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

Lesson 3: Area of Triangles

In this Grade 6 lesson from Pengi Math Chapter 6: Geometry, students learn to calculate the area of triangles by recognizing that a triangle is half of a parallelogram. The lesson covers identifying the base and corresponding height in acute, obtuse, and right triangles, including cases where the height falls outside the triangle. Students also practice solving for missing dimensions using the area formula.

Section 1

Deriving the Triangle Area Formula from Parallelograms

Property

A triangle is exactly half of a parallelogram with the same base and height. To find its area, choose any side of the triangle as its base (length bb), and let hh be the perpendicular distance from the base to its opposing vertex.
The formula is:

Area=12bh\operatorname{Area} = \frac{1}{2} bh

Examples

  • A right triangle has legs (base and height) of 5 m and 8 m. Its area is 12×5×8=20\frac{1}{2} \times 5 \times 8 = 20 square meters.
  • A triangular sign has a base of 40 cm and a height of 25 cm. Its area is 12×40×25=500\frac{1}{2} \times 40 \times 25 = 500 square cm.
  • An obtuse triangle has a base of 10 inches and its corresponding height is 6 inches. The area is 12×10×6=30\frac{1}{2} \times 10 \times 6 = 30 square inches.

Explanation

Any triangle is exactly half of a parallelogram! If you clone a triangle, flip it, and join it to the original, you create a parallelogram. That's why the triangle's area formula is simply one-half of the base times the height.

Section 2

Right Triangle Area Using Legs

Property

The area of a right triangle equals half the product of its two legs:

Area=ab2\text{Area} = \frac{ab}{2}
where aa and bb are the lengths of the legs (the two sides that form the right angle).

Examples

Section 3

Solving for a Missing Dimension

Property

If you know the area of a triangle and one of its dimensions (base or height), you can rearrange the area formula to solve for the missing dimension.

Given the area formula A=12bhA = \frac{1}{2}bh:

  • To find the height:
    h=2Abh = \frac{2A}{b}
  • To find the base:
    b=2Ahb = \frac{2A}{h}

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 2

    Lesson 2: Area of Parallelograms

  3. Lesson 3Current

    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

Deriving the Triangle Area Formula from Parallelograms

Property

A triangle is exactly half of a parallelogram with the same base and height. To find its area, choose any side of the triangle as its base (length bb), and let hh be the perpendicular distance from the base to its opposing vertex.
The formula is:

Area=12bh\operatorname{Area} = \frac{1}{2} bh

Examples

  • A right triangle has legs (base and height) of 5 m and 8 m. Its area is 12×5×8=20\frac{1}{2} \times 5 \times 8 = 20 square meters.
  • A triangular sign has a base of 40 cm and a height of 25 cm. Its area is 12×40×25=500\frac{1}{2} \times 40 \times 25 = 500 square cm.
  • An obtuse triangle has a base of 10 inches and its corresponding height is 6 inches. The area is 12×10×6=30\frac{1}{2} \times 10 \times 6 = 30 square inches.

Explanation

Any triangle is exactly half of a parallelogram! If you clone a triangle, flip it, and join it to the original, you create a parallelogram. That's why the triangle's area formula is simply one-half of the base times the height.

Section 2

Right Triangle Area Using Legs

Property

The area of a right triangle equals half the product of its two legs:

Area=ab2\text{Area} = \frac{ab}{2}
where aa and bb are the lengths of the legs (the two sides that form the right angle).

Examples

Section 3

Solving for a Missing Dimension

Property

If you know the area of a triangle and one of its dimensions (base or height), you can rearrange the area formula to solve for the missing dimension.

Given the area formula A=12bhA = \frac{1}{2}bh:

  • To find the height:
    h=2Abh = \frac{2A}{b}
  • To find the base:
    b=2Ahb = \frac{2A}{h}

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 2

    Lesson 2: Area of Parallelograms

  3. Lesson 3Current

    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