Learn on PengiPengi Math (Grade 6)Chapter 6: Geometry

Lesson 2: Area of Parallelograms

In this Grade 6 lesson from Pengi Math Chapter 6: Geometry, students learn how to calculate the area of parallelograms by identifying the base and perpendicular height and applying the area formula. The lesson guides students through deriving the formula by rearranging parallelograms into rectangles, then comparing figures that share the same base and height. Students practice using the area formula to solve equations in real problem-solving contexts.

Section 1

Area of a Parallelogram

Property

A parallelogram can be rearranged into a rectangle with the same base and height. Choose any side of the parallelogram as the base (length bb), and let hh be the perpendicular distance between the base and the opposite side.
The area is given by the formula:

Area=bh\operatorname{Area} = bh

Examples

  • A parallelogram has a base of 12 cm and a height of 5 cm. Its area is 12×5=6012 \times 5 = 60 square cm.
  • A section of a patio is shaped like a parallelogram with a base of 8 feet and a height of 6 feet. The area is 8×6=488 \times 6 = 48 square feet.
  • Even if the slanted side is 9 inches, if the base is 15 inches and the height is 7 inches, the area is 15×7=10515 \times 7 = 105 square inches.

Explanation

Think of a parallelogram as a slanted rectangle. By slicing off a triangle from one end and moving it to the other, you create a perfect rectangle. This new rectangle has the same base and height, which is why the area formula works!

Section 2

Finding Area by Rearrangement

Property

The area of a shape remains the same when it is cut into pieces and rearranged into a new shape without any gaps or overlaps. This method transforms a complex shape into a simpler one, like a rectangle or square, to make calculating the area easier.

Examples

  • A parallelogram with base b=8b=8 and height h=5h=5 can be rearranged into a rectangle. By cutting a triangle from one side and moving it to the other, we form a rectangle with dimensions 8×58 \times 5. The area is 8×5=408 \times 5 = 40 square units.
  • A trapezoid with bases b1=6b_1=6 and b2=10b_2=10 and height h=4h=4 can be transformed. By making a cut through the midpoints of the non-parallel sides and rotating the top section, a parallelogram is formed with base b1+b2=16b_1+b_2=16 and height h2=2\frac{h}{2}=2. Or, more simply, it can be rearranged into a rectangle with length b1+b22=8\frac{b_1+b_2}{2} = 8 and width h=4h=4, giving an area of 8×4=328 \times 4 = 32 square units.

Explanation

This skill involves visually decomposing a shape and then recomposing the pieces into a more familiar figure. It's a powerful reasoning tool that shows area is conserved even when a shape's form changes. Instead of just adding or subtracting areas, you are transforming the shape itself. This method is particularly useful for deriving area formulas for shapes like parallelograms and trapezoids by relating them back to rectangles.

Section 3

Area Invariance with Constant Base and Height

Property

Parallelograms that share the same base (bb) and have the same perpendicular height (hh) have equal areas. This is a direct consequence of the area formula, A=b×hA = b \times h.

Examples

Book overview

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Chapter 6: Geometry

  1. Lesson 1

    Lesson 1: Polygons and Their Properties

  2. Lesson 2Current

    Lesson 2: Area of Parallelograms

  3. Lesson 3

    Lesson 3: Area of Triangles

  4. Lesson 4

    Lesson 4: Area of Trapezoids

  5. Lesson 5

    Lesson 5: Composite and Irregular Shapes

  6. Lesson 6

    Lesson 6: Polyhedra, Prisms, and Pyramids

  7. Lesson 7

    Lesson 7: Nets and Surface Area

  8. Lesson 8

    Lesson 8: Volume of Rectangular Prisms

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Area of a Parallelogram

Property

A parallelogram can be rearranged into a rectangle with the same base and height. Choose any side of the parallelogram as the base (length bb), and let hh be the perpendicular distance between the base and the opposite side.
The area is given by the formula:

Area=bh\operatorname{Area} = bh

Examples

  • A parallelogram has a base of 12 cm and a height of 5 cm. Its area is 12×5=6012 \times 5 = 60 square cm.
  • A section of a patio is shaped like a parallelogram with a base of 8 feet and a height of 6 feet. The area is 8×6=488 \times 6 = 48 square feet.
  • Even if the slanted side is 9 inches, if the base is 15 inches and the height is 7 inches, the area is 15×7=10515 \times 7 = 105 square inches.

Explanation

Think of a parallelogram as a slanted rectangle. By slicing off a triangle from one end and moving it to the other, you create a perfect rectangle. This new rectangle has the same base and height, which is why the area formula works!

Section 2

Finding Area by Rearrangement

Property

The area of a shape remains the same when it is cut into pieces and rearranged into a new shape without any gaps or overlaps. This method transforms a complex shape into a simpler one, like a rectangle or square, to make calculating the area easier.

Examples

  • A parallelogram with base b=8b=8 and height h=5h=5 can be rearranged into a rectangle. By cutting a triangle from one side and moving it to the other, we form a rectangle with dimensions 8×58 \times 5. The area is 8×5=408 \times 5 = 40 square units.
  • A trapezoid with bases b1=6b_1=6 and b2=10b_2=10 and height h=4h=4 can be transformed. By making a cut through the midpoints of the non-parallel sides and rotating the top section, a parallelogram is formed with base b1+b2=16b_1+b_2=16 and height h2=2\frac{h}{2}=2. Or, more simply, it can be rearranged into a rectangle with length b1+b22=8\frac{b_1+b_2}{2} = 8 and width h=4h=4, giving an area of 8×4=328 \times 4 = 32 square units.

Explanation

This skill involves visually decomposing a shape and then recomposing the pieces into a more familiar figure. It's a powerful reasoning tool that shows area is conserved even when a shape's form changes. Instead of just adding or subtracting areas, you are transforming the shape itself. This method is particularly useful for deriving area formulas for shapes like parallelograms and trapezoids by relating them back to rectangles.

Section 3

Area Invariance with Constant Base and Height

Property

Parallelograms that share the same base (bb) and have the same perpendicular height (hh) have equal areas. This is a direct consequence of the area formula, A=b×hA = b \times h.

Examples

Book overview

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

Continue this chapter

Chapter 6: Geometry

  1. Lesson 1

    Lesson 1: Polygons and Their Properties

  2. Lesson 2Current

    Lesson 2: Area of Parallelograms

  3. Lesson 3

    Lesson 3: Area of Triangles

  4. Lesson 4

    Lesson 4: Area of Trapezoids

  5. Lesson 5

    Lesson 5: Composite and Irregular Shapes

  6. Lesson 6

    Lesson 6: Polyhedra, Prisms, and Pyramids

  7. Lesson 7

    Lesson 7: Nets and Surface Area

  8. Lesson 8

    Lesson 8: Volume of Rectangular Prisms