Solid Geometry and Mensuration

Mensuration is the part of the MSTE paper where a single remembered formula turns a wordy solid into one clean number. The board expects you to know the standard areas, surface areas, and volumes cold, and to recognize when a general tool like the prismatoid formula or a theorem of Pappus collapses a hard-looking solid into a short calculation. This lesson lays out the formulas in the order the exam uses them, from flat figures up to spheres and solids of revolution, with the arithmetic worked all the way through so you can check your own habits against it.

Areas of plane figures

Before any solid, you need its faces. The triangle alone shows up three ways: base times height halved, Heron's formula from the three sides, and the two-sides-and-included-angle form. Keep the whole toolkit ready.

Figure	Area
Triangle (base b, height h)	(1/2) b h
Triangle (Heron, sides a, b, c)	sqrt(s(s - a)(s - b)(s - c)), where s = (a + b + c)/2
Triangle (two sides, included angle C)	(1/2) a b sin(C)
Parallelogram (base b, height h)	b h
Trapezoid (parallel sides b1, b2, height h)	(1/2)(b1 + b2) h
Circle (radius r)	pi r^2
Circular sector (radius r, central angle theta in radians)	(1/2) r^2 theta
Regular polygon (perimeter P, apothem a)	(1/2) P a

Worked example (Heron): A triangle has sides 13, 14, and 15. The semiperimeter is s = (13 + 14 + 15)/2 = 21, so the area is sqrt(21(21 - 13)(21 - 14)(21 - 15)) = sqrt(21 * 8 * 7 * 6) = sqrt(7056) = 84 square units. Worked example (sector): A sector of radius 6 cm subtends 60 degrees. Convert the angle: 60 degrees = pi/3 radians, so the area is (1/2)(6^2)(pi/3) = 18 * (pi/3) = 6 pi = 18.85 cm^2, and the arc length is r theta = 6 * (pi/3) = 2 pi = 6.28 cm. The classic slip is to forget the 1/2 and report 37.70, or to report the arc length when the area was asked.

Prisms and cylinders

A prism and a cylinder share one idea: a constant cross-section swept a distance h. The volume is always the base area times the height, and the lateral (side) area is the base perimeter times the height.

• Right prism: V = B h, where B is the base area. Lateral area = (perimeter of base) * h. Total surface = lateral + 2B.
• Cylinder: V = pi r^2 h. Lateral area = 2 pi r h. Total surface = 2 pi r h + 2 pi r^2 = 2 pi r (h + r).

Worked example (prism): A prism whose base is a right triangle with legs 6 and 8 has base area (1/2)(6)(8) = 24 cm^2. If the prism is 10 cm long, V = 24 * 10 = 240 cm^3. Using the pyramid rule (1/3) B h here, which gives 80, is a common trap; a prism does not carry the 1/3. Worked example (cylinder): A closed cylinder has r = 5 cm and h = 12 cm. Its volume is pi (5^2)(12) = 300 pi = 942.5 cm^3, its lateral area is 2 pi (5)(12) = 120 pi = 377.0 cm^2, and its total surface is 2 pi (5)(12 + 5) = 170 pi = 534.1 cm^2.

Pyramids and cones

Cap a base with an apex and the volume drops to one third of the matching prism or cylinder. The lateral surface of a cone uses the slant height L, not the vertical height h.

• Pyramid: V = (1/3) B h.
• Cone: V = (1/3) pi r^2 h. Slant height L = sqrt(r^2 + h^2). Lateral area = pi r L. Total surface = pi r L + pi r^2 = pi r (L + r).

Worked example (cone): A right circular cone has r = 3 cm and h = 4 cm. First the slant height, L = sqrt(3^2 + 4^2) = sqrt(25) = 5 cm. Then V = (1/3) pi (3^2)(4) = 12 pi = 37.70 cm^3, the lateral area is pi (3)(5) = 15 pi = 47.12 cm^2, and the total surface is pi (3)(5 + 3) = 24 pi = 75.40 cm^2. Two errors to avoid: dropping the 1/3 in the volume (which gives 113.1) and using h instead of the slant L in the lateral area (which gives 37.70).

Frustums of pyramids and cones

Slice the top off a pyramid or cone with a plane parallel to the base and what remains is a frustum. It has two parallel end areas, A1 (lower) and A2 (upper), separated by a perpendicular height h. The volume rule is the same for both shapes:

V = (h/3)(A1 + A2 + sqrt(A1 A2)).

For a cone frustum with lower radius R and upper radius r this becomes V = (pi h/3)(R^2 + R r + r^2). The slanted side of a cone frustum has slant height L = sqrt((R - r)^2 + h^2), and the lateral (curved) area is pi (R + r) L.

Worked example: A cone frustum has R = 6 cm, r = 3 cm, and h = 4 cm. The volume is (pi (4)/3)(6^2 + 6 * 3 + 3^2) = (4 pi/3)(36 + 18 + 9) = (4 pi/3)(63) = 84 pi = 263.9 cm^3. For the lateral area, first the slant height L = sqrt((6 - 3)^2 + 4^2) = sqrt(9 + 16) = 5 cm, then lateral area = pi (6 + 3)(5) = 45 pi = 141.4 cm^2. Dropping the cross term R r in the volume gives 188.5, the usual miss.

The sphere, zones, and spherical segments

A sphere of radius r has volume V = (4/3) pi r^3 and surface area S = 4 pi r^2. Two parallel planes slice the surface into a zone and slice the solid into a spherical segment.

Archimedes' remarkable result is that a zone's area depends only on its altitude h (the distance between the cutting planes), not on where along the sphere it sits: a zone of altitude h on a sphere of radius r has area 2 pi r h. A spherical segment of one base (a cap, with a single cutting plane) has volume V = (pi h^2/3)(3r - h), and a segment of two bases with base radii a and b has volume V = (pi h/6)(3a^2 + 3b^2 + h^2).

Worked example (sphere): A sphere of radius 6 cm has V = (4/3) pi (6^3) = (4/3) pi (216) = 288 pi = 904.8 cm^3 and S = 4 pi (6^2) = 144 pi = 452.4 cm^2. Worked example (zone): On a sphere of radius R = 10 cm, a zone of altitude h = 4 cm has area 2 pi (10)(4) = 80 pi = 251.3 cm^2. Worked example (cap volume): A one-base segment of the same sphere with h = 4 cm has V = (pi (4^2)/3)(3 * 10 - 4) = (16 pi/3)(26) = (416/3) pi = 435.6 cm^3. Forgetting to subtract h and using (16 pi/3)(30) = 502.7 is the standard error.

The prismatoid formula

A prismatoid is any solid whose vertices all lie in two parallel planes; prisms, pyramids, cones, frustums, and spheres are all special cases. Its volume comes from one formula (also called the prismoidal formula, the same shape as Simpson's rule):

V = (h/6)(A1 + 4 Am + A2),

where A1 and A2 are the two end areas, Am is the area of the cross-section taken midway between the ends, and h is the perpendicular distance between the end planes. The one point to nail is that Am is a genuine mid-section area, not the average of the two ends; the 4 Am term is what makes the rule exact for these solids.

Worked example: A frustum of a pyramid has a 6 by 6 square base (A1 = 36), a 2 by 2 square top (A2 = 4), and height h = 9. The mid-section is a 4 by 4 square, so Am = 16. Then V = (9/6)(36 + 4 * 16 + 4) = (1.5)(36 + 64 + 4) = 1.5 * 104 = 156 cm^3. The frustum rule confirms it: (9/3)(36 + 4 + sqrt(36 * 4)) = 3(36 + 4 + 12) = 156 cm^3. Forgetting the factor 4 on Am gives 84, the trap distractor.

The theorems of Pappus

The two theorems of Pappus turn a plane figure into a solid or surface of revolution using nothing but the figure and the travel of its centroid. In both, the axis of revolution must lie in the plane of the figure and must not cut through it.

• First theorem (surface): revolving a plane arc of length L generates a surface of area S = 2 pi d L, where d is the distance from the axis to the centroid of the arc.
• Second theorem (volume): revolving a plane region of area A generates a solid of volume V = 2 pi d A, where d is the distance from the axis to the centroid of the region.

The factor 2 pi d is simply the distance the centroid travels in one full revolution, so a partial sweep through angle theta (in radians) just replaces 2 pi with theta.

Worked example: A circular region of radius 3 cm has its center 5 cm from an axis in the same plane, and it is revolved fully to form a torus. The region's area is A = pi (3^2) = 9 pi cm^2 with centroid at d = 5 cm, so V = 2 pi (5)(9 pi) = 90 pi^2 = 888.3 cm^3. The outer surface uses the first theorem with the boundary length L = 2 pi (3) = 6 pi: S = 2 pi (5)(6 pi) = 60 pi^2 = 592.2 cm^2. Mixing the two theorems, using the circumference in the volume formula, is the classic error and yields 592.2 for a volume.

Solid formulas at a glance

Solid	Volume	Lateral or surface area
Right prism	B h	(perimeter) * h; total = lateral + 2B
Cylinder	pi r^2 h	lateral 2 pi r h; total 2 pi r (h + r)
Pyramid	(1/3) B h	(1/2)(base perimeter)(slant height) if regular
Cone	(1/3) pi r^2 h	lateral pi r L; total pi r (L + r), L = sqrt(r^2 + h^2)
Cone frustum	(pi h/3)(R^2 + R r + r^2)	lateral pi (R + r) L, L = sqrt((R - r)^2 + h^2)
Sphere	(4/3) pi r^3	4 pi r^2
Spherical segment (one base)	(pi h^2/3)(3r - h)	zone 2 pi r h
Prismatoid	(h/6)(A1 + 4 Am + A2)	depends on the faces

Exam-day strategy

• Pick the right triangle-area tool by the givens: three sides means Heron, two sides plus the included angle means (1/2) a b sin(C), and a base with its own height means (1/2) b h; do not mix them.
• Convert every sector or arc angle to radians before using (1/2) r^2 theta or s = r theta; a degree left in the formula is the most common area error.
• Remember the 1/3 belongs to the apex solids only: prisms and cylinders have no 1/3, while pyramids and cones do.
• For any cone or cone frustum, compute the slant height first, L = sqrt(r^2 + h^2) for a full cone and L = sqrt((R - r)^2 + h^2) for a frustum, then feed it into the lateral-area formula; never substitute the vertical height for L.
• When a solid is awkward but has flat parallel ends, reach for the prismatoid formula V = (h/6)(A1 + 4 Am + A2), and take Am as the true middle cross-section, not the average of the two ends.
• On a solid of revolution, check that the axis does not cut the figure, then apply Pappus: area of the region times 2 pi d for volume, length of the boundary times 2 pi d for surface, using the centroid distance d for each.