Area and volume — SSC CGL Quant Notes

Concepts and rules

Triangle: area = 1/2 × base × height. Heron: area = √(s(s − a)(s − b)(s − c)) with s = (a + b + c)/2. With 2 sides and the included angle C: area = 1/2 × ab × sin C.

Equilateral triangle of side a: height = √3a/2, area = √3a²/4, inradius = a/(2√3), circumradius = a/√3. Right triangle: area = 1/2 × product of legs.

Rectangle l × b: area = lb, perimeter = 2(l + b), diagonal = √(l² + b²). Square of side a: area = a² = d²/2, diagonal d = √2a.

Parallelogram: area = base × height. Rhombus with diagonals d₁, d₂: area = d₁d₂/2, and the diagonals bisect at right angles, so side² = (d₁/2)² + (d₂/2)².

Trapezium with parallel sides a, b and distance h between them: area = 1/2 × (a + b) × h.

Regular polygon with n sides: each interior angle = (n − 2) × 180°/n, each exterior angle = 360°/n, number of diagonals = n(n − 3)/2. Regular hexagon of side a = 6 equilateral triangles, area = 3√3a²/2.

Circle of radius r: area = πr², circumference = 2πr. Semicircle perimeter = πr + 2r. Ring between radii R and r: area = π(R² − r²) = π(R + r)(R − r).

Sector with angle θ: area = (θ/360) × πr², arc length = (θ/360) × 2πr, perimeter = arc + 2r. Use π = 22/7 when radii are multiples of 7, otherwise 3.14 or leave π.

Paths: a path of width w outside a rectangle l × b has area (l + 2w)(b + 2w) − lb; inside it has area lb − (l − 2w)(b − 2w). 2 cross roads of width w through the middle have area w(l + b) − w², because the central square is counted twice.

A wheel covers 1 circumference per revolution: revolutions = distance ÷ (2πr). A wire bent from one shape into another keeps its length (perimeter).

If every length of a figure increases by x%, its area increases by (2x + x²/100)%: for x = 10, area rises 21%.

Method and shortcuts

Step 1: Draw a rough sketch and mark every given length. Identify the basic pieces (triangle, rectangle, sector) the figure is built from.

Step 2: Find missing lengths with Pythagoras or the given ratio before using any area formula.

Step 3: Add the pieces, or subtract the cut-out pieces (outer minus inner for paths and rings). Convert units at the end: 1 m² = 10000 cm², 1 hectare = 10000 m².

Shortcut: outside path area = 2w(l + b + 2w); inside path area = 2w(l + b − 2w). These come from expanding the products, so they always hold.

Shortcut: Pythagorean triplets (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25) and their multiples save time with diagonals and rhombus sides.

Example 1: Heron's formula

Question: Find the area of a triangle with sides 13 cm, 14 cm and 15 cm.

Solution: s = (13 + 14 + 15)/2 = 21. Area = √(21 × 8 × 7 × 6) = √7056 = 84 cm².

Answer: 84 cm²

Example 2: rhombus from diagonals

Question: The diagonals of a rhombus are 24 cm and 10 cm. Find its area and perimeter.

Solution: Area = 24 × 10/2 = 120 cm². Half-diagonals are 12 and 5, so side = √(144 + 25) = 13 cm and perimeter = 4 × 13 = 52 cm.

Answer: 120 cm², 52 cm

Example 3: path around a garden

Question: A garden is 40 m × 30 m. A path 2.5 m wide runs around it on the outside. Find the cost of paving the path at ₹12 per m².

Solution: Outer rectangle = 45 × 35 = 1575 m². Path = 1575 − 1200 = 375 m² (shortcut 2 × 2.5 × (40 + 30 + 5) = 375). Cost = 375 × 12 = ₹4,500.

Answer: ₹4,500

Example 4: regular hexagon

Question: Find the area of a regular hexagon of side 6 cm. (a) 36√3 cm² (b) 54√3 cm² (c) 72√3 cm² (d) 108√3 cm²

Solution: The hexagon is 6 equilateral triangles of side 6. Area = 6 × √3 × 36/4 = 54√3 cm².

Answer: (b) 54√3 cm²

Example 5: incircle of an equilateral triangle

Question: The area of the incircle of an equilateral triangle is 154 cm². Find the perimeter of the triangle. (a) 21√3 cm (b) 14√3 cm (c) 49√3 cm (d) 42√3 cm

Solution: (22/7) × r² = 154 gives r² = 49, r = 7. For an equilateral triangle r = a/(2√3), so a = 2√3 × 7 = 14√3 cm.

Perimeter = 3 × 14√3 = 42√3 cm. Option (b) is only 1 side.

Answer: (d) 42√3 cm

Example 6: inside path with a ratio and diagonal

Question: The length and breadth of a rectangular lawn are in the ratio 4 : 3 and its diagonal is 25 m. A path 2 m wide runs along the edges inside the lawn. Find the area of the path. (a) 124 m² (b) 140 m² (c) 156 m² (d) 176 m²

Solution: Sides 4k and 3k give diagonal 5k = 25, so k = 5, length = 20 m, breadth = 15 m, area = 300 m².

Inner rectangle = (20 − 4) × (15 − 4) = 16 × 11 = 176 m². Path = 300 − 176 = 124 m² (shortcut 2 × 2 × (20 + 15 − 4) = 124). Option (d) is the inner lawn, not the path.

Answer: (a) 124 m²

Common mistakes

Subtracting only w (not 2w) from each side for an inside path. The path takes w from both ends of every side.

Using the full diagonals of a rhombus in Pythagoras. The right triangle has the half-diagonals as legs.

Forgetting the 2 radii in the perimeter of a semicircle or sector; the perimeter is the arc plus the straight edges.

Adding cross-road areas as w × l + w × b without removing the common square w².

Mixing units: a path in metres and a rate per cm², or distance in km with radius in cm. Convert everything first.

Practice questions

Q1. A wire bent into a square encloses 484 cm². The same wire is bent into a circle. Find the area of the circle (π = 22/7). (a) 484 cm² (b) 554 cm² (c) 616 cm² (d) 704 cm²

Q2. Find the area of a sector of radius 21 cm and angle 120° (π = 22/7). (a) 462 cm² (b) 231 cm² (c) 924 cm² (d) 441 cm²

Q3. A wheel of diameter 70 cm covers 1.1 km. How many revolutions does it make (π = 22/7)? (a) 250 (b) 400 (c) 1000 (d) 500

Answers

Q1. (c) Side = 22, wire = 88 cm. 2 × (22/7) × r = 88 gives r = 14, area = (22/7) × 196 = 616 cm².

Q2. (a) (120/360) × (22/7) × 441 = 1/3 × 1386 = 462 cm².

Q3. (d) Circumference = (22/7) × 70 = 220 cm; 1.1 km = 110000 cm; revolutions = 110000 ÷ 220 = 500.

Quick revision

Heron: √(s(s − a)(s − b)(s − c)); equilateral √3a²/4; hexagon 3√3a²/2.

Rhombus d₁d₂/2; trapezium (a + b)h/2; square d²/2.

Sector (θ/360)πr²; arc (θ/360)2πr; ring π(R + r)(R − r).

Outside path 2w(l + b + 2w); inside path 2w(l + b − 2w); cross roads w(l + b) − w².

Concepts and rules

Right prism: volume = base area × height; lateral surface = base perimeter × height; total surface = lateral surface + 2 × base area.

Cylinder (radius r, height h): volume = πr²h, curved surface = 2πrh, total surface = 2πr(r + h). Hollow cylinder with radii R and r: volume of material = πh(R² − r²).

Cone: slant height l = √(r² + h²); volume = 1/3 × πr²h; curved surface = πrl; total surface = πr(l + r).

Sphere: volume = 4/3 × πr³, surface area = 4πr². A sphere has no separate curved and total surface; both are 4πr².

Frustum of a cone (radii R and r, height h): volume = πh/3 × (R² + r² + Rr); slant l = √(h² + (R − r)²); curved surface = π(R + r)l.

Cylinder and cone on the same base and height: volume ratio 3 : 1. Sphere fitting exactly in a cylinder: volume ratio sphere : cylinder = 2 : 3, and their curved surfaces are equal.

Recasting: melting one solid into others keeps the total volume. Number of small solids = big volume ÷ small volume. Surface area is NOT conserved.

Sector folded into a cone: the sector radius becomes the slant height, and the arc length becomes the base circumference 2πr.

Scaling: if every length is multiplied by k, surface areas are multiplied by k² and volumes by k³. Units: 1 litre = 1000 cm³, 1 m³ = 1000 litres.

Method and shortcuts

Step 1: Identify the solid and list r, h, l. If 2 of r, h, l are given, get the third from l² = r² + h².

Step 2: Decide what is asked: volume (capacity, melting, filling), curved surface (painting the side, label, canvas) or total surface (closed solid).

Step 3: In recasting, write volume before = volume after and cancel π and common factors before multiplying.

Shortcut: percentage change questions are ratio questions. Radius × 1.5 means surface × 2.25 (125% more) and volume × 3.375 (237.5% more).

Shortcut: wire drawn from a solid is a long cylinder: length = volume ÷ (π × wire radius²).

Example 1: cylinder volume and curved surface

Question: A cylinder has radius 7 cm and height 10 cm. Find its volume and curved surface area (π = 22/7).

Solution: Volume = (22/7) × 49 × 10 = 1540 cm³. Curved surface = 2 × (22/7) × 7 × 10 = 440 cm².

Answer: 1540 cm³, 440 cm²

Example 2: cone from radius and height

Question: A cone has radius 6 cm and height 8 cm. Find its curved surface area and volume in terms of π.

Solution: l = √(36 + 64) = 10. Curved surface = π × 6 × 10 = 60π cm². Volume = 1/3 × π × 36 × 8 = 96π cm³.

Answer: 60π cm², 96π cm³

Example 3: sphere recast into a cone

Question: A metal sphere of radius 6 cm is melted into a cone of base radius 6 cm. Find the height of the cone.

Solution: 4/3 × π × 216 = 1/3 × π × 36 × h. Cancel π/3: 4 × 216 = 36h, so h = 24 cm.

Answer: 24 cm

Example 4: sphere drawn into wire

Question: A copper sphere of radius 9 cm is drawn into a wire of radius 0.3 cm. Find the length of the wire. (a) 54 m (b) 72 m (c) 108 m (d) 216 m

Solution: 4/3 × π × 729 = π × 0.09 × L. So L = 972 ÷ 0.09 = 10800 cm = 108 m.

Answer: (c) 108 m

Example 5: frustum-shaped bucket

Question: A bucket is a frustum with top radius 28 cm, bottom radius 7 cm and height 45 cm. Find its capacity (π = 22/7). (a) 48.51 litres (b) 32.34 litres (c) 145.53 litres (d) 16.17 litres

Solution: R² + r² + Rr = 784 + 49 + 196 = 1029. Volume = (22/7) × (45/3) × 1029 = 22 × 15 × 147 = 48510 cm³.

48510 cm³ ÷ 1000 = 48.51 litres. Option (c) forgets the 1/3 in the frustum formula.

Answer: (a) 48.51 litres

Example 6: sector folded into a cone

Question: A sector of radius 21 cm and angle 120° is folded to form a cone. Find the volume of the cone (π = 22/7). (a) 1078√2/3 cm³ (b) 2156/3 cm³ (c) 1078√2 cm³ (d) 2156√2/3 cm³

Solution: Arc = (120/360) × 2π × 21 = 14π = 2πr, so r = 7 cm. Slant l = 21 cm, height h = √(441 − 49) = √392 = 14√2 cm.

Volume = 1/3 × (22/7) × 49 × 14√2 = 1/3 × 22 × 7 × 14√2 = 2156√2/3 cm³ (about 1016 cm³).

Answer: (d) 2156√2/3 cm³

Common mistakes

Using the slant height in the cone volume. Volume needs the vertical height h; slant l is only for the curved surface.

Taking diameter as radius. Check the word: a 14 cm diameter means r = 7.

Expecting surface area to stay the same after recasting. Only volume is conserved; many small spheres have more total surface than 1 big one.

Treating percentage change in radius as the same change in volume. Radius up 50% gives volume up 237.5%, not 50% or 150%.

Forgetting the unit change: cm³ to litres divides by 1000, m³ to litres multiplies by 1000.

Practice questions

Q1. A cylinder and a cone have the same radius and height. The volume of the cylinder is 924 cm³. Find the volume of the cone. (a) 462 cm³ (b) 616 cm³ (c) 308 cm³ (d) 231 cm³

Q2. The radius of a sphere is increased by 50%. By what percent does its surface area increase? (a) 125% (b) 150% (c) 237.5% (d) 225%

Q3. A metal pipe 14 cm long has outer radius 5 cm and inner radius 4 cm. Find the volume of metal (π = 22/7). (a) 352 cm³ (b) 440 cm³ (c) 418 cm³ (d) 396 cm³

Answers

Q1. (c) Cone = 1/3 of the cylinder = 924 ÷ 3 = 308 cm³.

Q2. (a) Surface ∝ r²: 1.5² = 2.25, so the increase is 125%. (237.5% is the volume increase.)

Q3. (d) (22/7) × 14 × (25 − 16) = 44 × 9 = 396 cm³.

Quick revision

Cylinder πr²h, 2πrh; cone 1/3 πr²h, πrl; sphere 4/3 πr³, 4πr².

Frustum πh/3 (R² + r² + Rr); slant √(h² + (R − r)²).

Same base and height: cylinder : cone = 3 : 1; sphere in cylinder = 2 : 3.

Recasting conserves volume only; lengths × k ⇒ area × k², volume × k³.

Concepts and rules

A hemisphere is half a sphere. Volume = 1/2 × 4/3 πr³ = 2/3 πr³.

Curved surface = half of 4πr² = 2πr². A solid hemisphere also has a flat circular face πr², so total surface = 3πr².

A thin hollow bowl (open hemisphere) has only the curved surface 2πr². Its capacity is 2/3 πr³ using the inner radius.

Thick hemispherical shell with outer radius R and inner radius r: volume of material = 2/3 π(R³ − r³); total surface = outer 2πR² + inner 2πr² + rim ring π(R² − r²).

Cutting a sphere into 2 hemispheres keeps the volume but adds 2 flat faces: total surface rises from 4πr² to 6πr², i.e. by 2πr², which is 50%.

Hemisphere, cone and cylinder with the same radius r and height r: volumes 2/3 πr³ : 1/3 πr³ : πr³ = 2 : 1 : 3.

Combined solids: add the volumes of the parts. For surface area, add only the outer exposed surfaces; a joined face is hidden and is not counted.

Cone on a hemisphere (toy, same radius): volume = 1/3 πr²h + 2/3 πr³; surface = πrl + 2πr². Cylinder with hemispherical ends (capsule): volume = πr²h + 4/3 πr³; surface = 2πrh + 4πr², where h is the cylinder part only.

Largest hemisphere cut from a cube of side a has radius a/2; largest hemisphere on top of a cylinder of radius r has radius r.

Method and shortcuts

Step 1: Read whether the hemisphere is solid (3πr²), an open bowl (2πr²) or a thick shell (outer + inner + rim).

Step 2: For combined solids, list each part with its own height. The total height of a toy = cone height + hemisphere radius.

Step 3: Take πr² common before multiplying: e.g. capsule volume = πr²(h + 4r/3). This keeps numbers small.

Shortcut: with π = 22/7 and r = 7, πr² = 154, 2πr² = 308, 3πr² = 462, 2/3 πr³ = 2156/3. With r = 21, 2/3 πr³ = 19404 and 2πr² = 2772.

Shortcut: in filling problems, number of containers = volume of source ÷ volume of 1 container; take the floor if the question asks for full containers.

Example 1: volume and surfaces of a hemisphere

Question: Find the volume, curved surface and total surface of a solid hemisphere of radius 21 cm (π = 22/7).

Solution: Volume = 2/3 × (22/7) × 9261 = 19404 cm³. Curved = 2 × (22/7) × 441 = 2772 cm². Total = 3 × (22/7) × 441 = 4158 cm².

Answer: 19404 cm³, 2772 cm², 4158 cm²

Example 2: radius from total surface

Question: The total surface area of a solid hemisphere is 462 cm². Find its radius (π = 22/7).

Solution: 3 × (22/7) × r² = 462 gives r² = 462 × 7/66 = 49, so r = 7 cm.

Answer: 7 cm

Example 3: cutting a sphere

Question: A sphere of radius 7 cm is cut into 2 hemispheres. By how much does the total surface area increase, and by what percent?

Solution: 2 new flat faces: 2 × (22/7) × 49 = 308 cm². Original surface = 4 × 154 = 616 cm², so the increase is 308/616 = 50%.

Answer: 308 cm², 50%

Example 4: cone mounted on a hemisphere

Question: A toy is a cone of radius 7 cm mounted on a hemisphere of the same radius. The total height of the toy is 31 cm. Find its total surface area (π = 22/7). (a) 770 cm² (b) 858 cm² (c) 1012 cm² (d) 616 cm²

Solution: Cone height = 31 − 7 = 24, slant = √(49 + 576) = 25. Cone surface = (22/7) × 7 × 25 = 550; hemisphere curved surface = 2 × 154 = 308.

Total = 550 + 308 = 858 cm². The joined circle is hidden, so it is not added.

Answer: (b) 858 cm²

Example 5: thick hemispherical shell

Question: A hemisphere of radius 7 cm is scooped out of a solid hemisphere of radius 14 cm, both having the same centre and the same flat face. Find the total surface area of the shell (π = 22/7). (a) 1540 cm² (b) 2156 cm² (c) 1848 cm² (d) 2002 cm²

Solution: Outer curved = 2π × 196 = 392π; inner curved = 2π × 49 = 98π; rim ring = π(196 − 49) = 147π.

Total = (392 + 98 + 147)π = 637π = 637 × 22/7 = 2002 cm². Option (b) wrongly adds the full flat circle 196π instead of the ring.

Answer: (d) 2002 cm²

Example 6: capsule melted into spheres

Question: A solid is a cylinder of radius 7 cm and height 14 cm with a hemisphere of radius 7 cm fixed on each end. It is melted into spheres of radius 3.5 cm. How many spheres are made? (a) 16 (b) 18 (c) 20 (d) 24

Solution: Volume = π × 49 × 14 + 4/3 × π × 343 = π(686 + 1372/3) = 3430π/3. 1 small sphere = 4/3 × π × 42.875 = 171.5π/3.

Number = 3430 ÷ 171.5 = 20. The 2 hemispheres together make 1 full sphere of radius 7.

Answer: (c) 20

Common mistakes

Using 3πr² for an open bowl. A bowl has no flat top, so its surface is 2πr² (outer or inner).

Adding the joined circular face in a combined solid. Faces glued together are not on the outside.

Taking the total height of a toy as the cone height. Subtract the hemisphere radius first.

For a thick shell, computing 2/3 π(R − r)³ instead of 2/3 π(R³ − r³).

Writing hemisphere volume as 1/3 πr³ (that is the cone with h = r) or 4/3 πr³ (that is the full sphere).

Practice questions

Q1. The curved surface area of a hemisphere is 1232 cm². Find its total surface area (π = 22/7). (a) 1848 cm² (b) 2464 cm² (c) 1232 cm² (d) 1386 cm²

Q2. A hemisphere, a cone and a cylinder have equal radii, and the cone and cylinder have height equal to that radius. The ratio of their volumes is: (a) 1 : 2 : 3 (b) 3 : 1 : 2 (c) 2 : 1 : 3 (d) 2 : 3 : 1

Q3. A hemispherical bowl of inner radius 21 cm is full of oil. The oil is poured into cylindrical bottles of radius 3.5 cm and height 6 cm. How many bottles are filled? (a) 72 (b) 96 (c) 63 (d) 84

Answers

Q1. (a) 2πr² = 1232 gives πr² = 616, so total = 3πr² = 1848 cm².

Q2. (c) 2/3 πr³ : 1/3 πr³ : πr³ = 2 : 1 : 3.

Q3. (d) Bowl = 2/3 × (22/7) × 9261 = 19404 cm³. Bottle = (22/7) × 12.25 × 6 = 231 cm³. 19404 ÷ 231 = 84.

Quick revision

Hemisphere: volume 2/3 πr³, curved 2πr², solid total 3πr².

Shell: volume 2/3 π(R³ − r³); surface 2πR² + 2πr² + π(R² − r²).

Sphere cut in 2: surface rises by 2πr² (50%); hemisphere : cone : cylinder (h = r) = 2 : 1 : 3.

Combined solids: add volumes; add only exposed surfaces.

Concepts and rules

Cuboid with length l, breadth b, height h: volume = lbh; total surface = 2(lb + bh + hl); lateral surface (4 walls) = 2h(l + b); diagonal = √(l² + b² + h²).

Cube of side a: volume = a³, total surface = 6a², lateral surface = 4a², body diagonal = √3a, face diagonal = √2a.

Identity: (l + b + h)² = l² + b² + h² + 2(lb + bh + hl), i.e. (sum of edges/4)² = diagonal² + total surface. The sum of all 12 edges is 4(l + b + h).

If 3 adjacent faces have areas x, y, z, then xyz = (lb)(bh)(hl) = (lbh)², so volume = √(xyz).

The longest rod that fits in a room is its diagonal. A room's 4 walls = 2h(l + b); with the ceiling add lb. Subtract doors and windows when told.

Painted cube of side n units cut into n³ unit cubes (n ≥ 2): 3 faces painted = 8 (corners); 2 faces = 12(n − 2) (edges); 1 face = 6(n − 2)² (face centres); no face = (n − 2)³ (inside).

Open box from a sheet L × B by cutting squares of side x from each corner: box = (L − 2x) × (B − 2x) × x.

Water problems: rain volume = area × depth of rain; water flowing through a channel per hour = cross-section × speed. 1 m³ = 1000 litres, 1 litre = 1000 cm³.

Bricks in a wall = wall volume ÷ brick volume (same units). Largest equal cubes cut from a cuboid have side = HCF(l, b, h).

Method and shortcuts

Step 1: Write l, b, h in the same unit. Decide whether the question needs volume (fill, bricks, water), lateral surface (walls) or total surface (closed box, painting all faces).

Step 2: If only sums, diagonals or face areas are given, use the identities above instead of finding l, b, h separately.

Step 3: For cost, multiply the area or volume by the rate only after converting to the unit in the rate.

Shortcut: for painted cubes check that 8 + 12(n − 2) + 6(n − 2)² + (n − 2)³ = n³. For n = 4: 8 + 24 + 24 + 8 = 64.

Shortcut: if each edge of a cube is increased by x%, the surface increases by (2x + x²/100)% and the volume by ((1 + x/100)³ − 1) × 100%.

Example 1: diagonal, volume and surface

Question: A cuboid measures 12 cm × 4 cm × 3 cm. Find its diagonal, volume and total surface area.

Solution: Diagonal = √(144 + 16 + 9) = √169 = 13 cm. Volume = 12 × 4 × 3 = 144 cm³. Surface = 2(48 + 12 + 36) = 192 cm².

Answer: 13 cm, 144 cm³, 192 cm²

Example 2: volume from face areas

Question: The areas of 3 adjacent faces of a cuboid are 12 cm², 20 cm² and 15 cm². Find its volume.

Solution: Volume = √(12 × 20 × 15) = √3600 = 60 cm³. (The sides are 3, 4 and 5 cm.)

Answer: 60 cm³

Example 3: painted cube

Question: A cube of side 5 cm is painted on all faces and cut into 1 cm cubes. How many small cubes have exactly 2 faces painted, and how many have no face painted?

Solution: n = 5. 2 faces: 12 × (5 − 2) = 36. No face: (5 − 2)³ = 27. Check: 8 + 36 + 54 + 27 = 125.

Answer: 36 and 27

Example 4: painting the walls of a room

Question: A room is 10 m long, 8 m wide and 4 m high. Find the cost of painting its 4 walls at ₹25 per m². (a) ₹3,600 (b) ₹5,600 (c) ₹7,200 (d) ₹4,800

Solution: Walls = 2 × 4 × (10 + 8) = 144 m². Cost = 144 × 25 = ₹3,600. Option (b) adds the ceiling 80 m² as well.

Answer: (a) ₹3,600

Example 5: surface from edge sum and diagonal

Question: The sum of the length, breadth and height of a cuboid is 19 cm and its diagonal is 11 cm. Find its total surface area. (a) 120 cm² (b) 361 cm² (c) 240 cm² (d) 260 cm²

Solution: (l + b + h)² = diagonal² + total surface, so 361 = 121 + surface, giving surface = 240 cm².

Option (a) is lb + bh + hl, which is only half the total surface.

Answer: (c) 240 cm²

Example 6: rain from a roof into a tank

Question: Rain of 5 cm falls on a flat roof 22 m × 20 m. All the water flows into a cylindrical tank of diameter 2 m. By how much does the water level in the tank rise (π = 22/7)? (a) 3.5 m (b) 7 m (c) 14 m (d) 5 m

Solution: Rain volume = 22 × 20 × 0.05 = 22 m³. Tank radius = 1 m, so (22/7) × 1 × h = 22.

h = 7 m. Option (a) uses the diameter 2 m as the radius.

Answer: (b) 7 m

Common mistakes

Using total surface for the walls of a room. 4 walls = 2h(l + b); floor and ceiling are separate.

Forgetting to convert rain in cm into metres (5 cm = 0.05 m) before multiplying by an area in m².

For an open box, subtracting x instead of 2x from each side of the sheet.

Treating lb + bh + hl as the total surface; the total surface is twice this.

Using the face diagonal √2a for the longest rod in a cube; the longest rod is the body diagonal √3a.

Practice questions

Q1. An open box is made from a sheet 48 cm × 36 cm by cutting a square of side 8 cm from each corner. Find the capacity of the box. (a) 4.8 litres (b) 6.4 litres (c) 5.12 litres (d) 5.6 litres

Q2. The diagonal of a cube is 6√3 cm. Find its total surface area. (a) 216 cm² (b) 108 cm² (c) 144 cm² (d) 324 cm²

Q3. A wall 8 m long, 6 m high and 22.5 cm thick is built with bricks of 25 cm × 11.25 cm × 6 cm. How many bricks are needed? (a) 5600 (b) 6000 (c) 7200 (d) 6400

Answers

Q1. (c) Box = (48 − 16) × (36 − 16) × 8 = 32 × 20 × 8 = 5120 cm³ = 5.12 litres.

Q2. (a) √3a = 6√3 gives a = 6; surface = 6 × 36 = 216 cm².

Q3. (d) Wall = 800 × 600 × 22.5 = 10800000 cm³; brick = 25 × 11.25 × 6 = 1687.5 cm³; bricks = 10800000 ÷ 1687.5 = 6400.

Quick revision

Cuboid: lbh; 2(lb + bh + hl); walls 2h(l + b); diagonal √(l² + b² + h²).

Cube: a³; 6a²; body diagonal √3a; face diagonal √2a.

(l + b + h)² = diagonal² + total surface; volume = √(product of 3 face areas).

Painted cube: 8, 12(n − 2), 6(n − 2)², (n − 2)³.

Concepts and rules

A right pyramid has its apex directly above the centre of its base. Height h = apex to base centre; slant height l = apex to the midpoint of a base edge; lateral edge e = apex to a base vertex.

Volume of any pyramid = 1/3 × base area × height. Lateral surface (regular base) = 1/2 × base perimeter × slant height. Total surface = lateral surface + base area.

The 3 lengths form right triangles: l² = h² + ρ², where ρ = inradius of the base (centre to edge midpoint); e² = h² + R², where R = circumradius of the base (centre to vertex). Also e² = l² + (side/2)².

Square base of side a: ρ = a/2, R = a/√2, base area = a², perimeter = 4a. So l = √(h² + a²/4) and e = √(h² + a²/2).

Equilateral triangle base of side a: ρ = a/(2√3), R = a/√3, base area = √3a²/4, perimeter = 3a. So l = √(h² + a²/12) and e = √(h² + a²/3).

Regular tetrahedron (all 6 edges = a, all 4 faces equilateral): height = a√(2/3), volume = a³/(6√2), total surface = √3a².

Counting: a pyramid on an n-sided base has n + 1 faces, n + 1 vertices and 2n edges. A prism on an n-sided base has n + 2 faces, 2n vertices and 3n edges. Euler: faces + vertices − edges = 2.

A pyramid and a prism with the same base and height: volume ratio 1 : 3, the same as a cone and a cylinder.

If the base is not regular (e.g. a right triangle), the volume formula still holds, but lateral faces may have different slant heights, so compute each face separately.

Method and shortcuts

Step 1: Note which length is given: height, slant height or lateral edge. Do not mix them up.

Step 2: Use the right triangle h–ρ–l or h–R–e of the base shape to get the missing length.

Step 3: Volume needs h; lateral surface needs l. Add the base only for total surface.

Shortcut: for a square base, the triangle (h, a/2, l) is often a triplet: a = 10, h = 12 gives l = 13; a = 12, h = 8 gives l = 10; a = 6, h = 4 gives l = 5.

Shortcut: from the lateral surface S of a regular pyramid, slant = 2S ÷ base perimeter.

Example 1: square pyramid

Question: A right pyramid has a square base of side 10 cm and height 12 cm. Find its slant height, lateral surface, total surface and volume.

Solution: l = √(144 + 25) = 13 cm. Lateral = 1/2 × 40 × 13 = 260 cm². Total = 260 + 100 = 360 cm². Volume = 1/3 × 100 × 12 = 400 cm³.

Answer: 13 cm, 260 cm², 360 cm², 400 cm³

Example 2: faces, edges and vertices

Question: How many faces, edges and vertices does a pyramid on a pentagonal base have?

Solution: n = 5: faces = 6, edges = 10, vertices = 6. Euler check: 6 + 6 − 10 = 2.

Answer: 6 faces, 10 edges, 6 vertices

Example 3: pyramid on a right-triangle base

Question: A pyramid has a triangular base with sides 3 cm, 4 cm and 5 cm and a height of 10 cm. Find its volume.

Solution: 3² + 4² = 5², so the base is right-angled with area 1/2 × 3 × 4 = 6 cm². Volume = 1/3 × 6 × 10 = 20 cm³.

Answer: 20 cm³

Example 4: height from slant height

Question: A right pyramid has a square base of side 8 cm and slant height 5 cm. Find its volume. (a) 64 cm³ (b) 192 cm³ (c) 80 cm³ (d) 128 cm³

Solution: h = √(25 − 16) = 3 cm. Volume = 1/3 × 64 × 3 = 64 cm³. Option (b) forgets the 1/3, and (c) uses the slant height 5 in place of h.

Answer: (a) 64 cm³

Example 5: equilateral-triangle base

Question: A right pyramid has an equilateral triangle base of side 6√3 cm and height 4 cm. Find its total surface area. (a) 45√3 cm² (b) 27√3 cm² (c) 36√3 cm² (d) 72√3 cm²

Solution: Inradius of base = 6√3/(2√3) = 3, so slant l = √(16 + 9) = 5 cm. Lateral = 1/2 × 18√3 × 5 = 45√3 cm².

Base = √3/4 × 108 = 27√3 cm². Total = 45√3 + 27√3 = 72√3 cm². Option (c) is the volume value 1/3 × 27√3 × 4 = 36√3 in cm³.

Answer: (d) 72√3 cm²

Example 6: volume from the lateral edge

Question: A right pyramid has a square base of side 8 cm, and each lateral edge is 9 cm. Find its volume. (a) 448 cm³ (b) 448/3 cm³ (c) 512/3 cm³ (d) 224/3 cm³

Solution: Centre to vertex R = half the diagonal = 8√2/2 = 4√2, so R² = 32. h² = 81 − 32 = 49, h = 7 cm.

Volume = 1/3 × 64 × 7 = 448/3 cm³. Using a/2 = 4 instead of R would give h = √65, a common trap.

Answer: (b) 448/3 cm³

Common mistakes

Using slant height in the volume formula. Volume always uses the vertical height h.

Confusing slant height with lateral edge. Slant goes to the midpoint of a base edge (uses ρ); lateral edge goes to a corner (uses R).

Forgetting the 1/3 in the volume, which gives the prism volume instead.

Adding the base to the lateral surface when only the lateral (slant) surface is asked.

For a triangular base, using a/2 as the centre-to-edge distance. For an equilateral triangle it is a/(2√3).

Practice questions

Q1. Find the volume of a regular tetrahedron of edge 6 cm. (a) 36√2 cm³ (b) 18√2 cm³ (c) 9√2 cm³ (d) 72 cm³

Q2. A right pyramid on a square base of side 12 cm has lateral surface area 240 cm². Find its volume. (a) 480 cm³ (b) 1152 cm³ (c) 384 cm³ (d) 288 cm³

Q3. A pyramid has 12 edges. How many faces does it have? (a) 6 (b) 8 (c) 12 (d) 7

Answers

Q1. (b) a³/(6√2) = 216/(6√2) = 36/√2 = 18√2 cm³.

Q2. (c) 1/2 × 48 × l = 240 gives l = 10; h = √(100 − 36) = 8; volume = 1/3 × 144 × 8 = 384 cm³.

Q3. (d) 2n = 12 gives n = 6 (hexagonal base), so faces = 6 + 1 = 7.

Quick revision

Volume = 1/3 × base × h; lateral = 1/2 × perimeter × l; total = lateral + base.

l² = h² + ρ²; e² = h² + R². Square: ρ = a/2, R = a/√2. Equilateral: ρ = a/(2√3), R = a/√3.

Tetrahedron: volume a³/(6√2), surface √3a².

n-gon pyramid: n + 1 faces, 2n edges, n + 1 vertices; prism: n + 2 faces, 3n edges, 2n vertices.