Mensuration and Geometry

Mensuration is the measurement of lengths, areas and volumes; the geometry it needs is a short list of facts about angles, triangles, circles and polygons. A figure or a description is given, and the task is a length, an area, a volume or a ratio of two of these. A scaling or a "melt and recast" step turns the task into a ratio calculation.

The method does not change: draw the figure, mark what is known, choose the formula whose inputs are known (or one step away, such as a slant height from Pythagoras), and keep everything exact in $\pi$ and surds until the last line.

Here “base” in an area formula means base area, not a length. Volume equals base area times perpendicular height for oblique prisms too; the displayed surface-area formula requires a right prism.

Exactness convention. Results are kept exact in $\pi$ and surds; decimals use $\pi\approx3.14159$ and are rounded to two places. With $\pi=22/7$, a radius divisible by 7 cancels a factor 7 and often simplifies arithmetic, but does not guarantee an integer answer. Check which value the question uses.

For points $(x_1, y_1)$ and $(x_2, y_2)$: distance $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, midpoint $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$, slope $\frac{y_2 - y_1}{x_2 - x_1}$ for $x_2\ne x_1$; a vertical line through distinct points has undefined slope. The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is

Heron's formula gives the area from the sides alone. For the 13, 14, 15 triangle, an independent route is the altitude: dropping the altitude $h$ to the side of 14 splits it into $x$ and $14 - x$ with $13^2 - x^2 = 15^2 - (14 - x)^2$, so $x = 5$ and $h = 12$; then the area is $\frac{1}{2} \times 14 \times 12 = 84$, the same as Heron.

Scaling. Area is a sum of tiny squares, each of whose sides scales by $k$, so each square scales by $k^2$; volume is a sum of tiny cubes, each scaling by $k^3$. A 20% increase in every length ($k = 1.2$) increases area by $1.44 - 1 = 44\%$ and volume by $1.728 - 1 = 72.8\%$.

Cone and cylinder. A cone has one third of the volume of the cylinder with the same base and height; a sphere has two thirds of the volume of the cylinder that just encloses it (radius $r$, height $2r$).

Example 1 (Heron). Sides 13, 14, 15: $s = 21$, area $\sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84$. The altitude to the side of 14 is $2 \times 84 / 14 = 12$ (Figure 1).

Example 2 (equilateral triangle). Side 6: area $\frac{\sqrt{3}}{4} \times 36 = 9\sqrt{3} \approx 15.59$; height $3\sqrt{3} \approx 5.20$.

Example 3 (sector). Radius 14, angle $90^\circ$: area $\frac{90}{360}\pi \times 196 = 49\pi \approx 153.94$ (exactly 154 with $\pi = 22/7$); arc length $\frac{90}{360} \times 2\pi \times 14 = 7\pi \approx 21.99$ (22 with $22/7$).

Example 4 (shaded region). A circle is inscribed in a square of side 10, so its radius is 5. The area inside the square but outside the circle is $100 - 25\pi \approx 21.46$ (Figure 1).

Example 5 (rhombus). Diagonals 10 and 24: area $\frac{1}{2} \times 10 \times 24 = 120$. The diagonals bisect each other at right angles, so the side is $\sqrt{5^2 + 12^2} = 13$.

Example 6 (cuboid). $12 \times 4 \times 3$: volume $144$; total surface area $2(48 + 12 + 36) = 192$; space diagonal $\sqrt{144 + 16 + 9} = 13$.

Example 7 (cylinder). $r = 7$, $h = 10$: volume $490\pi \approx 1539.38$ (1540 with $22/7$); curved surface $140\pi \approx 439.82$.

Example 8 (cone). $r = 3$, $h = 4$: slant $l = 5$; volume $\frac{1}{3}\pi \times 9 \times 4 = 12\pi \approx 37.70$; curved surface $15\pi$; total surface $24\pi$.

Example 9 (melt and recast). A solid sphere of radius 6 is melted and recast into cones of radius 3 and height 4. Sphere volume: $\frac{4}{3}\pi \times 216 = 288\pi$. Each cone: $12\pi$. Number of cones: $288\pi / 12\pi = 24$.

Example 10 (scaling). Each edge of a cube grows by 20%. Volume grows by $1.2^3 - 1 = 72.8\%$; surface area by $1.2^2 - 1 = 44\%$.

Example 11 (polygons). A regular octagon has each exterior angle $360/8 = 45^\circ$, so each interior angle is $135^\circ$. A regular polygon with interior angle $150^\circ$ has exterior angle $30^\circ$, so it has $360/30 = 12$ sides.

Example 12 (similar triangles). Two similar triangles have corresponding sides in the ratio $2 : 3$. Their areas are in the ratio $4 : 9$. If the smaller has area 20, the larger has area $20 \times 9/4 = 45$.

Example 13 (coordinates). The triangle with vertices $(0, 0)$, $(4, 0)$, $(1, 3)$ has area $\frac{1}{2}|0(0 - 3) + 4(3 - 0) + 1(0 - 0)| = 6$ (base 4, height 3 agrees). The distance from $(1, 2)$ to $(4, 6)$ is $\sqrt{9 + 16} = 5$.