1
Name the outer shape and the hole.
Inscribed means the circle touches all four sides — so the side equals the diameter.
diameter = 2(7) = 14
so the square's side is 14
Read the fit
Inscribed circle: side = diameter. Circle drawn around the square instead:
the square's diagonal is the diameter. Different setups, different numbers.
Circles in a row fix the rectangle. Two equal circles fit exactly inside a rectangle 20 long: each diameter is 20 ÷ 2 = 10, so the radius is 5 and the rectangle is 20 by 10. Shaded = 200 − 2 × π(5)² = 200 − 50π. The circles give both sides of the rectangle — its width is one diameter.
2
Find both areas.
square = 14 × 14 = 196
the whole
circle = π(7)² = 49π
keep π for now
shaded = 196 − 49π
exact form
≈ 196 − 153.9 = 42.1
cm², if a decimal is wanted
196 − 49π ≈ 42.1 cm²
Leave π in place until the last line. Answer choices are often written as
196 − 49π, and rounding early makes it hard to match.
3
The three patterns.
| Picture | Shaded is |
|---|
| shape with a hole | outer − inner |
|---|
| a fraction of a shape | use the angle or the symmetry |
|---|
| two shapes overlapping | both − the overlap counted twice |
|---|
| shapes joined edge to edge | add the pieces — a rectangle with a semicircle on top is rectangle + ½πr² |
|---|
Deciding which pattern you are looking at is most of the work. The arithmetic
that follows is simple.
A worked pair. A 10 by 7 rectangle with a 3 by 2 rectangle cut from one corner: area 70 − 6 = 64. A 10 by 4 rectangle with a semicircle on its 4 side: the radius is 2, so area 40 + ½π(2)² = 40 + 2π. The same 4 side with the semicircle cut out instead gives 40 − 2π.
Perimeter is not area. Cutting a rectangle from a corner removes area but not perimeter: the two edges lost are replaced by two edges of the same total length. The 10 by 7 rectangle with its corner cut still has perimeter 2(10 + 7) = 34 — check: 10 + 5 + 3 + 2 + 7 + 7 = 34. Adding the cut-out’s own perimeter is the trap.
4
Quarter circles in a square.
A square of side 10, with a quarter circle of radius 10 at each corner.
four quarters = one whole circle
4 × ¼
= π(10)² ÷ 4 × 4 = 100π ÷ 4 × 4
each quarter is 25π
total covered = 100π ≈ 314
more than the square itself
Overlaps mean you cannot just add
314 exceeds the square's 100, so the quarters overlap heavily. Whenever the parts
total more than the whole, an overlap has been counted twice.
For a cleaner version, use radius 5 — then the four quarters make one circle of area
25π, with no overlap, and the uncovered region left in the middle of the square is
100 − 25π ≈ 21.5.
5
Use symmetry to shrink the problem.
the four corners are identical
by symmetry
one corner = (196 − 49π) ÷ 4
≈ 10.5 cm²
If a question asks for just one corner, do the whole thing and divide. Trying to
compute a single corner region directly is far harder.
The other habit worth having: shade the picture with a pencil before calculating.
Half the errors in this topic are misreadings of which region is meant.
6
Variations: the square inside the circle, a ring, an overlap.
A square is inscribed in a circle of radius 5. Find the area inside the circle but outside the square.
circle = π(5)² = 25π
now the circle is the outer shape
square’s diagonal = diameter = 10
corners touch the circle
square = 10² ÷ 2 = 50
d² ÷ 2
Different from step 1: the roles swap. The diameter is the square’s diagonal, not its side.
Two circles share a centre, with radii 7 and 3. Find the area of the ring between them.
π(7)² − π(3)² = 49π − 9π
outer minus inner
Subtract areas, not radii
The ring is 4 wide, but π(7 − 3)² = 16π is the area of a small circle, not the ring.
Square each radius first, then subtract.
In a square of side 6, quarter circles of radius 6 are drawn from two opposite corners. Find the area of the leaf where they overlap.
each quarter = ¼ × π(6)² = 9π
two of them: 18π
18π = square + leaf = 36 + leaf
the leaf is covered twice
Different: the overlap is what is asked. Add the pieces, then take away the shape they cover once — what is left over is the part counted twice.
7
Between a chord and its arc: sector minus triangle.
Sector OAB has radius 6 and central angle 60°. Find the area of the region between chord AB and the arc.
sector = 60360 × π(6)² = 6π
one sixth of 36π
△OAB is equilateral, side 6
two radii with 60° between them
△OAB = (√34) × 36 = 9√3
equilateral area
| Central angle | The triangle OAB | Its area (radius r) |
|---|
| 90° | right isosceles | ½r² — radius 4 gives 4π − 8 |
|---|
| 60° | equilateral | (√34)r² |
|---|
The triangle needs its own formula
Using ½ × 6 × 6 = 18 for the 60° triangle treats it as right-angled. Only the 90°
sector has the two radii as base and height. And a square drawn inside a circle has
the diameter as its diagonal, so its area is d² ÷ 2, not d².
On the exam
Shaded-region items are almost always “whole minus pieces”: circles in a rectangle
(the circles fix the rectangle’s sides), a square in a circle, or a segment cut off by
a chord. Get every length from the fit first, keep π and surds exact, and match the
choice’s form. Ten-second check: the shaded area must be positive and smaller than
the outer shape.
✓
Check: is the size believable?
shaded vs square
42 out of 196
about 21% ✓
circle vs square
153.9 out of 196
about 79% — the usual ratio
one corner
10.5 cm²
small, as the picture suggests ✓
An inscribed circle always fills about 78.5% of its square — a useful sanity check.
Remember the flow
Name both shapes · Areas · Subtract · Keep π
Parts totalling more than the whole means an overlap.
