]
[illustration: cone with slant height labeled s=8dm and radius labeled r=4dm]
[illustration: cube with edge labeled 12 cm]
[illustration: cylinder with radius labeled r = 2m and height labeled h = 6m]

B. Solve the following problems. Write your answer in your answer sheet. Use π = 3.14

A boy is playing with a ball with radius 20 cm. Find the surface area of the ball.
A milk can has a radius of 5 cm and a height of 12 cm. How much tin was used to make it?

Lesson 1: Solving Word Problems Involving Measurement of Surface Area

In the previous lessons, you have learned different characteristics of solid figures and how to solve their surface area. This time, we will focus on how to solve word problems involving surface area of solid objects.

What’s In

A. On a separate sheet of paper, write T if the statement is true and F if the statement is false.

________1. The faces of rectangular pyramid are rectangle.

________2. A cone has two circular bases.

________3. The faces of the cube are triangles.

________4. Surface area refers to the sum of the faces of solid figures.

________5. A cylinder is a three-dimensional figure with two congruent circular bases that are parallel.

B. Match the solid figure in column A to its surface area formula on column B. Write your answer on a separate sheet of paper.

Column A	Column B
1. Cylinder	A. S.A = 2 (lw + lh + wh)
2. Pyramid	B. S.A = πrs + πr²
3. Cube	C. S.A = 6s²
4. Cone	D. S.A = 2πrh + 2πr²
5. Rectangular Prism	E. S.A = 4πr²
6. Sphere	F. S.A. = l x w x h
	G. S.A = L.A + B

What’s New

Read and study the problem below.

Mary wants to wrap her rectangular gift with a dimension of 15 cm by 6 cm by 2 cm. What is the least amount of wrapper can she use to wrap the gift?

Can you help Mary to find the least amount of wrapper to wrap her gift?

What is It

To solve the problem above, we will follow the following steps:

Understand:
a. What is asked?
Answer: The least amount of wrapper she can use to wrap the gift?
b. What are the given?
Answer: dimension: 15 cm by 6 cm by 2 cm
Plan: What strategy can we use to solve the problem?
Answer: The problem asked for the wrapper to be used around the gift.
We can solve the problem by finding the surface area of a rectangular prism.
Solution: Use the formula to find the surface area of a rectangular prism.

S.A. = 2 (lw + lh + wh)
     = 2 (15 cm x 6 cm + 15 cm x 2 cm + 6 cm x 2 cm)
     = 2 (90 cm² + 30 cm² + 12 cm²)
     = 2 (132 cm²)
S.A. = 264 cm²

4. Check:
To check our answer, we will find the area using another formula:

S.A. = 2lw + 2lh + 2wh
= 2 (15 cm x 6 cm) + 2 (15 cm x 2 cm) + 2 (6 cm x 2 cm)
= 2 (90cm2) + 2 (30cm2) + 2 (12 cm2)
= 180 cm2 + 60 cm2 + 24 cm2
S.A. = 264 cm2

We have observed that the answer is the same even if we are using different process. So, 264 cm2 is the amount of wrapper needed to wrap the gift.

Problem 2:

Find the surface area of a can of milk whose diameter is 12 inches and 20 inches high.

Let's analyze the problem:

1. Understand:

a. What is asked?
Answer: The surface area of a can of milk.

b. What are the given?
Answer: diameter is 12 inches, 20 inches high

2. Plan: What strategy can we use to solve the problem?

Answer: Find the surface area of a cylinder because a can of milk cylinder.

3. Solution: Use the formula in finding the surface area of a cylinder.
S. A. = 2πrh + 2πr2
= 2 (3.14) x 6 in x 20 in + 2 (3.14) (6in)2
= 6.28 x 120 in2 + 6.28 x 36 in2
= 753.6 in2 + 226.08 in2
S.A. = 979.68 in2

### What's More

4. Check
To check our answer, we will find the area using another formula:

Step 1: Solve the lateral area.
L.A. = 2πrh
= 2(3.14) x (6 in) x (20 in)
= 6.28 x 120 in2
= 753.6 in2

Step 2: Find the surface area.
S.A = L. A. + 2B (Substitute the formula in finding the area of circle)
= L.A. + 2πr2 (Substitute the value to the formula)
= 753.6 in2 + 2 (3.14) (6 in)2 (Multiply: 2 x 3.14 and 3 to itself)
= 753.6 in2 + 6.28 (36 in2)
= 753.6 in2 + 226.08 in2
S.A.= 979.68 in2

We have observed that the answer is the same even if we are using a different process. So, 979.68 in2 is the surface area of the can of milk.

. Read and solve the following problem using the steps illustrated in the previous section. **Use** π = 3.14

1. Find the surface area of the basketball with a radius of 4.25 cm. Round your answer to the nearest tenths.

2. An ice cream cone has a circular base of radius 2 cm and a slant height of 5 cm. What is the surface area of the cone?

### What I Have Learned

The following are the steps to solve a problem:

1. Understand:

a. What is asked?
b. What are given?

2. Plan: What formula are you going to use?

3. Solution: Show your computation.

4. Check:

❖ Surface area (S.A) is the total areas of the surface of solid object.

❖ Surface Area of Prism

The surface area of a rectangular prism is the sum of all the faces.
S.A = 2(lw) + 2(lh) + 2(wh) or SA = 2 (lw + lh + wh)

The surface area of a cube is the sum of the area of all the faces.
S.A = 6s2

❖ Surface Area of a Cylinder

The lateral area of a cylinder is the product of the circumference of the base and the height.

The surface of a cylinder is the sum of the lateral area and the areas of the two bases.
S.A = L.A + 2B or S.A = 2πrh + 2πr2

❖ Surface Area of a Pyramid

The surface area of a pyramid is the sum of the lateral area and the area of the base.
S.A = L.A + B

❖ Surface Area of a Cone

The surface area of a cone is the sum of the lateral area and the area of its base.
S.A = L.A + B or S.A = πrs + πr2

❖ Surface Area of a Sphere

The surface area of a sphere with radius (r) is **S.A = 4πr2**

### What I Can Do

Read and solve the problems using the steps solving problem. Write your answer on a separate sheet of paper. **Use** π = 3.14

1. A wooden cabinet needs repainting. How much surface area needs to be painted if the box has a length of 8m, width of 5 m and height of 6 m?

2. What is the surface area of a sphere whose radius is 4dm?

### Assessment

On a separate sheet of paper, solve following the steps on how to solve word problems. Use π = 3.14

1. Peter and Eric pitched a tent in a shape of a pyramid. The base of the tent is 3 meters wide by 4 meters long. The tent had a slanted height of 5 meters. What was the surface area of the tent?

2. Jason is wrapping a present. The box he is using is a rectangular prism with a length of 8 dm, a width of 5 dm and a height of 4 dm. Find how much paper he needs to wrap the entire box.

3. A cylinder-shaped water pitcher has a radius of 5 inches and a height of 15 inches. Find the surface area of the pitcher.

4. Find the amount of tin needed to make a milk can that has a diameter of 4 cm and a height of 5 cm.

5. A music box has the shape of a cube. Each edge of the music box is 12 centimeters long. What is the surface area of the box?

### Additional Activities

Read and solve the following problems. Use π = 3.14

1. The side of the cube measures 7 cm. Find its surface area.

2. The regular pyramid has a square base whose edge of the base measures 4 m and the slanted height measures 10m. Find the surface area of the pyramid?

3. A box of shoes has dimensions of 20 cm by 12 cm by 9 cm. What is the surface area of the box?

4. A cosmetics company that makes small cylindrical bars of soap. Find the surface area of a bar of soap if the diameter is 4 cm and the height is 9 cm.

5. Gerald own an antique shop. He is applying a coating to a rectangular jewelry box whose dimension is 5 m by 3m by 4m. The can of varnish is enough to cover 80 m2. Is there enough varnish left in the can to coat the jewelry box? Explain your answer.

## Answer Key

| What I Know | What I Can Do |
|---|---|
| 1. 615.44 cm2<br>2. 152 cm2<br>3. 208 ft2<br>4. 150.72 dm2<br>5. 864 cm2<br>6. 100.48 m2<br>7. 5,024 cm2<br>8. 533.8 cm2 | 1. 236 m2<br>2. 200.96 dm2 |

| What's In | Assessment |
|---|---|
| A.<br>1. F<br>2. F<br>3. F<br>4. F<br>5. T<br>B.<br>1. D<br>2. G<br>3. C<br>4. B<br>5. A<br>6. E | 1. 47 m2<br>2. 184 dm2<br>3. 628 m2<br>4. 87.92 cm2<br>5. 864 cm2 |

| What's More | Additional Activities |
|---|---|
| 1. 226.9 cm2<br>2. 31 cm2 | 1. 294 cm2<br>2. 96 m2<br>3. 1,056 cm2<br>4. 138.16 cm2<br>5. The varnish is not enough because the surface area of the jewelry box is 94 m2 and the varnish left could cover only an area of 80 m2. |

## References:

Lumbre, A. et al. (2016) 21st Century MATHletes Textbook for Grade 5: Vibal Group Inc.

Perez, M. et al. (2016) 21st Century MATHletes 6. Philippines: Vibal Group Inc.

Tan, A. et al. (2007) Hypermath 6.Philippines:Trinitas Publishing, Inc.

Most Essential Learning Competencies (MELC) in Mathematics 6

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