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Right Prisms: Triangular and Other Bases

Use V = Bh and SA = ph + 2B for prisms with triangle, trapezoid and other bases, and work backwards to find a missing dimension.

Practice this skill 30 min Right prisms Leer en español
Key idea

What you'll learn

A prism is any solid with two identical, parallel ends (the bases) joined by rectangles. A box is one kind of prism. In this lesson you will handle prisms whose bases are triangles, trapezoids and other polygons: tents, wedges, ramps, roof attics, water troughs, Toblerone-style boxes and sheds.

You will learn to:

  • find the volume with V = Bh,
  • find the surface area with SA = ph + 2B,
  • find a missing dimension when the volume or a side is given, sometimes using the Pythagorean theorem.

Why it matters on the GED: The GED formula sheet gives one pair of formulas for all right prisms. The challenge is not the formula; it is finding the right B (the area of the base shape) and the right h (the length of the prism). The test likes to turn a triangular prism on its side so the base is not on the bottom. Once you can spot the two matching ends, every prism becomes the same problem.

Words to know

  • Right prism — a prism whose side faces are rectangles standing at right angles to the bases (all GED prisms are right prisms).
  • Bases of a prism — the two identical, parallel faces that give the prism its name (a triangular prism has two triangle bases).
  • B — the area of one base. For a triangle with base 6 and height 4: B = \tfrac{1}{2}(6)(4) = 12.
  • Height of the prism (h) — the distance between the two bases; the "length" of the prism. A tent 10\text{ ft} long has h = 10\text{ ft}.
  • Height of the triangle — the perpendicular distance from the triangle's base to its top vertex. It lives inside the base shape and is different from the prism's height.
  • Lateral faces — the rectangles around the sides; a triangular prism has 3, a pentagonal prism has 5.
  • Perimeter of the base (p) — the distance around the base shape (3-4-5 triangle: p = 12).
  • Trapezoid — a four-sided shape with one pair of parallel sides b_1 and b_2: A = \tfrac{1}{2}h(b_1 + b_2).
  • Legs and hypotenuse — in a right triangle, the legs make the right angle; the hypotenuse is the longest side across from it.
  • Net — the flat pattern of a prism unfolded: two bases plus a strip of rectangles.

All Math words in the glossary

Find the bases first

Triangular prism. The two parallel, identical triangles are the bases (B). The prism's height h is the distance between the two triangles, measured along the length of the prism. The triangle's own base b and height are measured on the triangle itself. triangle base b prism height h triangle height

Every prism has two identical ends that are parallel. Those ends are the bases, no matter how the prism is sitting.

  • A tent sitting on its rectangular floor: the bases are the triangles at the front and back.
  • A wedge or doorstop lying on a rectangle: the bases are the two right triangles on the ends.
  • A shed: the bases are the house-shaped pentagons on the front and back.
Three prisms with their two bases shaded: a tent with triangle bases on the front and back, a wedge (doorstop) with right-triangle bases on the ends, and a shed with house-shaped pentagon bases on the front and back. tent wedge shed shaded faces = the bases (two matching ends)

Once the bases are found, h is the edge that connects them: the length of the prism.

Area of the base shape (all formulas are on the GED sheet):

Base shapeArea BExample
triangle\tfrac{1}{2}bhb = 6, h = 4: B = 12
rectanglelw5 \times 3: B = 15
trapezoid\tfrac{1}{2}h(b_1 + b_2)b_1 = 2, b_2 = 4, h = 1.5: B = 4.5
composite (rectangle + triangle)add the pieces80 + 20 = 100

The two heights. A triangular prism has two different "heights":

  1. the triangle's height, used inside B = \tfrac{1}{2}bh;
  2. the prism's height (length), used in V = Bh.

Label them clearly before you plug in.

Triangular prism showing its two different heights: the height of the triangle, 4 in, measured inside the triangular base from the 6 in base to the top vertex; and the height (length) of the prism, 10 in, the distance between the two triangular bases. triangle base 6 in prism height (length) 10 in triangle height 4 in

Right triangles: the two legs are perpendicular, so one leg is the base and the other is the height. The hypotenuse is never the height of the triangle.

Volume: V = Bh for every prism

The volume of any right prism is the area of a base times the distance between the bases:

V = Bh

Example (triangular prism): each triangle has base 6\text{ in} and height 4\text{ in}; the prism is 10\text{ in} long.

  1. B = \tfrac{1}{2}(6)(4) = 12\text{ in}^2.
  2. V = Bh = 12 \times 10 = 120\text{ in}^3.

Notice that this is half of the 6 \times 4 \times 10 = 240 box that would surround it. A triangular prism with a right-triangle base is exactly half of a box.

Triangular prism with right-triangle bases (legs 6 in and 4 in) and length 10 in, drawn inside the dashed 6 by 4 by 10 box that surrounds it. B = 12, V = 120 cubic inches, half of the box volume 240. b = 6 in h = 4 in 10 in B = ½(6)(4) = 12; V = 12 × 10 = 120 in³ (box: 240)

Example (trapezoid prism): a water trough has trapezoid ends with b_1 = 2\text{ ft}, b_2 = 4\text{ ft}, height 1.5\text{ ft}, and is 8\text{ ft} long.

  1. B = \tfrac{1}{2}(1.5)(2 + 4) = \tfrac{1}{2}(1.5)(6) = 4.5\text{ ft}^2.
  2. V = 4.5 \times 8 = 36\text{ ft}^3.
Water trough shaped like a prism with trapezoid ends: bottom 2 ft, top 4 ft, height 1.5 ft, and 8 ft long. B = 4.5 square feet and V = 36 cubic feet. b₁ = 2 ft b₂ = 4 ft 1.5 ft 8 ft B = ½(1.5)(2 + 4) = 4.5 ft²; V = 4.5 × 8 = 36 ft³

Example (other base): if a problem tells you a hexagon base has area 24\text{ cm}^2 and the prism is 7\text{ cm} tall, then V = 24 \times 7 = 168\text{ cm}^3. You don't need a hexagon formula when B is given.

Hexagonal prism standing on a hexagon base with area 24 square cm and height 7 cm. V = 24 × 7 = 168 cubic cm. B = 24 cm² h = 7 cm V = 24 × 7 = 168 cm³

Missing side using Pythagoras: if the triangle's height is not shown, you may have to find it. An isosceles triangle with sides 10, 10, 12: drop the height to the middle of the 12 side, making two right triangles with legs 6 and h and hypotenuse 10. Then h = \sqrt{10^2 - 6^2} = \sqrt{64} = 8, and B = \tfrac{1}{2}(12)(8) = 48.

Isosceles triangle with sides 10, 10 and 12. The height drawn to the middle of the 12 side splits it into two right triangles with legs 6 and h and hypotenuse 10, so h = 8 and B = 48. 10 10 6 6 h = 8 h = √(10² − 6²) = √64 = 8; B = ½(12)(8) = 48

Surface area: SA = ph + 2B

Net of a triangular prism with right-triangle bases (legs 3 and 4, hypotenuse 5) and length 10: three rectangles 3 by 10, 4 by 10 and 5 by 10 side by side, with a triangle base B attached at the top and at the bottom. The three rectangles together form one long rectangle whose width is the perimeter 3 + 4 + 5 = 12. 3×10 4×10 5×10 B B perimeter p = 3 + 4 + 5 = 12

Unfold a triangular prism and you get two triangles and three rectangles. All three rectangles have the same length (the prism's h), and their widths are the three sides of the triangle. Laid side by side, they form one long rectangle whose width is the perimeter of the triangle. So:

SA = ph + 2B

Example: right-triangle bases with legs 3 and 4 and hypotenuse 5; length 10.

  1. p = 3 + 4 + 5 = 12.
  2. Lateral area: ph = 12 \times 10 = 120.
  3. B = \tfrac{1}{2}(3)(4) = 6, so 2B = 12.
  4. SA = 120 + 12 = 132 square units.

Check with the net: rectangles 3 \times 10 + 4 \times 10 + 5 \times 10 = 30 + 40 + 50 = 120, plus two triangles of 6: 132 ✓.

Net of a triangular prism with right-triangle bases 3, 4, 5 and length 10: three rectangles 3 by 10, 4 by 10 and 5 by 10 (areas 30, 40, 50) form one strip 12 wide, plus two triangles of area 6 each. Surface area 132. 3 × 10 = 30 4 × 10 = 40 5 × 10 = 50 6 6 10 SA = 30 + 40 + 50 + 6 + 6 = 132

The perimeter needs every side. If a side is missing (often the hypotenuse), find it with a^2 + b^2 = c^2 first. Legs 6 and 8: c = \sqrt{36 + 64} = 10.

Only cover what is there. A tent with no floor leaves out one rectangle; an open trough leaves out the top rectangle. Add up only the faces described.

Formula

The formulas (on the GED formula sheet)

On the GED formula sheet, under right prism:

V = Bh \qquad\qquad SA = ph + 2B
  • B = area of one base (the shape on the two matching ends).
  • p = perimeter of that base.
  • h = height of the prism = distance between the two bases.

Also on the formula sheet: triangle A = \tfrac{1}{2}bh, trapezoid A = \tfrac{1}{2}h(b_1 + b_2), rectangle A = lw, and the Pythagorean theorem a^2 + b^2 = c^2.

Memorize the idea, not more formulas: find the two matching ends, find their area, multiply by the distance between them.

Worked example

Example 1 (basic): triangular prism volume

Each triangular end of a prism has a base of 8\text{ cm} and a height of 5\text{ cm}. The prism is 12\text{ cm} long. Find the volume.

Triangular prism whose triangular ends have base 8 cm and height 5 cm; the prism is 12 cm long. B = 20 square cm, V = 240 cubic cm. 8 cm 12 cm 5 cm
  1. B = \tfrac{1}{2}bh = \tfrac{1}{2}(8)(5) = 20\text{ cm}^2.
  2. V = Bh = 20 \times 12 = 240.

Answer: 240\text{ cm}^3

Check: the surrounding box is 8 \times 5 \times 12 = 480, and the prism is half of it: 480 \div 2 = 240 ✓.

Worked example

Example 2: surface area with a right-triangle base

A triangular prism has right-triangle bases with legs 6\text{ in} and 8\text{ in} and hypotenuse 10\text{ in}. It is 15\text{ in} long. Find its surface area.

Triangular prism with right-triangle bases, legs 6 in and 8 in and hypotenuse 10 in, and length 15 in. 8 in 10 in 6 in 15 in
  1. Perimeter: p = 6 + 8 + 10 = 24\text{ in}.
  2. Lateral area: ph = 24 \times 15 = 360\text{ in}^2.
  3. B = \tfrac{1}{2}(6)(8) = 24\text{ in}^2, so 2B = 48\text{ in}^2.
  4. SA = 360 + 48 = 408.
Net of the 6-8-10 triangular prism 15 in long: rectangles 6 by 15, 8 by 15 and 10 by 15 (90, 120, 150 square inches) plus two right triangles of 24 square inches each. Surface area 408 square inches. 6 × 15 = 90 8 × 15 = 120 10 × 15 = 150 24 24 15 in SA = 90 + 120 + 150 + 24 + 24 = 408 in²

Answer: 408\text{ in}^2

Check: rectangles 6 \cdot 15 + 8 \cdot 15 + 10 \cdot 15 = 90 + 120 + 150 = 360; plus two triangles 24 + 24: 408 ✓.

Worked example

Example 3 (real world): a pool with a sloped bottom

A pool is 15\text{ ft} wide. Seen from the side, it is a trapezoid: 30\text{ ft} long, 3\text{ ft} deep at the shallow end and 9\text{ ft} deep at the deep end. How many gallons does it hold? (1\text{ ft}^3 \approx 7.5 gallons.)

Pool drawn as a prism lying on its side: the side view is a trapezoid 30 ft long, 3 ft deep at the shallow end and 9 ft deep at the deep end; the pool is 15 ft wide. B = 180 square feet and V = 2,700 cubic feet. 30 ft 3 ft 9 ft 15 ft wide B = ½(30)(3 + 9) = 180 ft²; V = 180 × 15 = 2,700 ft³
  1. The trapezoid side is the base. Its parallel sides are the depths 3 and 9; the distance between them is 30.
  2. B = \tfrac{1}{2}(30)(3 + 9) = \tfrac{1}{2}(30)(12) = 180\text{ ft}^2.
  3. The prism's height is the width of the pool: V = 180 \times 15 = 2{,}700\text{ ft}^3.
  4. Gallons: 2{,}700 \times 7.5 = 20{,}250.

Answer: about 20{,}250 gallons

Check: the average depth is (3 + 9) \div 2 = 6\text{ ft}, and 30 \times 15 \times 6 = 2{,}700\text{ ft}^3 ✓.

Worked example

Example 4 (work backwards): missing triangle height

A chocolate box is a triangular prism 12\text{ in} long with a volume of 180\text{ in}^3. Each triangular end has a base of 6\text{ in}. What is the height of the triangle?

Chocolate box shaped like a triangular prism 12 in long; each triangular end has a 6 in base and an unknown height h. Volume 180 cubic inches, so B = 15 and h = 5 in. 6 in 12 in h = ?
  1. V = Bh: 180 = B \times 12, so B = 180 \div 12 = 15\text{ in}^2.
  2. B = \tfrac{1}{2}bh: 15 = \tfrac{1}{2}(6)(h) = 3h.
  3. h = 15 \div 3 = 5.

Answer: the triangle is 5\text{ in} tall.

Check: \tfrac{1}{2}(6)(5) = 15 and 15 \times 12 = 180 ✓.

Worked example

Example 5 (GED-hard): attic volume with Pythagoras

An attic is a triangular prism 40\text{ ft} long. Each end is an isosceles triangle with a base of 24\text{ ft} and two sloped sides of 13\text{ ft}. What is the volume of the attic?

Attic shaped like a triangular prism 40 ft long. Each end is an isosceles triangle with a 24 ft base and two 13 ft sloped sides; the triangle height h is unknown (it is 5 ft). 24 ft 13 ft 40 ft h
  1. The triangle's height is not given. It splits the 24\text{ ft} base into two halves of 12\text{ ft}, forming right triangles with hypotenuse 13.
  2. Pythagoras: h^2 + 12^2 = 13^2, so h^2 = 169 - 144 = 25 and h = 5\text{ ft}.
  3. B = \tfrac{1}{2}(24)(5) = 60\text{ ft}^2.
  4. V = Bh = 60 \times 40 = 2{,}400.
End of the attic: an isosceles triangle with base 24 ft and sides 13 ft. The height splits the base into two 12 ft halves, making right triangles 5-12-13, so h = 5 ft and B = 60 square feet. 13 ft 13 ft 12 ft 12 ft h = 5 ft h² + 12² = 13² → h² = 25, h = 5; B = ½(24)(5) = 60 ft²

Answer: 2{,}400\text{ ft}^3

Check: 5-12-13 is a known right triangle: 25 + 144 = 169 ✓; and 60 \times 40 = 2{,}400 ✓.

Calculator tip

On the TI-30XS MultiView

  • Triangle base area: 1 n/d 2 ▶ (gives \tfrac{1}{2}), then × 6 × 4 enter → 12. Or simply divide by 2: 6 × 4 ÷ 2.
  • Volume in one line: 0.5 × 6 × 4 × 10 enter → 120.
  • Trapezoid: use parentheses for the sum of the bases: 0.5 × 1.5 × ( 2 + 4 ) × 8 enter → 36.
  • Missing hypotenuse: 2nd x² (√) then 6 x² + 8 x² enter → 10.
  • toggle switches an answer like \tfrac{45}{2} to 22.5.

When not to use it: with small whole numbers, halving in your head (\tfrac{1}{2} \times 6 = 3) is quicker and avoids typing errors. Save the calculator for decimals and square roots.

Common mistake

Common mistakes

  • Forgetting the \tfrac{1}{2}. ✗ B = 6 \times 4 = 24. ✓ A triangle is half a rectangle: B = \tfrac{1}{2}(6)(4) = 12. The volume comes out twice too big otherwise.
  • Using the hypotenuse as the height. ✗ B = \tfrac{1}{2}(12)(13). ✓ The height is perpendicular to the base: for a right triangle use the legs, \tfrac{1}{2}(12)(5) = 30.
  • Treating the bottom rectangle as the base. When a triangular prism lies on a rectangle, B is still the triangle, not the floor rectangle. ✗ Using the 12 \times 20 floor as B. ✓ B = \tfrac{1}{2}(12)(5) = 30, so V = 30 \times 20 = 600.
  • Leaving out a side in the perimeter. ✗ p = 6 + 8 = 14 when the hypotenuse is not drawn. ✓ Find it first: c = 10, p = 24.
  • One base instead of two. ✗ SA = ph + B for a closed prism. ✓ SA = ph + 2B (unless the problem says an end is open).
  • Using a pyramid formula. ✗ V = \tfrac{1}{3}Bh. That is for pyramids and cones. Prisms use V = Bh.
Test strategy

GED test strategy

  • Shade the two matching ends (mentally or on the erasable note board). That shape is B, wherever it sits.
  • Write two lines: first "B = \ldots" with the area formula for the base shape, then "V = B \times h" or "SA = ph + 2B." Keeping B separate prevents mix-ups between the two heights.
  • Estimate with the surrounding box. A triangular prism holds about half of the box around it. If your answer is bigger than the box, something is wrong.
  • Look for 3-4-5, 5-12-13, 6-8-10, 9-12-15 right triangles. They save time when a side is missing.
  • Plug answers back in on work-backwards questions: test each choice in V = Bh.
  • Read for open or missing faces: tents without floors, troughs without tops, ramps sitting on the ground.

Check yourself

Right prisms

What is the volume of the triangular prism shown?

Triangular prism. Each triangular base has a base of 6 inches and a height of 4 inches. The prism is 10 inches long. 6 in 10 in 4 in
Answer choices
Hint

V = Bh. Here B is the area of the triangle, and h is the length of the prism.

Check yourself

Right prismsFill in

A glass paperweight is a prism with a hexagon-shaped base. The area of the hexagon is 24\text{ cm}^2, and the prism is 7\text{ cm} tall.

What is the volume of the paperweight?

\text{cm}^3

Type your answer in the box. Fractions like 3/4 are fine.

Hint

You don't need a hexagon formula. The base area B is already given; use V = Bh.

Summary

  • Every right prism uses the same two formulas from the formula sheet: V = Bh and SA = ph + 2B.
  • B is the area of the two matching ends: triangle \tfrac{1}{2}bh, trapezoid \tfrac{1}{2}h(b_1 + b_2), or the sum of simpler shapes.
  • h is the distance between the two bases (the prism's length), not the triangle's height.
  • Use the legs (not the hypotenuse) as base and height of a right triangle; use a^2 + b^2 = c^2 to find a missing side.
  • To work backwards: B = V \div h, then solve the base-area formula for the missing side.
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