Areas of combined circular shapes

A running track with rounded ends, a coin-shaped brooch with a decorative border, a window that is a rectangle topped by a semicircle — none of these is a pure circle or a pure polygon. Each is a composite figure: circles, sectors, or segments fused with squares, rectangles, and triangles. To find the area (or the area of a "leftover" or "shaded" part), you do not need any new formula. You need to see the simple pieces hiding inside the compound shape, find each piece's area with tools you already have — $A=\pi r^2$ for a circle, $A_{\text{sector}}=\frac{\theta}{360^\circ}\pi r^2$ for a sector — and then add or subtract them with care. This article sharpens exactly that skill: spotting the pieces, and getting the bookkeeping of addition versus subtraction right every time.

The core insight

Break a composite circular figure into circles, sectors, or segments plus straight-edged shapes, compute each known area, and add or subtract them to match exactly the region the question asks for.

The core idea

Composite-area problems always reduce to the same three moves: identify the simple shapes buried in the figure, compute each one's area from a known formula, and combine them by addition or subtraction to match the region asked for — the whole figure, a leftover region, or a shaded region. This is exactly the same three moves used earlier in the chapter to find the area of a single sector or segment; the only new demand here is that a figure now contains more than one circular piece, or a circular piece fused onto a straight-edged one, so the bookkeeping step matters far more than any new formula. It helps to think of every composite figure as a small accounting problem. Every simple shape you can spot is either "in" the final region (add it) or "cut out" of the final region (subtract it), and a figure can nest several layers of this — a square with a circle cut out of it, and then a smaller circle added back inside that circle, for instance. Reading the figure carefully, and marking with a pencil which parts are shaded before touching any formula, avoids most of the errors students make on this topic. A handful of configurations recur so often in exam figures that it pays to recognise them on sight.

Circle inscribed in a square of side $a$ . The circle touches all four sides, so its diameter equals $a$ and its radius is $r=\frac{a}{2}$ . The area left over between the square and the circle is $a^2-\pi r^2$ .
Four quarter-circles at the corners of a square, each of radius $r$ . Their central angles add to $4\times90^\circ=360^\circ$ , so together they are exactly one full circle of radius $r$ — combined area $\pi r^2$ , with no separate sector calculation needed.
Semicircle sitting on a side of length $s$ . If that side is the diameter, the radius is $r=\frac{s}{2}$ and the semicircle's area is $\frac{1}{2}\pi r^2$ .
Two or more sectors sharing a figure. Compute each sector's area from $\frac{\theta}{360^\circ}\pi r^2$ and combine, watching carefully for any region counted by more than one sector.

A_{\text{sector}}=\frac{\theta}{360^\circ}\pi r^2, \qquad A_{\text{segment}}=A_{\text{sector}}-A_{\triangle}

where: $\theta$ is the sector's central angle in degrees, $r$ is the radius, and $A_{\triangle}$ is the area of the triangle formed by the two radii and the chord, subtracted from the sector to leave only the segment.

Figure

Four shaded quarter circles at the corners of a square combine into one full circle of the same radius

Area between a square and its inscribed circle

Derive the general area left over when a circle of the largest possible size is inscribed in a square of side $a$ , so the result can be reused instantly for any side length instead of recomputing the square and circle areas from scratch each time.

Step 1. The circle is inscribed in the square, meaning it touches all four sides. The largest circle that fits this way has its diameter exactly equal to the side of the square, so the radius is

r=\frac{a}{2}

Step 2. The area of the square is simply

A_{\text{square}}=a^2

Step 3. The area of the inscribed circle follows from substituting $r=\frac{a}{2}$ into $A=\pi r^2$ :

A_{\text{circle}}=\pi\left(\frac{a}{2}\right)^2=\frac{\pi a^2}{4}

Step 4. The leftover region — inside the square but outside the circle — is the square's area minus the circle's area:

A_{\text{left over}}=a^2-\frac{\pi a^2}{4}=a^2\left(1-\frac{\pi}{4}\right)

where: $a$ is the side of the square and $r$ is the radius of the inscribed circle. This single boxed result replaces the square-then-circle-then-subtract routine with one substitution whenever a circle is inscribed in a square, and it is worth noticing that the leftover fraction of the square, $1-\frac{\pi}{4}\approx0.2146$ , does not depend on $a$ at all — it is always about $21.5\%$ of the square's area, whether the square has side $7\text{ cm}$ or $70\text{ m}$ , which is a fast way to sanity-check an answer before committing to it.

Worked example

A window is shaped like a rectangle $20\text{ cm}$ long and $14\text{ cm}$ wide, topped by a semicircle whose diameter is the shorter side. Find the total area of the window. Use $\pi=\frac{22}{7}$ .

Step 1. The area of the rectangular part is

A_{\text{rect}}=20\times14=280\text{ cm}^2

Step 2. The semicircle's diameter is the shorter side, $14\text{ cm}$ , so its radius is $r=\frac{14}{2}=7\text{ cm}$ .

Step 3. The semicircle's area is half the area of a full circle of that radius:

A_{\text{semi}}=\frac{1}{2}\pi r^2=\frac{1}{2}\times\frac{22}{7}\times7^2=\frac{1}{2}\times154=77\text{ cm}^2

Step 4. The total area of the window is the rectangle plus the semicircle, since the semicircle sits on top of the rectangle and adds new area rather than overlapping it:

A_{\text{window}}=A_{\text{rect}}+A_{\text{semi}}=280+77=357\text{ cm}^2

The window has a total area of $357\text{ cm}^2$ .

Competitive stretch

If a metal strip is fitted along the outer boundary of the window (the two vertical sides, the bottom, and the curved top — not the straight top edge, since the semicircle replaces it), the strip length is $14+14+20+\pi r=14+14+20+22=70\text{ cm}$ , using the semicircular arc length $\pi r=\frac{22}{7}\times7=22\text{ cm}$ . Notice that the straight top edge of the rectangle, $14\text{ cm}$ long, is not part of the boundary at all once the semicircle is glued on — it becomes an internal line, invisible from outside the window. Forgetting to drop it, and adding it into the perimeter anyway, is a common slip when a curved piece is attached to a straight-edged one; always trace the actual outer edge of the finished composite shape with a finger before writing down which lengths belong in the perimeter.

Common traps

Using the square's side directly as the circle's radius instead of halving it — the side is the diameter of an inscribed or side-fitting circle, so it must be halved to get $r$ .
Adding areas that actually overlap, which double-counts the shared region; always check whether the circular pieces touch, overlap, or sit cleanly beside each other before combining.
Losing track of which region the question wants — "the shaded part," "the remaining part," and "the total figure" are three different answers from the same picture.
Missing the "four quarters equal one circle" shortcut and instead computing four separate $90^\circ$ sector areas, which only wastes time since $4\times90^\circ=360^\circ$ already.

Ask the tutor: Not sure whether to add or subtract the pieces in a shaded-region figure — ask the AI tutor to walk through your specific diagram step by step.

Why do the four corner quarter-circles always add up to exactly one full circle?

Each quarter-circle is a $90^\circ$ sector, and four of them together sweep $4\times90^\circ=360^\circ$ — the full angle around a point, which is exactly what defines one complete circle. As long as all four quarters share the same radius $r$ , their combined area is simply $\pi r^2$ , with no need to compute and add four separate sector areas.

How do I know whether to add or subtract the areas in a composite figure?

Read the question for the exact region asked for. If it wants the total boundary figure, you usually add the pieces (like rectangle plus semicircle). If it wants the leftover or unshaded part after removing a circular piece from a straight-edged shape (like a square minus an inscribed circle), you subtract. Sketching the figure and shading the requested region before calculating prevents most sign errors.

Can I mix different values of pi, like 22/7 for one part and 3.14 for another, in the same problem?

No — always use the value of $\pi$ given in the question consistently across every circular piece in the figure. Mixing $\frac{22}{7}$ and $3.14$ in the same composite calculation introduces small inconsistencies that can throw off the final subtraction or addition, especially when the answer is expected to come out as a clean number.