Measuring Space: Perimeter and Area

The perimeter (circumference) of a circle of radius \(r\) is \(2\pi r\).

The radius is 7 cm.

Use \(C = 2\pi r\) with \(\pi = \frac{22}{7}\).

(i) \(r = 7\) cm:

(ii) \(r = 10\) cm:

(iii) \(r = 12\) cm:

An arc is the fraction \(\dfrac{\theta}{360^\circ}\) of the whole circumference:

(i) \(r = 3.5\) cm, \(\theta = 60^\circ\):

(ii) \(r = 6.3\) m, \(\theta = 120^\circ\):

A sector is bounded by two radii and one arc, so its perimeter is \(2r + \ell\).

The arc.

The two radii.

Total perimeter.

Reading the figures: each shape is built from straight pieces and circular arcs. The trick every time is to work out the radius of each arc and what fraction of a full circle it is, then add.

(i) Stadium shape — a rectangle \(80\) m long with a semicircular cap on each end; the \(60\) m dashed line is the width, so each cap has diameter \(60\) m. The two caps together make one full circle.

(ii) Half-ring — an outer semicircle of diameter \(12\) cm, an inner semicircle of diameter \(8\) cm, and two flat ends each \(\frac{12-8}{2} = 2\) cm long.

(iii) Four-lobed shape — a square of side \(10\) cm with a semicircle of diameter \(10\) cm bulging outwards on each of the four sides. The square's sides are dashed, so they are not part of the perimeter.

(iv) Three-lobed shape — an equilateral triangle of side \(12\) cm with a semicircle drawn outwards on each side as diameter.

(v) Rounded cross — the dashed grid is \(3\times 3\) squares of side \(14\) cm. The boundary is four quarter-circles of radius \(14\) cm at the corners, plus four semicircles of diameter \(14\) cm on the four arms.

(vi) Big arch on a wavy base — the \(28\) cm base is split into four equal parts of \(7\) cm. On top is one semicircle of diameter \(28\) cm; along the base are four semicircles of diameter \(7\) cm, alternating below and above.

(vii) Semicircles on a right triangle — the legs are \(8\) cm and \(6\) cm, so the dashed hypotenuse is \(\sqrt{8^2+6^2} = 10\) cm. A semicircle sits on each of the three sides.

(viii) Arch on three bumps — base \(4+4+4 = 12\) cm. One semicircle of diameter \(12\) cm on top, three semicircles of diameter \(4\) cm below it.

(ix) Comma shape — base \(10 + 10 = 20\) cm. One semicircle of diameter \(20\) cm above, one semicircle of diameter \(10\) cm above the left half, one semicircle of diameter \(10\) cm below the right half.

In one revolution a wheel rolls forward exactly one circumference.

(i) With \(d = 56\) cm:

The car travels 176 cm (1.76 m) per revolution.

(ii) Convert the journey to centimetres first:

So the tyre makes about 5682 revolutions — strictly, 5681 complete revolutions and a little over four-fifths of another.

(i) Square, side 14 cm. Each arc is centred at the midpoint of a side, and its radius is half the side, \(7\) cm. Each such arc is a semicircle drawn inside the square, and there are four of them — one per side. Every bit of every semicircle lies on the edge of some petal, so the total petal perimeter is just the total length of the four semicircles.

(ii) Regular hexagon, side 42 cm. Each arc is centred at a vertex. In a regular hexagon the distance from a vertex to the centre equals the side, so an arc of radius \(42\) cm centred at a vertex passes through both neighbouring vertices and through the centre of the hexagon.

Take a vertex \(A\), the hexagon centre \(O\), and a neighbouring vertex \(B\). Then \(AO = OB = AB = 42\) cm, so \(\triangle AOB\) is equilateral and the arc from \(O\) to \(B\) subtends

Each such arc therefore has length

There are six petals and each petal is bounded by two of these arcs, so there are \(12\) arcs in all.

Let the radii be \(r_1\) and \(r_2\). The perimeters are \(2\pi r_1\) and \(2\pi r_2\).

The factor \(2\pi\) cancels from top and bottom:

So the radii are also in the ratio 5 : 4.

Reading the figure: \(ABCD\) is a rectangle with \(A\) top-left, \(B\) top-right, \(C\) bottom-right and \(D\) bottom-left. The width is \(10\) cm and the height is \(8\) cm. The point \(E\) lies on the right-hand side \(BC\).

Take \(AD\) — the left-hand side — as the base of \(\triangle ADE\):

The height is the perpendicular distance from \(E\) to the line \(AD\). Since \(E\) is on the opposite side of the rectangle, that distance is the full width:

The area is 40 cm² — exactly half the rectangle.

The trapezium is isosceles. Drop a perpendicular from each end of the shorter parallel side to the longer one. This cuts the longer side into three pieces: two equal end-pieces and a middle piece equal to the shorter side.

Each end piece, the height \(h\) and a slant side of \(26\) cm form a right triangle:

Now use the trapezium formula:

The area is 720 cm². (The right triangle here is a \(10\text{-}24\text{-}26\) triangle, which is the \(5\text{-}12\text{-}13\) triple doubled.)

Step 1: the third side.

Step 2: Heron's formula. With \(a=8\), \(b=11\), \(c=13\):

The area is \(8\sqrt{30}\) cm², about 43.8 cm².

Step 1: the sides. Let the sides be \(3k\), \(5k\), \(7k\).

Step 2: Heron's formula.

The area is \(1500\sqrt{3}\) m², about 2598 m².

The area of a rhombus is half the product of its diagonals. Let the shorter diagonal be \(d\); then the longer one is \(2d\).

The shorter diagonal is \(8\sqrt{2}\) cm, about 11.3 cm; the longer one is \(16\sqrt{2} \approx 22.6\) cm.

Both triangles are built on the same base \(CD\).

In a parallelogram \(AB \parallel CD\), so every point of \(AB\) is the same perpendicular distance from the line \(CD\). That distance is the height \(h\) of the parallelogram. Since \(P\) and \(Q\) both lie on \(AB\), both triangles have the same height.

The two areas are equal, whatever the positions of \(P\) and \(Q\) on \(AB\). Each is exactly half the area of the parallelogram.

Both triangles share the base \(PO\), which lies along the diagonal \(PR\). So it is enough to show that \(Q\) and \(S\) are the same perpendicular distance from the line \(PR\).

Why they are. Let the diagonals \(PR\) and \(QS\) meet at \(M\). In a parallelogram the diagonals bisect each other, so:

Drop perpendiculars \(QX\) and \(SY\) from \(Q\) and \(S\) to the line \(PR\). Compare \(\triangle QXM\) and \(\triangle SYM\):

By AAS congruence \(\triangle QXM \cong \triangle SYM\), so by CPCT:

Now both triangles have base \(PO\) and equal heights:

Let the 4-gon be \(ABCD\) and let \(P\), \(Q\), \(R\), \(S\) be the midpoints of \(AB\), \(BC\), \(CD\), \(DA\). Draw the diagonal \(AC\).

Step 1: \(PQRS\) really is a parallelogram. In \(\triangle ABC\), \(P\) and \(Q\) are midpoints of two sides, so by the midpoint theorem:

In \(\triangle ADC\), the same theorem gives:

So \(PQ\) and \(SR\) are equal and parallel, which makes \(PQRS\) a parallelogram.

Step 2: the area. Cutting off the four corner triangles is what removes the other half. Take \(\triangle APS\): it is similar to \(\triangle ABD\) with ratio \(\frac{1}{2}\), so its area is \(\frac{1}{4}\) of \(\triangle ABD\). The same holds at each corner:

Adding the first pair and then the second pair:

The four corner triangles together take up \(\frac{1}{4}+\frac{1}{4} = \frac{1}{2}\) of the 4-gon, so what is left is the other half:

Step 1: the median halves the triangle. \(\triangle ABD\) and \(\triangle ACD\) have equal bases \(BD = DC\) and the same apex \(A\), so the same height:

Step 2: the median of the small triangle too. \(\triangle PBD\) and \(\triangle PCD\) also have equal bases \(BD = DC\), now with common apex \(P\):

Step 3: subtract. Since \(P\) lies on \(AD\), the triangle \(ABP\) is what is left of \(ABD\) after removing \(PBD\), and similarly on the other side:

The two right-hand sides are equal term by term, so:

Let the square have side \(a\). Put \(P\) at perpendicular distances \(h_1\) from \(AB\) and \(h_2\) from the opposite side \(CD\). Since \(AB\) and \(CD\) are opposite sides of the square:

The red pair. Both triangles have base \(a\):

The green pair. Let \(k_1\), \(k_2\) be the distances from \(P\) to \(BC\) and \(DA\); these are also opposite sides, so \(k_1+k_2 = a\) and the identical computation gives:

Both regions are half the square, so:

The red and green regions have equal area, wherever \(P\) is placed inside the square.

Step 1: use the parallel lines. \(\triangle QDP\) and \(\triangle CDP\) stand on the same base \(DP\), and their apexes \(Q\) and \(C\) lie on the line \(QC\), which is parallel to \(DP\). Triangles on the same base and between the same parallels are equal in area:

Step 2: build up \(\triangle BPQ\). The segment \(DP\) cuts \(\triangle BPQ\) into \(\triangle BDP\) and \(\triangle QDP\):

Replace \([QDP]\) using Step 1:

But \(\triangle BDP\) and \(\triangle CDP\) together make up exactly \(\triangle BDC\), since \(P\) lies on \(BC\):

Step 3: \(D\) is a midpoint. In \(\triangle ABC\), the segment \(CD\) is a median from \(C\), so it halves the area:

A sector is the fraction \(\dfrac{\theta}{360^\circ}\) of the whole disc:

The sector has area \(\frac{77}{3}\) cm², about 25.7 cm².

Step 1: the radius.

Step 2: a quadrant is a quarter of the disc (a \(90^\circ\) sector).

The quadrant has area 38.5 cm².

The minute hand goes right round the clock — \(360^\circ\) — in \(60\) minutes. In \(10\) minutes it turns through:

The region swept is a sector of radius \(7\) cm and angle \(60^\circ\):

The hand sweeps about 25.7 cm².

Here \(r = 10\) cm and \(\pi \approx 3.14\), so the whole disc has area \(3.14 \times 100 = 314\) cm².

(i) Minor sector, \(90^\circ\):

(ii) Major sector, \(270^\circ\):

Check. The two sectors must fill the circle exactly:

A segment is what is left of a sector once you cut off the triangle formed by the two radii and the chord.

Step 1: the sector.

Step 2: the triangle. The two radii are equal and the angle between them is \(60^\circ\), so the triangle is equilateral with side \(15\) cm:

Step 3: the minor segment.

Step 4: the major segment is the rest of the circle.

Each blade sweeps a sector of radius \(28\) cm and angle \(120^\circ\).

The wipers do not overlap, so the areas simply add:

The two blades clean about 1642.67 cm² in each sweep.

Step 1: the sector. A \(60^\circ\) sector is one-sixth of the disc:

Step 2: the triangle. The two bounding radii are equal and the angle between them is \(60^\circ\), so the triangle is equilateral with side \(r\):

Step 3: the minor segment.

Numerically \(\frac{\pi}{6} \approx 0.5236\) and \(\frac{\sqrt{3}}{4} \approx 0.4330\), so the segment is about \(0.0906\,r^2\) — a thin sliver, as the picture demands.

Step 1: the side of the triangle. Join the centre \(O\) to two vertices \(A\), \(B\). Since the three vertices divide the circle equally, the central angle is:

Drop the perpendicular from \(O\) to \(AB\); it bisects both the chord and the angle, giving a right triangle with hypotenuse \(r\) and angle \(60^\circ\):

Step 2: the area of the triangle.

Step 3: the ratio.

The triangle covers only about 41% of the circle — an equilateral triangle is a poor fit inside a circle.

Step 1: the side of the square. Each angle of a square is \(90^\circ\), so each diagonal subtends a right angle at the opposite vertex — which means every diagonal is a diameter.

For a square of side \(a\), the diagonal is \(a\sqrt{2}\):

Step 2: the areas.

Step 3: the ratio.

The square fills about 64% of the circle — a better fit than the triangle.

Step 1: split the hexagon. Join the centre to all six vertices. The central angles are:

Each of the six triangles has two sides equal to \(r\) and the angle between them \(60^\circ\), so each is equilateral with side \(r\).

Step 2: the ratio.

Why it is exactly double. Compare the two inscribed shapes directly:

Geometrically: the inscribed equilateral triangle uses alternate vertices of the inscribed regular hexagon. Joining those three vertices cuts the hexagon into the central triangle plus three corner triangles, and each corner triangle is congruent to one third of the central one — so the hexagon is exactly twice the triangle. Dividing both by the same circle area preserves that factor of \(2\).

How the given picture works. A square of side \(a+b\) is cut by one horizontal and one vertical line into four rectangles: an \(a\times a\) square, a \(b\times b\) square and two \(a\times b\) rectangles. The whole area equals the sum of the parts:

(i) A figure for \((a+b)(a-b) = a^2-b^2\).

Start with a square of side \(a\), area \(a^2\). From one corner cut away a small square of side \(b\), leaving an L-shape of area \(a^2-b^2\).

Now cut the L-shape into two rectangles: one measuring \(a \times (a-b)\) and one measuring \(b \times (a-b)\). Slide the smaller one round and join it to the longer edge of the other. The two together form a single rectangle of width \(a+b\) and height \(a-b\):

(ii) A figure for \((a+b+c)^2\).

Draw a square of side \(a+b+c\). Mark off lengths \(a\), \(b\), \(c\) along the top edge and the same along the left edge, then draw two horizontal and two vertical lines through those marks. The square is cut into \(3\times 3 = 9\) rectangles:

Three of the nine cells are squares and the other six pair up:

Step 1: the base.

Step 2: the height. In an isosceles triangle the altitude to the base bisects it, giving a right triangle with hypotenuse \(15\) and one leg \(5\):

Step 3: the area.

The area is \(50\sqrt{2}\) cm², about 70.7 cm². (Heron's formula with \(s=20\) gives \(\sqrt{20\cdot5\cdot5\cdot10} = \sqrt{5000} = 50\sqrt2\) — the same answer.)

Step 1: the height.

Step 2: the equal side. The altitude bisects the base, so each half is \(5\) cm:

Each of the equal sides is 13 cm — another \(5\text{-}12\text{-}13\) triangle.

Step 1: the other leg. For a right triangle the two legs are base and height:

Step 2: the hypotenuse.

Step 3: the perimeter.

The perimeter is 36 cm. (The sides \(9, 12, 15\) are the \(3\text{-}4\text{-}5\) triple tripled.)

Step 1: the sides.

Step 2: Heron's formula.

The area is \(\frac{75\sqrt{15}}{4}\) cm², about 72.6 cm².

Method 1 — spot the right angle. Test the three sides:

They agree, so by the converse of the Baudhāyana–Pythagoras theorem the triangle is right-angled, with the two shorter sides as legs:

Method 2 — Heron's formula.

Both methods give 84 cm².

One rotation carries the bicycle forward one circumference.

Converting: \(18857.14\) cm \(= 188.57\) m. The cyclist travels about 188.6 m.

Step 1: the radius.

Step 2: the quadrant.

The quadrant has area 86.625 cm².

Step 1: one turn.

So one complete turn carries the car 176 cm, that is \(1.76\) m.

Step 2: a journey of 1 km.

The wheel turns about 568 times in one kilometre.

Yes — they must be congruent.

Let the first rectangle have sides \(a\), \(b\) and the second have sides \(c\), \(d\). Equal perimeter and equal area give:

Call these common values \(S\) and \(P\). Then \(a\) and \(b\) are the two roots of the quadratic

and \(c\), \(d\) are the roots of the same quadratic, because it is built from the same \(S\) and \(P\). A quadratic has at most two roots, so the pair \(\{c,d\}\) must be the pair \(\{a,b\}\).

Either \(c=a\) and \(d=b\), or \(c=b\) and \(d=a\) — and the second case is just the first rectangle turned through \(90^\circ\). Either way the two rectangles have the same pair of side lengths, so they are congruent.

Reading the figure: the trapezium has parallel sides \(a\) (top) and \(b\) (bottom) with height \(h\), and it has been cut by a line from the top-right vertex down to the bottom edge, splitting it into a parallelogram on the left and a triangle on the right.

The parallelogram. Its base is the part of \(b\) that lies directly below the top side, which has length \(a\), and its height is \(h\):

The triangle. Its base is the remaining part of the bottom edge, \(b-a\), and its height is also \(h\):

Add them.

which is the required formula.

Let the trapezium be \(ABCD\) with \(AB \parallel DC\), \(AB = a\), \(DC = b\) and height \(h\). Draw the diagonal \(AC\); it cuts the trapezium into \(\triangle ABC\) and \(\triangle ACD\).

The first triangle. Take \(AB\) as its base. Its apex is \(C\), which lies on the line \(DC\), and the distance between the two parallel lines is \(h\):

The second triangle. Take \(DC\) as its base. Its apex is \(A\), on the line \(AB\), again at distance \(h\):

Add them.

The construction. Take the trapezium \(ABCD\) with \(AB = a\) parallel to \(DC = b\) and height \(h\). Make an identical copy, rotate the copy through \(180^\circ\) (a half-turn), and place it against the original along one of the slanting sides.

The half-turn sends the short parallel side to the position of the long one, so along each of the two horizontal edges of the new figure you now have one piece of length \(a\) followed by one of length \(b\). The result is a parallelogram with:

Reading off the formula. The parallelogram is made of exactly two copies of the trapezium, so:

Let the kite be \(ABCD\) with \(AB = AD\) and \(CB = CD\). Its diagonals \(AC\) and \(BD\) meet at \(M\); write \(AC = p\) and \(BD = q\).

A standard property of a kite is that the axis \(AC\) is the perpendicular bisector of the other diagonal, so:

(i) Using algebra. The diagonal \(BD\) cuts the kite into \(\triangle ABD\) and \(\triangle CBD\). Both have base \(BD = q\); their heights are \(AM\) and \(MC\), which add up to \(p\).

(ii) Using geometry. Draw the rectangle whose sides pass through the four vertices of the kite and run parallel to the diagonals. Its sides measure \(p\) and \(q\), so its area is \(pq\).

The kite cuts this rectangle into four right triangles, one in each corner. Each corner triangle has a matching partner inside the kite: the diagonals divide the kite into four right triangles, and each is congruent to the corner triangle sitting directly outside it (same two legs, right angle between them). So the kite covers exactly half of the rectangle:

(i) Rectangles.

So \([PQRS] = 4[ABCD]\). And yes — the copies really do fit. Lay them in a \(2\times 2\) block: two side by side make a \(2a \times b\) strip, and two such strips stacked give exactly \(2a \times 2b\).

(ii) Triangles, scale factor 2. The triangles are similar, and Heron's formula scales cleanly: doubling every side doubles \(s\) and each of \(s-a\), \(s-b\), \(s-c\), so the product under the root is multiplied by \(2^4 = 16\) and the square root by \(4\).

Again the copies fit. Join the midpoints of the sides of \(\triangle PQR\); this cuts it into four triangles, each with sides \(a\), \(b\), \(c\). Three of them point the same way as \(PQR\) and the middle one is upside down — but all four are congruent to \(\triangle ABC\).

(iii) Triangles, scale factor 3. By the same reasoning the product under Heron's root is multiplied by \(3^4 = 81\), so the area is multiplied by \(9\).

The copies fit here too. Divide each side of \(\triangle PQR\) into three equal parts and draw all the lines parallel to the sides through those points. This tiles \(\triangle PQR\) with \(9\) small triangles, each congruent to \(\triangle ABC\) — six pointing one way and three the other.

(i) The triangle (Fig. 6.43).

Reading the figure: in \(\triangle ABC\) the left side \(AB\) carries a single marked point with matching ticks on both halves — so it is the midpoint \(M\). The right side \(AC\) carries two marked points with three equal ticks — so \(AC\) is trisected at \(N\) (nearer \(A\)) and \(P\). The shaded region is the quadrilateral \(BMNP\).

Write \([ABC] = S\). Two triangles sharing the angle at \(A\) have areas in the ratio of the products of the sides around that angle.

The shaded quadrilateral is what remains of \(\triangle ABP\) once \(\triangle AMN\) is removed:

So exactly one half of the triangle is shaded.

(ii) The square (Fig. 6.44).

Reading the figure: every side of the square carries its midpoint (equal ticks on both halves), and four lines are drawn, each joining a vertex to the midpoint of a non-adjacent side, all turning the same way round. They enclose a small tilted square in the middle.

Put coordinates on it: let the square be \((0,0)\), \((1,0)\), \((1,1)\), \((0,1)\). The four lines are then

Solving them in pairs gives the four corners of the inner square:

Its side is the distance between two neighbouring corners:

So exactly one fifth of the square is shaded.

In both pictures the circles are equal, they touch each other and they touch the top and bottom of the rectangle. Let the radius be \(r\).

(i) Three circles. The rectangle's height is one diameter and its length is three diameters:

(ii) Four circles.

Both give the same answer:

About 78.5% of each rectangle is covered — and remarkably, the number of circles makes no difference.

The conjecture. However many circles you use, the fraction of the rectangle they cover is always \(\dfrac{\pi}{4}\), roughly \(78.5\%\).

Testing it. With \(n\) circles of radius \(r\) in a row, the rectangle measures \(2nr\) by \(2r\):

The proof. Take \(n\) circles of radius \(r\), each touching its neighbours and both long sides.

The \(n\) cancels, which is exactly why the answer never changes. The cleanest way to see it: each circle sits in its own \(2r \times 2r\) square, and a circle fills \(\frac{\pi r^2}{4r^2} = \frac{\pi}{4}\) of its square. Lining up \(n\) identical squares cannot change that proportion.

Reading the figure: the large rectangle has two rows. The top row holds 4 small rectangles lying on their long sides, and the bottom row holds 5 standing on their short sides. Four plus five is the nine rectangles.

Let each small rectangle be \(L\) long and \(W\) wide, with \(L > W\).

Step 1: match the widths. The top row spans \(4L\); the bottom row spans \(5W\). Both equal the width of the large rectangle:

Step 2: use the area. The large rectangle is \(4L\) wide and \(W + L\) tall:

Step 3: the perimeter.

Each small rectangle has perimeter \(\frac{18\sqrt{10}}{5}\) cm, about 11.4 cm.

Check. Each small rectangle has area \(LW = \sqrt{10}\times\frac{4\sqrt{10}}{5} = 8\) cm², and \(9 \times 8 = 72\) cm². ✓

Reading the figure: \(\triangle ABC\) has its base \(BC\) divided into three equal parts by two points \(X\) and \(Y\). Lines \(AX\) and \(AY\) are drawn from the apex, cutting the triangle into three smaller triangles: \(ABX\) (blue), \(AXY\) (white) and \(AYC\) (red).

Why the areas are equal. All three small triangles have their apex at \(A\), so all three have the same height — the perpendicular distance from \(A\) to the line \(BC\). Call it \(h\). Their bases are the three equal pieces of \(BC\):

So the blue and the red triangle are equal — each is a third of the whole.

Cutting one to cover the other. Two pieces are enough.

Let \(M\) be the midpoint of the blue triangle's side \(AX\), and let \(N\) be the midpoint of \(BX\). Cut the blue triangle \(ABX\) along \(MN\). This produces a small triangle \(MXN\) and a trapezium \(ABNM\).

Now rotate the small triangle \(MXN\) through \(180^\circ\) about the point \(M\). The half-turn carries \(X\) to \(A\), and the triangle lands alongside the trapezium so that the two pieces together form a parallelogram of base \(\frac{BC}{3}\) and height \(\frac{h}{2}\).

Doing exactly the same to the red triangle \(AYC\) produces a parallelogram with the same base and the same height. Two parallelograms with equal base and equal height are related by a shear, and a shear can be realised by one straight cut and a slide. So blue can be made to cover red in three pieces — and if the two triangles happen to be congruent, one piece is enough.

Let the square have side \(s\).

Step 1: the quarter circle. Its centre is a vertex and it passes through the two neighbouring vertices, so its radius is the full side \(s\):

Step 2: the two semicircles. Each is drawn on a side as diameter, so each has radius \(\frac{s}{2}\):

Step 3: notice they are equal.

Step 4: subtract what they share. Let \(X\) be the region belonging to both the quarter circle and the semicircles. Then, reading the picture:

The two left-hand sides are equal, so the right-hand sides are too. Cancelling the common \(X\):

The two shaded regions have equal area, and the result never needed the actual value of either one.

The square has side \(2\), so each semicircle has diameter \(2\) and radius \(1\), centred at a side's midpoint and drawn inwards.

The perimeter. Every one of the four semicircular arcs lies entirely on the boundary of the flower — each arc forms one side of two neighbouring petals. So the flower's perimeter is the total length of the four arcs:

The area. Work with one quarter of the square — a \(1\times 1\) corner square. Inside it sits one petal-half, formed where two of the semicircles overlap.

Take the corner square with corners at \((0,0)\), \((1,0)\), \((1,1)\), \((0,1)\). The two arcs crossing it are quarter circles of radius \(1\) centred at \((1,0)\) and \((0,1)\). The region inside both is a classic lens, and its area is

(quarter circle plus quarter circle, minus the unit square they both sit in). There are four such corner squares, so:

Check. The whole square has area \(4\), and \(2\pi-4 \approx 2.28\) is about \(57\%\) of it — which matches the picture, where the flower covers rather more than half the square.

Let the larger circle have radius \(R\) and the smaller one radius \(r\). The green region is an annulus:

The key step: find \(R^2-r^2\). The chord \(BC\) touches the inner circle at \(A\), so \(OA\) is perpendicular to \(BC\) at \(A\) — a radius is perpendicular to the tangent at the point of contact.

But \(OA\) is also the perpendicular from the centre to the chord \(BC\) of the outer circle, and such a perpendicular bisects the chord:

Now apply the Baudhāyana–Pythagoras theorem in the right triangle \(OAB\), whose hypotenuse is the outer radius \(OB = R\):

Substitute back.

Let the right triangle have legs \(a\) and \(b\) and hypotenuse \(c\), with the semicircles \(A\), \(B\) on the legs and \(C\) on the hypotenuse. Each semicircle is drawn on its side as a diameter, so its radius is half that side.

Add the first two:

By the Baudhāyana–Pythagoras theorem, \(a^2+b^2 = c^2\):

Let the centres be \(A\) and \(B\). Each circle passes through the other's centre, so:

The circles cross at two points, \(C\) and \(D\) — the shaded lens in the figure. Look at \(\triangle ABC\): all three of \(AB\), \(AC\), \(BC\) are radii, so it is equilateral, and

Step 1: the sector. The lens is made of two circular segments, one from each circle. Take the sector \(CAD\) of the circle centred at \(A\), of angle \(120^\circ\):

Step 2: the triangle. \(\triangle ACD\) has \(AC = AD = r\) with a \(120^\circ\) angle between them:

Step 3: one segment, then two.

So the shaded lens has area \(r^2\left(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\right)\).

If instead you want the whole region covered by the two circles (their union), add the two discs and subtract the double-counted lens:

Reading the figure: the rectangle has a corner \(O\) at the bottom-left. An interior point \(M\) is joined to \(O\); from \(M\) a vertical segment runs up to a point \(T\) on the top edge, and a horizontal segment runs right to a point \(R\) on the right edge. The three triangles are \(A = \triangle OTM\), \(C = \triangle TRM\) and \(B = \triangle ORM\).

Put coordinates on it: let the rectangle be \(w\) wide and \(h\) tall with \(O\) at the origin, and let \(M = (p, q)\). Then \(T = (p, h)\) and \(R = (w, q)\).

Step 1: the three areas.

Step 2: the two sums simplify beautifully.

Step 3: multiply and divide.

and \(wh\) is exactly the area of the rectangle, as required.

Reading the figure: \(AC\) is the diameter of a large semicircle with centre \(O\), and \(B\) is the point of that semicircle directly above \(O\). The chord \(AB\) is drawn, and a second, smaller semicircle is drawn on \(AB\) as diameter (centre \(D\), the midpoint of \(AB\)), bulging away from \(O\) and passing through \(E\). The two shaded regions are the crescent between the small semicircle and the big arc, and the triangle \(AOB\).

Let \(OA = OB = OC = r\). Since \(\angle AOB = 90^\circ\):

Step 1: the quarter circle. The sector \(AOB\) of the big circle has angle \(90^\circ\):

Step 2: the small semicircle. Its diameter is \(AB = r\sqrt2\), so its radius is \(\frac{r\sqrt2}{2}\):

Step 3: they are equal.

Step 4: remove the shared piece. Let \(S\) be the circular segment of the big circle cut off by the chord \(AB\) — the sliver between chord \(AB\) and the big arc. Reading the picture:

The two left-hand sides are equal by Step 3, so cancelling \(S\) from both:

The two shaded regions are equal. Their common value is easy to write down, since \(\triangle AOB\) is right-angled with legs \(r\):