面積と体積

The core area formulae are as follows. Rectangle: $A=\text{base}\times \text{height}$. Triangle: $A=\frac{1}{2}\times \text{base}\times \text{height}$. Parallelogram: $A=\text{base}\times \text{height}$, where the height is the perpendicular distance to the base. Trapezium: $A=\frac{1}{2}(a+b)h$, where $a$ and $b$ are the parallel sides and $h$ is the distance between them. For triangles and parallelograms, remember the height is the perpendicular distance, not the slanted side. Area is always measured in square units (cm${}^2$, etc.).

With radius $r$ and diameter $d=2r$, the distance around a circle (the circumference) is $C=2\pi r=\pi d$, and the area of the circle is $A=\pi r^2$. Here $\pi$ (pi) is a constant of about $3.142$; use the value the question specifies, or the calculator's $\pi$ otherwise. Take care not to confuse radius and diameter, and remember that for the area you square the radius first, then multiply by $\pi$. In this topic we use $\pi =3.142$ for numerical work.

A prism is a solid with the same cross-section all the way along its length. Its volume is always 'area of cross-section $\times$ length (height)'. A cuboid is a prism with a rectangular cross-section, so $V=\text{length}\times \text{width}\times \text{height}$. A cylinder is a prism with a circular cross-section, so we multiply the cross-sectional area $\pi r^2$ by the height $h$ to get $V=\pi r^{2}h$. Volume is always measured in cubic units (cm${}^3$, etc.).

The surface of a closed cylinder (like a solid can) is made of two circular ends and a curved side which unrolls into a rectangle. The two circles together have area $2\pi r^2$. The curved (lateral) surface is a rectangle whose width is the circumference $2\pi r$ and whose height is $h$, giving area $2\pi rh$. So the total surface area is $A=2\pi r^2+2\pi rh$. For an open tube or a container with no lid, be careful to add only the faces that are actually present.

A sector is the part of a circle cut off by an angle $\theta$ at the centre. The arc length is $\frac{\theta }{360}$ of the circumference, and the sector area is $\frac{\theta }{360}$ of the area of the circle. The 0580 List of formulas gives only the circumference $2\pi r$ and the area $\pi r^2$, so you multiply by the fraction of the full circle yourself.

The perimeter of a sector is the arc length plus two radii, $2r$. Answers may be asked for in terms of $\pi$; then leave $\pi$ in the answer instead of turning it into a decimal.

Scope: in 0580 (2025–2027), Core uses only sector angles that are factors of $360^{\circ }$ (such as $90^{\circ }$, $60^{\circ }$, $45^{\circ }$). Extended can use any angle and includes major sectors, whose angle is greater than $180^{\circ }$.

The formulas for spheres, cones and pyramids are on the List of formulas printed on page 2 of the question paper. Volume of a sphere $V=\frac{4}{3}\pi r^3$, surface area of a sphere $A=4\pi r^2$, volume of a cone $V=\frac{1}{3}\pi r^{2}h$, curved surface area of a cone $A=\pi rl$, and volume of a pyramid $V=\frac{1}{3}Ah$, where $A$ is the base area.

For a cone, $h$ is the perpendicular height and $l$ is the slant height (the sloping edge from the vertex to the edge of the base); they are different lengths. If you only know one, use Pythagoras' theorem, $l^2=r^2+h^2$ (this is not on the List of formulas). The total surface area of a cone is the curved surface $\pi rl$ plus the circular base $\pi r^2$.

For a pyramid, find the base and each triangular face separately and add them. The height of a triangular face is measured up the face, not the vertical height of the pyramid.

In 0580 (2025–2027), spheres, cones and pyramids are in both Core and Extended (C5.4 / E5.4).

The 0580 syllabus (E5.5) also covers the surface area and volume of compound solids and parts of solids; its example is a frustum. Find the volume of a frustum as 'large cone minus the small cone removed'. A hemisphere is half a sphere, so its volume is $\frac{2}{3}\pi r^3$, and the total surface area of a solid hemisphere is the curved part $2\pi r^2$ plus the flat circle $\pi r^2$, which is $3\pi r^2$.

Check on the diagram which faces are on the outside; faces that are joined together and hidden are not part of the surface area.