Middle School Math Quiz: Solve Area And Volume Problems

20 questions · exam conditions

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Solve Area And Volume ProblemsQuestion 1 of 20

A rectangular box has dimensions $3\text{ in} \times 4\text{ in} \times 5\text{ in}$ . What is the total surface area of the box (all 6 faces)?

47\text{ in}^2

94\text{ in}^2

60\text{ in}^2

110\text{ in}^2

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Middle School Math Quiz

Middle School Math Quiz: Solve Area And Volume Problems

Practice Solve Area And Volume Problems in Middle School Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solve Area And Volume Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A rectangular box has dimensions $3\text{ in} \times 4\text{ in} \times 5\text{ in}$ . What is the total surface area of the box (all 6 faces)?

$47\text{ in}^2$
$94\text{ in}^2$ (correct answer)
$60\text{ in}^2$
$110\text{ in}^2$

Explanation: This problem tests solving surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Surface area: sum all face areas (rectangular prism 3×4×5 has faces: two 3×4=12, two 3×5=15, two 4×5=20, total: 2(12+15+20)=94). The box has 6 faces in 3 pairs: two 3×4 faces (area = 12 each), two 3×5 faces (area = 15 each), two 4×5 faces (area = 20 each), so total = 2(12) + 2(15) + 2(20) = 24 + 30 + 40 = 94 in². The correct surface area is 94 in². Common errors include counting only 3 faces instead of 6 (giving 47), or making arithmetic mistakes. Steps: (1) identify all 6 faces of the rectangular box, (2) calculate areas of the 3 different face types: 3×4=12, 3×5=15, 4×5=20, (3) multiply each by 2 (opposite faces), (4) sum: 2(12)+2(15)+2(20)=24+30+40=94, (5) verify units (in²). Remember that a rectangular box has 6 faces, not 3—each dimension pair creates 2 opposite faces.

Question 2

A school display is shaped like a "house": a rectangle with a triangle on top. The rectangle is $8\text{ in}$ wide and $5\text{ in}$ tall. The triangle on top has the same base as the rectangle ( $8\text{ in}$ ) and height $3\text{ in}$ . What is the total area of the display?

$64\text{ in}^2$
$52\text{ in}^2$ (correct answer)
$40\text{ in}^2$
$76\text{ in}^2$

Explanation: This problem tests solving area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). The house-shaped display has: rectangle area = 8 × 5 = 40 in², triangle area = (1/2) × 8 × 3 = 12 in², so total area = 40 + 12 = 52 in². The correct total area is 52 in². Common errors include forgetting the (1/2) in the triangle formula (using 8 × 3 = 24, giving total 64), or arithmetic mistakes in addition. Steps: (1) identify composite structure (rectangle with triangle on top), (2) calculate rectangle area (8 × 5 = 40), (3) calculate triangle area using A = (1/2)bh = (1/2) × 8 × 3 = 12, (4) add areas (40 + 12 = 52), (5) verify units (in²). The "house" shape is a common composite figure—remember the triangle on top uses the same base width as the rectangle below.

Question 3

A triangular prism has a triangular base with base $9\text{ m}$ and height $4\text{ m}$ , and the prism length is $5\text{ m}$ . What is the volume of the prism?

$180\text{ m}^3$
$90\text{ m}^3$ (correct answer)
$72\text{ m}^3$
$45\text{ m}^3$

Explanation: This problem tests solving volume problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). For the triangular prism: triangular base area = (1/2) × 9 × 4 = 18 m², volume = 18 × 5 = 90 m³. The correct volume is 90 m³. Common errors include forgetting the (1/2) in the triangle area formula (using 9 × 4 = 36, giving volume 180), or confusing the prism length with other dimensions. Steps: (1) identify the shape (triangular prism), (2) calculate triangular base area using A = (1/2)bh = (1/2) × 9 × 4 = 18 m², (3) multiply base area by prism length: V = 18 × 5 = 90 m³, (4) verify units (m³). The triangular prism volume formula is (triangular base area) × length—don't forget the (1/2) factor in the triangle area.

Question 4

An L-shaped classroom floor needs new carpet. The floor can be seen as a large rectangle $10\text{ m}\times 8\text{ m}$ with a rectangular storage cutout $4\text{ m}\times 3\text{ m}$ removed from one corner. What is the area of the floor to be carpeted?

$92\text{ m}^2$
$56\text{ m}^2$
$80\text{ m}^2$
$68\text{ m}^2$ (correct answer)

Explanation: This question tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). Surface area: sum all face areas (rectangular prism 3×4×5 has faces: two 3×4=12, two 3×5=15, two 4×5=20, total: 2(12+15+20)=94). For this L-shaped floor, decompose as large rectangle 10 m × 8 m = 80 m² minus cutout 4 m × 3 m = 12 m², resulting in 68 m²; alternatively, two rectangles: one 10 m × 5 m = 50 m² and one 6 m × 3 m = 18 m² (assuming the cutout leaves an L with those dimensions), total 68 m². Common errors include calculating the large rectangle only (80 m²), adding instead of subtracting the cutout (92 m²), or wrong decomposition like treating as single shape without adjustment. Steps: (1) identify composite structure (L-shape with cutout), (2) decompose into standard shapes (large rectangle minus small rectangle), (3) calculate each component (apply formulas: A=lw), (4) combine (subtract cutout), (5) verify units (area m²). Decomposition choice: two rectangles OR large-minus-small (both valid, should give same answer—good check).

Question 5

A poster is shaped like a $10\text{ in}\times 7\text{ in}$ rectangle with a $3\text{ in}\times 4\text{ in}$ rectangle cut out of one corner for a logo space. What is the area of the poster that remains?

$82\text{ in}^2$
$46\text{ in}^2$
$70\text{ in}^2$
$58\text{ in}^2$ (correct answer)

Explanation: This question tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). Surface area: sum all face areas (rectangular prism 3×4×5 has faces: two 3×4=12, two 3×5=15, two 4×5=20, total: 2(12+15+20)=94). For this poster, large rectangle 10 in ×7 in=70 in² minus cutout 3 in ×4 in=12 in², remaining 58 in². Common errors include adding cutout (70+12=82 in²), or wrong area (10×7=70, but cutout 3×4=12; mistake like 10×4=40). Steps: (1) identify composite structure (rectangle with cutout), (2) decompose into standard shapes (large minus small rectangle), (3) calculate each component (apply A=lw), (4) combine (subtract), (5) verify units (area in²). Decomposition choice: large-minus-small (valid).

Question 6

An L-shaped classroom floor needs new carpet. The floor can be seen as a large rectangle $10\text{ m} \times 8\text{ m}$ with a rectangular corner cut out that is $4\text{ m} \times 3\text{ m}$ . What is the area of the L-shaped floor?

$68\text{ m}^2$ (correct answer)
$92\text{ m}^2$
$56\text{ m}^2$
$80\text{ m}^2$

Explanation: Tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). For this L-shaped floor, we can use the large-minus-cutout method: total rectangle area is 10×8=80 m², cutout area is 4×3=12 m², so L-shape area is 80-12=68 m². Alternatively, decompose into two rectangles: one 10×5=50 m² and one 6×3=18 m², giving 50+18=68 m² (both methods yield the same result, confirming our answer). Common error would be just using the large rectangle area (80 m²) without subtracting the cutout, or incorrect decomposition leading to wrong dimensions. Steps: (1) identify composite structure (L-shape from rectangle with corner cutout), (2) decompose (large rectangle minus small rectangle), (3) calculate each component (10×8=80, 4×3=12), (4) combine (80-12=68), (5) verify units (area in m²). The answer 68 m² correctly accounts for the cutout, while 80 m² ignores it, 92 m² adds instead of subtracts, and 56 m² likely has calculation errors.

Question 7

A storage container is made by attaching two rectangular prisms side-by-side (no overlap). Prism 1 is $7\text{ ft}\times 3\text{ ft}\times 2\text{ ft}$ and Prism 2 is $4\text{ ft}\times 3\text{ ft}\times 2\text{ ft}$ . What is the total volume of the container?

$90\text{ ft}^3$
$30\text{ ft}^3$
$42\text{ ft}^3$
$66\text{ ft}^3$ (correct answer)

Explanation: This question tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). Surface area: sum all face areas (rectangular prism 3×4×5 has faces: two 3×4=12, two 3×5=15, two 4×5=20, total: 2(12+15+20)=94). For this container, decompose into two prisms: 7 ft ×3 ft ×2 ft=42 ft³ and 4 ft ×3 ft ×2 ft=24 ft³, total 66 ft³ (no overlap). Common errors include adding dimensions instead of volumes (7+4=11 ×3×2=66, coincidental), or wrong volume (7×3×2=42, 4×3×2=24, but sum 60 if arithmetic error). Steps: (1) identify composite structure (two attached prisms), (2) decompose into standard shapes (two prisms), (3) calculate each component (apply V=lwh), (4) combine (add volumes), (5) verify units (volume ft³). Mistakes: double-counting overlapping areas (but none here), arithmetic errors.

Question 8

A composite solid is made by stacking a smaller rectangular prism on top of a larger one. The larger prism is $9\text{ m} \times 5\text{ m} \times 2\text{ m}$ . The smaller prism is $4\text{ m} \times 5\text{ m} \times 3\text{ m}$ . What is the total volume of the composite solid?

$180\text{ m}^3$
$90\text{ m}^3$
$210\text{ m}^3$
$150\text{ m}^3$ (correct answer)

Explanation: Tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). For this composite solid with stacked prisms, calculate volumes separately: larger prism volume is 9×5×2=90 m³, smaller prism volume is 4×5×3=60 m³, so total volume is 90+60=150 m³. The prisms are stacked without overlap, so we simply add their volumes. Common error would be multiplying dimensions incorrectly or confusing which dimensions belong to which prism. Steps: (1) identify composite structure (two rectangular prisms stacked), (2) note stacking means no overlap, (3) calculate each volume (larger: 9×5×2=90, smaller: 4×5×3=60), (4) combine by addition (90+60=150), (5) verify units (volume in m³). When prisms are stacked, their volumes add directly without any subtraction.

Question 9

A right triangular prism has a triangular base with legs of 5 cm and 12 cm, and the prism has a height of 8 cm. If the prism is cut by a plane parallel to its triangular base at a height of 3 cm from the bottom, what is the volume of the smaller piece?

90 cubic centimeters (correct answer)
120 cubic centimeters
150 cubic centimeters
180 cubic centimeters

Explanation: The triangular base has area ½ × 5 × 12 = 30 sq cm. The smaller piece has the same base area but height of 3 cm. Volume = base area × height = 30 × 3 = 90 cubic cm. Choice B uses height of 4 cm (8-4 error). Choice C uses height of 5 cm (leg length confusion). Choice D uses height of 6 cm (8-2 error).

Question 10

A toy block is formed by joining two rectangular prisms. Prism 1 is $8\text{ cm} \times 3\text{ cm} \times 4\text{ cm}$ . Prism 2 is $4\text{ cm} \times 3\text{ cm} \times 4\text{ cm}$ . They are joined along a full $3\text{ cm} \times 4\text{ cm}$ face (so there is no overlap). What is the total volume of the combined block?

$96\text{ cm}^3$
$144\text{ cm}^3$ (correct answer)
$192\text{ cm}^3$
$48\text{ cm}^3$

Explanation: Tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). For this toy block made of two joined prisms, calculate volumes separately: Prism 1 volume is 8×3×4=96 cm³, Prism 2 volume is 4×3×4=48 cm³, and since they're joined without overlap, total volume is 96+48=144 cm³. The key phrase "no overlap" means we simply add the volumes without any subtraction. Common error would be somehow subtracting or miscalculating individual volumes. Steps: (1) identify composite structure (two rectangular prisms joined), (2) note "no overlap" means simple addition, (3) calculate each volume (8×3×4=96, 4×3×4=48), (4) combine by addition (96+48=144), (5) verify units (volume in cm³). When solids are joined without overlap, their volumes simply add together.

Question 11

Use the table showing the dimensions of three different rectangular rooms. Room C measures 14 feet by 8 feet. If each room needs flooring that costs 4 dollars per square foot, and Room C also needs crown molding that costs 8 dollars per linear foot around its perimeter, what is the total cost for all materials needed for Room C?

$448
$800 (correct answer)
$352
$736

Explanation: Room C measures 14 feet by 8 feet, so its floor area is 14 x 8 = 112 square feet, and its perimeter is 2 x (14 + 8) = 44 feet. The flooring costs 112 x $4 = $448, and the crown molding costs 44 x $8 = $352. Adding both material costs together gives $448 + $352 = $800 for all the materials needed, matching Choice B. Choice A only includes the flooring cost and leaves out the crown molding entirely. Choice C only includes the molding cost and leaves out the flooring. Choice D comes from using an incorrect perimeter of 36 feet instead of 44 feet.

Question 12

A science class builds a triangular prism model. The triangular base has base $6\text{ in}$ and height $4\text{ in}$ . The length of the prism is $10\text{ in}$ . What is the volume of the triangular prism?

$240\text{ in}^3$
$96\text{ in}^3$
$60\text{ in}^3$
$120\text{ in}^3$ (correct answer)

Explanation: This question tests solving volume problems for prisms by applying formulas to the base and length. For a triangular prism, decompose by finding the triangular base area A = (1/2)bh, then V = base area × length; example: triangle with b=6 in, h=4 in (area 12 in²), length 10 in, volume 120 in³. Formulas include triangle A = (1/2)bh and prism V = base area × length. For example, base (1/2)×6×4=12, times 10=120; or another prism with base 12 in² and length 10 in giving 120 in³. The correct calculation is base area (1/2)×6×4=12 in², volume 12×10=120 in³. Common errors include forgetting (1/2) for triangle (using 24 in², volume 240 in³), using wrong dimensions, or arithmetic errors (12×5=60). Steps: (1) identify the prism and base triangle, (2) calculate base area with (1/2)bh, (3) multiply by length, (4) verify units in in³. Mistakes: missing the half in triangle area, confusing area with volume units, or double-counting.

Question 13

A badge design is a rectangle with a triangle attached on top. The rectangle is $8\text{ cm}$ wide and $5\text{ cm}$ tall. The triangle has the same base as the rectangle ( $8\text{ cm}$ ) and a height of $3\text{ cm}$ . What is the total area of the badge?

$64\text{ cm}^2$
$28\text{ cm}^2$
$40\text{ cm}^2$
$52\text{ cm}^2$ (correct answer)

Explanation: This question tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). Surface area: sum all face areas (rectangular prism 3×4×5 has faces: two 3×4=12, two 3×5=15, two 4×5=20, total: 2(12+15+20)=94). For this badge, decompose into rectangle 8 cm ×5 cm=40 cm² and triangle (1/2)×8 cm×3 cm=12 cm², total 52 cm². Common errors include missing (1/2) for triangle (using 24 cm², total 64 cm²), or wrong base (using height as base). Steps: (1) identify composite structure (rectangle with triangle top), (2) decompose into standard shapes (rectangle and triangle), (3) calculate each component (apply A=lw, A=(1/2)bh), (4) combine (add areas), (5) verify units (area cm²). Mistakes: forgetting factors (1/2 in formula), wrong decomposition.

Question 14

A garden is shaped like an L. You can decompose it into two rectangles: one rectangle is $9\text{ m}\times 4\text{ m}$ and the other is $5\text{ m}\times 3\text{ m}$ . What is the total area of the garden?

$36\text{ m}^2$
$51\text{ m}^2$ (correct answer)
$45\text{ m}^2$
$27\text{ m}^2$

Explanation: This question tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). Surface area: sum all face areas (rectangular prism 3×4×5 has faces: two 3×4=12, two 3×5=15, two 4×5=20, total: 2(12+15+20)=94). For this L-shaped garden, decompose into two rectangles: 9 m ×4 m=36 m² and 5 m ×3 m=15 m², total 51 m². Common errors include multiplying dimensions wrong (9×4=36, but 5×3=15, sum 51; mistake like 9×5=45 total), or assuming overlap and subtracting. Steps: (1) identify composite structure (L-shape), (2) decompose into standard shapes (two rectangles), (3) calculate each component (apply A=lw), (4) combine (add areas), (5) verify units (area m²). Decomposition choice: two rectangles (valid, no overlap assumed).

Question 15

A shipping box is a rectangular prism with dimensions $7\text{ in} \times 5\text{ in} \times 3\text{ in}$ . What is the total surface area of the box?

$105\text{ in}^2$
$71\text{ in}^2$
$142\text{ in}^2$ (correct answer)
$210\text{ in}^2$

10\times5=50

and

6\times3=18

, sum:

68

; or as large minus cutout:

10\times8=80

minus

4\times3=12

, difference:

68

, equivalent). Surface area: sum all face areas (rectangular prism 3×4×5 has faces: two

3\times4=12

, two

3\times5=15

, two

4\times5=20

, total:

2(12+15+20)=94

). For this 7×5×3 box, calculate areas of three pairs of faces: two

7\times5=35

in² faces, two

7\times3=21

in² faces, two

5\times3=15

in² faces, giving total surface area

2(35+21+15)=2(71)=142

in². Common error would be calculating volume (

7\times5\times3=105

) instead of surface area, which appears as option B. Steps: (1) identify all 6 faces of rectangular prism, (2) group into 3 pairs of identical opposite faces, (3) calculate area of each type (

7\times5=35

7\times3=21

5\times3=15

), (4) sum with factor of 2:

2(35+21+15)=142

, (5) verify units (surface area in in²). Surface area calculation requires identifying and summing all face areas, not just multiplying dimensions.

Question 16

A storage container is made by stacking a rectangular prism and a rectangular pyramid on top. The prism has dimensions $5\text{ ft} \times 4\text{ ft} \times 6\text{ ft}$ . The pyramid on top has the same base $5\text{ ft} \times 4\text{ ft}$ and height $3\text{ ft}$ . What is the total volume of the container?

$180\text{ ft}^3$
$120\text{ ft}^3$
$140\text{ ft}^3$ (correct answer)
$160\text{ ft}^3$

Explanation: Tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). For this composite container, calculate prism volume: V=5×4×6=120 ft³, then pyramid volume: V=(1/3)×(5×4)×3=(1/3)×20×3=20 ft³, so total volume is 120+20=140 ft³. Common error would be forgetting the (1/3) factor for pyramid volume, giving 120+60=180 ft³, or arithmetic mistakes. Steps: (1) identify composite structure (rectangular prism with pyramid on top), (2) decompose into standard shapes (prism and pyramid), (3) calculate each component (prism: 5×4×6=120, pyramid: (1/3)×5×4×3=20), (4) combine (120+20=140), (5) verify units (volume in ft³). The pyramid formula V=(1/3)Bh is crucial—without the (1/3), you'd get 180 ft³ instead of the correct 140 ft³.

Question 17

A composite solid is made of a rectangular prism and a rectangular pyramid on top.

Rectangular prism: $4\text{ ft} \times 3\text{ ft} \times 5\text{ ft}$
Rectangular pyramid on top: base $4\text{ ft} \times 3\text{ ft}$ and height $6\text{ ft}$

What is the total volume of the solid?

$84\text{ ft}^3$ (correct answer)
$96\text{ ft}^3$
$132\text{ ft}^3$
$60\text{ ft}^3$

Explanation: This problem tests solving volume problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). The composite solid has: rectangular prism volume = 4 × 3 × 5 = 60 ft³, pyramid base area B = 4 × 3 = 12 ft², pyramid volume = (1/3) × 12 × 6 = 24 ft³, so total = 60 + 24 = 84 ft³. The correct total volume is 84 ft³. Common errors include forgetting the (1/3) factor for the pyramid (calculating 12 × 6 = 72 instead of 24, giving total 132), or arithmetic mistakes. Steps: (1) identify composite structure (prism plus pyramid), (2) calculate prism volume (4 × 3 × 5 = 60), (3) calculate pyramid volume using V = (1/3)Bh where B = 4 × 3 = 12 and h = 6, giving (1/3) × 12 × 6 = 24, (4) add volumes (60 + 24 = 84), (5) verify units (ft³). Remember the pyramid volume formula requires the (1/3) factor.

Question 18

A storage container is a composite solid: a rectangular prism with dimensions $5\text{ cm} \times 4\text{ cm} \times 6\text{ cm}$ and a rectangular pyramid on top that has the same $5\text{ cm} \times 4\text{ cm}$ base and height $3\text{ cm}$ . What is the total volume of the container?

$140\text{ cm}^3$ (correct answer)
$180\text{ cm}^3$
$120\text{ cm}^3$
$130\text{ cm}^3$

Explanation: This problem tests solving volume problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). The container has a rectangular prism base: V₁ = 5 × 4 × 6 = 120 cm³, and a pyramid on top with base area B = 5 × 4 = 20 cm² and height 3 cm: V₂ = (1/3) × 20 × 3 = 20 cm³, so total volume = 120 + 20 = 140 cm³. The correct calculation gives 140 cm³. Common errors include forgetting the (1/3) factor for pyramid volume (calculating 20 × 3 = 60 instead of 20), or arithmetic mistakes like 120 + 20 = 130. Steps: (1) identify composite structure (prism plus pyramid), (2) calculate prism volume (5 × 4 × 6 = 120), (3) calculate pyramid volume using V = (1/3)Bh where B = 5 × 4 = 20 and h = 3, giving (1/3) × 20 × 3 = 20, (4) add volumes (120 + 20 = 140), (5) verify units (cm³). The pyramid formula V = (1/3)Bh is crucial—missing the (1/3) is a common mistake.

Question 19

An L-shaped classroom floor needs new carpet. The floor can be seen as a $10\text{ m} \times 8\text{ m}$ rectangle with a $4\text{ m} \times 3\text{ m}$ rectangular storage area cut out of one corner. What is the area of the carpeted floor?

$80\text{ m}^2$
$68\text{ m}^2$ (correct answer)
$92\text{ m}^2$
$56\text{ m}^2$

Explanation: This problem tests solving area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). The L-shaped floor is a 10m × 8m rectangle with a 4m × 3m cutout, so we calculate: large rectangle area = 10 × 8 = 80 m², cutout area = 4 × 3 = 12 m², carpeted area = 80 - 12 = 68 m². The correct answer is 68 m². Common errors include calculating only the large rectangle (80 m²) without subtracting the cutout, or making arithmetic mistakes like 80 - 12 = 56. Steps: (1) identify composite structure (rectangle with rectangular cutout), (2) calculate large rectangle area (10 × 8 = 80), (3) calculate cutout area (4 × 3 = 12), (4) subtract cutout from large rectangle (80 - 12 = 68), (5) verify units (m²). Alternative method: decompose into two rectangles that form the L-shape, which should give the same answer as a check.

Question 20

A tent stake is shaped like a triangular prism. The triangular base has base $6\text{ cm}$ and height $4\text{ cm}$ , and the prism length is $10\text{ cm}$ . What is the volume of the tent stake?

$120\text{ cm}^3$ (correct answer)
$240\text{ cm}^3$
$60\text{ cm}^3$
$100\text{ cm}^3$

Explanation: This question tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). Surface area: sum all face areas (rectangular prism 3×4×5 has faces: two 3×4=12, two 3×5=15, two 4×5=20, total: 2(12+15+20)=94). For this triangular prism, base area (1/2)×6 cm×4 cm=12 cm², then volume 12 cm² ×10 cm=120 cm³. Common errors include missing (1/2) for triangle (using 24 cm² ×10=240 cm³), or treating as rectangular prism (6×4×10=240 cm³). Steps: (1) identify composite structure (triangular prism), (2) decompose into base triangle and length, (3) calculate each component (apply formulas: A=(1/2)bh, V=base area × length), (4) combine (multiply), (5) verify units (volume cm³). Mistakes: forgetting factors (1/2 in formula), arithmetic errors, units wrong.