MathBored

Virginia SOL Mathematics Textbook

Grade 8 Workbook — Chapter 11: Surface Area and Volume: Pyramids and Cones

SOL 8.MG.2 · Companion to Textbook Chapter 11

Each page below is one Canva page. Headings are sized for direct paste: page title as H1, section labels as H2. Figures referenced by filename live in ../figures/. Every numbered item is the same problem as in the textbook, so one answer key serves both. Item numbers run continuously from 1 to 150.

Pi: use π3.14\pi \approx 3.14 everywhere in this chapter, not a calculator's stored π\pi. Give every answer containing π\pi exactly first, then approximately.

Units: volume in cubic units (cm3^3, in3^3, ft3^3, m3^3); surface area in square units (cm2^2, in2^2, ft2^2, m2^2).

Formula reminders: 2=h2+(s2)2\ell^2 = h^2 + \left(\frac{s}{2}\right)^2 · SA=s2+2sSA = s^2 + 2s\ell · LA=2sLA = 2s\ell · V=13s2hV = \frac{1}{3}s^2h · V=13πr2hV = \frac{1}{3}\pi r^2 h


PAGE 1 — Chapter opener

Chapter 11 · Surface Area and Volume: Pyramids and Cones

Standard 8.MG.2

In this chapter you will:

Words to know: square-based pyramid · base · apex · lateral face · base edge · height · slant height · net · surface area · lateral area · cone · radius · diameter · volume · cubic units · square units · base area BB

Every pyramid in this chapter has a square base. Surface area is found for pyramids only — for cones this chapter finds volume and nothing else.


PAGE 2 — The parts of a square-based pyramid

11.1 Pyramids, Cones, and Two Different Heights

FIGURE: fig1-pyramid-parts.png (full width)

Fill in the blanks.

The point where all four triangles meet is the ____________.

The four triangular faces are called ____________ faces.

The height hh runs from the apex to the ____________ of the base.

The slant height \ell runs from the apex to the ____________ of a base edge.

The longer of the two is always the ____________ ____________.

Volume uses the ____________. Surface area uses the ____________ ____________.

  1. Name the shape of the base and the shape of each lateral face of a square-based pyramid.

    Base: _______________ Each lateral face: _______________

  2. A square-based pyramid has ______ faces, ______ edges, and ______ vertices.

  3. Which labeled segment in the figure above runs from the apex to the center of the base? ______

    Which runs from the apex to the midpoint of a base edge? ______


PAGE 3 — The net, and the cone

Nets and Cones

FIGURE: fig2-pyramid-net.png (full width)

  1. In the net of a square-based pyramid there are ______ square(s) and ______ triangle(s).

FIGURE: fig3-cone-parts.png (full width)

  1. A cone has base diameter 1414 cm and height 99 cm.

    Radius: ______ cm Which given number is the height? ______ cm

  2. Describe the net of a square-based pyramid, and say what the height of each triangle in the net represents.




PAGE 4 — Finding the slant height

The Pythagorean Theorem Inside a Pyramid

FIGURE: fig4-slant-height-right-triangle.png (full width)

2=h2+(s2)2h2=2(s2)2\ell^2 = h^2 + \left(\frac{s}{2}\right)^2 \qquad\qquad h^2 = \ell^2 - \left(\frac{s}{2}\right)^2

Warning: use half the base edge, never the whole base edge.

  1. Base edge 66 cm, height 44 cm.

    2=2+2=\ell^2 = \underline{\hspace{1.5cm}}^2 + \underline{\hspace{1.5cm}}^2 = \underline{\hspace{1.5cm}}, so =\ell = \underline{\hspace{1.5cm}} cm.

  2. Base edge 1010 in, height 1212 in. Slant height: ______ in

    WORK SPACE: 1.2 in tall, full width

  3. Why can the slant height never equal the height?



PAGE 5 — Independent practice 11.1

Practice · Heights and Slant Heights

  1. Find the slant height of each square-based pyramid.
Base edge Height Slant height
a) 88 cm 33 cm
b) 2424 m 55 m
c) 1212 ft 88 ft
  1. Base edge 1818 in, height 1212 in. Slant height: ______ in

  2. Base edge 1616 cm, slant height 1717 cm. Height: ______ cm

  3. Base edge 1212 m, slant height 1010 m. Height: ______ m

  4. Base edge 1010 ft, slant height 1313 ft. Height: ______ ft

WORK SPACE: 2.0 in tall, full width

  1. Find the error. For a pyramid with base edge 66 cm and height 44 cm, a student wrote 2=42+62=52\ell^2 = 4^2 + 6^2 = 52, so =527.2\ell = \sqrt{52} \approx 7.2 cm.

    What went wrong? _______________________________________________

    Correct slant height: ______ cm

  2. Apply it. A glass paperweight is a square-based pyramid with base edge 1616 mm and height 1515 mm.

    Slant height: ______ mm

  3. Explain. Why are all four lateral faces congruent isosceles triangles?



PAGE 6 — Exit ticket 11.1

Exit Ticket · Lesson 11.1

Name: ________________________ Date: ____________

  1. Base edge 1414 cm, height 2424 cm. Slant height: ______ cm

  2. Base edge 3030 in, slant height 2525 in. Height: ______ in

  3. In a net of a square-based pyramid, the height of each triangular face is called the ____________ ____________.

  4. Explain the difference between the height and the slant height, and say which is longer.




PAGE 7 — Building the surface area formula

11.2 Surface Area of a Square-Based Pyramid

FIGURE: fig5-surface-area-pieces.png (full width)

Build it from the net.

The base is a square with side ss, so its area is ____________.

Each lateral face is a triangle with base ss and height ____________, so its area is ____________.

There are ______ lateral faces.

  1. SA=s2+4(12s)=s2+SA = s^2 + 4\left(\frac{1}{2}s\ell\right) = s^2 + \underline{\hspace{2cm}}

SA=s2+2sLA=2sSA = s^2 + 2s\ell \qquad\qquad LA = 2s\ell

Lateral area only is what you want when the base is not covered: a tent with no floor, shingles on a roof, paint on the four sloped sides.


PAGE 8 — Guided practice 11.2

Practice · Surface Area

Item Base edge Slant height Surface area
23. 66 cm 55 cm
24. 88 in 55 in
25. 44 m 66 m
26. 1010 ft 1313 ft
  1. Lateral area only: base edge 66 cm, slant height 88 cm. LA=LA = ______ cm2^2

  2. Base edge 66 m and height 44 m.

    Step 1 — slant height: ______ m Step 2 — surface area: ______ m2^2

    WORK SPACE: 1.4 in tall, full width

  3. Base edge 1212 in, slant height 1010 in. SA=SA = ______ in2^2


PAGE 9 — Independent practice 11.2

Practice · Surface Area and Lateral Area

  1. Find each surface area.
Base edge Slant height Surface area
a) 55 cm 88 cm
b) 99 m 1212 m
c) 33 in 44 in
  1. Base edge 1414 ft, slant height 2525 ft. SA=SA = ______ ft2^2

  2. Base edge 2020 cm, slant height 1515 cm. SA=SA = ______ cm2^2

  3. Lateral area: base edge 1818 m, slant height 1515 m. LA=LA = ______ m2^2

These three give the height. Find the slant height first.

  1. Base edge 1616 in, height 1515 in. =\ell = ______ in SA=SA = ______ in2^2

  2. Base edge 1212 cm, height 88 cm. =\ell = ______ cm SA=SA = ______ cm2^2

  3. Base edge 2424 ft, height 55 ft. =\ell = ______ ft SA=SA = ______ ft2^2

WORK SPACE: 2.2 in tall, full width

  1. Base edge 77 m, surface area 189189 m2^2. Slant height: ______ m

PAGE 10 — Applications and errors 11.2

Apply and Check

  1. Apply it. A tent is a square-based pyramid with base edge 66 ft and slant height 55 ft. Four fabric walls, no floor.

    Volume or surface area? ____________ Is the base included? ______

    Fabric needed: ______ ft2^2

  2. Apply it. A gift box is a square-based pyramid with base edge 1010 cm and height 1212 cm. Paper covers the entire outside.

    Slant height: ______ cm Paper needed: ______ cm2^2

    WORK SPACE: 1.4 in tall, full width

  3. Find the error. For a pyramid with base edge 66 cm and height 44 cm, a student wrote SA=62+2(6)(4)=84SA = 6^2 + 2(6)(4) = 84 cm2^2.

    What went wrong? _______________________________________________

    Why must the student's answer be too small? _______________________________________________

    Correct surface area: ______ cm2^2

  4. Explain. Why does the formula contain 2s2s\ell and not 4s4s\ell, when there are four lateral faces?



PAGE 11 — Exit ticket 11.2

Exit Ticket · Lesson 11.2

Name: ________________________ Date: ____________

  1. Base edge 55 cm, slant height 66 cm. SA=SA = ______ cm2^2

  2. Base edge 22 m, slant height 55 m. SA=SA = ______ m2^2

  3. Base edge 1616 in, height 66 in. SA=SA = ______ in2^2

  4. Lateral area: base edge 1212 ft, slant height 1313 ft. LA=LA = ______ ft2^2

  5. Why can the pyramid's height not be substituted into the surface area formula, and what do you do when a problem gives you the height?




PAGE 12 — The one-third relationship

11.3 Why One Third?

FIGURE: fig6-cone-in-cylinder.png (half width, left)

FIGURE: fig7-pyramid-in-prism.png (half width, right)

FIGURE: fig8-three-pours.png (full width)

Fill in the blanks.

A cylinder holds V=πr2hV = \pi r^2 h. A cone with the same base and the same height holds ____________ of that.

A prism with a square base holds V=s2hV = s^2h. A square-based pyramid with the same base and the same height holds ____________ of that.

One formula covers both: V=BhV = \underline{\hspace{2cm}}Bh, where BB is the ____________ of the base.

It takes ______ cone-fulls to fill the cylinder.

  1. A cone holds one third of a cylinder when the two solids have the same ____________ and the same ____________.

  2. How many cone-fulls fill a cylinder with the same base and the same height? ______

  3. In the pouring figure, what fraction is filled after two pours? ______


PAGE 13 — Guided practice 11.3

Practice · Reasoning Up and Down

Item Given Same base and height Answer
48. cylinder holds 6060 cm3^3 cone holds
49. cone holds 2525 in3^3 cylinder holds
50. prism holds 8181 ft3^3 pyramid holds
51. pyramid holds 1414 m3^3 prism holds
  1. Explain what "the same base and the same height" means for a cone and a cylinder. Name the two measurements that must match.




PAGE 14 — Independent practice 11.3

Practice · Cones with Cylinders, Pyramids with Prisms

  1. Cylinder: radius 66 cm, height 1010 cm. V=V = ______ Cone with the same base and height: ______ , or about ______

  2. Cylinder: radius 22 in, height 66 in. V=V = ______ Cone: ______ , or about ______

  3. Prism: square base side 99 m, height 44 m. V=V = ______ m3^3 Pyramid: ______ m3^3

  4. Prism: square base side 1212 ft, height 55 ft. V=V = ______ ft3^3 Pyramid: ______ ft3^3

  5. A cone holds 30π30\pi cm3^3. The cylinder with the same base and height holds ______ , or about ______

  6. A pyramid holds 5050 in3^3. The prism with the same base and height holds ______ in3^3

WORK SPACE: 2.0 in tall, full width


PAGE 15 — Reasoning 11.3

Explain It

  1. Explain. A cube with edge 66 cm cuts into three congruent square-based pyramids, each with a face of the cube as its base and height 66 cm.

    Volume of the cube: ______ cm3^3 Volume of one pyramid: ______ cm3^3

    Check with 13s2h\frac{1}{3}s^2h: _______________

  2. Explain. Why does a cone hold less than half of its cylinder? Use horizontal cross-sections.



  3. Explain. A cone and a cylinder have the same height, but the cone's radius is twice the cylinder's radius RR. Is the cone one third of the cylinder?

    Cone volume in terms of RR and hh: _______________ Cylinder volume: _______________

    Answer and why: _______________________________________________

  4. Apply it. A cylindrical container holds 900900 mL. A cone with the same base and height holds ______ mL

  5. Find the error. A student says a pyramid is half of its prism "because a triangle is half of a rectangle."

    Why does the two-dimensional fact not carry over? _______________________________________________

    Correct fraction: ______

  6. Explain. Why does V=13BhV = \frac{1}{3}Bh work for both a cone and a pyramid? What is BB for each?

    Cone: B=B = ____________ Pyramid: B=B = ____________


PAGE 16 — Exit ticket 11.3

Exit Ticket · Lesson 11.3

Name: ________________________ Date: ____________

  1. A cylinder holds 48π48\pi m3^3. The cone with the same base and height holds ______ , or about ______

  2. A pyramid holds 2121 cm3^3. The prism with the same base and height holds ______ cm3^3

  3. How many pyramid-fulls fill a prism with the same square base and height? ______

  4. Explain why a cone holds exactly one third of its cylinder. Use both "same base" and "same height."




PAGE 17 — The volume formulas

11.4 Volume of Square-Based Pyramids and Cones

V=13s2hV=13πr2hV = \frac{1}{3}s^2h \qquad\qquad V = \frac{1}{3}\pi r^2 h

The hh is the height — never the slant height.

Checklist before you compute. Square first, then multiply. Keep the 13\frac{1}{3}. Halve a diameter. Write cubic units. Approximate π\pi only at the last step.

The question asks for The formula needs If you were given the other one
surface area of a pyramid slant height \ell 2=h2+(s2)2\ell^2 = h^2 + \left(\frac{s}{2}\right)^2
volume of a pyramid height hh h2=2(s2)2h^2 = \ell^2 - \left(\frac{s}{2}\right)^2
  1. Base edge 33 cm, height 55 cm: V=13()2()=V = \frac{1}{3}(\underline{\hspace{1.5cm}})^2(\underline{\hspace{1.5cm}}) = \underline{\hspace{1.5cm}} cm3^3

PAGE 18 — Guided practice 11.4

Practice · Volume

Pyramids.

Item Base edge Height Volume
72. 66 in 1010 in
73. 55 m 66 m
74. 1010 ft 99 ft

Cones. Give each volume exactly in terms of π\pi, then approximately with π3.14\pi \approx 3.14.

Item Radius Height Exact Approximate
75. 33 cm 44 cm
76. 55 in 99 in
77. 44 m 33 m
78. d=10d = 10 ft 66 ft

WORK SPACE: 2.0 in tall, full width


PAGE 19 — Independent practice 11.4

Practice · Volume, Mixed

  1. Find each pyramid volume.
Base edge Height Volume
a) 44 cm 66 cm
b) 88 in 33 in
c) 1515 m 88 m
  1. Pyramid: base edge 77 ft, height 66 ft. V=V = ______ ft3^3

  2. Pyramid: base edge 2020 cm, height 66 cm. V=V = ______ cm3^3

  3. Cone: radius 66 cm, height 44 cm. V=V = ______ , or about ______

  4. Cone: diameter 88 in, height 66 in. r=r = ______ in V=V = ______ , or about ______

  5. Cone: radius 99 m, height 55 m. V=V = ______ , or about ______

  6. Cone: radius 11 ft, height 99 ft. V=V = ______ , or about ______

WORK SPACE: 2.2 in tall, full width


PAGE 20 — Working backward, and applications 11.4

Work Backward and Apply

  1. Pyramid: base edge 1212 ft, volume 240240 ft3^3. Height: ______ ft

    WORK SPACE: 1.0 in tall, full width

  2. Cone: radius 55 in, volume 100π100\pi in3^3. Height: ______ in

  3. Pyramid: base edge 66 m, slant height 55 m.

    Step 1 — height: ______ m Step 2 — volume: ______ m3^3

  4. Cone: base area 27π27\pi cm2^2, height 55 cm. Using V=13BhV = \frac{1}{3}Bh: ______ , or about ______

  5. Apply it. A cone-shaped paper cup has radius 33 cm and height 1010 cm. It holds ______ , or about ______

  6. Apply it. A candle mold is a square-based pyramid with base edge 1010 cm and height 1212 cm. It holds ______ cm3^3

  7. Find the error. For a cone with radius 66 cm and height 1010 cm, a student wrote V=π(6)2(10)=360πV = \pi(6)^2(10) = 360\pi cm3^3.

    What went wrong? _______________________________________________

    Correct volume: ______ , or about ______


PAGE 21 — Exit ticket 11.4

Exit Ticket · Lesson 11.4

Name: ________________________ Date: ____________

  1. Pyramid: base edge 1111 cm, height 33 cm. V=V = ______ cm3^3

  2. Pyramid: base edge 22 m, height 99 m. V=V = ______ m3^3

  3. Cone: radius 66 in, height 99 in. V=V = ______ , or about ______

  4. Cone: diameter 66 ft, height 88 ft. V=V = ______ , or about ______

  5. Where does the 13\frac{1}{3} in both formulas come from, and why is it the height and not the slant height in them?




PAGE 22 — Deciding what a problem wants

11.5 Problems in Context

Volume is about the inside: fill, hold, pour, capacity, how much sand, water, wax, grain. → cubic units

Surface area is about the outside: cover, wrap, paint, tile, how much fabric, paper, glass. → square units

Then ask: is the base included?

Four moves for every context problem. 1. Decide. 2. Collect (halve a diameter; find \ell or hh if needed). 3. Substitute. 4. Report with the right unit.

  1. Volume or surface area?
Situation Volume or surface area?
a) painting the four sloped sides of a pyramid monument
b) filling a cone with popcorn
c) gift wrap for a closed pyramid box
d) water in a cone-shaped paper cup

PAGE 23 — Guided practice 11.5

Practice · In Context

  1. A sandbox is a square-based pyramid with base edge 1212 cm and height 88 cm.

    Volume or surface area? ____________ Answer: ______ cm3^3

  2. An ornament is a square-based pyramid with base edge 1010 cm and height 1212 cm. Foil covers all five faces.

    Slant height: ______ cm Answer: ______ cm2^2

  3. A snow cone cup has radius 33 cm and height 88 cm. It holds ______ , or about ______

  4. A paperweight is a square-based pyramid with base edge 44 cm and slant height 66 cm. A label covers the four lateral faces only.

    Answer: ______ cm2^2

  5. A cone-shaped cup has radius 22 in and height 99 in. It holds ______ , or about ______

WORK SPACE: 2.0 in tall, full width


PAGE 24 — Independent practice 11.5, part 1

Apply It

  1. Apply it. A roof is a square-based pyramid with base edge 1818 ft and slant height 1515 ft. Shingles cover the four sloped faces only. ______ ft2^2

  2. Apply it. A gravel pile is a cone with radius 1010 ft and height 66 ft. ______ , or about ______

  3. Apply it. A display case is a square-based pyramid with base edge 1616 in and height 1515 in, glass on all five faces.

    Slant height: ______ in Glass: ______ in2^2

  4. Apply it. A cone-shaped filter has base diameter 1212 cm and height 77 cm. r=r = ______ cm It holds ______ , or about ______

  5. Apply it. A candle is a square-based pyramid with base edge 66 cm and height 44 cm. Wax has mass 0.90.9 g per cm3^3.

    Volume: ______ cm3^3 Mass: ______ g

WORK SPACE: 2.2 in tall, full width


PAGE 25 — Independent practice 11.5, part 2

Apply It (continued)

  1. Apply it. Which holds more?
Container Volume
cone, radius 66 cm, height 66 cm
square-based pyramid, base edge 1010 cm, height 66 cm
 Which holds more, and by about how much? _______________________________________________
  1. Apply it. A cone-shaped funnel has radius 55 cm and height 1212 cm and drains at 2020 cm3^3 per second.

    Volume: ______ , or about ______ Time to empty: about ______ seconds

  2. Apply it. A tent is a square-based pyramid with base edge 88 ft and height 33 ft, four walls and no floor.

    Slant height: ______ ft Fabric: ______ ft2^2

  3. Apply it. A cone-shaped hole has radius 33 ft and depth 44 ft. Mulch costs $3 per cubic foot.

    Volume: ______ , or about ______ ft3^3 Cost: $______

  4. Explain. A problem gives a pyramid's base edge and slant height and asks for volume. What must you find first, and why?


  5. Find the error. For wrapping paper on a closed pyramid box with base edge 1010 in and height 1212 in, a student wrote SA=102+2(10)(12)=340SA = 10^2 + 2(10)(12) = 340 in2^2.

    What went wrong? _______________________________________________

    Correct surface area: ______ in2^2

  6. Explain. How does the unit on your answer tell you which kind of problem you solved?



PAGE 26 — Exit ticket 11.5

Exit Ticket · Lesson 11.5

Name: ________________________ Date: ____________

  1. A cone-shaped cup has radius 44 cm and height 99 cm. It holds ______ , or about ______

  2. A square-based pyramid has base edge 2424 in and height 55 in. Sand fills it: ______ in3^3

  3. A square-based pyramid has base edge 2020 cm and slant height 2626 cm. Paper covering all five faces: ______ cm2^2

  4. Cardboard for a closed pyramid box with base edge 66 cm and slant height 88 cm: ______ cm2^2

  5. How do you decide whether a context problem wants volume or surface area?




PAGE 27 — Chapter 11 review, Part A

Chapter 11 Review

Part A · Surface area of square-based pyramids (8.MG.2a)

  1. Describe the net of a square-based pyramid: how many pieces of each shape, and what the height of each triangle represents.



  2. Base edge 4848 cm, height 77 cm. Slant height: ______ cm

  3. Find each surface area.

Base edge Slant height Surface area
a) 77 cm 1010 cm
b) 1111 m 99 m
  1. Base edge 3030 ft, height 2020 ft. =\ell = ______ ft SA=SA = ______ ft2^2

  2. Lateral area: base edge 1414 in, slant height 2525 in. LA=LA = ______ in2^2

  3. Base edge 2424 in, surface area 12001200 in2^2. Slant height: ______ in

  4. Find the error. For base edge 66 cm and slant height 55 cm, a student wrote 36+3(15)=8136 + 3(15) = 81 cm2^2.

    What went wrong? _______________________________________________ Correct: ______ cm2^2

  5. Explain. How does the net show that SA=s2+2sSA = s^2 + 2s\ell? Account for every piece and for the 22.




PAGE 28 — Chapter 11 review, Part B

Chapter 11 Review (continued)

Part B · Volume of cones and square-based pyramids (8.MG.2b)

Give every cone volume exactly, then approximately with π3.14\pi \approx 3.14.

  1. Find each pyramid volume.
Base edge Height Volume
a) 66 cm 44 cm
b) 1212 in 55 in
c) 1010 m 1212 m
  1. Pyramid: base edge 1616 ft, height 1515 ft. V=V = ______ ft3^3

  2. Find each cone volume.

Radius Height Exact Approximate
a) 33 cm 1010 cm
b) 88 in 33 in
  1. Cone: diameter 1010 m, height 1212 m. V=V = ______ , or about ______

  2. Cone: radius 1212 ft, height 55 ft. V=V = ______ , or about ______

  3. Pyramid: base edge 1515 m, volume 600600 m3^3. Height: ______ m

  4. Cone: height 44 cm, volume 48π48\pi cm3^3. Radius: ______ cm

  5. Pyramid: base edge 1212 cm, slant height 1010 cm. h=h = ______ cm V=V = ______ cm3^3

WORK SPACE: 2.2 in tall, full width


PAGE 29 — Chapter 11 review, Part C

Chapter 11 Review (continued)

Part C · Explaining the volume relationships (8.MG.2c)

  1. Cylinder: radius 55 cm, height 99 cm. V=V = ______ Cone with the same base and height: ______ , or about ______

  2. Prism: square base side 1010 in, height 99 in. V=V = ______ in3^3 Pyramid: ______ in3^3

  3. A cone holds 16π16\pi ft3^3. The cylinder with the same base and height holds ______ ft3^3

  4. Explain. Use the dissection of a cube into three congruent square-based pyramids to explain the one-third relationship.




  5. Explain. Describe the pouring experiment and say exactly what it shows. Why are both "same base" and "same height" required?



  6. Explain. A classmate writes, "A cone is one third of any cylinder." Correct the statement, then give a counterexample with numbers.

    Correction: _______________________________________________

    Counterexample: cone r=r = ______ , h=h = ______ , V=V = ______ ; cylinder r=r = ______ , h=h = ______ , V=V = ______


PAGE 30 — Chapter 11 review, Part D

Chapter 11 Review (continued)

Part D · Problems in context (8.MG.2d)

Give every cone volume exactly, then approximately with π3.14\pi \approx 3.14.

  1. Apply it. A cone-shaped cup has radius 44 cm and height 66 cm. It holds ______ , or about ______

  2. Apply it. A gift box is a square-based pyramid with base edge 1818 cm and height 1212 cm. Paper for the whole outside: =\ell = ______ cm, SA=SA = ______ cm2^2

  3. Apply it. A sand pile is a cone with base diameter 1414 ft and height 33 ft. r=r = ______ ft V=V = ______ , or about ______

  4. Apply it. A monument is a square-based pyramid with base edge 3030 m and slant height 2525 m. Paint covers the four sloped faces only: ______ m2^2

  5. Apply it. A cone-shaped funnel has radius 66 cm and height 1010 cm.

    Volume: ______ , or about ______ Full 100100 cm3^3 jars it can fill: ______

  6. Apply it. A planter is a square-based pyramid with base edge 1212 in and height 88 in.

    a) Soil it holds: ______ in3^3

    b) Liner for the four lateral faces only: =\ell = ______ in, LA=LA = ______ in2^2

WORK SPACE: 2.0 in tall, full width

  1. Explain. A square-based pyramid has base edge 66 cm and height 44 cm.

    Volume: ______ cm3^3 Surface area: ______ cm2^2

    Why do the two answers carry different units? _______________________________________________

  2. Write your own. Write a context problem about a cone that must be solved with volume, and one about a square-based pyramid that must be solved with surface area. Then solve both.

    WORK SPACE: 3.5 in tall, full width