Mensuration

Recasting solids (melting and moulding)

Recasting means melting a solid and moulding the same metal into a new shape. The governing principle is conservation of volume: the volume of metal before melting equals the total volume of the new objects, provided no metal is lost. Surface area is not conserved, so a recast object usually has a different surface area from the original. Keep these two quantities separate, because one recasting problem may ask for a volume-based quantity and a surface-based quantity.

The standard formulas you need are:

• Sphere: volume \frac{4}{3}\pi R^3, surface area 4\pi R^2.
• Hemisphere: volume \frac{2}{3}\pi R^3, curved surface area 2\pi R^2.
• Cylinder: volume \pi r^2 h, curved surface area 2\pi r h.
• Cone: volume \frac{1}{3}\pi r^2 h.
• Cube of edge a: volume a^3, surface area 6a^2.
• Cuboid: volume lbh, surface area 2(lb+bh+hl).

The working method is always the same. Find the volume of the original solid, equate it to the sum of the volumes of the new pieces, and solve for the unknown dimension. If the unknown is an edge or radius inside a cube or sphere, take a cube root, so that r^3 = 27 gives r = 3. When a large cube is cut into several smaller cubes, the leftover volume after the known cubes are removed equals the volume of the remaining cube, and its edge is the cube root of that volume.

A sphere stretched into a wire is a volume equation with the wire regarded as a cylinder: \frac{4}{3}\pi R^3 = \pi r^2 L, where r is the wire radius and L its length. For example, a sphere of radius 2 cm drawn into a wire of length 96 cm has \frac{32\pi}{3} = \pi r^2 (96), so r^2 = \frac{1}{9} and r = \frac{1}{3} cm. Check whether the question asks for the radius or the diameter, since the diameter is twice the radius.

Units must be made consistent before equating volumes. Note that 1 \text{ m}^3 = 10^6 \text{ cm}^3 and 1 \text{ litre} = 1000 \text{ cm}^3. A coin or a thin plate is a cylinder whose height is its thickness, and a thickness given in millimetres must be converted to centimetres before use. The number of identical objects is the total volume divided by the volume of one object.

A hemispherical bowl filled with liquid and poured into identical cylindrical bottles gives the number of bottles as the bowl volume divided by one bottle volume. For a bowl of radius 6 cm and bottles of radius 2 cm and height 3 cm, the bowl volume is 144\pi cm^3 and each bottle holds 12\pi cm^3, so 12 bottles are needed.

Some solids combine into a known solid. A cone of base radius x and height 2x has volume \frac{2}{3}\pi x^3, and a hemisphere of radius x has volume \frac{2}{3}\pi x^3. Together they occupy \frac{4}{3}\pi x^3, which is the volume of a sphere of radius x. Such equivalences are decided by comparing volume expressions, not dimensions.

If several spheres are melted into one, the radii satisfy R^3 = r_1^3 + r_2^3 + \cdots, because volumes add and volume of a sphere is proportional to the cube of its radius. A resulting radius that is not a whole number is found by a cube-root estimate, and the estimate must be checked against the given range of options.

For n equal spheres of radius r melted into one sphere of radius R, we have R = n^{1/3} r. The ratio of the large sphere's surface area to the total surface area of the small spheres is \frac{4\pi R^2}{n \cdot 4\pi r^2} = n^{-1/3}. For example, 8 spheres of radius 1 cm melted into one sphere give R = 2 cm, so the total surface area falls from 32\pi to 16\pi.

The percentage diminution of surface area is \frac{\text{original} - \text{new}}{\text{original}} \times 100\%. Melting several spheres into one always reduces the total surface area because a single sphere has the smallest surface for a given volume.

For a cuboid, the ratio problems are solved by writing each dimension as a multiple of one parameter. A cuboid of 16 cm \times 2 cm \times 2 cm has volume 64 cm^3, so the cube has edge 4 cm. The surface areas are 2(32+4+32) = 136 cm^2 for the cuboid and 96 cm^2 for the cube, giving the ratio \frac{17}{12}.

A sphere that touches the base and the lateral surface of a cone is the inscribed sphere of the axial triangle. Its radius is the inradius r = \frac{A}{s}, where A is the area and s the semiperimeter of that triangle. The slant height of the cone comes from Pythagoras, l = \sqrt{R^2+h^2}.

A thin circular plate of radius r and thickness t has volume \pi r^2 t. If a sector of angle \theta is cut off, the remaining metal has volume \frac{360^\circ - \theta}{360^\circ}\pi r^2 t. Thickness and radius must be expressed in the same unit before the remaining volume is equated to a sphere's volume.

Sector, arc and semicircle measures

A sector is the region enclosed by two radii and the arc between them. Its measures depend on the central angle \theta and the radius r. The angle must be in radians when used in the arc-length formula, with \theta_{\text{rad}} = \theta^\circ \cdot \frac{\pi}{180}.

The key formulas are:

• Arc length: l = r\theta, with \theta in radians.
• Sector area: \frac{\theta}{360^\circ}\pi r^2 = \frac{1}{2} r^2 \theta = \frac{1}{2} r l.
• Sector perimeter: 2r + l.
• Semicircle perimeter: \pi r + 2r = (\pi + 2) r; semicircle area: \frac{1}{2}\pi r^2.
• Circumference: 2\pi r, and circumference minus diameter equals d(\pi - 1) for diameter d.

The relation between arc and area gives a shortcut: if the arc length and the radius are known, the area is \frac{1}{2} r \times l, with no need to find the angle. Conversely, if the sector perimeter and the radius are known, the arc is the perimeter minus 2r. For example, a sector of radius 5 cm and perimeter 14 cm has arc 4 cm and area 10 cm^2.

A semicircle is half of a circle, so its area is half the full-circle area with the same radius. When two regions have different radii, compare the squares of the radii, since areas scale with r^2.

The minute hand of a clock turns 6^\circ per minute, so a stretch of 20 minutes sweeps 120^\circ. The area swept by a hand of length r is the sector area for that angle. For a hand of length 14 cm sweeping 90 minutes-equivalent angle, the same method applies.

A semicircular track of radius 7 m with \pi = \frac{22}{7} has perimeter 22 + 14 = 36 m and area 77 m^2. When the perimeter is given, first write it as (\pi+2)r and solve for r.

Equal arcs of two circles satisfy r_1\theta_1 = r_2\theta_2. The radii are therefore inversely proportional to the central angles, so the ratio of radii equals the inverse of the ratio of angles.

For a circle, the difference between the circumference and the diameter is d(\pi-1). Solving d(\pi-1) = k gives d = \frac{k}{\pi-1}, which is the standard way to find the diameter from such a difference.

A pendulum of length L swinging through angle \theta describes an arc L\theta with \theta in radians. Thus L = \frac{\text{arc}}{\theta_{\text{rad}}}. For an angle of 60^\circ, \theta_{\text{rad}} = \frac{\pi}{3}, so an arc of 11 cm corresponds to L = \frac{33}{\pi} cm.

A chord of length c in a circle of radius r subtends angle \theta at the centre with c = 2r\sin\frac{\theta}{2}. Its distance from the centre is \sqrt{r^2 - (c/2)^2}. Parallel chords that are on the same side of the centre are separated by the difference of these distances, and the longer chord is always nearer to the centre.

A statement-sufficiency question asks whether a given piece of information alone fixes the answer. Each statement must be tested separately. Here, a single arc length with its angle determines the radius, and a single chord with its angle also determines the radius through c = 2r\sin\frac{\theta}{2}, so either piece of information can be sufficient on its own.

When a set of sectors divides a circle, the ratio parts sum to a total T, and each sector's angle is \frac{\text{part}}{T}\times 360^\circ. Using the sector area formula, each area is \frac{\theta}{360^\circ}\pi r^2, so the smallest and largest sectors can be found directly from their parts.

The perimeter of a sector with a small central angle is 2r plus a short arc. Compute the arc from the fraction of the full circumference, \frac{\theta}{360^\circ}\cdot 2\pi r, then add 2r.

A quick reference list:

• radian — the angle subtended by an arc equal in length to the radius; use \theta = \frac{l}{r}. Example: an arc of length 6 cm on a radius of 3 cm subtends 2 radians.
• sector — the region between two radii and an arc; its area is \frac{\theta}{360^\circ}\pi r^2.
• semicircle — half a circle; its perimeter includes the diameter once.

Cylinder volume and curved surface area

A right circular cylinder of radius r and height h has the following measures:

• Volume: \pi r^2 h.
• Curved (lateral) surface area: 2\pi r h.
• Total surface area: 2\pi r (r + h), which is the curved surface plus two circular bases.

When a ratio is given between radius and height, write them as r = ak and h = bk, substitute into the volume, and solve for k by cube root. The curved surface area is then found from k. For a ratio of radius to height of 2:3 and a volume equal to 12\pi k^3, the value of k is found first, and only then the curved surface area 2\pi r h = 12\pi k^2.

A solid made of two cylinders on top of each other has volume equal to the sum of the two volumes. Its total surface area is not the sum of the separate surfaces, because the joined faces are hidden. Convert the total volume to a mass or density unit when required, using 1 \text{ m}^3 = 10^6 \text{ cm}^3. For example, a cylinder of volume 3000 cm^3 plus one of volume 400 cm^3 gives 3400 cm^3 = 0.0034 m^3, and a density of 10000 kg per m^3 gives a mass of 34 kg.

Water flowing through a pipe has volume per unit time equal to the cross-sectional area times the speed of flow, \pi r^2 v. Because a litre is 1000 cm^3, the volume flow must be converted to cm^3 per minute before finding the length of water column per minute. Then convert the length to km per hour.

A tank is emptied by removing spherical droplets one at a time. The number of droplets needed is the cylinder volume divided by the droplet volume, and the time is that number times the interval between droplets. Convert seconds to hours by dividing by 3600.

A cone with base radius R and height h has slant height l = \sqrt{R^2 + h^2}, and its lateral surface area is \pi R l. A sphere that just fits inside the cone touching its sides has radius r = \frac{Rh}{R+l}. The ratio of the cone's lateral area to the sphere's surface area 4\pi r^2 is then a number independent of the scale.

If the total surface area of a cylinder is \frac{2}{3} of... more precisely, when a stated multiple of the total surface area equals a multiple of the curved surface area, substitute \text{TSA} = 2\pi r^2 + 2\pi r h and \text{CSA} = 2\pi r h, then cancel 2\pi r from both sides. The resulting linear relation gives the required ratio of height to radius. For example, the relation 4\pi r^2 = 2\pi r h reduces to h = 2r.

If the sum of radius and height is known together with the curved surface area, let r + h = s and rh = \frac{\text{CSA}}{2\pi}. Then r and h are the roots of t^2 - st + rh = 0. Choose the root that satisfies the condition on which of r and h is larger, and compute the volume \pi r^2 h.

Quick reference list:

• curved surface area (CSA) — the lateral area 2\pi r h, excluding the two bases.
• total surface area (TSA) — 2\pi r(r+h), including both bases.
• slant height — the distance from the apex to a point on the base rim of a cone, \sqrt{R^2+h^2}.

Water level change in cylindrical vessel

When a solid is lowered into a liquid, or liquid is poured from one container into another, the volume of liquid is conserved, so the change in level follows from a volume balance. The basic relation is: volume of liquid added or displaced = base area of the container \times change in level. For a right circular cylinder of radius R, the base area is \pi R^2, so a volume V raises the level by h = \dfrac{V}{\pi R^2}.

When a full cone is emptied into a cylinder, the water volume is the cone volume \frac{1}{3}\pi r^2 H, so the new level is h = \dfrac{r^2 H}{3R^2}. For example, a cone with r = 2 cm and H = 6 cm emptied into a cylinder with R = 4 cm raises the water to h = 0.5 cm.

A fully submerged sphere of radius a displaces exactly its own volume, \frac{4}{3}\pi a^3, so the rise in a cylinder of radius R is h = \dfrac{4a^3}{3R^2}. For a = 2 cm and R = 4 cm, the rise is \frac{2}{3} cm.

Only the part of a body that lies below the liquid surface displaces liquid, and the displaced volume must always be compared with the base area of the vessel. When many identical balls are immersed, the number needed to produce a given rise is the rise volume \pi R^2 h divided by the volume of one ball.

When a container is already full, an immersed body makes water overflow equal to its own volume. The water remaining is therefore the container volume minus the volume of the body. For a cylinder of radius a and height 2a holding a sphere of radius a, the water left is 2\pi a^3 - \frac{4}{3}\pi a^3 = \frac{2}{3}\pi a^3. Since 1\ \text{cm}^3 = 1\ \text{mL}, the remaining volume can be read directly in millilitres when the dimensions are in centimetres.

Frustum of a cone

A frustum is the part of a cone lying between its base and a plane cut parallel to the base. Its two circular ends have radii R (larger) and r (smaller), vertical height h, and slant height l. The slant height comes from the right triangle formed by the height and the difference of the radii: l = \sqrt{h^2 + (R-r)^2}. For R = 10 cm, r = 6 cm and h = 3 cm, the slant height is \sqrt{9+16} = 5 cm.

The volume of a frustum is V = \frac{\pi h}{3}(R^2 + Rr + r^2), and its curved surface area is \pi(R+r)l. A convenient method is to subtract the small cone removed above the cut from the whole cone.

Because the small cone and the whole cone are similar, linear dimensions scale by a ratio k and volumes by k^3. If the small cone has \frac{1}{27} of the volume, then k = \frac{1}{3}. For a whole cone of height 30 cm, the small cone has height 10 cm, so the frustum has height 30 - 10 = 20 cm. Note that the frustum height is the difference, not the height of the small cone.

Spheres placed inside a cone are solved with apex geometry. If a sphere of radius \rho touches the lateral surface and its centre is at distance D from the apex, then D\sin\theta = \rho, where \theta is the half vertex angle, so D = \rho/\sin\theta. For two tangent spheres of radii R and r stacked on the axis with the larger one resting on the table, comparing their centre distances gives \sin\theta = \frac{R-r}{R+r} and a cone height H = \frac{2R^2}{R-r}. Once H is known, the base radius is \rho_b = H\tan\theta = \frac{2R^2\tan\theta}{R-r}.

For a frustum bounded by two parallel circles, the radii are proportional to their distances from the apex. The radii are in the same ratio as the circumferences, so if the circumferences are in ratio 3:4, then r:R = 3:4. If the apex is x from the smaller end and the frustum height is 6, then \frac{x}{x+6} = \frac{3}{4}, which gives x = 18.

Area of circular segment

A chord divides a circle into two segments. The minor segment is the region between the chord and the smaller arc. For a central angle \theta < 180^\circ, it equals the sector minus the triangle formed by the two radii and the chord: A_{\text{minor}} = \frac{\theta}{360^\circ}\pi r^2 - \frac{1}{2}r^2\sin\theta. For r = 6 and \theta = 60^\circ, the segment is 6\pi - 9\sqrt{3} \approx 3.26 square units.

The major segment is the rest of the disc, so A_{\text{major}} = \pi r^2 - A_{\text{minor}}. For r = 6 and the same chord, the major segment is 30\pi + 9\sqrt{3} \approx 109.8 square units.

When the central angle is 90^\circ, the triangle is isosceles right with legs r and chord l = r\sqrt{2}. Hence r^2 = \frac{l^2}{2}, and the minor segment is \frac{\pi r^2}{4} - \frac{r^2}{2} = \frac{r^2}{4}(\pi - 2). In terms of the chord this becomes \frac{l^2}{4}\left(\frac{\pi}{2} - 1\right). The major segment written in the same way is \pi r^2 - \frac{r^2}{4}(\pi-2) = \frac{r^2}{4}(3\pi+2) = \frac{l^2}{4}\left(\frac{3\pi}{2}+1\right).

If a chord is 2\sqrt{3}\,h long, where h is the height of the segment, its half-length is \sqrt{3}h. The intersecting chords relation c^2 = h(2r-h) then gives 3h^2 = h(2r-h), so r = 2h. The central angle satisfies \sin\frac{\theta}{2} = \frac{\sqrt{3}}{2}, so \theta = 120^\circ. Comparing the sector area \frac{4\pi h^2}{3} with the segment area gives a ratio of sector to segment of about 1.7 when \pi = 3.14 and \sqrt{3} = 1.7.

Rectangle area and perimeter word problems

The perimeter of a rectangle is P = 2(l+b) and its area is A = lb, where l is the length and b the breadth. Most word problems reduce to a fixed perimeter with varying sides, or a fixed area with a trade-off between sides.

• breadth — the shorter side, or width, of a rectangle. Example: a rectangle with length 9 m and breadth 4 m has perimeter 26 m.
• plinth — the built-up base area of a building. Example: a plinth of 50 m^2 is the floor area used for the calculation.
• hectare — a unit of area equal to 10000\ \text{m}^2.

For a square of side s and a rectangle with sides s+d and s-d, the perimeters are equal, and the area difference is (s^2) - (s^2 - d^2) = d^2. For example, if the area differs by 4 cm^2, then d = 2 and the length exceeds the breadth by 4 cm.

If two rectangles have equal areas and their lengths and breadths change in opposite directions, the sum l+b changes by the difference of the two changes. The perimeter changes by twice that amount. For lengths differing by 6 and breadths by 4 in opposite directions, the sum changes by 2, so the perimeters differ by 2 \times 2 = 4 units.

For work involving areas and speeds, convert units first. Since 1\ \text{hectare} = 10000\ \text{m}^2, a square of area A has side \sqrt{A} and boundary 4\sqrt{A}. Time equals distance divided by speed. For example, a square of 4 hectares has a side of 200 m and a boundary of 800 m, which takes about 9.6 minutes at 5 km/hr.

If a border of width d is added on each side of a rectangle, each dimension increases by 2d. The new area is (l+2d)(b+2d), and the percentage increase is \frac{\text{new} - \text{old}}{\text{old}} \times 100. For a plinth of area 72 m^2 with ratio 2:1, the breadth is 6 m and the length is 12 m. Adding 1 m on each side gives 14 \times 8 = 112 m^2, an increase of about 55.6\%.

When a rectangle is split into two equal squares and two equal rectangles, the squares' side determines the other dimensions. A statement that uses 'always' is false if one arrangement gives a counterexample. Different arrangements of the same rectangle can make one piece larger than another, so a single valid arrangement can decide the truth of a claim.

Volume of cone and hemisphere

The volume of a cone with base radius r and height h is V = \frac{1}{3}\pi r^2 h, and the volume of a hemisphere of radius r is V = \frac{2}{3}\pi r^3. The phrase 'equal bases' means the two solids have the same base radius.

If a cone and a hemisphere have equal base radii and equal volumes, then \frac{1}{3}\pi r^2 h = \frac{2}{3}\pi r^3, so h = 2r. The ratio of the cone's height to the hemisphere's radius is 2:1.

When a cone is mounted on a hemisphere of the same radius, the common circular face is internal, so the volume of the combined solid is simply the sum of the two volumes. For a cone of radius 3 and height 4 on a hemisphere of radius 3, the cone volume is 12\pi and the hemisphere volume is 18\pi, so the total is 30\pi.

When radius and height are given as a ratio, write r = ak and h = bk. The volume then becomes proportional to k^3, so a given volume determines k through a cube root. For r:h = 3:7, the volume is \frac{1}{3}\pi(3k)^2(7k) = 21\pi k^3, which equals 66k^3 with \pi = \frac{22}{7}. A known volume therefore gives k and then the height h = 7k.