Chapter 6: Measuring Space: Perimeter and Area (Perimeter and Area)

$$2\pi r=44\Rightarrow 2\times \frac{22}{7}\times r=44\Rightarrow r=\frac{44\times 7}{44}=7\text{cm}$$

The radius is 7 cm.

$$\text{(i) }2\times \frac{22}{7}\times 7=44.0\text{cm}$$

$$\text{(ii) }2\times \frac{22}{7}\times 10=\frac{440}{7}=62.857\dots \approx 62.9\text{cm}$$

$$\text{(iii) }2\times \frac{22}{7}\times 12=\frac{528}{7}=75.428\dots \approx 75.4\text{cm}$$

(i) 44.0 cm (ii) 62.9 cm (iii) 75.4 cm

$$\text{(i) }\frac{60}{360}\times 2\times \frac{22}{7}\times 3.5=\frac{1}{6}\times 22=\frac{11}{3}\approx 3.67\text{cm}$$

$$\text{(ii) }\frac{120}{360}\times 2\times \frac{22}{7}\times 6.3=\frac{1}{3}\times 39.6=13.2\text{m}$$

(i) $\frac{11}{3}$ cm ≈ 3.67 cm (ii) 13.2 m

$$\text{Arc}=\frac{75}{360}\times 2\times \frac{22}{7}\times 14=\frac{75}{360}\times 88=\frac{55}{3}\approx 18.33\text{cm}$$

The two straight portions are radii: $14+14=28$ cm.

$$\text{Perimeter}=\frac{55}{3}+28=\frac{139}{3}\approx 46.33\text{cm}$$

The perimeter of the sector is about 46.33 cm.

A semicircle of diameter d has length $\frac{1}{2}\pi d$, and a quarter circle of radius r has length $\frac{1}{2}\pi r$. Only the outer boundary counts, not the dashed lines.

(i) A 80 m × 60 m rectangle with a semicircle (diameter 60 m) on each 60 m side. The boundary is the two 80 m sides and two semicircles, i.e. one full circle of diameter 60 m:

$$2\times 80+\frac{22}{7}\times 60=160+188.57=348.57\text{m}$$

(ii) A semicircular ring: outer diameter 12 cm, inner diameter 8 cm, and two straight ends of $\frac{12-8}{2}=2$ cm each:

$$\frac{1}{2}\pi (12)+\frac{1}{2}\pi (8)+2+2=10\pi +4=\frac{220}{7}+4\approx 35.43\text{cm}$$

(iii) A 10 cm square with a semicircle on every side: four semicircles of diameter 10 cm = two full circles:

$$2\times \frac{22}{7}\times 10=\frac{440}{7}\approx 62.86\text{cm}$$

(iv) An equilateral triangle of side 12 cm with a semicircle on each side: three semicircles of diameter 12 cm:

$$3\times \frac{1}{2}\pi \times 12=18\pi =\frac{396}{7}\approx 56.57\text{cm}$$

(v) A plus-shape of 14 cm squares. The boundary is four semicircles of diameter 14 cm (at the ends of the arms) and four quarter circles of radius 14 cm (at the corners between the arms), and no straight parts:

$$4\times \frac{1}{2}\pi (14)+4\times \frac{1}{2}\pi (14)=28\pi +28\pi =56\pi =176\text{cm}$$

(vi) A large semicircle of diameter 28 cm; the base is replaced by four small semicircles, each of diameter $28÷4=7$ cm:

$$\frac{1}{2}\pi (28)+4\times \frac{1}{2}\pi (7)=14\pi +14\pi =28\pi =88\text{cm}$$

(vii) A right triangle with legs 8 cm and 6 cm has hypotenuse $\sqrt{64+36}=10$ cm. The boundary is three semicircles on the sides:

$$\frac{1}{2}\pi (8+6+10)=12\pi =\frac{264}{7}\approx 37.71\text{cm}$$

(viii) A semicircle of diameter 12 cm with three semicircles of diameter 4 cm along its base:

$$\frac{1}{2}\pi (12)+3\times \frac{1}{2}\pi (4)=6\pi +6\pi =12\pi \approx 37.71\text{cm}$$

(ix) A semicircle of diameter 20 cm on top, and two semicircles of diameter 10 cm (one inside on the left, one below on the right):

$$\frac{1}{2}\pi (20)+\frac{1}{2}\pi (10)+\frac{1}{2}\pi (10)=20\pi =\frac{440}{7}\approx 62.86\text{cm}$$

(i) 348.57 m (ii) 35.43 cm (iii) 62.86 cm (iv) 56.57 cm (v) 176 cm (vi) 88 cm (vii) 37.71 cm (viii) 37.71 cm (ix) 62.86 cm

(i) In one revolution the car moves one circumference:

$$\pi d=\frac{22}{7}\times 56=176\text{cm}$$

(ii) 10 km $=10\times 1000\times 100=1000000$ cm.

$$\frac{1000000}{176}\approx 5681.8$$

(i) 176 cm (ii) about 5682 revolutions (5681 complete revolutions).

(i) Each arc is part of a semicircle of radius 7 cm (half the side) drawn inside the square on a side as diameter; each semicircle passes through the centre of the square. The four semicircles make up all the petal edges:

$$4\times \frac{1}{2}\times 2\pi \times 7=28\pi =88\text{cm}$$

(ii) In a regular hexagon of side 42 cm, each vertex is 42 cm from its two neighbouring vertices and from the centre. So the arc with centre at a vertex has radius 42 cm and runs from one neighbouring vertex, through the centre, to the other. It turns through the interior angle of the hexagon, 120°:

$$\text{One arc}=\frac{120}{360}\times 2\times \frac{22}{7}\times 42=\frac{1}{3}\times 264=88\text{cm}$$

There are 6 such arcs (one for each vertex), and together they form all the petal edges:

$$6\times 88=528\text{cm}$$

(i) 88 cm (ii) 528 cm

$$\frac{2\pi r_1}{2\pi r_2}=\frac{5}{4}\Rightarrow \frac{r_1}{r_2}=\frac{5}{4}$$

Since the circumference is a fixed multiple ($2\pi$) of the radius, the ratio does not change.

The radii are in the ratio 5 : 4.

Take AD as the base: $AD=BC=8$ cm. The height of △ADE on base AD is the perpendicular distance from E to AD. Since E is on BC and BC ∥ AD, this distance is $AB=10$ cm, wherever E is on BC.

$$\text{Area}=\frac{1}{2}\times 8\times 10=40{\text{cm}}^2$$

(This is half the rectangle, $\frac{1}{2}\times 80$.)

Area of △ADE = 40 cm².

Drop perpendiculars from the ends of the 20 cm side onto the 40 cm side. Since the trapezium is isosceles, each overhang is $\frac{40-20}{2}=10$ cm.

$$h=\sqrt{26^2-10^2}=\sqrt{676-100}=\sqrt{576}=24\text{cm}$$

$$\text{Area}=\frac{1}{2}(40+20)\times 24=30\times 24=720{\text{cm}}^2$$

720 cm²

Third side $=32-8-11=13$ cm, and $s=16$ cm. By Heron's formula:

$$\sqrt{16(16-8)(16-11)(16-13)}=\sqrt{16\times 8\times 5\times 3}=\sqrt{1920}=8\sqrt{30}\approx 43.82{\text{cm}}^2$$

$8\sqrt{30}\approx 43.82$ cm²

Let the sides be 3k, 5k, 7k. Then $15k=300$, so $k=20$ and the sides are 60 m, 100 m and 140 m. $s=150$ m.

$$\sqrt{150\times 90\times 50\times 10}=\sqrt{6750000}=1500\sqrt{3}\approx 2598{\text{m}}^2$$

$1500\sqrt{3}\approx 2598$ m²

Area of a rhombus $=\frac{1}{2}{d}_1{d}_2$. Let the diagonals be d and 2d.

$$\frac{1}{2}\times d\times 2d=128\Rightarrow d^2=128\Rightarrow d=\sqrt{128}=8\sqrt{2}\approx 11.31\text{cm}$$

The shorter diagonal is $8\sqrt{2}\approx 11.31$ cm.

Both triangles have the same base CD. Their third vertices P and Q lie on AB, which is parallel to CD, so both are at the same distance (the height of the parallelogram) from CD. Same base and same height give the same area (each is half the parallelogram).

area (△PCD) : area (△QCD) = 1 : 1.

The diagonal PR divides the parallelogram into two congruent triangles PQR and PSR of equal area. Both have base PR, so their heights are equal: Q and S are at the same distance h from the line PR.

△PQO and △PSO have the common base PO (on line PR) and equal heights h (from Q and from S). Hence

$$\text{area}(△PQO)=\frac{1}{2}\times PO\times h=\text{area}(△PSO)$$

Q and S are equidistant from PR, and the triangles share the base PO, so their areas are equal.

Let ABCD be the 4-gon (convex), and P, Q, R, S the midpoints of AB, BC, CD, DA.

Corner triangle APS: AP = ½AB and AS = ½AD, so it has half the base and half the height of △ABD (measuring from A). Hence $\text{area}(APS)=\frac{1}{4}\text{area}(ABD)$. Similarly $\text{area}(CRQ)=\frac{1}{4}\text{area}(CDB)$. Adding:

$$\text{area}(APS)+\text{area}(CRQ)=\frac{1}{4}[\text{area}(ABD)+\text{area}(CDB)]=\frac{1}{4}\text{area}(ABCD)$$

In the same way, using diagonal AC: $\text{area}(BQP)+\text{area}(DSR)=\frac{1}{4}\text{area}(ABCD)$.

The four corner triangles together have area $\frac{1}{2}\text{area}(ABCD)$, so what remains, PQRS, has the other half.

The four corner triangles total $\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$ of ABCD, so the midpoint parallelogram PQRS has exactly half the area of ABCD.

A median divides a triangle into two triangles of equal area (equal bases BD = DC, same height from the apex).

• In △ABC, AD is a median: $\text{area}(ABD)=\text{area}(ACD)$.
• In △PBC, PD is a median: $\text{area}(PBD)=\text{area}(PCD)$.

Subtracting the second from the first:

$$\text{area}(ABD)-\text{area}(PBD)=\text{area}(ACD)-\text{area}(PCD)\Rightarrow \text{area}(ABP)=\text{area}(ACP)$$

area(ABD) = area(ACD) and area(PBD) = area(PCD); subtracting gives area(ABP) = area(ACP).

Let the side be s. Let P be at distances $h_1$ from AB and $h_2$ from CD; since AB ∥ CD, $h_1+h_2=s$.

$$\text{Red}=\frac{1}{2}s{h}_1+\frac{1}{2}s{h}_2=\frac{1}{2}s(h_1+h_2)=\frac{1}{2}{s}^2$$

So the red region is half the square, and the green region is the other half.

Red : Green = 1 : 1, wherever P is inside the square.

△PDQ and △PDC have the same base PD, and Q and C lie on the line CQ, which is parallel to PD. So they are at equal distances from PD, and

$$\text{area}(PDQ)=\text{area}(PDC)$$

Now

$$\text{area}(BPQ)=\text{area}(BPD)+\text{area}(PDQ)=\text{area}(BPD)+\text{area}(PDC)=\text{area}(BDC)$$

Since D is the midpoint of AB, CD is a median of △ABC, so $\text{area}(BDC)=\frac{1}{2}\text{area}(ABC)$.

area(BPQ) = area(BPD) + area(PDQ) = area(BPD) + area(PDC) = area(BDC) = ½ area(ABC).

$$\frac{60}{360}\times \frac{22}{7}\times 7^2=\frac{1}{6}\times 154=\frac{77}{3}\approx 25.67{\text{cm}}^2$$

$\frac{77}{3}\approx 25.67$ cm²

$2\pi r=44$ gives $r=7$ cm. A quadrant is a quarter of the circle:

$$\frac{1}{4}\times \frac{22}{7}\times 49=\frac{154}{4}=38.5{\text{cm}}^2$$

38.5 cm²

In 60 minutes the hand turns 360°, so in 10 minutes it turns 60°.

$$\frac{60}{360}\times \frac{22}{7}\times 7^2=\frac{154}{6}=\frac{77}{3}\approx 25.67{\text{cm}}^2$$

$\frac{77}{3}\approx 25.67$ cm²

$$\text{(i) }\frac{90}{360}\times 3.14\times 100=78.5{\text{cm}}^2$$

$$\text{(ii) }\frac{270}{360}\times 3.14\times 100=235.5{\text{cm}}^2(=314-78.5)$$

(i) 78.5 cm² (ii) 235.5 cm²

Minor segment = minor sector − △OAB. Since OA = OB and the angle is 60°, △OAB is equilateral with side 15 cm.

$$\text{Sector}=\frac{60}{360}\times 3.14\times 225=117.75{\text{cm}}^2$$

$$△OAB=\frac{\sqrt{3}}{4}\times 15^2=\frac{1.73\times 225}{4}=97.3125{\text{cm}}^2$$

$$\text{Minor segment}=117.75-97.3125=20.4375\approx 20.44{\text{cm}}^2$$

$$\text{Major segment}=\pi r^2-\text{minor segment}=706.5-20.4375=686.0625\approx 686.06{\text{cm}}^2$$

Minor segment ≈ 20.44 cm²; major segment ≈ 686.06 cm².

Each blade sweeps a sector of radius 28 cm and angle 120°:

$$\frac{120}{360}\times \frac{22}{7}\times 28^2=\frac{1}{3}\times 2464=\frac{2464}{3}{\text{cm}}^2$$

Two non-overlapping wipers:

$$2\times \frac{2464}{3}=\frac{4928}{3}\approx 1642.67{\text{cm}}^2$$

About 1642.67 cm².

The sector has area $\frac{60}{360}\pi r^2=\frac{\pi r^2}{6}$. The triangle formed by the chord and the two radii is equilateral (two sides r and the angle between them 60°), with area $\frac{\sqrt{3}}{4}{r}^2$. So

$$\text{Minor segment}=\frac{\pi r^2}{6}-\frac{\sqrt{3}}{4}{r}^2=r^2\left(\frac{\pi }{6}-\frac{\sqrt{3}}{4}\right)$$

Segment = sector − equilateral triangle = $r^2\left(\frac{\pi }{6}-\frac{\sqrt{3}}{4}\right)$.

Join the centre O to the three vertices. This makes three congruent triangles, each with two sides r and angle $\frac{360°}{3}=120°$ at O. The distance from O to each side is $\frac{r}{2}$ (see the note below), so half of a side is $\sqrt{r^2-\frac{r^2}{4}}=\frac{\sqrt{3}}{2}r$, and the side is $r\sqrt{3}$.

$$\text{Area of triangle}=\frac{\sqrt{3}}{4}(r\sqrt{3}{)}^2=\frac{3\sqrt{3}}{4}{r}^2$$

$$\frac{\text{Area of triangle}}{\text{Area of circle}}=\frac{\frac{3\sqrt{3}}{4}{r}^2}{\pi r^2}=\frac{3\sqrt{3}}{4\pi }\approx \frac{5.196}{12.566}\approx 0.413$$

(The distance from O to a side is $\frac{r}{2}$ because O is also the centroid, which divides each median 2 : 1, and the median is $r+\frac{r}{2}$.)

Side $=r\sqrt{3}$, area $=\frac{3\sqrt{3}}{4}{r}^2$, so the ratio is $\frac{3\sqrt{3}}{4\pi }\approx 0.413$.

The diagonals of the inscribed square are diameters, each of length 2r. Area of a square $=\frac{1}{2}(\text{diagonal}{)}^2$:

$$\text{Area of square}=\frac{1}{2}(2r{)}^2=2r^2,\frac{2r^2}{\pi r^2}=\frac{2}{\pi }\approx 0.637$$

Area of square = 2r², so the ratio is $\frac{2}{\pi }\approx 0.637$.

A regular hexagon inscribed in a circle is made of 6 equilateral triangles of side r (see Chapter 5, End-of-Chapter Q19):

$$\text{Area of hexagon}=6\times \frac{\sqrt{3}}{4}{r}^2=\frac{3\sqrt{3}}{2}{r}^2,\frac{\frac{3\sqrt{3}}{2}{r}^2}{\pi r^2}=\frac{3\sqrt{3}}{2\pi }\approx 0.827$$

Why twice: joining alternate vertices of the hexagon gives the inscribed equilateral triangle of Q8. Join O to the vertices A, C, E of that triangle. The hexagon splits into 3 rhombuses OABC, OCDE, OEFA (each made of 2 equilateral triangles of side r). Each side of the big triangle (AC, CE, EA) is a diagonal of one rhombus and cuts it into two equal halves, and the big triangle contains exactly one half of each rhombus. So the hexagon has exactly twice the area of the triangle.

Hexagon area = $\frac{3\sqrt{3}}{2}{r}^2$, ratio $\frac{3\sqrt{3}}{2\pi }\approx 0.827$; it is twice the triangle's ratio because the hexagon has exactly twice the area of the triangle on its alternate vertices.

In Fig. 6.41 the square of side a + b is split into a square $a^2$, a square $b^2$ and two rectangles ab, so $(a+b{)}^2=a^2+2ab+b^2$.

$(a+b)(a-b)=a^2-b^2$: from a square of side a, cut away a square of side b at one corner. The L-shaped remainder has area $a^2-b^2$. Cut the L into two rectangles, $a\times (a-b)$ and $b\times (a-b)$, and place them end to end: they form one rectangle $(a+b)\times (a-b)$.

$(a+b+c{)}^2$: divide each side of a square of side a + b + c into parts a, b, c. The square splits into 9 pieces: three squares $a^2,b^2,c^2$ along the diagonal and the rectangles ab, bc, ca twice each.

See the two area models above: an L-shape of area a² − b² rearranges into an (a + b) × (a − b) rectangle; a square of side a + b + c splits into a², b², c² and two each of ab, bc, ca.

Base $=40-30=10$ cm. The altitude to the base bisects it, so

$$h=\sqrt{15^2-5^2}=\sqrt{200}=10\sqrt{2}\text{cm},\text{Area}=\frac{1}{2}\times 10\times 10\sqrt{2}=50\sqrt{2}\approx 70.71{\text{cm}}^2$$

$50\sqrt{2}\approx 70.71$ cm²

$$\frac{1}{2}\times 10\times h=60\Rightarrow h=12\text{cm}$$

The altitude bisects the base, so each equal side is

$$\sqrt{5^2+12^2}=\sqrt{169}=13\text{cm}$$

Each equal side is 13 cm.

$$\frac{1}{2}\times 12\times x=54\Rightarrow x=9\text{cm},\text{hypotenuse}=\sqrt{144+81}=15\text{cm}$$

$$\text{Perimeter}=12+9+15=36\text{cm}$$

36 cm

$9k=45$, so $k=5$: the sides are 10, 15 and 20 cm, and $s=22.5$.

$$\sqrt{22.5\times 12.5\times 7.5\times 2.5}=\sqrt{\frac{45\times 25\times 15\times 5}{16}}=\frac{\sqrt{84375}}{4}=\frac{75\sqrt{15}}{4}\approx 72.62{\text{cm}}^2$$

$\frac{75\sqrt{15}}{4}\approx 72.62$ cm²

Method 1: $7^2+24^2=49+576=625=25^2$, so it is a right triangle with legs 7 and 24.

$$\text{Area}=\frac{1}{2}\times 7\times 24=84{\text{cm}}^2$$

Method 2 (Heron): $s=28$.

$$\sqrt{28\times 21\times 4\times 3}=\sqrt{7056}=84{\text{cm}}^2$$

84 cm² (both ways).

$$100\times \pi d=100\times \frac{22}{7}\times 60=\frac{132000}{7}\approx 18857\text{cm}\approx 188.57\text{m}$$

About 188.57 m.

$$2\times \frac{22}{7}\times r=66\Rightarrow r=10.5\text{cm},\frac{1}{4}\times \frac{22}{7}\times 10.5^2=\frac{346.5}{4}=86.625{\text{cm}}^2$$

86.625 cm²

$$2\times \frac{22}{7}\times 28=176\text{cm},\frac{100000}{176}\approx 568.2$$

176 cm per turn; about 568 turns in 1 km.

Yes. Let a rectangle have sides l and b with perimeter P and area A. Then $l+b=\frac{P}{2}$ and $lb=A$. So

$$(l-b{)}^2=(l+b{)}^2-4lb=\frac{P^2}{4}-4A$$

fixes $|l-b|$; together with $l+b$, this fixes the two side lengths (the larger is $\frac{1}{2}\left(\frac{P}{2}+|l-b|\right)$ and the smaller is $\frac{1}{2}\left(\frac{P}{2}-|l-b|\right)$). So both rectangles have the same pair of sides, and rectangles with the same sides are congruent.

Yes. Equal perimeter and area fix l + b and lb, which fix the two sides, so the rectangles are congruent.

From one end of the shorter side a, draw a line parallel to the opposite slanting side. This cuts the trapezium into a parallelogram with base a and height h, and a triangle with base $b-a$ and height h.

$$\text{Area}=ah+\frac{1}{2}(b-a)h=\frac{2ah+bh-ah}{2}=\frac{1}{2}(a+b)h$$

Parallelogram (ah) + triangle (½(b − a)h) = ½(a + b)h.

Draw a diagonal. It splits the trapezium into two triangles: one with base a (the top side) and one with base b (the bottom side). Both have height h, the distance between the parallel sides.

$$\text{Area}=\frac{1}{2}ah+\frac{1}{2}bh=\frac{1}{2}(a+b)h$$

The diagonal gives triangles of areas ½ah and ½bh, total ½(a + b)h.

Take a second copy of the trapezium, turn it upside down (rotate it through 180°) and place it against the first along a slanting side. The top side a of one copy now lies in line with the bottom side b of the other, so the combined figure has both pairs of opposite sides parallel: it is a parallelogram with base $a+b$ and height h.

$$\text{Area of 2 trapeziums}=(a+b)h\Rightarrow \text{Area of one trapezium}=\frac{1}{2}(a+b)h$$

Two copies (one rotated) make a parallelogram of base (a + b) and height h, so one trapezium has area ½(a + b)h.

Let the kite be ABCD with AB = AD and CB = CD. The diagonal AC (the axis of symmetry, length $d_1$) is the perpendicular bisector of the other diagonal BD (length $d_2$); they meet at O with $BO=OD=\frac{d_2}{2}$.

(i) Algebra: AC splits the kite into △ABC and △ADC, each with base AC = $d_1$ and height $\frac{d_2}{2}$:

$$\text{Area}=\frac{1}{2}{d}_1\cdot \frac{d_2}{2}+\frac{1}{2}{d}_1\cdot \frac{d_2}{2}=\frac{1}{2}{d}_1{d}_2$$

(ii) Geometry: draw the rectangle through A, B, C, D with sides parallel to the diagonals; its sides are $d_1$ and $d_2$. The diagonals cut the rectangle into 4 smaller rectangles, and each is cut in half by a side of the kite. So the kite is exactly half the rectangle:

$$\text{Area of kite}=\frac{1}{2}\times d_1\times d_2$$

Area of a kite = ½ × d₁ × d₂ (two triangles of base d₁ and height d₂/2; or half of the d₁ × d₂ rectangle).

(i) Area of PQRS $=2a\times 2b=4ab=4\times$ area of ABCD. Yes, 4 copies fit, in a 2 × 2 arrangement.

(ii) By Heron's formula, doubling every side doubles s and each of $s-a$, $s-b$, $s-c$:

$$\sqrt{2s\cdot 2(s-a)\cdot 2(s-b)\cdot 2(s-c)}=\sqrt{16}\sqrt{s(s-a)(s-b)(s-c)}=4\times \text{area}(ABC)$$

Yes, 4 copies fit: join the midpoints of the sides of △PQR. This makes 4 triangles, each with sides a, b, c (by the midpoint theorem); three are upright copies and the middle one is a copy turned upside down.

(iii) Tripling every side multiplies each factor by 3, so the area is multiplied by $\sqrt{3^4}=9$. Yes, 9 copies fit: divide each side of △PQR into three equal parts and draw lines parallel to the sides through these points. This gives 9 small triangles with sides a, b, c: 6 upright and 3 upside down.

(i) 4ab = 4 × ab; 4 copies fit (2 × 2). (ii) Area × 4; 4 copies fit (midpoint triangles). (iii) Area × 9; 9 copies fit (6 upright + 3 inverted).

Fig. 6.43: In △ABC (A at the top, B and C at the base), P is the midpoint of AB, and Q, R divide AC into three equal parts (AQ = QR = RC). The shaded region is the quadrilateral BPQR.

Take area(ABC) = 1.

• △ABR has the same height from B as △ABC and base AR $=\frac{2}{3}AC$, so area(ABR) $=\frac{2}{3}$.
• △APQ: AP $=\frac{1}{2}AB$ and AQ $=\frac{1}{3}AC$. Halving one side halves the area, and taking a third of the other takes a third of that: area(APQ) $=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$.

$$\text{Shaded}=\text{area}(ABR)-\text{area}(APQ)=\frac{2}{3}-\frac{1}{6}=\frac{1}{2}$$

Fig. 6.44: Each vertex of the square is joined to the midpoint of a side, making a small tilted square in the middle. The four lines meet at the corners of the inner square. We use a square of side 5 to get whole-number coordinates; the fraction does not depend on the size.

With the square from (0, 0) to (5, 5), the lines are $y=5-2x$, $y=2.5+\frac{x}{2}$, $y=10-2x$ and $y=\frac{x}{2}$. Solving them in pairs gives the corners of the inner square: (1, 3), (3, 4), (4, 2), (2, 1). Its side is

$$\sqrt{(3-1{)}^2+(4-3{)}^2}=\sqrt{5},\text{area}=5=\frac{5}{25}=\frac{1}{5}\text{ of the square}$$

Fig. 6.43: $\frac{1}{2}$ of the triangle is shaded. Fig. 6.44: $\frac{1}{5}$ of the square is shaded.

Let each circle have radius r. The circles touch each other and the sides of the rectangle, so the rectangle is $2r$ high, and $6r$ long for 3 circles ($8r$ for 4 circles).

$$\text{Fig. 6.45: }\frac{3\pi r^2}{6r\times 2r}=\frac{3\pi r^2}{12r^2}=\frac{\pi }{4},\text{Fig. 6.46: }\frac{4\pi r^2}{8r\times 2r}=\frac{\pi }{4}$$

In both cases the circles cover $\frac{\pi }{4}\approx 0.785$ (about 78.5%) of the rectangle.

Conjecture: however many circles are fitted in a row in this way, they always cover $\frac{\pi }{4}$ of the rectangle.

Tests: for 10 circles, $\frac{10\pi r^2}{20r\times 2r}=\frac{\pi }{4}$; for 20 circles, $\frac{20\pi r^2}{40r\times 2r}=\frac{\pi }{4}$; for 50 circles, $\frac{50\pi r^2}{100r\times 2r}=\frac{\pi }{4}$.

Proof: for n circles of radius r, the rectangle is $2nr$ long and $2r$ high.

$$\frac{n\pi r^2}{2nr\times 2r}=\frac{n\pi r^2}{4n{r}^2}=\frac{\pi }{4}$$

The n cancels: the rectangle is just n copies of a $2r\times 2r$ square with one circle in each, and a circle covers $\frac{\pi r^2}{4r^2}=\frac{\pi }{4}$ of its square.

The fraction is always $\frac{\pi }{4}\approx 0.785$, for any number of circles.

Let each small rectangle have length L and width W (L > W). In the figure the top row has 4 rectangles lying flat (each L wide, W tall) and the bottom row has 5 rectangles standing up (each W wide, L tall). Both rows have the same total width:

$$4L=5W\Rightarrow L=\frac{5W}{4}$$

The nine rectangles make up the large one:

$$9LW=72\Rightarrow LW=8\Rightarrow \frac{5W^2}{4}=8\Rightarrow W^2=\frac{32}{5}\Rightarrow W=\frac{4\sqrt{10}}{5}\approx 2.53\text{cm}$$

So $L=\frac{5W}{4}=\sqrt{10}\approx 3.16$ cm, and

$$\text{Perimeter}=2(L+W)=2\times \frac{9W}{4}=\frac{9W}{2}=\frac{18\sqrt{10}}{5}\approx 11.38\text{cm}$$

Check: $L\times W=\sqrt{10}\times \frac{4\sqrt{10}}{5}=8$ ✓

Each small rectangle is about 3.16 cm × 2.53 cm, with perimeter $\frac{18\sqrt{10}}{5}\approx 11.38$ cm.

Equal areas: the base is divided into three equal parts of length $\frac{b}{3}$. The blue and red triangles both have base $\frac{b}{3}$ on the same line and the same apex, so the same height h. Each has area $\frac{1}{2}\cdot \frac{b}{3}\cdot h=\frac{bh}{6}$.

Cutting and rearranging (one way):

• Cut the blue triangle along the line joining the midpoints of its two slanting sides (this line is parallel to the base, at height $\frac{h}{2}$). Turn the small top triangle through 180° about the midpoint of one slanting side. The two pieces form a parallelogram with base $\frac{b}{3}$ and height $\frac{h}{2}$.
• Do the same to the red triangle, in your mind: it is also equivalent to a parallelogram with base $\frac{b}{3}$ and height $\frac{h}{2}$.
• Two parallelograms with equal bases and equal heights can be turned into each other: cut a right triangle off one end of the first and slide it to the other end to make a rectangle $\frac{b}{3}\times \frac{h}{2}$; then cut that rectangle the same way to make the second parallelogram.
• Finally reverse the first step for the red triangle: the pieces now cover the red triangle exactly.

Both triangles have base b/3 and the same height, so each has area bh/6. Cutting the blue one at half-height gives a parallelogram (base b/3, height h/2), which can be cut and slid into the matching parallelogram of the red triangle, and then re-formed into the red triangle.

Let the side of the square be s. The quarter circle has radius s; the two semicircles (on the two sides through the centre of the quarter circle) each have diameter s, i.e. radius $\frac{s}{2}$.

$$\text{Quarter circle}=\frac{1}{4}\pi s^2,\text{Two semicircles}=2\times \frac{1}{2}\pi {\left(\frac{s}{2}\right)}^2=\frac{1}{4}\pi s^2$$

The two semicircles lie inside the quarter circle and overlap in region A. Region B is the part of the quarter circle not covered by either semicircle. So

$$\text{Quarter circle}=(\text{semicircle 1}+\text{semicircle 2}-A)+B$$

Since the quarter circle and the two semicircles have the same total area, $-A+B=0$, i.e. $A=B$.

Quarter circle = sum of the two semicircles (both ¼πs²), and quarter = semicircles − A + B; hence A = B.

Each semicircle has its diameter on a side of the square, so its radius is 1 unit, and it passes through the centre of the square.

Perimeter: the edges of the petals are made up of exactly the four semicircular arcs:

$$4\times \frac{1}{2}\times 2\pi \times 1=4\pi \approx 12.57\text{units}$$

Area: the four semicircles together cover the whole square, and the petals are exactly the parts covered twice. So

$$\text{(sum of semicircles)}=\text{area of square}+\text{area of petals}$$

$$4\times \frac{1}{2}\pi (1{)}^2=4+\text{petals}\Rightarrow \text{petals}=2\pi -4\approx 2.28\text{sq. units}$$

Perimeter = 4π ≈ 12.57 units; area = 2π − 4 ≈ 2.28 square units.

Let the radii be R (large) and r (small). The chord BC touches the small circle at A, so OA = r and OA ⊥ BC. The perpendicular from the centre bisects the chord, so $AB=\frac{l}{2}$. In right △OAB:

$$R^2=r^2+{\left(\frac{l}{2}\right)}^2\Rightarrow R^2-r^2=\frac{l^2}{4}$$

$$\text{Green area}=\pi R^2-\pi r^2=\pi (R^2-r^2)=\frac{1}{4}\pi l^2$$

$R^2-r^2={\left(\frac{l}{2}\right)}^2$, so the ring has area $\pi (R^2-r^2)=\frac{1}{4}\pi l^2$.

Let the legs be a, b and the hypotenuse c, so $a^2+b^2=c^2$. A semicircle on a side of length x has area $\frac{\pi x^2}{8}$, so

$$\text{semicircle}(a)+\text{semicircle}(b)=\frac{\pi (a^2+b^2)}{8}=\frac{\pi c^2}{8}=\text{semicircle}(c)$$

The semicircle on the hypotenuse passes through the right-angle vertex (angle in a semicircle) and is made of the triangle C plus two segments, $S_1$ (on side a) and $S_2$ (on side b). Each leg's semicircle is made of a crescent plus the same segment: semicircle(a) = A + $S_1$ and semicircle(b) = B + $S_2$. So

$$(A+S_1)+(B+S_2)=C+S_1+S_2\Rightarrow A+B=C$$

The two leg semicircles together equal the hypotenuse semicircle (Pythagoras); removing the common segments leaves A + B = C.

Let the centres be A and B (AB = r) and let the circles meet at C and D. Then AC = BC = AB = r, so △ABC is equilateral; similarly △ABD. Hence ∠CAD = 120° and ∠CBD = 120°.

The common region is two equal segments, one cut from each circle by the chord CD, each with central angle 120°.

$$\text{One segment}=\frac{120}{360}\pi r^2-\text{area}(△ACD)$$

△ACD has two sides r with 120° between them. The chord CD is bisected by AB at right angles, so CD $=2\times \frac{\sqrt{3}}{2}r=\sqrt{3}r$ and the distance from A to CD is $\frac{r}{2}$, so area(ACD) $=\frac{1}{2}\times \sqrt{3}r\times \frac{r}{2}=\frac{\sqrt{3}}{4}{r}^2$.

$$\text{Common region}=2\left(\frac{\pi r^2}{3}-\frac{\sqrt{3}}{4}{r}^2\right)=\frac{2\pi r^2}{3}-\frac{\sqrt{3}}{2}{r}^2=r^2\left(\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\right)\approx 1.23r^2$$

$r^2\left(\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\right)\approx 1.23r^2$

Place the rectangle with its bottom-left corner P at (0, 0), width w and height h. From the figure: T = (t, h) is on the top side, R = (w, k) is on the right side, and X = (t, k) is directly below T and level with R. The triangles are A = △PTX, B = △PXR and C = △TXR.

$$C=\frac{1}{2}(w-t)(h-k),A=\frac{1}{2}t(h-k),B=\frac{1}{2}(w-t)k$$

(A has the vertical base TX = h − k and height t; B has the horizontal base XR = w − t and height k.) Then

$$A+C=\frac{1}{2}(h-k)w,B+C=\frac{1}{2}(w-t)h$$

$$\frac{2(A+C)(B+C)}{C}=\frac{2\cdot \frac{1}{2}(h-k)w\cdot \frac{1}{2}(w-t)h}{\frac{1}{2}(w-t)(h-k)}=wh$$

which is the area of the rectangle.

With the coordinates above, A + C = ½w(h − k), B + C = ½h(w − t), C = ½(w − t)(h − k), and the expression simplifies to wh.

In Fig. 6.55, O is the centre of the semicircle ABC of radius r, and OB ⊥ AC. Arc AFB is the quarter circle (centre O). On AB as diameter (centre D, the midpoint of AB) a semicircle AEB is drawn. The shaded regions are the crescent AEBF and the triangle AOB.

Since OA = OB = r and ∠AOB = 90°, $AB=r\sqrt{2}$.

$$\text{Semicircle on }AB=\frac{1}{2}\pi {\left(\frac{r\sqrt{2}}{2}\right)}^2=\frac{\pi r^2}{4}$$

$$\text{Segment AFB}=\text{quarter circle}-△AOB=\frac{\pi r^2}{4}-\frac{r^2}{2}$$

$$\text{Crescent AEBF}=\text{semicircle on }AB-\text{segment AFB}=\frac{\pi r^2}{4}-\left(\frac{\pi r^2}{4}-\frac{r^2}{2}\right)=\frac{r^2}{2}$$

And $\text{area}(△AOB)=\frac{1}{2}\times r\times r=\frac{r^2}{2}$.

Both shaded regions have area $\frac{r^2}{2}$: the crescent equals the semicircle on AB (πr²/4) minus the segment (πr²/4 − r²/2).