Textbook Exercises & Solutions

Given

A circle has circumference 44 cm. Use π≈$\frac{22}{7}$ unless stated otherwise in this exercise.

To Find

Circle radius

Reasoning

C = 2πr.

Solution

1. r = 44 ÷ (2 × $\frac{22}{7}$) = 7.

Result

7 cm using the prescribed approximation.

Given

Find circumferences to three significant figures for radii (i) 7 cm, (ii) 10 cm, (iii) 12 cm, using π≈$\frac{22}{7}$.

To Find

Three rounded circumferences

Reasoning

C = 2πr.

Solution

1. (i) 2 × $\frac{22}{7}$ × 7 = 44, written 44.0 cm to three significant figures.

2. (ii) $\frac{440}{7}$ = 62.857… cm, rounding to 62.9 cm.

3. (iii) $\frac{528}{7}$ = 75.428… cm, rounding to 75.4 cm.

Result

44.0 cm; 62.9 cm; 75.4 cm.

Given

Find arc lengths: (i) r = 3.5 cm, θ = 60°; (ii) r = 6.3 m, θ = 120°. Use π≈$\frac{22}{7}$.

To Find

The two arc lengths

Reasoning

Arc length = (θ/360) × 2πr.

Solution

1. (i) ($\frac{60}{360}$) × 2 × $\frac{22}{7}$ × 3.5 = $\frac{11}{3}$ cm.

2. (ii) ($\frac{120}{360}$) × 2 × $\frac{22}{7}$ × 6.3 = 13.2 m.

Result

(i) $\frac{11}{3}$ cm; (ii) 13.2 m with the prescribed π approximation.

Given

Find the full perimeter of a radius-14 cm sector with angle 75°, using π≈$\frac{22}{7}$.

To Find

Sector perimeter including its straight edges

Reasoning

Perimeter = arc length + two radii.

Solution

1. Arc = ($\frac{75}{360}$) × 88 = $\frac{55}{3}$ cm.

2. Add 28 cm for the two radii: $\frac{55}{3}$+28 = $\frac{139}{3}$ cm.

Result

$\frac{139}{3}$ cm, approximately 46.33 cm.

Given

Find the nine perimeters in Fig. 6.14, using the stated quarter/half-circle arcs and π≈$\frac{22}{7}$.

To Find

Perimeters in source figure order (i)–(ix)

Reasoning

Add boundary arcs and exposed straight edges only. Dashed construction lines are not boundary.

Solution

1. (i) Two straight lengths 80 m and two semicircles of diameter 60 m: P = 160+60π≈$\frac{2440}{7}$ m.

   The source repeats equality ticks on sides labelled 60 and 80

   the numerical dimensions, not those inconsistent ticks, are used here.

2. (ii) Semicircular band with outer diameter 12 cm and inner diameter 8 cm: P = 6π+4π+(12−8) = 10π+4≈$\frac{248}{7}$ cm.

3. (iii) Four outward semicircles on a side-10 cm square: P = 4 × 5π = 20π≈$\frac{440}{7}$ cm.

4. (iv) Three semicircles of diameter 12 cm: P = 18π≈$\frac{396}{7}$ cm.

5. (v) Write s for each equal marked grid interval.

   Four radius-s quarter circles and four diameter-s semicircles give P = 2πs+2πs = 4πs.

   Reading the 14 cm label as one marked interval gives P = 56π≈176 cm.

   This dimension convention must be stated because the source has no dimension-arrow endpoints for the label.

6. (vi) One diameter-28 cm semicircle plus four diameter-7 cm semicircles: P = 14π+4 × 3.5π = 28π≈88 cm.

7. (vii) The perpendicular sides 6,8 give hypotenuse 10.

   Three semicircles have these diameters: P = (6+8+10)$\frac{\pi }{2}$ = 12π≈$\frac{264}{7}$ cm.

8. (viii) Diameter-12 cm outer semicircle plus three diameter-4 cm semicircles: P = 6π+3 × 2π = 12π≈$\frac{264}{7}$ cm.

9. (ix) Diameter-20 cm outer semicircle plus two diameter-10 cm semicircles: P = 10π+5π+5π = 20π≈$\frac{440}{7}$ cm.

Result

In order: $\frac{2440}{7}$ m; $\frac{248}{7}$,$\frac{440}{7}$,$\frac{396}{7}$,176,88,$\frac{264}{7}$,$\frac{264}{7}$,$\frac{440}{7}$ cm, with the stated source conventions.

Given

A car tyre has diameter 56 cm. Find travel in one revolution and revolutions in 10 km. Use π≈$\frac{22}{7}$ and rolling without slipping.

To Find

Rolling distance and rotation count

Reasoning

Distance = number of turns × circumference.

Solution

1. One revolution = πd = ($\frac{22}{7}$) × 56 = 176 cm = 1.76 m.

2. 10 km = 1000000 cm.

   Turns = $\frac{1000000}{176}$ = $\frac{62500}{11}$≈5681.818.

3. This is 5681 complete revolutions plus $\frac{9}{11}$ of another

   do not round it to an exact integer count.

Result

176 cm per turn; $\frac{62500}{11}$ turns over 10 km.

Given

Find total petal perimeters in Fig. 6.15: (i) square side 14 cm, arc centres at side midpoints; (ii) the depicted regular hexagon of side 42 cm, arc centres at vertices.

To Find

Total curved boundary lengths

Reasoning

Sum the arcs around all petals; use π≈$\frac{22}{7}$.

Solution

1. (i) Each of four petals has two quarter-circle arcs of radius 7.

   Total = 8 × ($\frac{1}{4}$) × 2π × 7 = 28π≈88 cm.

2. (ii) Under the depicted regular-hexagon construction, each of six petals has two 60° arcs of radius 42.

   Total = 12 × ($\frac{1}{6}$) × 2π × 42 = 168π≈528 cm.

3. Equal side marks alone do not prove an arbitrary hexagon regular.

   The second result uses the regular arrangement shown

   without its angle/symmetry condition, side length alone does not fix the arc angles.

Result

(i) 88 cm. (ii) 528 cm for the stated regular-hexagon arrangement.

Given

Two circle perimeters are in the ratio 5:4.

To Find

Ratio of their radii

Reasoning

C = 2πr, with the same constant factor for both circles.

Solution

1. C1/C2 = $\frac{\left(2\mathrm{\pi r}1\right)}{\left(2\mathrm{\pi r}2\right)}$ = r1/r2.

Result

5:4.

Given

In rectangle ABCD of width 10 cm and height 8 cm, E lies on BC (Fig. 6.31). Find area of △ADE.

To Find

Area of △ADE

Reasoning

Triangle area = half base × perpendicular height.

Solution

1. Take AD = 8 cm as base.

   The perpendicular distance from E to AD is the rectangle width 10 cm.

2. Area = 8 × $\frac{10}{2}$ = 40 ${\mathrm{cm}}^2$, regardless of E’s position on BC.

Result

40 ${\mathrm{cm}}^2$.

Given

An isosceles trapezium has parallel sides 40 cm and 20 cm and legs 26 cm.

To Find

Trapezium area

Reasoning

A perpendicular from each end of the shorter base leaves equal overhangs.

Solution

1. Each overhang = $\frac{\left(40-20\right)}{2}$ = 10 cm.

   Height h satisfies $h^2$ = ${26}^2$−${10}^2$ = 576, so h = 24 cm.

2. Area = (40+20) × $\frac{24}{2}$ = 720 ${\mathrm{cm}}^2$.

Result

720 ${\mathrm{cm}}^2$.

Given

A triangle has sides 8 cm and 11 cm and perimeter 32 cm.

To Find

Triangle area

Reasoning

Heron’s formula A = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$.

Solution

1. Third side = 32−8−11 = 13 cm

   semiperimeter s = 16 cm.

2. Area = $\sqrt{16\times 8\times 5\times 3}$ = $\sqrt{1920}$ = 8$\sqrt{30}$ ${\mathrm{cm}}^2$.

Result

8$\sqrt{30}$ ${\mathrm{cm}}^2$.

Given

A triangular plot has side ratio 3:5:7 and perimeter 300 m.

To Find

Plot area

Reasoning

Scale the ratio, then apply Heron’s formula.

Solution

1. One ratio unit = $\frac{300}{15}$ = 20 m

   sides are 60,100,140 m.

2. s = 150 m.

   A = $\sqrt{150\times 90\times 50\times 10}$ = $\sqrt{6750000}$ = 1500$\sqrt{3}$ $m^2$.

Result

1500$\sqrt{3}$ $m^2$.

Given

A rhombus has area 128 ${\mathrm{cm}}^2$ and one diagonal is twice the other.

To Find

Shorter diagonal length

Reasoning

Rhombus area = half the product of its diagonals.

Solution

1. Let the shorter diagonal be d > 0 and the longer be 2d.

2. 128 = d × 2$\frac{d}{2}$ = $d^2$, giving d = $\sqrt{128}$ = 8$\sqrt{2}$ cm.

Result

8$\sqrt{2}$ cm.

Given

P and Q lie on AB of parallelogram ABCD. Compare areas △PCD and △QCD.

To Find

Ratio of the two triangle areas

Reasoning

Common base CD and equal perpendicular heights between parallel lines AB and CD.

Solution

1. Both triangle areas equal CD × $\frac{h}{2}$.

Result

1:1.

Given

O lies on diagonal PR of parallelogram PQRS. Prove equal areas for △PSO and △PQO.

To Prove

Equality of the two areas

Reasoning

A parallelogram diagonal divides its area in half; triangles on the same line have area proportional to base.

Proof

1. △PSR and △PQR each have half the parallelogram area.

2. Their subtriangles PSO and PQO have common base PO on line PR.

   Each is the fraction $\frac{\mathrm{PO}}{\mathrm{PR}}$ of its respective larger triangle.

3. Multiplying equal areas by the same fraction gives equal areas.

   If O = P, both areas are zero.

Result

Area(△PSO) = Area(△PQO).

Given

Join consecutive side midpoints of a quadrilateral. Prove the resulting parallelogram has half the original area.

To Prove

Midpoint-parallelogram area relationship

Reasoning

Midpoint theorem and area decomposition for a simple quadrilateral.

Proof

1. For convex ABCD let E,F,G,H be consecutive side midpoints.

   In triangles ABC and ADC, EF and HG are each parallel to AC and half its length.

   Similarly FG and HE are parallel and equal, giving a parallelogram.

2. Each corner triangle is one-quarter of the triangle made by that corner and its two neighbouring vertices.

   The sum of those four larger triangle areas is twice area(ABCD).

3. The removed corner triangles therefore total half the original area.

   The inner parallelogram occupies the remaining half.

4. For a simple concave quadrilateral the same identity follows with signed areas: substituting E = $\frac{\left(A+B\right)}{2}$, F = $\frac{\left(B+C\right)}{2}$, G = $\frac{\left(C+D\right)}{2}$, H = $\frac{\left(D+A\right)}{2}$ in the polygon area sum gives exactly half the original signed sum.

   Take absolute values.

   Self-crossing figures require a different convention for area.

Result

The midpoint parallelogram has half the area of the simple quadrilateral.

Given

D is the midpoint of BC in △ABC; P lies on median AD. Prove area(△ABP) = area(△ACP).

To Prove

Equality of the areas on opposite sides of AP

Reasoning

Equal bases on line BC give equal areas for triangles with common apex.

Proof

1. BD = DC, so area(ABD) = area(ACD) and area(PBD) = area(PCD).

2. Subtract the second equality from the first.

   The remainders are ABP and ACP.

   Endpoint cases follow as well.

Result

Area(△ABP) = area(△ACP).

Given

P lies inside square ABCD. Compare the combined areas PAB+PCD with PBC+PDA.

To Find

Ratio of the two opposite-pair area sums

Reasoning

Distances from an interior point to opposite sides add to the square side.

Solution

1. Let side length be s.

   If distances from P to AB and CD are h and s−h, their triangle areas sum to s$\frac{\left(h+s-h\right)}{2}$ = $\frac{s^2}{2}$.

2. The other opposite pair also sums to $\frac{s^2}{2}$.

Result

1:1.

Given

D is the midpoint of AB in △ABC. P lies on BC; draw CQ parallel to PD to meet line AB at Q. Prove area(BPQ) = area$\frac{\left(\mathrm{ABC}\right)}{2}$, and check the source’s requirement that Q lie on side AB.

To Prove

Area identity and valid placement of Q

Reasoning

Similar triangles BPD and BCQ give proportional sides.

Proof

1. For P≠B, $\frac{\mathrm{BP}}{\mathrm{BC}}$ = $\frac{\mathrm{BD}}{\mathrm{BQ}}$.

   Since BD = $\frac{\mathrm{BA}}{2}$, ($\frac{\mathrm{BP}}{\mathrm{BC}}$)($\frac{\mathrm{BQ}}{\mathrm{BA}}$) = $\frac{1}{2}$.

2. Triangles BPQ and BAC have the same included angle at B, so their area ratio is that product, $\frac{1}{2}$.

3. To keep Q on segment AB, BQ ≤ BA requires BP ≥ $\frac{\mathrm{BC}}{2}$.

   If P is nearer B than the midpoint of BC, Q lies on AB extended beyond A.

   At P = B the construction is undefined.

4. Thus the source’s “any point” and “Q on AB” need this qualification

   the area identity remains valid for the permitted construction, including Q on the extension.

Result

Area(BPQ) = area$\frac{\left(\mathrm{ABC}\right)}{2}$ for P≠B; Q lies on side AB only when BP ≥ $\frac{\mathrm{BC}}{2}$.

Given

Find area of a sector with radius 7 cm and angle 60°, using π≈$\frac{22}{7}$.

To Find

Sector area

Reasoning

A = (θ/360)π$r^2$.

Solution

1. A = ($\frac{60}{360}$) × $\frac{22}{7}$ × 49 = $\frac{77}{3}$ ${\mathrm{cm}}^2$.

Result

$\frac{77}{3}$ ${\mathrm{cm}}^2$.

Given

Find area of a quadrant of a circle with circumference 44 cm, using π≈$\frac{22}{7}$.

To Find

Quadrant area

Reasoning

r = $\frac{C}{\left(2\pi \right)}$; a quadrant is one-quarter of the circle.

Solution

1. r = 7 cm.

   Area = π × $\frac{49}{4}$ = 38.5 ${\mathrm{cm}}^2$ using $\frac{22}{7}$.

Result

38.5 ${\mathrm{cm}}^2$.

Given

A clock minute hand is 7 cm long. Find the area it sweeps in 10 minutes, using π≈$\frac{22}{7}$.

To Find

Swept sector area

Reasoning

The hand completes 360° in 60 minutes.

Solution

1. Angle = 360 × $\frac{10}{60}$ = 60°.

2. Area = ($\frac{1}{6}$) × $\frac{22}{7}$ × 49 = $\frac{77}{3}$ ${\mathrm{cm}}^2$.

Result

$\frac{77}{3}$ ${\mathrm{cm}}^2$.

Given

A radius-10 cm circle has a chord subtending 90°. Find minor and major sector areas using π≈3.14.

To Find

The two sector areas

Reasoning

Sectors occupy $\frac{1}{4}$ and $\frac{3}{4}$ of the whole circle.

Solution

1. Whole area = 3.14 × 100 = 314 ${\mathrm{cm}}^2$.

2. Minor sector = 78.5 ${\mathrm{cm}}^2$

   major sector = 235.5 ${\mathrm{cm}}^2$.

Result

78.5 ${\mathrm{cm}}^2$ and 235.5 ${\mathrm{cm}}^2$.

Given

In a radius-15 cm circle, a chord subtends 60°. Find minor and major segment areas using π≈3.14 and $\sqrt{3}$≈1.73.

To Find

Minor and major circular-segment areas

Reasoning

Subtract the equilateral triangle from the 60° sector.

Solution

1. Sector area = ($\frac{1}{6}$) × 3.14 × 225 = 117.75 ${\mathrm{cm}}^2$.

2. Triangle area = ($\frac{\sqrt{3}}{4}$) × 225≈97.3125 ${\mathrm{cm}}^2$.

3. Minor segment ≈117.75−97.3125 = 20.4375 ${\mathrm{cm}}^2$.

4. Whole circle ≈706.5 ${\mathrm{cm}}^2$, so major segment ≈706.5−20.4375 = 686.0625 ${\mathrm{cm}}^2$.

Result

Approximately 20.4375 ${\mathrm{cm}}^2$ and 686.0625 ${\mathrm{cm}}^2$ with the prescribed approximations.

Given

Two non-overlapping wipers each sweep 120° with a 28 cm blade. Find the total area cleaned using π≈$\frac{22}{7}$.

To Find

Combined swept area

Reasoning

Use the source model of two radius-28 cm sectors with blades extending from their pivots.

Solution

1. One sector = ($\frac{120}{360}$) × $\frac{22}{7}$ × ${28}^2$ = $\frac{2464}{3}$ ${\mathrm{cm}}^2$.

2. Two disjoint sweeps give $\frac{4928}{3}$ ${\mathrm{cm}}^2$.

   If a blade starts away from the pivot, an inner radius is needed instead

   this answer uses the source sector model.

Result

$\frac{4928}{3}$ ${\mathrm{cm}}^2$ in the stated model.

Given

Explain the area identity (a+b)${}^2$ and draw area models for (a+b)(a−b) = $a^2$−$b^2$ and (a+b+c)${}^2$ = $a^2$+$b^2$+$c^2$+2ab+2bc+2ca.

To Find

Geometric decompositions for the identities

Reasoning

Areas add when a region is partitioned without gaps or overlap.

Solution

1. The side-(a+b) square splits at lengths a and b in both directions into $a^2$,$b^2$ and two ab rectangles.

2. For a ≥ b > 0, remove a side-b square from a side-a square at a corner.

   Split the L-shaped remainder into rectangles a × (a−b) and b × (a−b)

   place them together to form (a+b) × (a−b).

3. For the third identity, split each side of the (a+b+c) square into lengths a,b,c.

   The nine cells have areas $a^2$,$b^2$,$c^2$,ab,ab,bc,bc,ca,ca.

4. The geometric constructions use non-negative lengths

   the algebraic identities themselves extend beyond that restriction.

Result

The decomposed areas give exactly the three stated expansions.

Given

An isosceles triangle has perimeter 40 cm and equal sides 15 cm.

To Find

Triangle area

Reasoning

Its altitude bisects the base.

Solution

1. Base = 40−30 = 10 cm.

   Half-base = 5 cm.

2. Height = $\sqrt{{15}^2-5^2}$ = $\sqrt{200}$ = 10$\sqrt{2}$ cm.

3. Area = 10 × $\frac{10\sqrt{2}}{2}$ = 50$\sqrt{2}$ ${\mathrm{cm}}^2$.

Result

50$\sqrt{2}$ ${\mathrm{cm}}^2$.

Given

An isosceles triangle has base 10 cm and area 60 ${\mathrm{cm}}^2$.

To Find

Length of each equal side

Reasoning

Find the height from area, then apply the right-triangle theorem.

Solution

1. 60 = 10$\frac{h}{2}$ gives h = 12 cm.

   Half-base = 5 cm.

2. Equal side = $\sqrt{{12}^2+5^2}$ = 13 cm.

Result

13 cm each.

Given

A right triangle has area 54 ${\mathrm{cm}}^2$ and one leg 12 cm.

To Find

Triangle perimeter

Reasoning

A = $\frac{\mathrm{ab}}{2}$, then $c^2$ = $a^2$+$b^2$.

Solution

1. Other leg = $\frac{108}{12}$ = 9 cm.

   Hypotenuse = $\sqrt{144+81}$ = 15 cm.

2. Perimeter = 12+9+15 = 36 cm.

Result

36 cm.

Given

Triangle sides are in ratio 2:3:4 and perimeter is 45 cm.

To Find

Triangle area

Reasoning

Heron’s formula.

Solution

1. Sides are 10,15,20 cm

   semiperimeter is $\frac{45}{2}$ cm.

2. Area = $\sqrt{\left(\frac{45}{2}\right)\left(\frac{25}{2}\right)\left(\frac{15}{2}\right)\left(\frac{5}{2}\right)}$ = $\frac{\sqrt{84375}}{4}$ = $\frac{75\sqrt{15}}{4}$ ${\mathrm{cm}}^2$.

Result

$\frac{75\sqrt{15}}{4}$ ${\mathrm{cm}}^2$.

Given

Find the area of a triangle with sides 7,24,25 cm in two ways.

To Find

Area by right-triangle and Heron methods

Reasoning

$7^2$+${24}^2$ = 49+576 = 625 = ${25}^2$, so the converse establishes a right angle. Method 1: A = 7 × $\frac{24}{2}$ = 84 ${\mathrm{cm}}^2$. Method 2: s = 28; A = $\sqrt{28\times 21\times 4\times 3}$ = $\sqrt{7056}$ = 84 ${\mathrm{cm}}^2$.

Solution

1. The two independent methods agree.

Result

84 ${\mathrm{cm}}^2$.

Given

A bicycle wheel has diameter 60 cm and makes 100 turns without slipping. Use π≈$\frac{22}{7}$.

To Find

Travel distance

Reasoning

Distance = turns × πd.

Solution

1. 100 × π × 60 cm = 6000π cm = 60π m≈$\frac{1320}{7}$ m.

Result

$\frac{1320}{7}$ m ≈188.57 m.

Given

Find the quadrant area when the circle circumference is 66 cm. Use π≈$\frac{22}{7}$.

To Find

Quadrant area

Reasoning

r = $\frac{C}{\left(2\pi \right)}$; quadrant area = $\frac{{\mathrm{\pi r}}^2}{4}$.

Solution

1. r = 66 × $\frac{7}{44}$ = 10.5 cm.

2. A = ($\frac{22}{7}$) × $\frac{{10.5}^2}{4}$ = 86.625 ${\mathrm{cm}}^2$.

Result

86.625 ${\mathrm{cm}}^2$.

Given

A car wheel has outer radius 28 cm. Find distance per turn and turns in 1 km, using π≈$\frac{22}{7}$ and no slipping.

To Find

Rolling distance and turns

Reasoning

C = 2πr; turns = distanc$\frac{e}{C}$.

Solution

1. C = 2 × $\frac{22}{7}$ × 28 = 176 cm = 1.76 m.

2. 1 km = 100000 cm

   turns = $\frac{100000}{176}$ = $\frac{6250}{11}$≈568.182.

Result

176 cm per turn; $\frac{6250}{11}$ turns in 1 km.

Given

Use the parallelogram in Fig. 6.42 to derive the trapezium area formula for parallel sides a,b and height h.

To Find

Area of the trapezium

Reasoning

Complete the trapezium to a parallelogram and remove the excess triangle.

Solution

1. Extend the shorter top base to length b, forming a parallelogram with the same base b and height h.

2. The added triangle has base b−a and height h.

3. Trapezium area = bh−(b−a)$\frac{h}{2}$ = (a+b)$\frac{h}{2}$.

Result

A = (a+b)$\frac{h}{2}$.

Given

Divide a trapezium into two triangles to derive its area formula.

To Find

Area from a diagonal partition

Reasoning

The two triangles have bases a and b and the common perpendicular height h between parallel sides.

Solution

1. Draw either diagonal.

   Triangle areas are $\frac{\mathrm{ah}}{2}$ and $\frac{\mathrm{bh}}{2}$.

2. Add them: A = (a+b)$\frac{h}{2}$.

Result

A = (a+b)$\frac{h}{2}$.

Given

Fit two identical trapezia together to form a parallelogram and derive the area formula.

To Find

Area of one trapezium

Reasoning

A half-turn of a copy permits a shared non-parallel edge.

Solution

1. Join a trapezium to its rotated copy along a non-parallel side.

   The resulting parallelogram has base a+b and height h.

2. Its area is (a+b)h, twice one trapezium’s area.

Result

A = (a+b)$\frac{h}{2}$.

Given

Show algebraically and geometrically that a kite has half the product of its diagonals as its area.

To Find

Kite area formula

Reasoning

The symmetry diagonal bisects the other at right angles.

Solution

1. Let perpendicular diagonals be d1 and d2, with the first divided into lengths x,y, so x+y = d1.

2. Algebra: the two triangles with base d2 have heights x,y.

   Their total area = d2$\frac{\left(x+y\right)}{2}$ = d1d$\frac{2}{2}$.

3. Geometry: split along both diagonals into four right triangles.

   Duplicate the kite

   pair each matching triangle across its hypotenuse into a rectangle.

   The four rectangles have total area d1d2, so one kite has half that area.

4. This argument uses a convex kite and its perpendicular, internally intersecting diagonals.

Result

A = d1d$\frac{2}{2}$.

Given

Show the area scale and whether congruent copies fit: (i) rectangle sides a,b enlarged to 2a,2b; (ii) triangle sides a,b,c doubled; (iii) the triangle sides tripled (continued page 151).

To Find

Area ratios and actual dissections

Reasoning

Uniform scaling multiplies area by the square of the length factor; a tiling needs a construction as well.

Solution

1. (i) Area (2a)(2b) = 4ab.

   Bisect both sides to make a 2 × 2 grid of four original rectangles.

2. (ii) The doubled triangle has area four times the original by similarity.

   Join its three side midpoints: four triangles have side lengths a,b,c, so SSS makes all four congruent to the original.

3. (iii) The tripled triangle has nine times the area.

   Trisect each side and draw lines parallel to the other sides through the trisection points.

   The triangular grid contains nine small triangles (six one way, three inverted), each with sides a,b,c.

4. Thus the equal area totals are accompanied by explicit gap-free tilings, rather than used alone to infer a fit.

Result

Four rectangle copies; four triangle copies; nine triangle copies.

Given

Find the covered-area fraction for the rows of three and four congruent tangent circles fitted exactly in their rectangles (Figs. 6.45–6.46).

To Find

Circle-covered fraction in each rectangle

Reasoning

For n radius-r circles in one row, the tight rectangle is 2nr by 2r.

Solution

1. Three circles: area 3π$r^2$

   rectangle area (6r)(2r) = $12r^2$.

   Ratio = $\frac{\pi }{4}$.

2. Four circles: area 4π$r^2$

   rectangle area (8r)(2r) = $16r^2$.

   Ratio = $\frac{\pi }{4}$.

Result

Both fractions are $\frac{\pi }{4}$, approximated by $\frac{11}{14}$ with π≈$\frac{22}{7}$.

Given

Conjecture and prove the fraction covered by equal circles in a tight rectangular row. Test 10,20 and 50 circles.

To Prove

General fraction and three test cases

Reasoning

Area ratios cancel the common number of circles and squared radius.

Proof

1. For 10 circles: circle area 10π$r^2$, rectangle area $40r^2$

   ratio $\frac{\pi }{4}$.

2. For 20: 20π$r^2$/$80r^2$ = $\frac{\pi }{4}$.

   For 50: 50π$r^2$/$200r^2$ = $\frac{\pi }{4}$.

3. Generally, n tangent equal circles in a tight single row have area nπ$r^2$ and rectangle area (2nr)(2r) = $4{\mathrm{nr}}^2$.

   Their ratio is $\frac{\pi }{4}$ for n ≥ 1,r > 0.

4. The same calculation holds for a tight rectangular grid of equal circles.

   It is not a claim about arbitrary packings, margins or unequal radii.

Result

The covered fraction is always $\frac{\pi }{4}$ for the specified arrangement.