Area of a semicircle

A semicircle is exactly half of a circle — the shape you get when a diameter slices a circle into two equal, mirror-image pieces. Once you know the area of a full circle, finding the area of a semicircle is a short, satisfying step, and it shows up everywhere: arched windows, running-track ends, protractors, and semicircular plots of land. In the ICSE curriculum this idea is formally developed in Class 9 Mensuration, right alongside the full circle-area formula $A = \pi r^2$ and its circumference formula $C = 2\pi r$ . Only the semicircle and quarter-circle are studied here as "special fractions" of a circle; the general sector formula for an arbitrary angle comes later. This page builds only on ideas you already have — the area of a circle and the idea of taking a half — and sharpens the one distinction students trip on most: area and perimeter of a semicircle do not behave the same way.

The core insight

A semicircle's area is exactly half a circle's area, $A = \tfrac{1}{2}\pi r^2$ , but its perimeter is NOT half the circumference — it also needs the straight diameter edge.

The core idea

Start from the area of a full circle of radius $r$ .

A_{\text{circle}} = \pi r^2

where: $r$ is the radius of the circle and $\pi \approx 3.14159$ , commonly taken as $\dfrac{22}{7}$ or $3.14$ for calculation. A semicircle is exactly one half of that circle, cut along a diameter, so its area is exactly one half of the circle's area.

A_{\text{semicircle}} = \frac{1}{2}\pi r^2

If a problem gives the diameter $d$ instead of the radius, use $r = \dfrac{d}{2}$ to rewrite the formula directly in terms of $d$ .

A_{\text{semicircle}} = \frac{1}{2}\pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{8}

where: $d$ is the diameter of the full circle (the straight edge that closes the semicircle). It also helps to keep the perimeter of a semicircle in your toolkit, because problems often ask for area and boundary length together, and the two formulas look deceptively similar but are not interchangeable. The boundary has two parts: the curved arc, which is half the circle's circumference, and the straight diameter that closes the shape.

P_{\text{semicircle}} = \pi r + 2r = r(\pi + 2)

where: $\pi r$ is the length of the curved arc and $2r$ is the length of the straight diameter edge. For calculations, use $\pi = \dfrac{22}{7}$ when the radius is a multiple of 7 (it cancels neatly), and $\pi = 3.14$ otherwise.

Figure

A circle split by a diameter into two shaded semicircles labelled with radius r, curved arc pi r and straight edge 2r

Derivation AArea of a semicircle from the area of a full circle

Show that halving the area formula for a full circle of radius $r$ gives the exact area enclosed by a semicircle, and that this area can be rewritten equally correctly in terms of the diameter $d$ , which matters because textbook problems mix both forms freely and students must be able to move between them without introducing an error.

Step 1. Begin with the standard area formula for a full circle of radius $r$ , which every semicircle result is built from: $A_{\text{circle}} = \pi r^2$ . This formula itself comes from the circle-area derivation done earlier in mensuration and is taken here as the known starting point.

Step 2. A diameter is a straight line through the centre of a circle, and it divides the circular region into two congruent pieces that are exact mirror images of each other — this is the definition of a semicircle. Because the two halves are congruent, they must have equal area, and together they must add up to the full circle's area. So each half's area is $\dfrac{1}{2}$ of the whole.

Step 3. Divide the full circle's area by 2 to get the semicircle's area: $A_{\text{semicircle}} = \dfrac{1}{2}\times \pi r^2 = \dfrac{1}{2}\pi r^2$ . This is the working formula whenever the radius is given directly.

Step 4. To rewrite it in terms of the diameter, substitute $r = \dfrac{d}{2}$ into the formula from Step 3: $A_{\text{semicircle}} = \dfrac{1}{2}\pi \left(\dfrac{d}{2}\right)^2 = \dfrac{1}{2}\pi \times \dfrac{d^2}{4} = \dfrac{\pi d^2}{8}$ . Both forms describe the identical area — choose whichever matches the quantity the problem actually gives you, and never square the diameter and forget to divide by the extra factor of 4 that comes from squaring $\tfrac{d}{2}$ .

Worked example

A circular metal plate has radius 7 cm. It is cut exactly along a diameter into two equal semicircular halves. Find the area of one semicircular half. (Use $\pi = \dfrac{22}{7}$ .)

Step 1. Write down what is known: radius $r = 7$ cm.

Step 2. Apply the semicircle-area formula: $A_{\text{semicircle}} = \dfrac{1}{2}\pi r^2 = \dfrac{1}{2}\times\dfrac{22}{7}\times 7^2$ .

Step 3. Simplify: $= \dfrac{1}{2}\times\dfrac{22}{7}\times 49 = \dfrac{1}{2}\times 22 \times 7 = \dfrac{1}{2}\times 154 = 77$ .

Answer: the area of one semicircular half is $77\ \text{cm}^2$ .

Competitive stretch

A semicircular flower bed has a diameter of 20 m. Find (a) the area of the flower bed, and (b) the length of fencing needed to go all the way around its boundary. (Use $\pi = 3.14$ .)

Step 1. Find the radius: diameter $d = 20$ m, so radius $r = \dfrac{20}{2} = 10$ m.

Step 2. Find the area: $A = \dfrac{1}{2}\pi r^2 = \dfrac{1}{2}\times 3.14 \times 10^2 = \dfrac{1}{2}\times 314 = 157$ , so the area is $157\ \text{m}^2$ .

Step 3. Find the perimeter (fencing length), remembering it needs both the curved arc and the straight diameter edge: $P = \pi r + 2r = (3.14 \times 10) + (2\times 10) = 31.4 + 20 = 51.4$ .

Answer: area $= 157\ \text{m}^2$ ; fencing needed $= 51.4$ m. Notice how different these two answers look even though they came from the same shape — that is exactly the trap discussed below.

Common traps

Forgetting to halve. The most common mistake is computing the full circle's area, $\pi r^2$ , and forgetting the extra $\div 2$ that turns it into a semicircle's area.
Squaring the diameter instead of the radius. If a question gives the diameter, halve it to get the radius before squaring — squaring the diameter directly and then halving gives a wrong, too-large answer unless the correct $\pi d^2/8$ form is used.
Mixing up area and perimeter. The perimeter of a semicircle is not half the circle's circumference — the straight diameter edge must be added too. Halving the circumference alone gives only the length of the curved arc.
Using the wrong value of pi. Using $\dfrac{22}{7}$ when the radius is not a nice multiple of 7 creates messy fractions and invites arithmetic slips; switch to $3.14$ in those cases.

Ask the tutor: Still mixing up when to add the diameter and when not to? Ask your tutor to sketch a semicircle, trace the curved arc in one colour and the straight edge in another, and walk through both the area and perimeter calculations side by side on the same figure.

Why is the area formula half of pi r squared but the perimeter formula is not half of 2 pi r?

Area measures the flat region enclosed by the shape, and a semicircle genuinely encloses exactly half the flat region of the full circle, so halving works cleanly. Perimeter measures the total boundary you would walk around, and a semicircle's boundary is not just half the circle's rim — after tracing the curved arc, you still have to walk back along the straight diameter to close the shape. That extra straight edge is why $P = \pi r + 2r$ , not just $\pi r$ .

If I am given the diameter, should I convert to radius first or use the pi d squared over 8 formula directly?

Either is correct and gives the identical answer, since $\dfrac{\pi d^2}{8}$ is simply $\dfrac{1}{2}\pi r^2$ with $r=\dfrac{d}{2}$ substituted in. Most students find it safer to convert to radius first, because it avoids the easy slip of squaring the whole diameter and forgetting the extra division by 4 that the substitution introduces. Use whichever method you can execute without arithmetic errors under exam conditions.

Does the semicircle formula work for any half of a circle, or only the one cut along a diameter?

The formula $A = \tfrac{1}{2}\pi r^2$ works only for a true semicircle, meaning the circle is cut exactly along a diameter through its centre. A curved region bounded by any other chord, not passing through the centre, is called a circular segment, and it needs a different, more advanced formula that is not part of this topic. Always check that the straight edge in the figure passes through the centre before applying the semicircle formula.

Area of a semicircle

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads