Mini-Quizzes with Answers: Practice Sets for Perimeter, Area, and Volume

How to Use These Mini-Quizzes

This chapter is a practice workshop: short, focused quizzes with answers so you can check your work immediately. Each set targets a specific skill (perimeter, area, volume, and mixed applications). You will see real-life contexts, numbers chosen to be realistic, and step-by-step solutions when a multi-step method is needed.

Guidelines for using the practice sets:

• Work each problem without looking at the answer first.
• Write down the formula you plan to use before calculating.
• Include units in every intermediate step (ft, m, in, cm; and square/cubic units).
• After checking the answer, redo any missed problem using the solution steps as a template.

Mini-Quiz Set 1: Perimeter Practice (Real Boundaries)

Focus: adding side lengths, handling missing sides, and interpreting “border length” in practical scenarios.

1) Baseboard for a rectangular room (door opening excluded)

A room is 14 ft by 11 ft. You will install baseboard around the room, but there is a 3 ft wide doorway where no baseboard is needed. How many feet of baseboard should you buy (not counting waste)?

Answer (step-by-step):

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• Perimeter of rectangle = 2(L + W) = 2(14 + 11) = 2(25) = 50 ft
• Exclude doorway section: 50 − 3 = 47 ft
• Baseboard needed: 47 ft

2) Fencing a garden with one side already fenced

A rectangular garden is 9 m by 6 m. One 9 m side is already fenced by a wall, so you only need fencing for the other three sides. How many meters of fencing are needed?

Answer:

• Total perimeter = 2(9 + 6) = 30 m
• Subtract the wall side (9 m): 30 − 9 = 21 m
• Fencing needed: 21 m

3) Perimeter of a composite outline (simple “step” shape)

A patio outline is shaped like a step: start with a 10 ft by 6 ft rectangle, and a 4 ft by 2 ft rectangle is attached to the right side, centered vertically (so it sticks out). What is the perimeter of the combined shape?

Answer (step-by-step):

• Perimeter of 10×6 rectangle alone would be 2(10+6)=32 ft, but attachment changes the boundary.
• Think in edges: the attached 4×2 adds new outer edges but removes the shared edge.
• Shared edge length equals the height of the attachment where it touches: 2 ft.
• Added perimeter from attachment if separate: 2(4+2)=12 ft.
• When attached, the shared edge is counted twice in the separate perimeters, but becomes interior, so subtract 2×(shared edge) = 2×2 = 4 ft.
• Combined perimeter = 32 + 12 − 4 = 40 ft.
• Perimeter: 40 ft

4) Missing side from perimeter information

A rectangular banner has a perimeter of 86 in. One side is 24 in. What is the other side length?

Answer:

• Perimeter P = 2(L + W) = 86
• L + W = 43
• If L = 24, then W = 43 − 24 = 19 in
• Other side: 19 in

5) Border trim around a circular table

A round table has a diameter of 48 in. You want to add trim around the edge. How many inches of trim are needed? Use π ≈ 3.14.

Answer:

• Circumference C = πd = 3.14 × 48 = 150.72 in
• Trim needed: 150.72 in (about 151 in)

Mini-Quiz Set 2: Area Practice (Coverage and Layout)

Focus: choosing the correct area formula, subtracting cutouts, and converting “coverage” into quantities.

1) Flooring for a rectangular room

A room is 12 ft by 15 ft. What is the floor area in square feet?

Answer:

• Area = L × W = 12 × 15 = 180 ft²
• Area: 180 ft²

2) Paintable wall area (subtracting a door and a window)

A wall is 18 ft wide and 9 ft tall. There is a door 3 ft by 7 ft and a window 4 ft by 3 ft. What is the paintable area?

Answer (step-by-step):

• Total wall area = 18 × 9 = 162 ft²
• Door area = 3 × 7 = 21 ft²
• Window area = 4 × 3 = 12 ft²
• Paintable area = 162 − 21 − 12 = 129 ft²
• Paintable area: 129 ft²

3) Tile count from area (with tile size)

A kitchen floor area is 96 ft². You are using square tiles that are 12 in by 12 in. How many tiles are needed (not counting waste)?

Answer:

• 12 in = 1 ft, so each tile is 1 ft × 1 ft = 1 ft²
• Number of tiles = 96 ft² ÷ 1 ft² per tile = 96 tiles
• Tiles needed: 96

4) Area of a circular rug

A circular rug has a radius of 5 ft. What is its area? Use π ≈ 3.14.

Answer:

• Area A = πr² = 3.14 × 5² = 3.14 × 25 = 78.5 ft²
• Area: 78.5 ft²

5) Fabric panel with a cutout

A rectangular fabric panel is 2.4 m by 1.6 m. A rectangular cutout of 0.5 m by 0.4 m is removed for a pocket opening. What is the remaining fabric area?

Answer:

• Total area = 2.4 × 1.6 = 3.84 m²
• Cutout area = 0.5 × 0.4 = 0.20 m²
• Remaining area = 3.84 − 0.20 = 3.64 m²
• Remaining area: 3.64 m²

Mini-Quiz Set 3: Volume Practice (Capacity and Fill)

Focus: rectangular prisms, cylinders, and interpreting “how much fits” or “how much is needed.”

1) Storage bin volume

A storage bin is 22 in long, 16 in wide, and 14 in tall. What is its volume in cubic inches?

Answer:

• Volume V = L × W × H = 22 × 16 × 14
• 22 × 16 = 352
• 352 × 14 = 4928
• Volume: 4928 in³

2) Concrete for a slab (rectangular prism)

A small slab is 10 ft by 6 ft and 4 in thick. Find the volume in cubic feet.

Answer (step-by-step):

• Convert thickness to feet: 4 in = 4/12 ft = 1/3 ft
• Volume = 10 × 6 × (1/3) = 60 × (1/3) = 20 ft³
• Volume: 20 ft³

3) Cylindrical water tank

A cylindrical tank has radius 0.75 m and height 1.8 m. What is its volume in cubic meters? Use π ≈ 3.14.

Answer:

• Volume V = πr²h = 3.14 × (0.75²) × 1.8
• 0.75² = 0.5625
• 3.14 × 0.5625 = 1.76625
• 1.76625 × 1.8 = 3.17925
• Volume: 3.18 m³ (approx.)

4) How many small boxes fit by volume (idealized packing)

A large box has interior dimensions 60 cm by 40 cm by 30 cm. Small boxes are 10 cm by 10 cm by 10 cm. Assuming perfect packing with no wasted space, how many small boxes fit?

Answer:

• Large box volume = 60 × 40 × 30 = 72,000 cm³
• Small box volume = 10 × 10 × 10 = 1,000 cm³
• Count = 72,000 ÷ 1,000 = 72
• Small boxes: 72

5) Liquid in a rectangular aquarium

An aquarium is 80 cm long, 35 cm wide, and filled to a water height of 28 cm. What is the water volume in cubic centimeters?

Answer:

• Volume = 80 × 35 × 28
• 80 × 35 = 2800
• 2800 × 28 = 78,400
• Water volume: 78,400 cm³

Mini-Quiz Set 4: Mixed Perimeter–Area–Volume (Choose the Right Tool)

Focus: reading what the question actually asks, combining steps, and separating “edge length” from “coverage” from “capacity.”

1) Garden edging and mulch (perimeter and area)

A rectangular garden bed is 5.5 m by 3 m. You want edging around the outside and mulch covering the entire bed to a uniform depth of 6 cm. Find (a) the edging length in meters and (b) the mulch volume in cubic meters.

Answer (step-by-step):

• (a) Edging length is perimeter: P = 2(5.5 + 3) = 2(8.5) = 17 m
• (b) Mulch volume = area × depth
• Area = 5.5 × 3 = 16.5 m²
• Depth: 6 cm = 0.06 m
• Volume = 16.5 × 0.06 = 0.99 m³
• (a) 17 m edging, (b) 0.99 m³ mulch

2) Wrapping a gift box (surface coverage idea using area)

A gift box is 9 in by 6 in by 4 in. You want to cover the top only with decorative paper (not the sides). How much paper area is needed for the top?

Answer:

• Top is a rectangle: 9 × 6 = 54 in²
• Paper area for top: 54 in²

3) Picture frame: outer border length and inner opening area

A picture frame is a rectangle 16 in by 20 in on the outside. The opening (visible picture area) is 12 in by 16 in. Find (a) the outer perimeter and (b) the opening area.

Answer:

• (a) Outer perimeter = 2(16 + 20) = 72 in
• (b) Opening area = 12 × 16 = 192 in²
• (a) 72 in, (b) 192 in²

4) Circular patio: border stones and surface coverage

A circular patio has diameter 4.2 m. You want stones around the edge and sealant over the surface. Find (a) the border length and (b) the surface area. Use π ≈ 3.14.

Answer (step-by-step):

• Radius r = 4.2/2 = 2.1 m
• (a) Border length is circumference: C = πd = 3.14 × 4.2 = 13.188 m ≈ 13.19 m
• (b) Area: A = πr² = 3.14 × (2.1²)
• 2.1² = 4.41
• A = 3.14 × 4.41 = 13.8474 m² ≈ 13.85 m²
• (a) 13.19 m, (b) 13.85 m²

5) Shelf liner: roll length needed (area to length)

You have a roll of shelf liner that is 45 cm wide. You need to cover two shelves, each 90 cm long and 35 cm deep. How many centimeters of roll length do you need (ignoring waste and assuming you cut strips along the roll length)?

Answer (step-by-step):

• Each shelf area = 90 × 35 = 3150 cm²
• Two shelves area = 2 × 3150 = 6300 cm²
• Roll width = 45 cm, so length needed = area ÷ width = 6300 ÷ 45 = 140 cm
• Roll length needed: 140 cm

Mini-Quiz Set 5: Word Problems That Hide the Geometry

Focus: translating everyday language into a perimeter/area/volume plan, then executing cleanly.

1) “How much trim?” (perimeter with an interruption)

A rectangular mirror is 30 in by 22 in. You will add trim around it, but you will leave a 6 in gap at the bottom for a mounting clip. How much trim is needed?

Answer:

• Perimeter = 2(30 + 22) = 104 in
• Subtract gap: 104 − 6 = 98 in
• Trim needed: 98 in

2) “How much material?” (area with a non-covered region)

A tabletop is 1.2 m by 0.7 m. A decorative inlay in the center is a rectangle 0.4 m by 0.2 m and will not be covered. What area must be covered?

Answer:

• Total area = 1.2 × 0.7 = 0.84 m²
• Inlay area = 0.4 × 0.2 = 0.08 m²
• Covered area = 0.84 − 0.08 = 0.76 m²
• Area to cover: 0.76 m²

3) “How many bags?” (volume from area × depth)

You are filling a rectangular planter that is 120 cm by 45 cm. You want soil depth of 25 cm. What volume of soil is needed in cubic centimeters?

Answer:

• Volume = 120 × 45 × 25
• 120 × 45 = 5400
• 5400 × 25 = 135,000
• Soil volume: 135,000 cm³

4) “How much label?” (circumference from diameter)

A jar is cylindrical with diameter 7 cm. A label wraps once around the jar. What label length is needed (ignoring overlap)? Use π ≈ 3.14.

Answer:

• Length needed = circumference = πd = 3.14 × 7 = 21.98 cm
• Label length: 21.98 cm (about 22.0 cm)

5) “How much space inside?” (volume with a missing dimension)

A rectangular box has volume 2,160 cm³. Its base is 18 cm by 12 cm. What is the height?

Answer:

• Volume = base area × height
• Base area = 18 × 12 = 216 cm²
• Height = 2160 ÷ 216 = 10 cm
• Height: 10 cm

Mini-Quiz Set 6: Timed Drill (10 Quick Checks)

These are designed to be fast. Try to answer each in under 45 seconds. Answers are listed immediately after.

Problems

• 1) Rectangle perimeter: L = 8 m, W = 3.5 m.
• 2) Rectangle area: L = 9 ft, W = 4 ft.
• 3) Circle circumference: d = 10 in, π ≈ 3.14.
• 4) Circle area: r = 6 cm, π ≈ 3.14.
• 5) Box volume: 5 cm × 4 cm × 9 cm.
• 6) Paint area: wall 12 ft × 8 ft minus window 3 ft × 4 ft.
• 7) Perimeter after removing a 2 m gate from a 24 m fence loop.
• 8) Area of a triangle: base 14 m, height 5 m.
• 9) Volume: slab 7 m × 2 m × 0.15 m.
• 10) Missing side: rectangle perimeter 50 in, one side 12 in.

Answers

• 1) P = 2(8 + 3.5) = 23 m
• 2) A = 9 × 4 = 36 ft²
• 3) C = 3.14 × 10 = 31.4 in
• 4) A = 3.14 × 6² = 3.14 × 36 = 113.04 cm²
• 5) V = 5 × 4 × 9 = 180 cm³
• 6) 12×8 − 3×4 = 96 − 12 = 84 ft²
• 7) 24 − 2 = 22 m
• 8) A = (1/2)bh = 0.5 × 14 × 5 = 35 m²
• 9) V = 7 × 2 × 0.15 = 2.1 m³
• 10) L + W = 25, so other side = 25 − 12 = 13 in

Answer-Checking Templates (Reusable Step Patterns)

When you miss a problem, it usually helps to rewrite it using a consistent template. Use these patterns to diagnose where the mistake happened: formula choice, unit handling, arithmetic, or interpreting what is included/excluded.

Template A: Perimeter with an excluded section

1) Compute full perimeter of the shape (include all sides/edges on the outside boundary). 2) Identify excluded lengths (doors, gates, gaps). 3) Subtract excluded lengths from full perimeter. 4) State final answer with linear units.

Template B: Area with cutouts

1) Compute total area of the full region. 2) Compute area of each cutout/opening. 3) Subtract cutouts from total. 4) State final answer with square units.

Template C: Volume from a footprint and depth/height

1) Compute base/footprint area. 2) Convert depth/height into matching units. 3) Multiply area × depth/height. 4) State final answer with cubic units.

Template D: “How many items?” from area or volume

1) Compute total area/volume needed. 2) Compute area/volume per item. 3) Divide: total ÷ per-item. 4) If items must be whole, round up and note why.