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Mensuration for CUET General Aptitude

About 15 min to learn · 68 questions

CUET mensuration uses the Class 8–10 formulas for lengths, areas and volumes: rectangles and triangles, circles and sectors, cuboids, cylinders, cones and spheres, and solids that are melted down or joined together. The formulas are few. What costs marks is a diameter used as a radius, a slant height used as a height, or centimetres mixed with metres. This lesson gives every formula in a table, with worked examples and the traps beside them.

Syllabus area
Reasoning
On this page · 10 sections
  1. 01The core idea
  2. 02Plane figures
  3. 03Circles, rings and sectors
  4. 04Cubes and cuboids
  5. 05Cylinders, cones and spheres
  6. 06Melting, recasting and combined solids
  7. 07Exam shortcuts
  8. 08Common mistakes
  9. 09Quick revision
  10. 10Check your understanding

The core idea

Every mensuration question asks for one of three things: a length (m), an area (m²) or a volume (m³). Decide which one before you pick a formula.

Three habits save most marks. Write the unit next to every number. Convert everything to one unit before you multiply. And when a question says to take π=227\pi = \tfrac{22}{7}, expect the dimensions to be multiples of 7 so that the 7 cancels. If your working leaves an ugly decimal, check the formula before you check the arithmetic.

Plane figures

What are the perimeter and area formulas for rectangles, squares and triangles?

Perimeter is the length of the boundary, and area is the space inside. For a rectangle of length ll and breadth bb, the perimeter is 2(l+b)2(l + b) and the area is lblb. For a triangle, the area is half the base times the height, where the height is perpendicular to the base.

FigurePerimeterArea
Rectangle, ll by bb2(l+b)2(l + b)lblb
Square, side aa4a4aa2a^2
TriangleSum of the three sides12×base×height\tfrac{1}{2} \times \text{base} \times \text{height}
Right triangle, legs pp and qqp+q+hypotenusep + q + \text{hypotenuse}12pq\tfrac{1}{2}pq
Equilateral triangle, side aa3a3a34a2\tfrac{\sqrt{3}}{4}a^2

A rectangle's diagonal is l2+b2\sqrt{l^2 + b^2}.

Example 1 · Perimeter and area

A badminton court is 18 m long and 11 m wide. Find its perimeter and area.

  1. Perimeter: 2×(18+11)=582 \times (18 + 11) = 58 m.

  2. Area: 18×11=19818 \times 11 = 198 m².

Answer

58 m and 198 m²

Exam shortcut

Fencing, a border or a walk all round needs the perimeter. Carpeting, tiling or painting a floor needs the area. A cost question adds one step: cost = quantity × rate.

How do you find the area of a triangle when only its three sides are given?

Use Heron's formula. Find the semi-perimeter ss, half the sum of the sides, then take the square root of s(s−a)(s−b)(s−c)s(s - a)(s - b)(s - c).

Heron's formula

Area=s(s−a)(s−b)(s−c),s=a+b+c2\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}, \qquad s = \frac{a + b + c}{2}

Example 2 · Heron's formula

Find the area of a triangle with sides 9 cm, 10 cm and 17 cm.

  1. The semi-perimeter is s=(9+10+17)÷2=18s = (9 + 10 + 17) \div 2 = 18.

  2. Then s−as - a, s−bs - b and s−cs - c are 9, 8 and 1.

  3. Area: 18×9×8×1=1296=36\sqrt{18 \times 9 \times 8 \times 1} = \sqrt{1296} = 36.

Answer

36 cm²

Watch out

Check first whether the triangle is right-angled, as with 5, 12, 13 or 8, 15, 17. If it is, the area is simply half the product of the two shorter sides. If it is not, you cannot multiply two sides and halve.

Exam shortcut

For an isosceles triangle, drop the height to the unequal side. It cuts that side in half, and Pythagoras gives the height.

What are the area formulas for a parallelogram, a rhombus and a trapezium?

A parallelogram's area is base × height. A rhombus's area is half the product of its diagonals. A trapezium's area is half the sum of the parallel sides times the distance between them.

FigureAreaUseful fact
Parallelogrambase×height\text{base} \times \text{height}Either pair of sides can be the base, each with its own height
Rhombus, diagonals d1d_1 and d2d_212d1d2\tfrac{1}{2}d_1d_2Side =(d12)2+(d22)2= \sqrt{(\tfrac{d_1}{2})^2 + (\tfrac{d_2}{2})^2}
Trapezium, parallel sides aa and bb, height hh12(a+b)h\tfrac{1}{2}(a + b)hThe height is never a slanted side

Key idea

A parallelogram has one area but two heights. Base × height gives the same area with either pair of sides, so the longer sides are closer together.

How do you find the area of a path around or inside a rectangle?

Subtract the smaller area from the bigger one. A path of width ww outside a rectangle adds 2w2w to each dimension, not ww, because it runs along both sides. A path inside takes 2w2w off each dimension.

Example 3 · Path around a garden

A garden 10 m by 8 m has a path 1 m wide all round it on the outside. Find the area of the path.

  1. The outer rectangle is (10+2)(10 + 2) by (8+2)(8 + 2), so 12 m by 10 m, with area 120 m².

  2. The garden's area is 10×8=8010 \times 8 = 80 m².

  3. The path is the difference: 120−80=40120 - 80 = 40 m².

Answer

40 m²

Watch out

For two roads crossing in the middle of a park, add the two road areas, then subtract the square where they cross once. It was counted in both roads.

Circles, rings and sectors

How do you find the circumference and area of a circle, a sector and an arc?

For a circle of radius rr, the circumference is 2πr2\pi r and the area is πr2\pi r^2. A sector with angle θ\theta at the centre is the fraction θ360∘\tfrac{\theta}{360^\circ} of the circle, so its arc and its area are that fraction of the circumference and the area.

ShapeBoundaryArea
Circle2πr2\pi rπr2\pi r^2
Semicircleπr+2r\pi r + 2r12πr2\tfrac{1}{2}\pi r^2
Quadrant12πr+2r\tfrac{1}{2}\pi r + 2r14πr2\tfrac{1}{4}\pi r^2
Sector, angle θ\thetaarc+2r\text{arc} + 2rθ360∘×πr2\tfrac{\theta}{360^\circ} \times \pi r^2, or 12×arc×r\tfrac{1}{2} \times \text{arc} \times r
Ring, radii RR and rr2πR+2πr2\pi R + 2\pi rπ(R2−r2)\pi(R^2 - r^2)
Minor segmentarc+chord\text{arc} + \text{chord}Sector − triangle

Example 4 · Sector

A sector has radius 28 cm and an angle of 135° at the centre. Find its arc length and area. Take π=227\pi = \tfrac{22}{7}.

  1. For the full circle: circumference =2×227×28=176= 2 \times \tfrac{22}{7} \times 28 = 176 cm, and area =227×28×28=2464= \tfrac{22}{7} \times 28 \times 28 = 2464 cm².

  2. The sector is 135360=38\tfrac{135}{360} = \tfrac{3}{8} of the circle.

  3. Arc: 38×176=66\tfrac{3}{8} \times 176 = 66 cm. Area: 38×2464=924\tfrac{3}{8} \times 2464 = 924 cm².

Answer

Arc 66 cm, area 924 cm²

With π=227\pi = \tfrac{22}{7}, these values come up again and again. Knowing them means a circumference of 132 tells you r=21r = 21 at once.

RadiusCircumferenceArea
3.52238.5
744154
1488616
211321,386
281762,464
352203,850

Wheels, clock hands and tethered animals

  • One turn of a wheel covers one circumference, so turns = distance ÷ circumference, both in the same unit.
  • A minute hand turns 6° a minute, so in tt minutes it sweeps t60\tfrac{t}{60} of a circle.
  • An animal tied at the corner of a rectangular field grazes a quadrant. Tied halfway along a straight fence, it grazes a semicircle.

Watch out

When a wire is bent into a new shape, its length stays the same. The perimeter is kept, not the area. And for a ring, R2−r2R^2 - r^2 is not (R−r)2(R - r)^2: square each radius first.

Cubes and cuboids

What are the surface area and volume formulas for a cuboid and a cube?

For a cuboid with length ll, breadth bb and height hh, the volume is lbhlbh and the total surface area is 2(lb+bh+hl)2(lb + bh + hl). For a cube of edge aa, the volume is a3a^3 and the total surface area is 6a26a^2.

SolidVolumeTotal surfaceFour walls onlyDiagonal
Cuboidlbhlbh2(lb+bh+hl)2(lb + bh + hl)2h(l+b)2h(l + b)l2+b2+h2\sqrt{l^2 + b^2 + h^2}
Cubea3a^36a26a^24a24a^2a3a\sqrt{3}

Example 5 · Cuboid

A cuboid is 11 cm by 7 cm by 5 cm. Find its volume, its total surface area, and the area of its four side faces.

  1. Volume: 11×7×5=38511 \times 7 \times 5 = 385 cm³.

  2. Total surface: 2×(77+35+55)=2×167=3342 \times (77 + 35 + 55) = 2 \times 167 = 334 cm².

  3. Four side faces: 2×5×(11+7)=1802 \times 5 \times (11 + 7) = 180 cm².

Answer

385 cm³, 334 cm² and 180 cm²

Unit links you must know

  • 1 m³ = 1,000 litres, and 1 litre = 1,000 cm³.
  • 1 m = 100 cm, so 1 m² = 10,000 cm² and 1 m³ = 10,00,000 cm³.
  • Rain of depth dd on an area AA gives a volume A×dA \times d. Convert the depth to metres first.

Watch out

When packing small boxes into a big one, count how many fit along each edge, round each count down, and multiply. Dividing the volumes gives too many whenever an edge does not divide exactly.

Common variation: three face areas

If the three faces that meet at one corner have areas xx, yy and zz, then xyz=lb×bh×hl=(lbh)2xyz = lb \times bh \times hl = (lbh)^2. So the volume is xyz\sqrt{xyz}. For faces of 48, 60 and 80 cm², the volume is 230400=480\sqrt{230400} = 480 cm³.

Cylinders, cones and spheres

What are the formulas for a cylinder, a cone and a sphere?

A cylinder's volume is πr2h\pi r^2 h. A cone on the same base and height holds exactly one-third of that. A sphere's volume is 43πr3\tfrac{4}{3}\pi r^3 and its surface area is 4πr24\pi r^2. A cone's curved surface uses its slant height l=r2+h2l = \sqrt{r^2 + h^2}.

A cone with its vertical height h drawn as a dashed line from the tip to the centre of the base, the radius r along the base, and the slant height l along the sloping side.
The volume uses the vertical height h. The curved surface uses the slant height l, where l² = r² + h².
SolidCurved surfaceTotal surfaceVolume
Cylinder2πrh2\pi rh2πr(r+h)2\pi r(r + h)πr2h\pi r^2 h
Hollow cylinder, radii RR and rr2πh(R+r)2\pi h(R + r)Add 2π(R2−r2)2\pi(R^2 - r^2) for the rimsπh(R2−r2)\pi h(R^2 - r^2)
Coneπrl\pi rlπr(l+r)\pi r(l + r)13πr2h\tfrac{1}{3}\pi r^2 h
Sphere4πr24\pi r^24πr24\pi r^243πr3\tfrac{4}{3}\pi r^3
Solid hemisphere2πr22\pi r^23πr23\pi r^223πr3\tfrac{2}{3}\pi r^3

Example 6 · Cone

A cone has radius 14 cm and height 48 cm. Find its curved surface area and its volume. Take π=227\pi = \tfrac{22}{7}.

  1. Slant height: l=142+482=2500=50l = \sqrt{14^2 + 48^2} = \sqrt{2500} = 50 cm.

  2. Curved surface: 227×14×50=2200\tfrac{22}{7} \times 14 \times 50 = 2200 cm².

  3. Volume: 13×227×14×14×48=9856\tfrac{1}{3} \times \tfrac{22}{7} \times 14 \times 14 \times 48 = 9856 cm³.

Answer

Curved surface 2,200 cm², volume 9,856 cm³

Exam shortcut

Slant heights often come from triplets: (7, 24, 25), (8, 15, 17), (20, 21, 29) and (12, 35, 37). Spotting one saves a square root.

How do areas and volumes change when a solid is scaled?

When every length is multiplied by kk, every area is multiplied by k2k^2 and every volume by k3k^3. So two spheres with surface areas in the ratio 4:254 : 25 have radii in the ratio 2:52 : 5 and volumes in the ratio 8:1258 : 125.

Watch out

Take the square root of an area ratio before you cube it. Cubing 4:254 : 25 directly, or squaring it, gives the wrong ratio, and both wrong answers are usually among the options.

Melting, recasting and combined solids

How do you solve melting, recasting and water-displacement questions?

Recasting keeps the volume the same. Set the volume of the old solid equal to the total volume of the new pieces, then solve. Surface area is not kept. When a solid sinks in water, the water rises by exactly the solid's volume.

Example 7 · Recasting

A solid cylinder of radius 4 cm and height 9 cm is melted and recast into cones of radius 2 cm and height 3 cm. How many cones are made?

  1. Volume of the cylinder: π×42×9=144π\pi \times 4^2 \times 9 = 144\pi cm³.

  2. Volume of one cone: 13×π×22×3=4π\tfrac{1}{3} \times \pi \times 2^2 \times 3 = 4\pi cm³.

  3. Number of cones: 144π÷4π=36144\pi \div 4\pi = 36.

Answer

36 cones

The π\pi cancels, so no value of π\pi is needed.

Set-upWhat to write
A solid into many small piecesNumber = big volume ÷ one small volume
A solid drawn into a wireWire length = volume ÷ (π×\pi \times wire radius²)
Earth from a well spread over a plotRise = volume ÷ area it is spread over
Water flowing through a pipe or channelVolume per hour = cross-section area × speed
Balls dropped into a cylinder of waterTotal ball volume = πR2×\pi R^2 \times rise in level

How do you find the surface area of a combined solid?

Add only the surfaces you can see from outside. Where two solids are joined, the joining faces are hidden, so leave them out. Where a hollow is cut, remove the face that was cut away and add the new inner surface.

  • Tent, a cylinder with a cone on top: the cylinder's curved surface plus the cone's curved surface. There is no floor unless the question says so.
  • Capsule, a cylinder with a hemisphere at each end: the cylinder's curved surface plus one full sphere, 4πr24\pi r^2.
  • Toy, a cone on a hemisphere: the cone's curved surface plus 2πr22\pi r^2.
  • Cylinder with a cone hollowed out: the outer wall, the remaining flat base, and the cone's curved surface inside.

Key idea

Volume is simpler than surface area for a combined solid: add the parts, or subtract the part removed.

Exam shortcuts

  • Let the 7 cancel. With π=227\pi = \tfrac{22}{7}, multiply by 22 only after cancelling a 7 from a radius.
  • Leave π\pi as π\pi in recasting. It cancels from both sides.
  • Learn the circle values. Circumference 44, 88, 132, 176 means radius 7, 14, 21, 28.
  • Scale with powers. Lengths by kk, areas by k2k^2, volumes by k3k^3.
  • Three faces at one corner. If their areas are xx, yy and zz, the volume is xyz\sqrt{xyz}.

Common mistakes

  • Diameter used as radius. It makes an area four times too big and a volume eight times too big.
  • Height or slant height? A cone's curved surface uses ll, and its volume uses hh. A trapezium or parallelogram needs the perpendicular height, never a slanted side.
  • Forgetting the 13\tfrac{1}{3} or the 12\tfrac{1}{2}. A cone is a third of a cylinder, a triangle half a rectangle, and a rhombus half of d1×d2d_1 \times d_2.
  • Mixed units. 1 m² is 10,000 cm², not 100.
  • Adding the path width once. A path on all sides changes each dimension by 2w2w.
  • Counting an overlap twice. Crossing roads share a square.
  • Total or curved surface? Open, without a lid, label, four walls and no floor all mean leaving faces out.
  • Counting hidden faces. Joined faces do not show, and a capsule's ends are hemispheres, not flat circles.
  • Keeping the area when a wire is reshaped. The length is kept.
  • Dividing volumes when boxes do not pack. Count whole boxes along each edge.

Quick revision

Plane figures

  • Rectangle: 2(l+b)2(l + b) and lblb. Triangle: 12×base×height\tfrac{1}{2} \times \text{base} \times \text{height}
  • Heron: s(s−a)(s−b)(s−c)\sqrt{s(s - a)(s - b)(s - c)}
  • Rhombus: 12d1d2\tfrac{1}{2}d_1d_2. Trapezium: 12(a+b)h\tfrac{1}{2}(a + b)h
  • Path of width ww: each dimension changes by 2w2w

Circles

  • 2πr2\pi r and πr2\pi r^2
  • Sector: θ360∘\tfrac{\theta}{360^\circ} of the circle, or 12×arc×r\tfrac{1}{2} \times \text{arc} \times r
  • Ring: π(R2−r2)\pi(R^2 - r^2)
  • Wheel turns = distance ÷ circumference

Cuboids and cubes

  • Cuboid: lbhlbh, 2(lb+bh+hl)2(lb + bh + hl), diagonal l2+b2+h2\sqrt{l^2 + b^2 + h^2}
  • Cube: a3a^3, 6a26a^2, diagonal a3a\sqrt{3}
  • 1 m³ = 1,000 litres

Round solids

  • Cylinder: πr2h\pi r^2 h, curved surface 2πrh2\pi rh
  • Cone: 13πr2h\tfrac{1}{3}\pi r^2 h, curved surface πrl\pi rl, with l2=r2+h2l^2 = r^2 + h^2
  • Sphere: 43πr3\tfrac{4}{3}\pi r^3, surface 4πr24\pi r^2. Hemisphere: 23πr3\tfrac{2}{3}\pi r^3, total surface 3πr23\pi r^2

Recasting and combined solids

  • Recasting keeps the volume, not the surface area
  • Combined solid: add only the visible surfaces
  • Scaling: areas by k2k^2, volumes by k3k^3

Right triangles and circle facts come from lines, triangles and circles, and scale factors from ratio and proportion.

Next · 09Quick revision

Check your understanding

68 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

Perimeter and areaQ1 of 68Q1

Perimeter and area

13 questions

Question 1

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Harpreet wants to fence her rectangular kitchen garden, which is 24 m long and 15 m wide. Fencing costs ₹35 per metre. What will the fencing cost?

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AnswerD · ₹2,730

  1. Fencing goes round the boundary, so find the perimeter: 2×(24+15)=782 \times (24 + 15) = 78 m.

  2. Cost = 78 × ₹35 = ₹2,730.

Worth noting

₹1,365 uses only the half-perimeter 24+1524 + 15. ₹1,890 and ₹2,205 fence only three sides of the garden.

Question 2

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The boundary of a square carrom board measures 52 cm in all. What is the area of the board?

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AnswerA · 169 cm²

  1. Side = perimeter ÷ 4=52÷4=134 = 52 \div 4 = 13 cm.

  2. Area = 132=16913^2 = 169 cm².

Worth noting

676 cm² comes from dividing the perimeter by 2 instead of 4 (side 26 cm), and 2,704 cm² squares the perimeter itself.

Question 3

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A rhombus-shaped kite has diagonals of 16 cm and 30 cm. What is its area?

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AnswerB · 240 cm²

  1. Area of a rhombus = 12×d1×d2\frac{1}{2} \times d_1 \times d_2.

  2. Area = 12×16×30=240\frac{1}{2} \times 16 \times 30 = 240 cm².

Worth noting

480 cm² forgets the half. 255 cm² multiplies the side (17 cm) by half a diagonal, which is not a formula for anything.

Question 4

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The floor of a storeroom is a rectangle 6 m long and 4.5 m wide. How many square tiles of side 30 cm are needed to cover it exactly?

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AnswerB · 300

  1. Work in centimetres: the floor is 600 cm × 450 cm = 2,70,000 cm².

  2. One tile covers 30×30=90030 \times 30 = 900 cm².

  3. Number of tiles = 2,70,000÷900=3002{,}70{,}000 \div 900 = 300. (Check: 20 tiles along the length and 15 along the width, 20×15=30020 \times 15 = 300.)

Worth noting

90 divides the area 27 m² by 0.3 instead of 0.32=0.090.3^2 = 0.09. Always convert before you divide areas.

Question 5

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A rectangular hall floor measures 32 m by 18 m. A square courtyard has exactly the same area. By how much does the perimeter of the hall floor exceed the perimeter of the courtyard?

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AnswerC · 4 m

  1. Area of the hall = 32×18=57632 \times 18 = 576 m², so the courtyard's side = 576=24\sqrt{576} = 24 m.

  2. Perimeters: hall 2×(32+18)=1002 \times (32 + 18) = 100 m, courtyard 4×24=964 \times 24 = 96 m.

  3. Difference = 100−96=4100 - 96 = 4 m.

Worth noting

Equal areas do not mean equal perimeters, so 0 m is wrong. For a fixed area, the square has the smallest perimeter of all rectangles.

Question 6

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A rectangular lawn 40 m long and 24 m wide has a gravel path 2.5 m wide running all round it on the outside. What is the area of the path?

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AnswerD · 345 m²

  1. The outer rectangle measures (40+2×2.5)(40 + 2 \times 2.5) by (24+2×2.5)=45(24 + 2 \times 2.5) = 45 m by 29 m, area 1,305 m².

  2. The lawn is 40×24=96040 \times 24 = 960 m².

  3. Path = 1,305−960=3451{,}305 - 960 = 345 m².

Worth noting

The width is added on BOTH sides, so each dimension grows by 5 m, not 2.5 m (that gives 166.25 m²). 295 m² is a path inside the lawn, and 320 m² (perimeter × width) misses the four corner squares.

Question 7

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A park is a rectangle 60 m by 40 m. Two concrete roads, each 3 m wide, cross at the centre: one runs parallel to the length and the other parallel to the breadth. What is the total area of the roads?

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AnswerD · 291 m²

  1. Road along the length: 60×3=18060 \times 3 = 180 m². Road along the breadth: 40×3=12040 \times 3 = 120 m².

  2. The 3 m×3m \times 3 m square where they cross has been counted twice, so subtract it once: 180+120−9=291180 + 120 - 9 = 291 m².

Worth noting

282 m² subtracts the crossing square twice; it must be taken away only once. 180 m² and 120 m² are single roads.

Question 8

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A triangular plot has sides 17 m, 25 m and 28 m. What is its area?

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AnswerA · 210 m²

  1. Use Heron's formula. Semi-perimeter s=(17+25+28)÷2=35s = (17 + 25 + 28) \div 2 = 35 m.

  2. Area = 35×18×10×7=44,100=210\sqrt{35 \times 18 \times 10 \times 7} = \sqrt{44{,}100} = 210 m².

Worth noting

The plot is not right-angled (172+252≠282)(17^2 + 25^2 \ne 28^2), so ½ × one side × another is wrong: that gives 212.5, 238 or 350 m², depending on which two sides are used.

Question 9

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A parallelogram has sides 20 cm and 12 cm. The distance between the two longer sides is 9 cm. What is the distance between the two shorter sides?

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AnswerD · 15 cm

  1. Area = longer side × distance between the longer sides = 20×9=18020 \times 9 = 180 cm².

  2. The same area = shorter side × distance between the shorter sides, so the distance = 180÷12=15180 \div 12 = 15 cm.

Worth noting

5.4 cm comes from pairing the base with the wrong height (12×9÷20)(12 \times 9 \div 20). The shorter sides are further apart, because the same area is spread over a shorter base.

Question 10

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The two equal sides of an isosceles triangle are 25 cm each and the third side is 14 cm. What is its area?

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AnswerA · 168 cm²

  1. The height to the 14 cm side bisects it into 7 cm and 7 cm.

  2. Height = 252−72=576=24\sqrt{25^2 - 7^2} = \sqrt{576} = 24 cm.

  3. Area = 12×14×24=168\frac{1}{2} \times 14 \times 24 = 168 cm².

Worth noting

175 cm² uses the slant side 25 cm as the height, and 350 cm² also forgets the half. The height must be perpendicular to the base.

Question 11

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A rectangular photo frame has a perimeter of 46 cm and a diagonal of 17 cm. What is its area?

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AnswerB · 120 cm²

  1. Let the sides be l and b. Then l+b=46÷2=23l + b = 46 \div 2 = 23 and l2+b2=172=289l^2 + b^2 = 17^2 = 289.

  2. (l+b)2=l2+b2+2lb(l + b)^2 = l^2 + b^2 + 2lb, so 529=289+2lb529 = 289 + 2lb, giving 2lb=2402lb = 240.

  3. Area = lb=120lb = 120 cm². (The sides are 15 cm and 8 cm.)

Worth noting

240 is 2lb; halve it. 132.25 cm² assumes the frame is a square with side 11.5 cm, but then the diagonal would not be 17 cm.

Question 12

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A path 1.5 m wide runs all round a square garden, just inside its boundary. The path covers 165 m². What is the length of a side of the garden?

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AnswerC · 29 m

  1. Let the side be s. The part inside the path is a square of side s−2×1.5=s−3s - 2 \times 1.5 = s - 3.

  2. Path area = s2−(s−3)2=6s−9s^2 - (s - 3)^2 = 6s - 9.

  3. 6s−9=1656s - 9 = 165, so 6s=1746s = 174 and s=29s = 29 m. Check: 292−262=841−676=16529^2 - 26^2 = 841 - 676 = 165.

Worth noting

26 m is the side of the inner square, and also what you get by putting the path outside. 55.75 m forgets that the width is taken off both sides.

Question 13

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A field is a trapezium whose parallel sides are 25 m and 11 m. Its two non-parallel sides are 13 m and 15 m. What is the area of the field?

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AnswerA · 216 m²

  1. Cut off a parallelogram of sides 11 m and 13 m. What remains is a triangle with sides 13 m, 15 m and 25−11=1425 - 11 = 14 m, with the same height as the trapezium.

  2. That triangle has s=21s = 21, so area = 21×8×6×7=84\sqrt{21 \times 8 \times 6 \times 7} = 84 m².

  3. Height = 2×84÷14=122 \times 84 \div 14 = 12 m.

  4. Trapezium area = 12×(25+11)×12=216\frac{1}{2} \times (25 + 11) \times 12 = 216 m².

Worth noting

234 m² and 270 m² use a slanted side (13 m or 15 m) as the height. 432 m² forgets the half.

Circles, rings and sectors

14 questions

Question 14

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A round dining-table top has a diameter of 42 cm. What is the area of the top? Take π=227\pi = \frac{22}{7}.

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AnswerB · 1,386 cm²

  1. Radius = 42÷2=2142 \div 2 = 21 cm.

  2. Area = πr2=227×21×21=1386\pi r^2 = \frac{22}{7} \times 21 \times 21 = 1386 cm².

Worth noting

5,544 cm² uses the diameter as the radius. Halve the diameter first.

Question 15

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The wheel of Tenzin's bicycle has a diameter of 56 cm. How far does the bicycle travel when the wheel turns 440 times? Take π=227\pi = \frac{22}{7}.

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AnswerD · 774.4 m

  1. One turn covers one circumference: πd=227×56=176\pi d = \frac{22}{7} \times 56 = 176 cm.

  2. 440 turns cover 440×176=77,440440 \times 176 = 77{,}440 cm.

  3. 77,440 cm = 774.4 m.

Worth noting

387.2 m uses the radius in place of the diameter (πr), and 77.44 m and 7.744 m slip zeros when changing centimetres to metres (100 cm = 1 m).

Question 16

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A window is shaped like a semicircle of radius 14 cm. What length of metal strip is needed to go all round its edge, including the straight edge? Take π=227\pi = \frac{22}{7}.

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AnswerC · 72 cm

  1. Curved edge = half the circumference = πr=227×14=44\pi r = \frac{22}{7} \times 14 = 44 cm.

  2. Straight edge = diameter = 28 cm.

  3. Total = 44+28=7244 + 28 = 72 cm.

Worth noting

44 cm leaves out the straight edge, and 88 cm is the full circle's circumference.

Question 17

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A slice of a round cake is a sector of radius 21 cm with an angle of 60° at the centre. What is the area of the top of the slice? Take π=227\pi = \frac{22}{7}.

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AnswerA · 231 cm²

  1. Sector area = θ360∘×πr2\frac{\theta}{360^\circ} \times \pi r^2.

  2. = 60360×227×21×21=16×1386=231\frac{60}{360} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{6} \times 1386 = 231 cm².

Worth noting

462 cm² uses 120° in place of 60°, 693 cm² is a semicircle and 1,386 cm² is the whole cake. A 60° slice is one-sixth of it.

Question 18

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A curved garden border is an arc of a circle of radius 35 m, and the arc makes an angle of 72° at the centre. How long is the border? Take π=227\pi = \frac{22}{7}.

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AnswerB · 44 m

  1. Arc length = θ360∘×2πr\frac{\theta}{360^\circ} \times 2\pi r.

  2. = 72360×2×227×35=15×220=44\frac{72}{360} \times 2 \times \frac{22}{7} \times 35 = \frac{1}{5} \times 220 = 44 m.

Worth noting

22 m uses πr instead of 2πr. 770 comes from the sector-area formula, and it would be square metres, not metres.

Question 19

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The circumference of a circle is 44 cm. Taking π=227\pi = \frac{22}{7}, what is its area?

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AnswerB · 154 cm²

  1. Find the radius from the circumference: 2πr=442\pi r = 44, so r=44×72×22=7r = \frac{44 \times 7}{2 \times 22} = 7 cm.

  2. Then apply the area formula: πr2=227×72\pi r^2 = \frac{22}{7} \times 7^2.

  3. 227×49=22×7=154\frac{22}{7} \times 49 = 22 \times 7 = 154 cm².

Worth noting

The radius is the bridge between the two formulae. Going straight from circumference to area without finding it is where this question is usually lost.

Question 20

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A circular running track lies between two circles with the same centre. The outer edge is 176 m round and the inner edge is 132 m round. What is the area of the track? Take π=227\pi = \frac{22}{7}.

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AnswerC · 1,078 m²

  1. Outer radius: 2πR=1762\pi R = 176 gives R=28R = 28 m. Inner radius: 2πr=1322\pi r = 132 gives r=21r = 21 m.

  2. Track area = π(R2−r2)=227×(784−441)=227×343=1078\pi (R^2 - r^2) = \frac{22}{7} \times (784 - 441) = \frac{22}{7} \times 343 = 1078 m².

Worth noting

154 m² squares the difference of the radii, π(R−r)2\pi(R - r)^2, which is not the same as π(R2−r2)\pi(R^2 - r^2).

Question 21

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A piece of wire is bent into a square that encloses 484 cm². The same wire is then straightened and bent into a circle. What area does the circle enclose? Take π=227\pi = \frac{22}{7}.

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AnswerC · 616 cm²

  1. The square's side is 484=22\sqrt{484} = 22 cm, so the wire is 4×22=884 \times 22 = 88 cm long.

  2. The circle's circumference is 88 cm: 2×227×r=882 \times \frac{22}{7} \times r = 88, so r=14r = 14 cm.

  3. Area = 227×14×14=616\frac{22}{7} \times 14 \times 14 = 616 cm².

Worth noting

The length of wire stays the same, not the area, so 484 cm² is wrong. For the same boundary a circle encloses more area than a square.

Question 22

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The minute hand of a tower clock is 21 cm long. What area does the hand sweep out between 6:00 and 6:45? Take π=227\pi = \frac{22}{7}.

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AnswerC · 1,039.5 cm²

  1. In 45 minutes the minute hand turns through 45/6045/60 = ¾ of a full turn, i.e. 270°.

  2. Full circle = 227×21×21=1386\frac{22}{7} \times 21 \times 21 = 1386 cm².

  3. Swept area = 34×1,386=1,039.5\tfrac{3}{4} \times 1{,}386 = 1{,}039.5 cm².

Worth noting

346.5 cm² is the quarter the hand has NOT yet swept. 1,386 cm² is a full hour.

Question 23

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In a circle of radius 14 cm, a chord makes a right angle at the centre. What is the area of the smaller segment cut off by the chord? Take π=227\pi = \frac{22}{7}.

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AnswerA · 56 cm²

  1. Quarter sector area = 14×227×14×14=154\frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154 cm².

  2. The triangle formed by the chord and the two radii has area 12×14×14=98\frac{1}{2} \times 14 \times 14 = 98 cm².

  3. Segment = sector − triangle = 154−98=56154 - 98 = 56 cm².

Worth noting

154 cm² is the whole sector and 98 cm² is only the triangle. The segment is the difference, not the sum (252 cm²).

Question 24

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A goat is tied at one corner of a rectangular field 30 m by 20 m with a rope 14 m long. What area of the field can the goat NOT graze? Take π=227\pi = \frac{22}{7}.

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AnswerC · 446 m²

  1. At a corner the rope sweeps a quarter circle, because the corner angle is 90°.

  2. Grazed area = 14×227×14×14=154\frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154 m².

  3. Not grazed = 600−154=446600 - 154 = 446 m².

Worth noting

154 m² is what the goat CAN graze. 292 m² takes a semicircle instead of a quarter circle.

Question 25

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Four circular coasters, each of radius 7 cm, lie flat on a table. Each touches the two next to it, and their centres form a square. What is the area of the space enclosed between the four coasters? Take π=227\pi = \frac{22}{7}.

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AnswerA · 42 cm²

  1. The centres are 7+7=147 + 7 = 14 cm apart, so they form a square of side 14 cm, area 196 cm².

  2. Each corner of the square holds a quarter of a coaster, so the four quarters make one full circle: 227×7×7=154\frac{22}{7} \times 7 \times 7 = 154 cm².

  3. Enclosed space = 196−154=42196 - 154 = 42 cm².

Worth noting

168 cm² uses a square of side 28 cm (joining the far edges instead of the centres). 154 cm² is the area of the four quarter-circles.

Question 26

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A car's wheel has a diameter of 70 cm. How many revolutions does each wheel make per minute when the car moves at 66 km/h? Take π=227\pi = \frac{22}{7}.

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AnswerB · 500

  1. One revolution = circumference = 227×70=220\frac{22}{7} \times 70 = 220 cm.

  2. 66 km/h=66,00,000h = 66{,}00{,}000 cm per hour = 1,10,000 cm per minute.

  3. Revolutions per minute = 1,10,000÷220=5001{,}10{,}000 \div 220 = 500.

Worth noting

1,000 uses πr=110\pi r = 110 cm per turn, 250 takes 70 cm as the radius (440 cm per turn), and 30,000 is revolutions per HOUR.

Question 27

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A sector of a circle of radius 14 cm has a perimeter of 50 cm (two radii plus the arc). What is the area of the sector? Take π=227\pi = \frac{22}{7}.

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AnswerA · 154 cm²

  1. Arc length = 50−2×14=2250 - 2 \times 14 = 22 cm.

  2. The full circumference is 2×227×14=882 \times \frac{22}{7} \times 14 = 88 cm, so the sector is 22/8822/88 = ¼ of the circle.

  3. Area = 14×227×14×14=154\frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154 cm². (Shortcut: area = ½ × arc × radius = 12×22×14\tfrac{1}{2} \times 22 \times 14.)

Worth noting

350 cm² puts the whole perimeter, 50 cm, into ½ × arc × radius. Only the curved part is the arc.

Cubes and cuboids

15 questions

Question 28

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A cuboid measures 8 cm by 5 cm by 3 cm. What is its total surface area?

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AnswerB · 158 cm²

  1. A cuboid has three pairs of identical faces, so the total surface area is 2(lb+bh+lh)2(lb + bh + lh).

  2. Here lb=8×5=40lb = 8 \times 5 = 40, bh=5×3=15bh = 5 \times 3 = 15 and lh=8×3=24lh = 8 \times 3 = 24.

  3. Their sum is 40+15+24=7940 + 15 + 24 = 79, and doubling it gives 158158 cm².

Worth noting

The 120 on offer is the volume, 8×5×3=1208 \times 5 \times 3 = 120 cm³, which is a different quantity in different units. The 79 forgets to double.

Question 29

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A rectangular water tank on a school roof is 3 m long, 2 m wide and 1.5 m deep. How many litres does it hold when full? (1 m3=1,000m^3 = 1{,}000 litres)

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AnswerD · 9,000 litres

  1. Volume = 3×2×1.5=93 \times 2 \times 1.5 = 9 m³.

  2. 9 m³ = 9×1,000=9,0009 \times 1{,}000 = 9{,}000 litres.

Worth noting

1 m³ is 1,000 litres, not 100 (which gives 900) or 10 (which gives 90). 6,000 litres drops the depth of 1.5 m to 1 m.

Question 30

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A cube-shaped gift box has edges of 9 cm. Only its four side faces are to be covered with coloured paper. How much paper is needed?

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AnswerD · 324 cm²

  1. The four side faces make the lateral surface: 4 × edge² = 4×81=3244 \times 81 = 324 cm².

Worth noting

81 cm² is one face and 243 cm² is three. Four side faces means 4×814 \times 81.

Question 31

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A room is 12 m long, 9 m wide and 8 m high. What is the length of the longest straight pole that can fit inside it?

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AnswerC · 17 m

  1. The longest pole runs along the room's diagonal: l2+b2+h2\sqrt{l^2 + b^2 + h^2}.

  2. = 144+81+64=289=17\sqrt{144 + 81 + 64} = \sqrt{289} = 17 m.

Worth noting

15 m is only the diagonal of the floor. Tilting the pole up to the far top corner gives more room.

Question 32

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A warehouse is 40 m long, 25 m wide and 10 m high. How many cartons, each 2 m×1.25m \times 1.25 m×1m \times 1 m, can be stacked inside it?

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AnswerC · 4,000

  1. Along the length: 40÷2=2040 \div 2 = 20. Along the width: 25÷1.25=2025 \div 1.25 = 20. Up the height: 10÷1=1010 \div 1 = 10.

  2. Cartons = 20×20×10=4,00020 \times 20 \times 10 = 4{,}000. (The same as 10,000 m3÷2.5m^3 \div 2.5 m³, since the cartons fit exactly.)

Worth noting

400 and 40,000 are slips of a zero. Counting how many fit along each edge avoids them.

Question 33

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The four walls of a hall 18 m long, 12 m wide and 5 m high are to be plastered. What is the area to be plastered?

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AnswerA · 300 m²

  1. Area of four walls = 2×(l+b)×h2 \times (l + b) \times h.

  2. = 2×(18+12)×5=3002 \times (18 + 12) \times 5 = 300 m².

Worth noting

516 m² adds the floor, and 732 m² adds both the floor and the ceiling (the total surface area). 1,080 is the volume in m³.

Question 34

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The volume of a cube is 216 cm³. What is its total surface area?

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AnswerC · 216 cm²

  1. The volume of a cube is a3a^3, so a3=216a^3 = 216 and the edge is a=6a = 6 cm.

  2. A cube has six identical square faces, each of area a2=36a^2 = 36 cm².

  3. The total surface area is 6×36=2166 \times 36 = 216 cm².

Worth noting

The surface area comes out numerically equal to the volume. That is a coincidence of this particular edge length, not a rule, and the units differ: cm² against cm³. A cube of edge 6 is the only one for which it happens.

Question 35

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A packing box is 20 cm long, 14 cm wide and 8 cm high. How many whole cubes of edge 4 cm can be packed inside it?

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AnswerA · 30

  1. Along the length 20÷4=520 \div 4 = 5 cubes, along the height 8÷4=28 \div 4 = 2 cubes.

  2. Along the width 14÷4=3.514 \div 4 = 3.5, so only 3 whole cubes fit.

  3. Cubes = 5×3×2=305 \times 3 \times 2 = 30.

Worth noting

Dividing volumes (2,240÷64=35)(2{,}240 \div 64 = 35) assumes the cubes can be cut. When one edge does not divide exactly, count along each edge.

Question 36

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Rain falls to a depth of 2 cm on a flat roof 22 m long and 20 m wide, and all of it runs into a tank. How many litres of water reach the tank? (1 m3=1,000m^3 = 1{,}000 litres)

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AnswerD · 8,800 litres

  1. Depth of rain = 2 cm = 0.02 m.

  2. Volume = 22×20×0.02=8.822 \times 20 \times 0.02 = 8.8 m³.

  3. 8.8 m3=8,800m^3 = 8{,}800 litres.

Worth noting

4,400 litres uses 1 cm. 880 and 88 litres slip a zero or two in the conversion: 2 cm = 0.02 m and 1 m3=1,000m^3 = 1{,}000 litres.

Question 37

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Three faces of a cuboid that meet at one corner have areas 48 cm², 60 cm² and 80 cm². What is the volume of the cuboid?

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AnswerC · 480 cm³

  1. If the edges are l, b and hh, the face areas are lb, bh and hl.

  2. Their product is (lbh)², so (lbh)2=48×60×80=2,30,400(lbh)^2 = 48 \times 60 \times 80 = 2{,}30{,}400.

  3. Volume = 2,30,400=480\sqrt{2{,}30{,}400} = 480 cm³. (The edges are 6, 8 and 10 cm.)

Worth noting

188 is just the sum of the three areas. The product of the three face areas is the SQUARE of the volume.

Question 38

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Meenakshi cuts a 5 cm square from each corner of a metal sheet 40 cm by 30 cm and folds up the sides to make an open box. What is the volume of the box?

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AnswerB · 3,000 cm³

  1. Removing a 5 cm square at each end shortens each side by 10 cm: the base is 30 cm by 20 cm.

  2. The folded sides make the height 5 cm.

  3. Volume = 30×20×5=3,00030 \times 20 \times 5 = 3{,}000 cm³.

Worth noting

4,375 cm³ takes 5 cm off only one end of each side, and 6,000 cm³ uses the full sheet as the base.

Question 39

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A fish tank is 50 cm long, 40 cm wide and 30 cm high, with water up to 20 cm. When Joseph lowers a stone into it, the water rises to 23.5 cm without spilling. What is the volume of the stone?

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AnswerA · 7,000 cm³

  1. The stone pushes up water equal to its own volume.

  2. Rise in level = 23.5−20=3.523.5 - 20 = 3.5 cm.

  3. Stone volume = 50×40×3.5=7,00050 \times 40 \times 3.5 = 7{,}000 cm³.

Worth noting

40,000 cm³ is the water that was already there, 47,000 cm³ is water plus stone, and 13,000 cm³ is the space still empty above the water.

Question 40

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A room is 8 m long, 6 m wide and 3.5 m high. It has one door 2 m by 1 m and two windows each 1.5 m by 1 m. The four walls and the ceiling are to be painted at ₹25 per m². What will the painting cost?

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AnswerB · ₹3,525

  1. Four walls = 2×(8+6)×3.5=982 \times (8 + 6) \times 3.5 = 98 m². Ceiling = 8×6=488 \times 6 = 48 m².

  2. Door and windows = 2×1+2×1.5×1=52 \times 1 + 2 \times 1.5 \times 1 = 5 m².

  3. Area to paint = 98+48−5=14198 + 48 - 5 = 141 m².

  4. Cost = 141 × ₹25 = ₹3,525.

Worth noting

₹3,650 forgets to leave out the door and windows, ₹2,325 leaves out the ceiling, and ₹4,725 paints the floor too.

Question 41

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Water flows at 5 km/h through a rectangular channel 1.5 m wide and 0.8 m deep. How much area can it cover with water 8 cm deep in 30 minutes?

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AnswerB · 37,500 m²

  1. In one hour a length of 5 km = 5,000 m of water passes, so the flow is 1.5×0.8×5,000=6,0001.5 \times 0.8 \times 5{,}000 = 6{,}000 m³ per hour.

  2. In 30 minutes: 3,000 m³.

  3. Area = volume ÷ depth = 3,000÷0.08=37,5003{,}000 \div 0.08 = 37{,}500 m².

Worth noting

75,000 m² uses a full hour. 3,750 m² treats 8 cm as 0.8 m.

Question 42

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A wall 15 m long, 30 cm thick and 2 m high is built with bricks measuring 25 cm × 12.5 cm × 7.5 cm. Mortar fills one-tenth of the wall's volume. How many bricks are used?

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AnswerA · 3,456

  1. Wall volume = 1,500×30×200=90,00,0001{,}500 \times 30 \times 200 = 90{,}00{,}000 cm³.

  2. Bricks fill nine-tenths of it: 81,00,000 cm³.

  3. One brick = 25×12.5×7.5=2,343.7525 \times 12.5 \times 7.5 = 2{,}343.75 cm³.

  4. Bricks = 81,00,000÷2,343.75=3,45681{,}00{,}000 \div 2{,}343.75 = 3{,}456.

Worth noting

3,840 ignores the mortar, and 4,224 adds the mortar instead of taking it away. 38,400 slips a zero in the wall volume.

Cylinders, cones and spheres

13 questions

Question 43

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A cylindrical steel drum has a base radius of 7 cm and a height of 20 cm. What is its volume? Take π=227\pi = \frac{22}{7}.

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AnswerC · 3,080 cm³

  1. Volume of a cylinder = πr2h\pi r^2 h.

  2. = 227×7×7×20=3080\frac{22}{7} \times 7 \times 7 \times 20 = 3080 cm³.

Worth noting

880 is the curved surface area (2πrh), not a volume. 12,320 cm³ uses 14 cm as the radius.

Question 44

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A label is wrapped exactly once round the curved side of a cylindrical tin of radius 14 cm and height 10 cm, with no overlap. What is the area of the label? Take π=227\pi = \frac{22}{7}.

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AnswerA · 880 cm²

  1. The label covers the curved surface: 2πrh2\pi r h.

  2. = 2×227×14×10=8802 \times \frac{22}{7} \times 14 \times 10 = 880 cm².

Worth noting

1,496 cm² adds one circular end and 2,112 cm² adds both ends; a label covers only the curved side. 6,160 is the volume in cm³.

Question 45

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A globe has a radius of 21 cm. What is the area of its surface? Take π=227\pi = \frac{22}{7}.

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AnswerD · 5,544 cm²

  1. Surface area of a sphere = 4πr24\pi r^2.

  2. = 4×227×21×21=55444 \times \frac{22}{7} \times 21 \times 21 = 5544 cm².

Worth noting

1,386 cm² is πr², a flat circle of the same radius. 2,772 cm² is half the globe, and 4,158 cm² is the surface of a solid hemisphere (3πr²).

Question 46

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A hemispherical bowl has an inner radius of 21 cm. How much can it hold, in cubic centimetres? Take π=227\pi = \frac{22}{7}.

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AnswerC · 19,404 cm³

  1. Volume of a hemisphere = 23πr3\frac{2}{3}\pi r^3.

  2. = 23×227×21×21×21=19,404\frac{2}{3} \times \frac{22}{7} \times 21 \times 21 \times 21 = 19{,}404 cm³.

Worth noting

38,808 cm³ is a full sphere. A hemisphere is half of it.

Question 47

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A conical heap of grain has a base radius of 7 m and a height of 24 m. What is the volume of the heap? Take π=227\pi = \frac{22}{7}.

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AnswerA · 1,232 m³

  1. Volume of a cone = 13πr2h\frac{1}{3}\pi r^2 h.

  2. = 13×227×7×7×24=1232\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24 = 1232 m³.

Worth noting

3,696 m³ forgets the ⅓ (that is a cylinder), and 4,928 m³ uses 14 m as the radius.

Question 48

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A solid cone and a solid cylinder have bases of the same radius and have equal volumes. The cylinder is 8 cm tall. How tall is the cone?

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AnswerD · 24 cm

  1. Equal volumes: 13πr2H=πr2×8\frac{1}{3}\pi r^2 H = \pi r^2 \times 8.

  2. Cancel πr²: H÷3=8H \div 3 = 8, so H=24H = 24 cm.

Worth noting

A cone holds one-third of the cylinder with the same base and height, so it needs three times the height to match. 8/38/3 cm divides instead of multiplying.

Question 49

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A cylindrical pillar has a radius of 7 cm, and its curved surface area is 1,320 cm². What is its volume? Take π=227\pi = \frac{22}{7}.

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AnswerC · 4,620 cm³

  1. Curved surface: 2×227×7×h=44h=13202 \times \frac{22}{7} \times 7 \times h = 44h = 1320, so h=30h = 30 cm.

  2. Volume = 227×7×7×30=4620\frac{22}{7} \times 7 \times 7 \times 30 = 4620 cm³.

Worth noting

9,240 cm³ comes from using πrh=1,320\pi rh = 1{,}320 (dropping the 2), which doubles the height to 60 cm.

Question 50

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An iron pipe is 70 cm long. Its outer diameter is 14 cm and its inner diameter is 10 cm. What volume of iron is used in the pipe? Take π=227\pi = \frac{22}{7}.

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AnswerB · 5,280 cm³

  1. Outer radius 7 cm, inner radius 5 cm.

  2. Iron = π(R2−r2)h=227×(49−25)×70\pi (R^2 - r^2) h = \frac{22}{7} \times (49 - 25) \times 70.

  3. = 22×24×10=5,28022 \times 24 \times 10 = 5{,}280 cm³.

Worth noting

880 cm³ uses (R−r)2=4(R - r)^2 = 4 instead of R2−r2=24R^2 - r^2 = 24. 10,780 cm³ is the whole solid rod with no hole.

Question 51

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A conical tent has a base radius of 7 m and a height of 24 m. How many metres of canvas 2 m wide are needed to make its curved surface? Take π=227\pi = \frac{22}{7}.

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AnswerC · 275 m

  1. Slant height = 72+242=25\sqrt{7^2 + 24^2} = 25 m.

  2. Canvas area = πrl=227×7×25=550\pi r l = \frac{22}{7} \times 7 \times 25 = 550 m².

  3. Length of canvas = area ÷ width = 550÷2=275550 \div 2 = 275 m.

Worth noting

264 m uses the height 24 m instead of the slant height. 550 is the area in m², not the length.

Question 52

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The surface areas of two steel ball bearings are in the ratio 9:169 : 16. What is the ratio of their volumes?

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AnswerB · 27:6427 : 64

  1. Surface area goes as r², so r1:r2=9:16=3:4r_1 : r_2 = \sqrt{9} : \sqrt{16} = 3 : 4.

  2. Volume goes as r³, so the volumes are in the ratio 33:43=27:643^3 : 4^3 = 27 : 64.

Worth noting

9:169 : 16 copies the area ratio, and 81:25681 : 256 squares it. Take the square root first, then cube.

Question 53

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A solid wooden cone has a base radius of 7 cm and a height of 24 cm. What is its total surface area, including the base? Take π=227\pi = \frac{22}{7}.

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AnswerD · 704 cm²

  1. Slant height = 72+242=25\sqrt{7^2 + 24^2} = 25 cm.

  2. Total surface = πr(l+r)=227×7×(25+7)=22×32=704\pi r (l + r) = \frac{22}{7} \times 7 \times (25 + 7) = 22 \times 32 = 704 cm².

Worth noting

550 cm² is the curved part alone. 682 cm² uses the height 24 cm in place of the slant height 25 cm, and 528 cm² does both.

Question 54

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The curved surface of a cone is 4,070 cm² and the diameter of its base is 70 cm. What is its volume? Take π=227\pi = \frac{22}{7}.

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AnswerB · 15,400 cm³

  1. Radius = 35 cm. πrl=4070\pi r l = 4070 gives l=4070÷110=37l = 4070 \div 110 = 37 cm.

  2. Height = 372−352=144=12\sqrt{37^2 - 35^2} = \sqrt{144} = 12 cm.

  3. Volume = 13×227×35×35×12=15,400\frac{1}{3} \times \frac{22}{7} \times 35 \times 35 \times 12 = 15{,}400 cm³.

Worth noting

46,200 cm³ forgets the ⅓, and 61,600 cm³ uses 70 cm as the radius. The height is not the slant height: find it by Pythagoras.

Question 55

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For a closed cylindrical container, the radius and the height add up to 37 cm, and the total surface area is 3,256 cm². What is its volume? Take π=227\pi = \frac{22}{7}.

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AnswerB · 14,168 cm³

  1. Total surface = 2πr(r+h)=2×227×r×372\pi r (r + h) = 2 \times \frac{22}{7} \times r \times 37.

  2. So 16287r=3256\frac{1628}{7} r = 3256, giving r=14r = 14 cm and h=37−14=23h = 37 - 14 = 23 cm.

  3. Volume = 227×14×14×23=14,168\frac{22}{7} \times 14 \times 14 \times 23 = 14{,}168 cm³.

Worth noting

22,792 cm³ takes the height as 37 cm; the 37 is r+hr + h, not hh. 28,336 cm³ uses 2πr²h, mixing the 2 of the surface formula into the volume.

Melting, recasting and combined solids

13 questions

Question 56

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A solid silver cube of edge 12 cm is melted and recast into small cubes of edge 3 cm. How many small cubes are made?

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AnswerD · 64

  1. Volume is unchanged when a solid is recast.

  2. Number of cubes = 123÷33=1,728÷27=6412^3 \div 3^3 = 1{,}728 \div 27 = 64.

Worth noting

4 divides the edges (12÷3)(12 \div 3) and 16 divides the face areas. Recasting keeps the VOLUME, so divide the cubes of the edges.

Question 57

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A solid metal sphere of radius 9 cm is melted and recast into a solid cylinder of radius 6 cm. What is the height of the cylinder?

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AnswerD · 27 cm

  1. Volumes are equal: 43π×93=π×62×h\frac{4}{3}\pi \times 9^3 = \pi \times 6^2 \times h.

  2. π cancels: 972=36h972 = 36h, so h=27h = 27 cm. (No value of π is needed.)

Worth noting

20.25 cm drops the 4/34/3 from the sphere's volume, and 13.5 cm halves the answer. 9 cm wrongly assumes the height equals the sphere's radius.

Question 58

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Three identical cubes, each of edge 5 cm, are glued face to face in a row to form a cuboid. What is the total surface area of the cuboid?

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AnswerA · 350 cm²

  1. The cuboid measures 15 cm × 5 cm × 5 cm.

  2. Total surface area = 2×(15×5+5×5+15×5)=2×175=3502 \times (15 \times 5 + 5 \times 5 + 15 \times 5) = 2 \times 175 = 350 cm².

Worth noting

450 cm² adds the three cubes' surfaces (3×150)(3 \times 150) but the four glued faces are hidden. 400 cm² hides only two faces.

Question 59

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A well of diameter 7 m is dug 20 m deep. The earth taken out is spread evenly over a rectangular plot 70 m long and 22 m wide. By how much does the level of the plot rise? Take π=227\pi = \frac{22}{7}.

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AnswerC · 50 cm

  1. Earth dug out = 227×3.5×3.5×20=770\frac{22}{7} \times 3.5 \times 3.5 \times 20 = 770 m³.

  2. Plot area = 70×22=1,54070 \times 22 = 1{,}540 m².

  3. Rise = 770÷1,540=0.5770 \div 1{,}540 = 0.5 m=50m = 50 cm.

Worth noting

200 cm uses 7 m as the radius. 5 cm slips a decimal when changing 0.5 m into centimetres.

Question 60

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A block of metal 5.5 cm × 10 cm × 3.5 cm is melted and made into coins, each 1.4 cm in diameter and 0.2 cm thick. How many coins are made? Take π=227\pi = \frac{22}{7}.

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AnswerB · 625

  1. Block volume = 5.5×10×3.5=192.55.5 \times 10 \times 3.5 = 192.5 cm³.

  2. One coin is a short cylinder: 227×0.7×0.7×0.2=0.308\frac{22}{7} \times 0.7 \times 0.7 \times 0.2 = 0.308 cm³.

  3. Coins = 192.5÷0.308=625192.5 \div 0.308 = 625.

Worth noting

2,500 halves the diameter twice (radius 0.35 cm). 6,250 takes the thickness as 0.02 cm, and 125 takes it as 1 cm.

Question 61

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A solid hemisphere of radius 6 cm is melted and recast into a solid cone with a base radius of 12 cm. What is the height of the cone?

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AnswerB · 3 cm

  1. Hemisphere volume = 23π×63=144π\frac{2}{3}\pi \times 6^3 = 144\pi.

  2. Cone volume = 13π×122×h=48πh\frac{1}{3}\pi \times 12^2 \times h = 48\pi h.

  3. 48πh=144π48\pi h = 144\pi, so h=3h = 3 cm.

Worth noting

1 cm forgets the ⅓ in the cone formula, and 6 cm assumes the height equals the hemisphere's radius.

Question 62

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A tent is a cylinder of radius 7 m and height 3 m with a cone of the same radius on top. The cone's vertical height is 24 m. How much canvas is needed for the tent, with no floor? Take π=227\pi = \frac{22}{7}.

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AnswerC · 682 m²

  1. Cylinder wall = 2πrh=2×227×7×3=1322\pi r h = 2 \times \frac{22}{7} \times 7 \times 3 = 132 m².

  2. Cone: slant height 72+242=25\sqrt{7^2 + 24^2} = 25 m, curved surface πrl=22×25=550\pi r l = 22 \times 25 = 550 m².

  3. Canvas = 132+550=682132 + 550 = 682 m².

Worth noting

836 m² adds a floor (154 m²) that the tent does not have. 660 m² uses the height 24 m in place of the slant height.

Question 63

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A medicine capsule is a cylinder 10 mm long with a hemisphere of radius 7 mm fixed at each end. What is the total outer surface area of the capsule? Take π=227\pi = \frac{22}{7}.

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AnswerB · 1,056 mm²

  1. Cylinder wall = 2×227×7×10=4402 \times \frac{22}{7} \times 7 \times 10 = 440 mm².

  2. Two hemispheres make one sphere: 4×227×7×7=6164 \times \frac{22}{7} \times 7 \times 7 = 616 mm².

  3. Total = 440+616=1,056440 + 616 = 1{,}056 mm².

Worth noting

748 mm² treats each hemisphere end as a flat circle. 1,364 mm² adds two circular faces that are hidden inside the joins.

Question 64

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An ice-cream cone is filled completely and topped with a hemisphere of ice cream. The cone has radius 7 cm and height 10 cm, and the hemisphere has the same radius. What is the total volume of ice cream? Take π=227\pi = \frac{22}{7}.

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AnswerA · 1,232 cm³

  1. Cone = 13×227×7×7×10=15403\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 10 = \frac{1540}{3} cm³.

  2. Hemisphere = 23×227×7×7×7=21563\frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 = \frac{2156}{3} cm³.

  3. Total = 36963=1232\frac{3696}{3} = 1232 cm³.

Worth noting

2,258⅔ cm³ treats the cone as a cylinder (forgetting the ⅓), and 1,540 cm³ is that cylinder with the hemisphere left out. 1,950⅔ cm³ tops the cone with a full sphere instead of a hemisphere.

Question 65

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A copper ball of diameter 6 cm is melted and drawn into a wire of diameter 0.2 cm. How long is the wire?

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AnswerD · 36 m

  1. Ball: radius 3 cm, volume 43π×27=36π\frac{4}{3}\pi \times 27 = 36\pi cm³.

  2. Wire: a long cylinder of radius 0.1 cm, volume π×0.01×L\pi \times 0.01 \times L.

  3. 0.01L=360.01L = 36, so L=3,600L = 3{,}600 cm = 36 m.

Worth noting

9 m uses 0.2 cm as the wire's radius and 27 m forgets the 4/34/3. 3.6 m divides 3,600 cm by 1,000 instead of 100.

Question 66

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From a solid cylinder of radius 7 cm and height 24 cm, a cone of the same radius and height is hollowed out, with its base on the top face of the cylinder. What is the total surface area of the solid that remains? Take π=227\pi = \frac{22}{7}.

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AnswerC · 1,760 cm²

  1. Outside wall = 2×227×7×24=10562 \times \frac{22}{7} \times 7 \times 24 = 1056 cm². Bottom face = 227×49=154\frac{22}{7} \times 49 = 154 cm².

  2. The top face is now the cone's opening, so it is gone. Inside, the cone's curved surface shows: slant 72+242=25\sqrt{7^2 + 24^2} = 25, area 22×25=55022 \times 25 = 550 cm².

  3. Total = 1,056+154+550=1,7601{,}056 + 154 + 550 = 1{,}760 cm².

Worth noting

1,914 cm² keeps the top face, which has been cut away. 1,210 cm² forgets the new inner surface, and 1,738 cm² uses the height instead of the slant height for it.

Question 67

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A cylindrical glass beaker of diameter 7 cm contains water. How many spherical glass marbles, each of diameter 1.4 cm, must be dropped in to raise the water level by 5.6 cm? Take π=227\pi = \frac{22}{7}.

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AnswerB · 150

  1. Extra water level: π×3.52×5.6=68.6π\pi \times 3.5^2 \times 5.6 = 68.6\pi cm³.

  2. One marble: 43π×0.73=43π×0.343\frac{4}{3}\pi \times 0.7^3 = \frac{4}{3}\pi \times 0.343 cm³.

  3. Number = 68.6÷(43×0.343)=68.6×3÷1.372=15068.6 \div \left(\frac{4}{3} \times 0.343\right) = 68.6 \times 3 \div 1.372 = 150.

Worth noting

600 takes 7 cm as the beaker's radius. 200 forgets the 4/34/3 in the sphere's volume, and 50 uses 4 in place of 4/34/3.

Question 68

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Water flows at 15 km/h through a pipe of internal diameter 14 cm into a rectangular tank 50 m long and 44 m wide. How long will it take for the water level in the tank to rise by 21 cm? Take π=227\pi = \frac{22}{7}.

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AnswerC · 2 hours

  1. Pipe radius = 7 cm = 0.07 m. Each hour a 15,000 m length of water flows in: 227×0.07×0.07×15,000=231\frac{22}{7} \times 0.07 \times 0.07 \times 15{,}000 = 231 m³.

  2. Water needed = 50×44×0.21=46250 \times 44 \times 0.21 = 462 m³.

  3. Time = 462÷231=2462 \div 231 = 2 hours.

Worth noting

30 minutes uses 14 cm as the radius (four times the flow). 20 hours treats 21 cm as 2.1 m.

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