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How do you find the volume and surface area of solids using the reference sheet formulas?

Compute the volume and surface area of prisms, cylinders, cones, spheres, and pyramids using the reference sheet formulas, and solve real-world problems involving capacity and material.

A Grade 10 Math MCAS answer on volume and surface area of prisms, cylinders, cones, spheres, and pyramids, using the reference sheet formulas, and applying them to capacity and material problems with appropriate units.

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  1. What this topic is asking
  2. Volume formulas
  3. Surface area
  4. Real-world problems and units
  5. Scaling solids by a factor
  6. Working backward from a given volume
  7. Try this

What this topic is asking

The Geometry category requires you to compute volume and surface area of solids (the G-GMD standards). On the Grade 10 MCAS you handle prisms, cylinders, cones, spheres, and pyramids, applying the reference sheet formulas to capacity and material problems. All the volume formulas are on the reference sheet, so the credit is for choosing the right formula, substituting correctly, and reporting with appropriate units.

Volume formulas

Volume measures the space inside a solid, in cubic units. The reference sheet provides:

Vprism=Bh,Vcylinder=πr2h,Vsphere=43πr3,Vcone=13πr2h,Vpyramid=13Bh.V_{\text{prism}} = Bh, \quad V_{\text{cylinder}} = \pi r^2 h, \quad V_{\text{sphere}} = \frac{4}{3}\pi r^3, \quad V_{\text{cone}} = \frac{1}{3}\pi r^2 h, \quad V_{\text{pyramid}} = \frac{1}{3}Bh.

Here BB is the area of the base and hh the height. For a prism, find the base area first (a rectangle, triangle, or other polygon), then multiply by the height. The crucial pattern: a cone is one-third of a cylinder with the same base and height, and a pyramid is one-third of the matching prism, which is why those formulas carry the 13\frac{1}{3}. Forgetting that factor is the single most common error.

Surface area

Surface area is the total area of all the faces of a solid, in square units. For a solid with flat faces, add the areas of every face; the reference sheet provides the total surface area formulas alongside the volumes.

  • A rectangular prism has surface area 2(lw+lh+wh)2(lw + lh + wh): the six faces in three matching pairs.
  • A cylinder has surface area 2πr2+2πrh2\pi r^2 + 2\pi r h: the two circular ends plus the rectangle that wraps around (its width is the circumference).
  • A sphere has surface area 4πr24\pi r^2.

The distinction the MCAS tests is dimension: surface area is square units (covering material, paint, wrapping), while volume is cubic units (capacity, filling). Reading whether the problem asks for the amount to fill (volume) or to cover (surface area) decides which to compute.

Real-world problems and units

Capacity and material problems are volume and surface-area problems in context. "How much water fills the tank" is a volume in cubic units or a capacity (liters, gallons). "How much paint covers the box" is a surface area in square units. Always attach the correct unit, and when a decimal is required, use π3.14\pi \approx 3.14 from the reference sheet, rounding at the end.

For composite solids (a cylinder topped by a cone, say), find each part's volume separately and add them. Subtraction is used when a hole is removed.

Scaling solids by a factor

When a solid is scaled by a linear factor kk, its volume scales by k3k^3, because volume is a product of three lengths. So doubling every dimension of a box (k=2k = 2) makes it hold 23=82^3 = 8 times as much, not twice as much. This often appears as a "how many times bigger" question, and the answer is the cube of the linear factor. By the same logic, surface area scales by k2k^2. A common error is to scale volume by the linear factor or its square instead of its cube.

Working backward from a given volume

Sometimes the MCAS gives a volume and asks for a missing dimension. Substitute into the formula and solve. If a cylinder has volume 100π100\pi and radius 5, then π(5)2h=100π\pi(5)^2 h = 100\pi, so 25h=10025h = 100 and h=4h = 4. The same rearrangement works for any solid: write the formula, plug in the known volume and dimensions, and solve for the unknown, treating it like any equation.

Try this

Q1. A sphere has radius 3. Find its volume in terms of π\pi.

  • Cue. 43π(3)3=43π(27)=36π\frac{4}{3}\pi(3)^3 = \frac{4}{3}\pi(27) = 36\pi.

Q2. A rectangular prism is 2 by 3 by 5. Find its volume.

  • Cue. V=2×3×5=30V = 2 \times 3 \times 5 = 30 cubic units.

Exam-style practice questions

Practice questions written in the style of MA DESE exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Grade 10 Math MCAS (style)1 marksSelected-response. A cylinder has radius 3 and height 10. What is its volume, in terms of pi? (A) 30π30\pi (B) 90π90\pi (C) 60π60\pi (D) 300π300\pi
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The correct answer is (B).

Cylinder volume is V=πr2hV = \pi r^2 h (on the reference sheet): V=π(3)2(10)=π(9)(10)=90πV = \pi (3)^2 (10) = \pi (9)(10) = 90\pi. Choice (A) uses rr instead of r2r^2; choice (D) uses the diameter. Squaring the radius first is the key step.

Grade 10 Math MCAS (style)4 marksConstructed-response. A cone-shaped cup has radius 4 cm and height 9 cm. Find its volume in terms of pi, then state how much it holds to the nearest cubic centimeter (use pi about 3.14). Show your work.
Show worked answer →

A 4-point constructed-response: credit for the formula, the substitution, the exact value, and the rounded value.

Cone volume is V=13πr2hV = \frac{1}{3}\pi r^2 h (on the reference sheet): V=13π(4)2(9)=13π(16)(9)=13π(144)=48πV = \frac{1}{3}\pi (4)^2 (9) = \frac{1}{3}\pi (16)(9) = \frac{1}{3}\pi (144) = 48\pi cubic cm. Numerically, 48×3.14=150.7215148 \times 3.14 = 150.72 \approx 151 cubic cm. Forgetting the 13\frac{1}{3} (using the cylinder formula) is the most common error and triples the answer.

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