This section covers formulas for the perimeter, area, and volume of common geometric shapes - triangles, rectangles, circles, and rectangular prisms - as well as three-dimensional solids like cylinders, cones, and spheres. It also covers key circle concepts such as radius, diameter, arc length, sectors, tangents, and inscribed angles, all applied to real-world measurement problems.

Note: $s=$ arc length, $h=$ height, $r=$ radius, $b=$ base, $d=$ diameter, $l=$ length, $w=$ width.

Definitions

Radius

Distance from the center of a circle to its edge.

Diameter

A line segment that passes through the center of a circle; its length is twice the radius ($d=2r$).

Circumference of a circle

The perimeter of a circle ($C=2\pi r=\pi d$).

Area of a circle

The space enclosed by a circle ($A=\pi r^2$).

Perimeter

The total distance around a polygon.

Area of a rectangle

$A=l\times w$

Perimeter of a rectangle

$P=2(l+w)$

Perimeter of a triangle

$P=a+b+c$

Area of a triangle

$A=\frac{1}{2}bh$

Volume of a rectangular prism

$V=lwh$

Sector of a circle

A slice of a circle defined by a central angle $\theta$. Area: $A=\frac{\theta }{360^{\circ }}\cdot \pi r^2$; Arc length: $s=\frac{\theta }{360^{\circ }}\cdot 2\pi r$.

Circles: area and circumference

A circle is a two-dimensional shape where every point on the edge is the same distance from the center. That constant distance is the radius. The circumference is the distance around the circle (the circle’s perimeter), and the area is the amount of space inside the circle.

Most circle problems come down to moving between radius, diameter, circumference, and area. Use $d=2r$ to switch between radius and diameter, depending on what the problem gives you.

Example: Area from diameter

A circle has a diameter of $10\text{ cm}$. Find its area.

First, find the radius: $r=\frac{d}{2}=\frac{10}{2}=5\text{ cm}$

Then apply the area formula: $A=\pi r^2=\pi \cdot 25=25\pi \approx 78.54{\text{ cm}}^2$

Answer: $25\pi \approx 78.54{\text{ cm}}^2$ :::

Rectangles and triangles: perimeter and area

Rectangles and triangles are the standard starting point for perimeter and area.

• Perimeter is the total distance around a polygon. You find it by adding the side lengths. This often models fencing, framing, or outlining a region.
• Area measures how much flat (two-dimensional) space a shape covers. This comes up in tiling, painting, and covering surfaces. Area is measured in square units, while perimeter is measured in linear units.

Rectangles are especially direct: their right angles make the area simply length times width. For triangles, the area formula is always one-half the product of a base and its perpendicular height - don’t forget the $\frac{1}{2}$.

Pitfall: identifying the correct height The height must form a right angle with the base - it isn’t always a side of the triangle. In obtuse or scalene triangles, the height may fall outside the triangle entirely, but the formula $A=\frac{1}{2}bh$ still works - just use the perpendicular distance. For example, a base of $10\text{ cm}$ with an external height of $4\text{ cm}$ gives $A=\frac{1}{2}\cdot 10\cdot 4=20{\text{ cm}}^2$. Look for right-angle markings or dashed altitude lines; if none are shown, sketch the height yourself before applying the formula.

These ideas come up again in composite figures, coordinate geometry, and surface area.

Geometry often involves shapes that combine simpler figures. A composite figure is made up of two or more basic shapes joined together. To find its area, break it into the shapes you recognize, find each area separately, then add them together.

Example: Composite figure - rectangle and triangle

A shape is made of a rectangle $8\text{ ft}$ long and $4\text{ ft}$ wide, with a right triangle attached to one end. The triangle has a base of $4\text{ ft}$ and a height of $3\text{ ft}$. Find the total area.

• Rectangle: $A=8\cdot 4=32{\text{ ft}}^2$
• Triangle: $A=\frac{1}{2}\cdot 4\cdot 3=6{\text{ ft}}^2$
• Total: $32+6=38{\text{ ft}}^2$

Answer: $38{\text{ ft}}^2$

Example: Composite figure - rectangle and semicircle

A shape consists of a rectangle $10\text{ m}$ long and $4\text{ m}$ wide, with a semicircle attached to one of the short ends (diameter $=4\text{ m}$). Find the total area.

The semicircle’s radius is $r=\frac{4}{2}=2\text{ m}$.

• Rectangle: $A=10\cdot 4=40{\text{ m}}^2$
• Semicircle: $A=\frac{1}{2}\pi r^2=\frac{1}{2}\cdot \pi \cdot 4=2\pi \approx 6.28{\text{ m}}^2$
• Total: $40+2\pi \approx 46.28{\text{ m}}^2$

Answer: $40+2\pi \approx 46.28{\text{ m}}^2$ :::

Volume of rectangular prisms

Volume measures how much three-dimensional space an object occupies. A rectangular prism stacks rectangular layers, so you multiply length, width, and height: $V=lwh$. A cube is a special case where all sides are equal, giving $V=s^3$.

Pitfall: mixed units All three dimensions must be in the same unit before you multiply. Convert first, then apply $V=lwh$. Volume is always expressed in cubic units.

Sectors and arcs

Now that you know how to work with a full circle, you can work with a portion of one. A sector is a portion of a circle defined by a central angle measured at the center of the circle. Picture it as a “slice” of the circle - the larger the central angle, the bigger the slice.

Two common measurements come from a sector:

• Arc length: the curved distance along the outside edge of the sector
• Sector area: the area inside that slice

Both rely on the same idea: the central angle tells you what fraction of the whole circle you’re working with. That fraction is $\frac{\theta }{360^{\circ }}$. Apply it to:

• the full circumference to get arc length
• the full circle area to get sector area

This is why sector and arc problems are really just extensions of the circle formulas you already know.

To build intuition, start with a half-circle. A central angle of $180^{\circ }$ is exactly half of $360^{\circ }$, so the fraction is $\frac{1}{2}$. The arc length is half the circumference, and the sector area is half the circle area. Once that feels natural, the same logic works for any angle.

Example: Sector calculations

A sector has radius $6\text{ cm}$ and central angle $60^{\circ }$. Find the sector area and arc length.

Step 1 - Find the fraction of the circle: $\frac{60}{360}=\frac{1}{6}$

Step 2 - Apply to the full circle area: $A=\frac{1}{6}\cdot \pi \cdot 6^2=\frac{1}{6}\cdot 36\pi =6\pi \approx 18.85{\text{ cm}}^2$

Step 3 - Apply to the full circumference: $s=\frac{1}{6}\cdot 2\pi \cdot 6=\frac{1}{6}\cdot 12\pi =2\pi \approx 6.28\text{ cm}$

Answer: Area $\approx 18.85{\text{ cm}}^2$, Arc $\approx 6.28\text{ cm}$