Learn on PengienVision, Mathematics, Grade 6Chapter 7: Solve Area, Surface Area, and Volume Problems

Lesson 7: Find Surface Areas of Pyramids

In this Grade 6 enVision Mathematics lesson, students learn how to find the volume of rectangular prisms with fractional edge lengths by filling the prism with unit-fraction cubes and applying the formula V = ℓwh with mixed numbers and fractions. Students practice converting mixed numbers to improper fractions to calculate volume, including special cases like cubes using the formula V = s³. The lesson connects to Common Core standards 6.G.A.2 and 6.EE.A.2 within Chapter 7 on area, surface area, and volume problems.

Section 1

Surface Area of a Regular Square Pyramid

Property

The surface area (SASA) of a square pyramid is the sum of the area of its square base and the area of its four triangular lateral faces.

SA=Area of base+Area of 4 triangular facesSA = \text{Area of base} + \text{Area of 4 triangular faces}
SA=s2+4(12sl)=s2+2slSA = s^2 + 4 \left( \frac{1}{2} s l \right) = s^2 + 2sl

where ss is the side length of the square base and ll is the slant height of the pyramid.

Examples

  • A square pyramid has a base side length of 55 cm and a slant height of 88 cm. Its surface area is SA=52+2(5)(8)=25+80=105 cm2SA = 5^2 + 2(5)(8) = 25 + 80 = 105 \text{ cm}^2.
  • A square pyramid has a base with an area of 36 m236 \text{ m}^2 and a slant height of 1010 m. The side length is 36=6\sqrt{36} = 6 m. Its surface area is SA=36+2(6)(10)=36+120=156 m2SA = 36 + 2(6)(10) = 36 + 120 = 156 \text{ m}^2.

Explanation

To find the surface area of a square pyramid, you must calculate two components: the area of the square base and the total area of the four identical triangular faces. The area of the base is found by squaring its side length (s2s^2). The area of the four lateral faces is found by multiplying the area of one triangle (12×base×slant height\frac{1}{2} \times \text{base} \times \text{slant height}) by four. The total surface area is the sum of the base area and the lateral area.

Section 2

Surface Area of a Regular Triangular Pyramid

Property

The surface area (S.A.S.A.) of a triangular pyramid is the sum of the area of its triangular base (AbaseA_{base}) and the areas of its three triangular lateral faces.

S.A.=Abase+Area of Lateral FacesS.A. = A_{base} + \text{Area of Lateral Faces}

Examples

  • A triangular pyramid has a base with an area of 15 cm215 \text{ cm}^2 and three lateral faces each with an area of 12 cm212 \text{ cm}^2. The total surface area is S.A.=15+12+12+12=51 cm2S.A. = 15 + 12 + 12 + 12 = 51 \text{ cm}^2.
  • A regular triangular pyramid has an equilateral triangle base with a side length of 66 m and an area of 15.6 m215.6 \text{ m}^2. Its slant height is 88 m. The area of each lateral face is 12(6)(8)=24 m2\frac{1}{2}(6)(8) = 24 \text{ m}^2. The surface area is S.A.=15.6+3(24)=15.6+72=87.6 m2S.A. = 15.6 + 3(24) = 15.6 + 72 = 87.6 \text{ m}^2.

Explanation

To find the surface area of a triangular pyramid, you must calculate the area of four distinct triangles: the base and the three lateral faces. First, find the area of the triangular base. Next, calculate the area of each of the three triangular faces that meet at the apex, often using the slant height. Finally, add all four areas together to get the total surface area of the pyramid.

Book overview

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Chapter 7: Solve Area, Surface Area, and Volume Problems

  1. Lesson 1

    Lesson 1: Find Areas of Parallelograms and Rhombuses

  2. Lesson 2

    Lesson 2: Solve Triangle Area Problems

  3. Lesson 3

    Lesson 3: Find Areas of Trapezoids and Kites

  4. Lesson 4

    Lesson 4: Find Areas of Polygons

  5. Lesson 5

    Lesson 5: Represent Solid Figures Using Nets

  6. Lesson 6

    Lesson 6: Find Surface Areas of Prisms

  7. Lesson 7Current

    Lesson 7: Find Surface Areas of Pyramids

  8. Lesson 8

    Lesson 8: Find Volume with Fractional Edge Lengths

Lesson overview

Expand to review the lesson summary and core properties.

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Section 1

Surface Area of a Regular Square Pyramid

Property

The surface area (SASA) of a square pyramid is the sum of the area of its square base and the area of its four triangular lateral faces.

SA=Area of base+Area of 4 triangular facesSA = \text{Area of base} + \text{Area of 4 triangular faces}
SA=s2+4(12sl)=s2+2slSA = s^2 + 4 \left( \frac{1}{2} s l \right) = s^2 + 2sl

where ss is the side length of the square base and ll is the slant height of the pyramid.

Examples

  • A square pyramid has a base side length of 55 cm and a slant height of 88 cm. Its surface area is SA=52+2(5)(8)=25+80=105 cm2SA = 5^2 + 2(5)(8) = 25 + 80 = 105 \text{ cm}^2.
  • A square pyramid has a base with an area of 36 m236 \text{ m}^2 and a slant height of 1010 m. The side length is 36=6\sqrt{36} = 6 m. Its surface area is SA=36+2(6)(10)=36+120=156 m2SA = 36 + 2(6)(10) = 36 + 120 = 156 \text{ m}^2.

Explanation

To find the surface area of a square pyramid, you must calculate two components: the area of the square base and the total area of the four identical triangular faces. The area of the base is found by squaring its side length (s2s^2). The area of the four lateral faces is found by multiplying the area of one triangle (12×base×slant height\frac{1}{2} \times \text{base} \times \text{slant height}) by four. The total surface area is the sum of the base area and the lateral area.

Section 2

Surface Area of a Regular Triangular Pyramid

Property

The surface area (S.A.S.A.) of a triangular pyramid is the sum of the area of its triangular base (AbaseA_{base}) and the areas of its three triangular lateral faces.

S.A.=Abase+Area of Lateral FacesS.A. = A_{base} + \text{Area of Lateral Faces}

Examples

  • A triangular pyramid has a base with an area of 15 cm215 \text{ cm}^2 and three lateral faces each with an area of 12 cm212 \text{ cm}^2. The total surface area is S.A.=15+12+12+12=51 cm2S.A. = 15 + 12 + 12 + 12 = 51 \text{ cm}^2.
  • A regular triangular pyramid has an equilateral triangle base with a side length of 66 m and an area of 15.6 m215.6 \text{ m}^2. Its slant height is 88 m. The area of each lateral face is 12(6)(8)=24 m2\frac{1}{2}(6)(8) = 24 \text{ m}^2. The surface area is S.A.=15.6+3(24)=15.6+72=87.6 m2S.A. = 15.6 + 3(24) = 15.6 + 72 = 87.6 \text{ m}^2.

Explanation

To find the surface area of a triangular pyramid, you must calculate the area of four distinct triangles: the base and the three lateral faces. First, find the area of the triangular base. Next, calculate the area of each of the three triangular faces that meet at the apex, often using the slant height. Finally, add all four areas together to get the total surface area of the pyramid.

Book overview

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

Continue this chapter

Chapter 7: Solve Area, Surface Area, and Volume Problems

  1. Lesson 1

    Lesson 1: Find Areas of Parallelograms and Rhombuses

  2. Lesson 2

    Lesson 2: Solve Triangle Area Problems

  3. Lesson 3

    Lesson 3: Find Areas of Trapezoids and Kites

  4. Lesson 4

    Lesson 4: Find Areas of Polygons

  5. Lesson 5

    Lesson 5: Represent Solid Figures Using Nets

  6. Lesson 6

    Lesson 6: Find Surface Areas of Prisms

  7. Lesson 7Current

    Lesson 7: Find Surface Areas of Pyramids

  8. Lesson 8

    Lesson 8: Find Volume with Fractional Edge Lengths