Class 9 · MathsNCERT 2026 Revised Syllabus

Chapter 14: Math of Space: Surface Area and Volume (0904CH14)

Ganita Manjari Part II · 54 questions · 6 exercises

Chapter code: 0904ch14 · Book: Ganita Manjari Part II · Class 9 Maths

Find complete NCERT solutions for Class 9 Maths, Ganita Manjari Part II Chapter 14. Every exercise has a clear, step-by-step answer for chapter code 0904CH14.

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Exercise Set 14.1(9 questions)

The volume of a cube is 64 cm³. Find its total surface area.

School Level Solution

If a is the side, a³=64 ⇒ a=4 cm. Total surface area S=6a²=6(4²)=96 cm².

How many cubes of side 20 cm fit tightly in a cubical box of side 2 m?

School Level Solution

Convert 2 m to 200 cm. Along each edge, 200/20=10 small cubes fit. Total =10×10×10=1000 cubes.

A godown is 40 m×25 m×10 m and is filled with boxes measuring 2 m×1.25 m×1 m. Find the number of boxes.

School Level Solution

Godown volume=40×25×10=10,000 m³. One box volume=2×1.25×1=2.5 m³. Since each dimension divides the corresponding godown dimension exactly, the packed count is (40/2)(25/1.25)(10/1)=20×20×10=4000 boxes.

Two cubes, each of volume 125 cm³, are joined end to end. Find the surface area of the resulting cuboid.

School Level Solution

Each cube has side ∛125=5 cm. Joined face-to-face, the cuboid measures 10 cm×5 cm×5 cm. Its total surface area is 2(lw+lh+wh)=2(10×5+10×5+5×5)=2(50+50+25)=250 cm².

A cube of side 4 cm is cut into 1 cm cubes. Find the ratio of the original cube’s surface area to the sum of the surface areas of all the small cubes.

School Level Solution

Original surface area=6×4²=96 cm². There are 4³=64 small cubes; each has surface area 6×1²=6 cm², so their total surface area is 64×6=384 cm². Ratio=96:384=1:4. Volume is preserved, but cutting exposes new surfaces.

The three faces of a cuboid meeting at one corner have areas 6 cm², 15 cm² and 10 cm². Find its volume.

School Level Solution

Let its edges be l,w,h. The face areas give lw=6, wh=15 and lh=10. Multiplying, (lwh)²=(lw)(wh)(lh)=6×15×10=900. Since volume is positive, lwh=√900=30 cm³.

A 5 cm cube is painted on all faces and cut into 1 cm³ cubes. How many small cubes have (i) exactly three painted faces, (ii) exactly two, (iii) exactly one, (iv) no painted faces?

School Level Solution

(i) Exactly three painted faces occur only at the 8 corners: 8.(ii) On each of 12 edges, the non-corner cubes have two painted faces. There are 5−2=3 such cubes per edge, so 12×3=36.(iii) On each of 6 faces, the interior face cubes (not edges) have one painted face. Each face has (5−2)²=9, so 6×9=54.(iv) Fully internal cubes have no painted faces. Count=(5−2)³=3³=27. Check: 8+36+54+27=125=5³.

Find all cuboids with positive integer edge lengths and total surface area exactly 100 cm². Are there more than one? Show all cases.

School Level Solution

Let integer edges be a≤b≤c. From 2(ab+bc+ca)=100, we need ab+bc+ca=50. Since a≤b≤c, 3a²≤50, so a≤4.a=1: b+c+bc=50 ⇒ (b+1)(c+1)=51=3×17 (the only factor pair with b≥1 and c≥b is 3×17). Thus (b,c)=(2,16).a=2: 2b+2c+bc=50 ⇒ (b+2)(c+2)=54. With b≥2 and c≥b, the valid factor pair is 6×9, giving (b,c)=(4,7).a=3: (b+3)(c+3)=59. Since 59 is prime, no pair with b≥3 exists.a=4: (b+4)(c+4)=66, but both factors are at least 8; no factor pair of 66 has both factors at least 8. Hence no further cases. The two cuboids are 1×2×16 and 2×4×7. Their surface areas are respectively 2(2+16+32)=100 and 2(8+14+28)=100 cm².

Think and Reflect: compare school pool designs A=8 ft×50 ft×20 ft and B=6 ft×100 ft×15 ft. Which is appropriate and why? Use 1 ft³≈28.3 L.

School Level Solution

Option A: Volume = depth × length × width = 8 × 50 × 20 = 8,000 ft³. In litres, 8,000 × 28.3 = 226,400 L.Option B: Volume = 6 × 100 × 15 = 9,000 ft³. In litres, 9,000 × 28.3 = 254,700 L.Option B holds 9,000 − 8,000 = 1,000 ft³ more water, equivalent to 1,000 × 28.3 = 28,300 L. B is shallower but needs a 100 ft length; A is deeper and shorter. There is no universally correct choice: the answer should justify dimensions, site space, intended users, safety and cost.

Exercise Set 14.2(5 questions)

Cylinder B has twice the radius and half the height of cylinder A. Find CSA(A):CSA(B) and volume(A):volume(B).

School Level Solution

Let cylinder A have radius r and height h. Then B has radius 2r and height h/2.CSA(A)=2πrh. CSA(B)=2π(2r)(h/2)=2πrh, so the ratio is 1:1.V(A)=πr²h. V(B)=π(2r)²(h/2)=2πr²h, so V(A):V(B)=1:2.

The radii of two cylinders are in the ratio 2:3 and heights in the ratio 3:2. Find (a) volume ratio and (b) CSA ratio.

School Level Solution

Let r₁:r₂=2:3 and h₁:h₂=3:2.(a) Volumes are proportional to r²h, so V₁:V₂=(2²×3):(3²×2)=12:18=2:3.(b) Curved areas are proportional to rh, so CSA₁:CSA₂=(2×3):(3×2)=1:1.

A cube has edge r cm. A largest possible right circular cylinder is cut out of it. Find the cylinder’s volume.

School Level Solution

The cylinder’s height equals the cube’s edge, h=r. Its circular base is the largest circle that fits in a square face, so its diameter is r and radius is r/2. Therefore V=π(r/2)²(r)=πr³/4 cm³.

The radius of a cylinder is increased by 10% and its height is decreased by x%. Its volume remains unchanged. Find x.

School Level Solution

Let original volume be V=πr²h. New radius=1.1r and new height=(1−x/100)h. Equal volumes give 1.1²(1−x/100)=1. Thus 1−x/100=1/1.21=100/121. Hence x/100=21/121 and x=2100/121≈17.36%.

A metal cube of side 12 cm is melted and recast as cylindrical rods of radius 2 cm and height 12 cm. Find (i) cube volume, (ii) one rod volume, (iii) approximate number of complete rods using π≈22/7.

School Level Solution

(i) Cube volume=12³=1728 cm³.(ii) Rod volume=πr²h=π×2²×12=48π cm³. With π≈22/7, this is 1056/7≈150.86 cm³.(iii) Number of rods by volume=1728/(48π)=36/π≈11.46. Only whole rods can be made, so 11 complete rods can be formed.

Exercise Set 14.3(9 questions)

Find the total surface area of a cone with slant height 21 m and base diameter 24 m.

School Level Solution

Radius r=24/2=12 m and slant height l=21 m. Total surface area of a cone = πr(l+r) = π×12×(21+12)=396π m². Using π≈3.14159, 396π≈1244.07 m².

Find the curved surface area of a right circular cone with slant height 10 cm and radius 7 cm.

School Level Solution

CSA=πrl=π×7×10=70π cm². Using π=22/7, CSA=70×22/7=220 cm².

A cone has height 16 cm and base radius 12 cm. Find its curved and total surface areas.

School Level Solution

Slant height l=√(h²+r²)=√(16²+12²)=√400=20 cm.CSA=πrl=π×12×20=240π cm².TSA=πr(l+r)=π×12×(20+12)=384π cm².

A cone has height 15 cm and volume 1570 cm³. Find its base radius.

School Level Solution

V=(1/3)πr²h. Hence 1570=(1/3)πr²(15)=5πr². Thus r²=1570/(5π)=314/π. With π≈3.14, r²=100 and r=10 cm.

The curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) radius, (ii) total surface area.

School Level Solution

(i) CSA=πrl=308. With l=14 and π=22/7, 308=(22/7)×r×14=44r, so r=7 cm.(ii) Base area=πr²=(22/7)×49=154 cm². TSA=CSA+base area=308+154=462 cm².

A joker’s cap is a cone of radius 7 cm and height 24 cm. Find the sheet area needed to make 10 caps.

School Level Solution

Slant height l=√(7²+24²)=√625=25 cm. A cap is open at its base, so only curved area is needed. One cap: πrl=(22/7)×7×25=550 cm². Ten caps require 10×550=5500 cm².

A conical tent has height 8 m and base radius 6 m. Tarpaulin is 3 m wide. Find the length needed, including 20 cm extra for stitching and cutting waste.

School Level Solution

Slant height l=√(8²+6²)=√100=10 m. The conical tent has no base sheet, so tarpaulin area=CSA=πrl=π×6×10=60π m². With a 3 m wide roll, required length before allowance=(60π)/3=20π≈62.832 m. Add 0.20 m for stitching/waste: 20π+0.2≈63.03 m.

A right triangle with sides 6 cm, 8 cm and 10 cm is rotated 360° about its 8 cm side. Find the volume and curved surface area of the solid.

School Level Solution

The 8 cm side is the cone’s height, the other perpendicular side 6 cm becomes the base radius, and the hypotenuse 10 cm is the slant height. Thus V=(1/3)πr²h=(1/3)π×36×8=96π cm³. Curved area=πrl=π×6×10=60π cm².

A conical cup is filled with water to half its depth. What fraction of the cup’s volume does the water occupy?

School Level Solution

The water fills a smaller cone similar to the full cup. Its height is half the original, so every linear dimension (including radius) is scaled by 1/2. Volumes of similar solids scale as the cube of the linear scale factor: (1/2)³=1/8. Only one-eighth of the cup’s volume is occupied.

Exercise Set 14.4(8 questions)

A ball bearing has radius 0.7 cm. Find its surface area.

School Level Solution

Surface area of a sphere=4πr²=4π(0.7)²=1.96π. Using π=22/7 gives 1.96×22/7=6.16 cm².

Two solid spheres of the same metal weigh 5920 g and 740 g. Find the radius of the larger sphere if the smaller sphere has diameter 5 cm.

School Level Solution

For the same material, mass is proportional to volume. Volume ratio=5920:740=8:1. For spheres, volume ratio equals the cube of the radius ratio; thus (R/r)³=8 ⇒ R/r=2. The smaller sphere radius is 5/2=2.5 cm, so larger radius=2×2.5=5 cm.

The moon’s diameter is approximately one fourth of Earth’s diameter. Find the ratio of their surface areas.

School Level Solution

For spheres, surface area is proportional to the square of diameter. The diameter ratio (Moon:Earth)=1:4, so surface-area ratio=(1²):(4²)=1:16.

Find (i) the curved surface area and (ii) total surface area of a hemisphere of radius 21 cm.

School Level Solution

For a hemisphere, CSA=2πr² and TSA=3πr². With r=21 cm and π=22/7, CSA=2×(22/7)×441=2772 cm². TSA=3×(22/7)×441=4158 cm².

Metal spheres of radius 2 cm are packed in a box 16 cm×8 cm×8 cm. The box is then filled with preservative liquid. Find the liquid volume to the nearest integer.

School Level Solution

Box volume=16×8×8=1024 cm³. One sphere volume=(4/3)π(2³)=32π/3 cm³. Sixteen spheres occupy 16×32π/3=512π/3≈536.17 cm³. Liquid volume=1024−512π/3≈487.83 cm³, which rounds to 488 cm³.

Show that increasing the radius of a sphere by 10% increases its volume by approximately 33.1%.

School Level Solution

Let the original radius be r. The old volume is (4/3)πr³. New radius=1.1r, so new volume=(4/3)π(1.1r)³=1.331(4/3)πr³. The new volume is 133.1% of the old volume; therefore the increase is 33.1%.

The radius of a sphere is increased by x%, and its volume increases by 72.8%. Find x.

School Level Solution

The volume factor is 1+72.8/100=1.728. If the radius factor is 1+x/100, then (1+x/100)³=1.728. Since 1.2³=1.728, 1+x/100=1.2, hence x=20%.

A hemispherical dome has base circumference 35.2 m. Find the cost of painting its curved surface at ₹10 per 100 cm².

School Level Solution

Base circumference 2πr=35.2 m. Using π=22/7, r=35.2×7/44=5.6 m. Curved surface area of a hemisphere=2πr²=2×(22/7)×(5.6)²=197.12 m². Since 1 m²=10,000 cm², area=1,971,200 cm²=19,712 lots of 100 cm². Cost=19,712×₹10=₹1,97,120.

End-of-Chapter Exercises(22 questions)

Estimate how many scoops of ice cream can be obtained from a cuboidal container measuring 10 cm×15 cm×20 cm.

School Level Solution

Container volume=10×15×20=3000 cm³. Assume one average scoop is approximated by a sphere of radius 2.5 cm. Its volume is (4/3)π(2.5)³≈65.45 cm³. The theoretical quotient is 3000/65.45≈45.8, so the container contains about 45 full scoops under this assumption. Scoop size and trapped air change the real count, so the result is an estimate, not an exact universal number.

A cube of integer side length a is made of unit cubes. Give an expression for the number of unit cubes to be added to make a cube of side a+1.

School Level Solution

The existing cube contains a³ unit cubes. The larger cube contains (a+1)³. The extra cubes needed are (a+1)³−a³=a³+3a²+3a+1−a³=3a²+3a+1.

Estimate: (i) could one person drink enough water in a lifetime to fill a classroom? (ii) estimate the number of bricks used in the walls of a classroom.

School Level Solution

(i) Assume a person drinks 2 L/day for 80 years. Water consumed≈2×365×80=58,400 L=58.4 m³. A sample classroom of 8 m×7 m×3 m has volume 168 m³, nearly three times as much. Under these assumptions the answer is no. A smaller room or higher daily water amount changes the comparison.(ii) Assume the classroom is 8 m×7 m with 3 m wall height. Gross wall area=perimeter×height=2(8+7)×3=90 m². Subtract about 8 m² for doors and windows, leaving 82 m². Approximate exposed face area of one brick is 0.20×0.10=0.020 m², so estimated brick count=82/0.020=4100 bricks, before allowing for mortar and breakage. This is a transparent estimate, not a site measurement.

Compare: (A) one chocolate cube of side 50 mm; (B) 100 chocolate cubes of side 10 mm. Which gives more chocolate?

School Level Solution

Convert units: 50 mm=5 cm, so A volume=5³=125 cm³. Each cube in B has side 10 mm=1 cm and volume 1 cm³; 100 cubes have total volume 100 cm³. Option A gives 25 cm³ more chocolate.

If all people in the world were crowded together at one location, approximately how much ground area would they cover?

School Level Solution

For an estimate, take world population as about 8 billion and allocate 0.25 m² per person for very tightly packed standing people. Area=8×10⁹×0.25=2×10⁹ m². Since 1 km²=10⁶ m², this is 2000 km². This is equivalent to a square roughly √2000≈44.7 km on each side. The answer scales directly with the assumed space per person.

Milk is served in cylindrical glasses of diameter 7 cm and filled to height 12 cm. How many litres are needed for 1600 students? Use 1000 cm³=1 L.

School Level Solution

Radius=7/2=3.5 cm, height=12 cm. One glass volume=πr²h=π(3.5)²(12)=147π cm³. With π=22/7, this is 462 cm³. For 1600 students: 462×1600=739,200 cm³=739.2 L.

The surface area of a sphere of radius 5 cm is five times the curved surface area of a cone of radius 4 cm. Find the cone’s height and volume.

School Level Solution

Sphere surface area=4π(5²)=100π cm². This equals five times the cone’s CSA, so the cone CSA=20π cm². For radius 4, CSA=πrl=4πl; hence 4πl=20π ⇒ l=5 cm. Cone height h=√(l²−r²)=√(25−16)=3 cm. Volume=(1/3)πr²h=(1/3)π×16×3=16π cm³.

Take Earth radius 6370 km, Jupiter radius 69,900 km and Sun radius 695,700 km. Estimate (i) Jupiter/Earth volume ratio and (ii) Sun/Earth volume ratio.

School Level Solution

For spheres, V=(4/3)πr³, so the ratios are cubes of the radius ratios.(i) VJ/VE=(69,900/6370)³≈1321.34. Jupiter’s volume is about 1321 times Earth’s.(ii) VS/VE=(695,700/6370)³≈1,302,710. The Sun’s volume is about 1.30 million times Earth’s. The common factor (4/3)π cancels.

Show that the volume of a sphere equals 2/3 of the volume of the smallest cylinder that encloses it.

School Level Solution

Let the sphere radius be r. The smallest right circular cylinder enclosing it has radius r and height 2r. Cylinder volume=πr²(2r)=2πr³. Sphere volume=(4/3)πr³. Therefore Vsphere/Vcylinder=[(4/3)πr³]/(2πr³)=2/3, as required.

A string wraps tightly around the equator of a spherical body. A second string 1 m longer forms a concentric circle at a constant height above it. Find the height above the surface for Earth, Moon, Jupiter and a volleyball.

School Level Solution

Let the original radius be R and the added height be h. The circumference increases from 2πR to 2π(R+h), an increase of 2πh. Since the added string is 1 m longer, 2πh=1 ⇒ h=1/(2π)=0.15915 m≈15.9 cm. The original radius cancels, so the answer is the same for Earth, the Moon, Jupiter and a volleyball, assuming perfect spheres and a uniform gap.

A square pyramid is fitted inside a cylinder of radius r and height h. The square base lies on the cylinder’s base with its corners on the circular boundary; the apex lies on the top. Find volume(pyramid):volume(cylinder).

School Level Solution

The square base is inscribed in the circle of radius r. Its diagonal is the circle’s diameter 2r. If the square side is s, s√2=2r, so s²=2r². Pyramid volume=(1/3)(base area)(height)=(1/3)(2r²)h=(2/3)r²h. Cylinder volume=πr²h. Thus the ratio is [(2/3)r²h] : [πr²h] = 2/(3π):1.

Find the change in volume when: (i) the length of a cuboid l,w,h is increased by 1 unit; (ii) the radius of a cylinder r,h is increased by 1 unit; (iii) the radius of a sphere is decreased by 1 unit.

School Level Solution

(i) Old V=lwh; new V=(l+1)wh. Change=(l+1)wh−lwh=wh cubic units (option c).(ii) Old V=πr²h; new V=π(r+1)²h. Increase=πh[(r+1)²−r²]=πh(2r+1)=2πrh+πh cubic units (option e).(iii) For r≥1, new radius r−1. Decrease=(4/3)πr³−(4/3)π(r−1)³=(4/3)π[3r²−3r+1]=4πr²−4πr+4π/3 cubic units. The signed change (new minus old) is the negative of this quantity.

Given a cube of volume V, express its total surface area S in terms of V.

School Level Solution

If the cube side is a, then V=a³ ⇒ a=V^(1/3). Total surface area S=6a²=6(V^(1/3))²=6V^(2/3).

Given a cube with total surface area S, express its volume V in terms of S.

School Level Solution

For side a, S=6a², so a=√(S/6). Volume V=a³=[√(S/6)]³=(S/6)^(3/2).

A cylindrical glass has radius 4 cm and water depth 16 cm. How many marbles of radius 1 cm must be dropped in to raise the water to 20 cm?

School Level Solution

The rise is 20−16=4 cm. The additional volume needed is the cylinder’s base area times the rise: π(4²)(4)=64π cm³. One marble displaces its own volume (fully submerged): (4/3)π(1³)=4π/3 cm³. Number=64π÷(4π/3)=64×3/4=48 marbles. The final water level 20 cm is below the glass height 25 cm.

Estimate the volume of ink in a new ballpoint pen by making reasonable measurements and approximations.

School Level Solution

Use a sample measurement: assume the ink reservoir is approximately cylindrical, with internal radius 1 mm=0.1 cm and ink length 9 cm. Cylinder volume=π(0.1)²(9)=0.09π≈0.283 cm³. Allow roughly 0.02 cm³ for the small ink-filled tip, giving an estimate of about 0.30 cm³ (0.30 mL). For a particular pen, measure its actual inner radius and ink length; changing either dimension changes the result.

(i) A ball of chapati dough has radius 6 cm. Estimate how many chapatis can be made. (ii) Estimate edible flesh in half a coconut or muskmelon after taking measurements.

School Level Solution

(i) Dough volume=(4/3)π(6³)=288π≈904.8 cm³. Assume one chapati uses a dough volume equivalent to a disk of radius 12 cm and average thickness 0.2 cm: V≈π(12²)(0.2)=28.8π≈90.5 cm³. Number≈904.8/90.5≈10. Thus about 10 chapatis can be made; actual count depends on size and thickness.(ii) As an example, approximate a half-muskmelon by a hemisphere of radius 5 cm. Half-sphere volume=(2/3)π(5³)≈261.8 cm³. If about 70% is edible flesh after excluding rind/seeds, edible volume≈0.70×261.8≈183 cm³. Measure the actual fruit and substitute its radius; this is only a worked estimate.

(i) Would laying every page of this textbook side-by-side cover a classroom floor? (ii) Estimate the maximum number of textbooks that fit in an empty 15 ft×20 ft×30 ft storeroom.

School Level Solution

(i) Suppose the book has about 150 pages, each approximately 20 cm×27 cm=0.054 m². Total area≈150×0.054=8.1 m². A sample classroom floor 8 m×7 m=56 m² is much larger, so the pages would not cover it.(ii) Storeroom volume=15×20×30=9000 ft³. Since 1 ft³≈0.0283168 m³, that is about 254.85 m³. Approximate one book as 20 cm×28 cm×1.5 cm=840 cm³=0.00084 m³. Volume ratio=254.85/0.00084≈303,393, so the idealized upper bound is about 303,000 books. Real packing is lower because of gaps, orientation and access space.

Estimate the side length of a giant cube that could contain the entire human population.

School Level Solution

For an order-of-magnitude estimate, take population as 8×10⁹ and average human body volume as approximately 0.07 m³. Total body volume≈8×10⁹×0.07=5.6×10⁸ m³. If tightly packed into one cube, side s satisfies s³=5.6×10⁸, so s=∛(5.6×10⁸)≈824 m. Real packing would require extra space, making the cube larger.

Earth radius is about 6371 km and the estimated volume of surface water is 1.38 billion km³. If all this water formed a uniform spherical layer around Earth, (i) write an expression for thickness t; (ii) evaluate it.

School Level Solution

(i) Let R=6371 km. The water shell volume equals the difference of sphere volumes:(4/3)π[(R+t)³−R³]=1.38×10⁹.Solving for t gives t=[R³+3(1.38×10⁹)/(4π)]^(1/3)−R, with R=6371 km.(ii) Substitution gives t≈2.704 km, or about 2.7 km. The thin-shell approximation V≈4πR²t gives almost the same result.

Give the dimensions of a cuboid whose volume is halved when its surface area is doubled.

School Level Solution

The prompt does not state the starting cuboid, so the phrase “halved” and “doubled” needs a reference. To give a concrete example, take a unit cube as the reference: volume 1 unit³ and total surface area 6 unit². We seek a cuboid with dimensions a × a × c satisfying(a) a²c = 1/2(b) 2(a² + 2ac) = 12, so a² + 2ac = 6.From (a), c = 1/(2a²). Substitute into (b): a² + 1/a = 6, or a³ − 6a + 1 = 0. One positive root is approximately a = 2.361; then c = 1/(2a²) ≈ 0.0897.Thus one approximate example is 2.361 × 2.361 × 0.0897 units. Its volume is about 0.5 unit³ and its total surface area about 12 unit². Other dimensions may also work; the original wording is ambiguous without a specified reference object.

Project: estimate the volume of a house using reasonable dimensions and show the calculations.

School Level Solution

Divide the house into simple cuboids and add their volumes. One sample layout, with 3 m ceiling height, is:

Room	Assumed dimensions (m)	Volume
Living room	5×4×3	60 m³
Bedroom 1	4×3×3	36 m³
Bedroom 2	4×3×3	36 m³
Kitchen	3×3×3	27 m³
Bathroom	2×2×3	12 m³
Corridor	4×1.5×3	18 m³

Total=60+36+36+27+12+18=189 m³. This is an illustrative solution: replace the assumed dimensions with measurements of the actual house, and include or exclude wall thickness consistently.

In-text and Think and Reflect(1 questions)

Explain why the volume of a cuboid is area of base × height. Then write the total surface area and volume formulas for a cube of side a.

School Level Solution

If the base has length l and width w, one layer of unit-height cubes contains lw unit cubes. Stacking h such layers gives lwh cubes, so V = lwh = (lw)h = area of base × height. A cube has six square faces of area a² each, giving total surface area 6a². Its volume is a³.

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