Learn on PengiPengi Math (Grade 5)Chapter 2: Multi-Digit Multiplication and Division with Place Value

Lesson 2: Modeling Multiplication with Area Models and Partial Products

In this Grade 5 Pengi Math lesson, students learn to represent multi-digit multiplication using area models and the distributive property. They practice decomposing factors into place value parts and interpreting each partial product as a rectangular area. This lesson is part of Chapter 2: Multi-Digit Multiplication and Division with Place Value.

Section 1

Introduction: Representing Multiplication with an Area Model

Property

An area model represents a multi-digit multiplication problem as the area of a big rectangle. First, break both multi-digit numbers into their place-value parts. The total area of the big rectangle (which is the final product) is the sum of the areas of all the small rectangles made from these parts. These small areas are called partial products.
When you multiply two broken-down numbers like (a1+a2)×(b1+b2+b3)(a_1 + a_2) \times (b_1 + b_2 + b_3), you can find the total product by adding up all the partial products:

(a1×b1)+(a1×b2)+(a1×b3)+(a2×b1)+(a2×b2)+(a2×b3)(a_1 \times b_1) + (a_1 \times b_2) + (a_1 \times b_3) + (a_2 \times b_1) + (a_2 \times b_2) + (a_2 \times b_3)

Examples

  • To model 23×347\boldsymbol{23 \times 347}, you draw a rectangle with side lengths 2323 and 347347. Decompose 2323 into 20+320 + 3, and decompose 347347 into 300+40+7300 + 40 + 7. The partial products are the areas of the smaller rectangles: 20×300=600020 \times 300 = 6000, 20×40=80020 \times 40 = 800, 20×7=14020 \times 7 = 140, 3×300=9003 \times 300 = 900, 3×40=1203 \times 40 = 120, and 3×7=213 \times 7 = 21.
  • To model 256×347\boldsymbol{256 \times 347}, you draw a rectangle with side lengths 256256 and 347347. Decompose 256256 into 200+50+6200 + 50 + 6, and decompose 347347 into 300+40+7300 + 40 + 7. The partial products are the areas of the smaller rectangles: 200×300=60000200 \times 300 = 60000, 200×40=8000200 \times 40 = 8000, 200×7=1400200 \times 7 = 1400, 50×300=1500050 \times 300 = 15000, 50×40=200050 \times 40 = 2000, 50×7=35050 \times 7 = 350, 6×300=18006 \times 300 = 1800, 6×40=2406 \times 40 = 240, and 6×7=426 \times 7 = 42.

Section 2

Using the Distributive Property for Partial Products

Property

The distributive property allows us to multiply a sum by multiplying each addend separately and then adding the products.
When we decompose factors by place value, the results of these smaller multiplications are called partial products.

a×(b+c)=(a×b)+(a×c)a \times (b + c) = (a \times b) + (a \times c)

Examples

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Chapter 2: Multi-Digit Multiplication and Division with Place Value

  1. Lesson 1

    Lesson 1: Estimating Products Using Rounding and Compatible Numbers

  2. Lesson 2Current

    Lesson 2: Modeling Multiplication with Area Models and Partial Products

  3. Lesson 3

    Lesson 3: Connecting Partial Products to the Standard Algorithm

  4. Lesson 4

    Lesson 4: Understanding Division as Finding a Missing Factor

  5. Lesson 5

    Lesson 5: Dividing Multi-Digit Numbers Using Area Models and Partial Quotients

  6. Lesson 6

    Lesson 6: Applying Division Algorithms and Interpreting Remainders

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Introduction: Representing Multiplication with an Area Model

Property

An area model represents a multi-digit multiplication problem as the area of a big rectangle. First, break both multi-digit numbers into their place-value parts. The total area of the big rectangle (which is the final product) is the sum of the areas of all the small rectangles made from these parts. These small areas are called partial products.
When you multiply two broken-down numbers like (a1+a2)×(b1+b2+b3)(a_1 + a_2) \times (b_1 + b_2 + b_3), you can find the total product by adding up all the partial products:

(a1×b1)+(a1×b2)+(a1×b3)+(a2×b1)+(a2×b2)+(a2×b3)(a_1 \times b_1) + (a_1 \times b_2) + (a_1 \times b_3) + (a_2 \times b_1) + (a_2 \times b_2) + (a_2 \times b_3)

Examples

  • To model 23×347\boldsymbol{23 \times 347}, you draw a rectangle with side lengths 2323 and 347347. Decompose 2323 into 20+320 + 3, and decompose 347347 into 300+40+7300 + 40 + 7. The partial products are the areas of the smaller rectangles: 20×300=600020 \times 300 = 6000, 20×40=80020 \times 40 = 800, 20×7=14020 \times 7 = 140, 3×300=9003 \times 300 = 900, 3×40=1203 \times 40 = 120, and 3×7=213 \times 7 = 21.
  • To model 256×347\boldsymbol{256 \times 347}, you draw a rectangle with side lengths 256256 and 347347. Decompose 256256 into 200+50+6200 + 50 + 6, and decompose 347347 into 300+40+7300 + 40 + 7. The partial products are the areas of the smaller rectangles: 200×300=60000200 \times 300 = 60000, 200×40=8000200 \times 40 = 8000, 200×7=1400200 \times 7 = 1400, 50×300=1500050 \times 300 = 15000, 50×40=200050 \times 40 = 2000, 50×7=35050 \times 7 = 350, 6×300=18006 \times 300 = 1800, 6×40=2406 \times 40 = 240, and 6×7=426 \times 7 = 42.

Section 2

Using the Distributive Property for Partial Products

Property

The distributive property allows us to multiply a sum by multiplying each addend separately and then adding the products.
When we decompose factors by place value, the results of these smaller multiplications are called partial products.

a×(b+c)=(a×b)+(a×c)a \times (b + c) = (a \times b) + (a \times c)

Examples

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 2: Multi-Digit Multiplication and Division with Place Value

  1. Lesson 1

    Lesson 1: Estimating Products Using Rounding and Compatible Numbers

  2. Lesson 2Current

    Lesson 2: Modeling Multiplication with Area Models and Partial Products

  3. Lesson 3

    Lesson 3: Connecting Partial Products to the Standard Algorithm

  4. Lesson 4

    Lesson 4: Understanding Division as Finding a Missing Factor

  5. Lesson 5

    Lesson 5: Dividing Multi-Digit Numbers Using Area Models and Partial Quotients

  6. Lesson 6

    Lesson 6: Applying Division Algorithms and Interpreting Remainders