Measuring Space: Perimeter and Area

44 = 2(22/7)r.

Use C = 2(22/7)r and round only at the end to three significant figures.

Arc length = (θ/360)2πr.

(i) (60/360)×2×(22/7)×3.5 = 11/3 cm.

(ii) (120/360)×2×(22/7)×6.3 = 13.2 m.

The arc is (75/360)×2×(22/7)×14 = 55/3 cm.

Add both radii, totalling 28 cm.

(i) Two semicircles of diameter 60 plus two straight lengths 80: P = 60π + 160.

(ii) Outer and inner semicircles of diameters 12 and 8, plus two straight gaps totalling 4: P = 10π + 4.

(iii) Four semicircles of diameter 10: P = 20π.

(iv) Three outward semicircles on the 12 cm sides: P = 3×π×6 = 18π.

(v) The equal marked grid steps are 14 cm. Four quarter-circles of radius 14 and four semicircles of radius 7 give P = 56π.

(vi) One diameter-28 semicircle plus four diameter-7 semicircles gives P = 14π + 14π = 28π.

(vii) The right-triangle hypotenuse is √(8² + 6²) = 10; the three arcs have total length π(8 + 6 + 10)/2 = 12π.

(viii) Diameters are 12, 4, 4, 4: total 12π.

(ix) Diameters are 20, 10, 10: total 20π. Dotted lines are not boundary edges.

One turn covers πd = (22/7)56 = 176 cm.

10 km = 1,000,000 cm; divide by 176.

(i) Eight quarter-arcs of radius 7 form four petals: total 4π×7 = 88 cm.

(ii) Twelve 60° arcs of radius 42 form six petals: total 12×(60/360)×2π×42 = 168π.

C₁/C₂ = (2πr₁)/(2πr₂) = r₁/r₂.

Use AD = 8 cm as base. The perpendicular distance from E on BC to AD is 10 cm regardless of E’s location.

The horizontal offset of each slanted side is (40 − 20)/2 = 10.

Height = √(26² − 10²) = 24. Area = ½(40 + 20)24.

Third side = 32 − 8 − 11 = 13; semiperimeter s = 16.

Heron’s area = √(16×8×5×3).

Ratio sum 15 gives scale 300/15 = 20. Sides are 60, 100, 140 m; s = 150.

Area = √(150×90×50×10) = √6750000.

Let the shorter diagonal be d; the other is 2d.

Area = ½d(2d) = d² = 128.

Both triangles have base CD and their vertices P, Q lie on a line parallel to CD. Their heights are equal.

Write O on PR with PO/PR = t.

Triangles PSO and PSR share altitude from S to PR, so [PSO] = t[PSR]. Likewise [PQO] = t[PQR].

A parallelogram diagonal divides it into equal-area triangles: [PSR] = [PQR].

Draw diagonal AC. The midpoint theorem makes opposite midpoint edges parallel to AC or BD, so the inner figure is a parallelogram.

Each corner triangle has two adjacent sides halved, so its area is one quarter of its corresponding full triangle.

The two corners at B and D total ¼[ABCD]; those at A and C total another ¼[ABCD].

The median gives [ABD] = [ACD].

If P is on AD, triangles ABP and ACP are the same fraction AP/AD of ABD and ACD respectively, using AD as base line.

For square side s, distances from P to AB and CD sum to s. Thus [PAB] + [PCD] = ½s².

The distances to BC and DA also sum to s.

Since PD ∥ CQ, triangles BPD and BCQ are similar: BP/BC = BD/BQ.

[BPQ]/[ABC] = (BP/BC)(BQ/BA). Substitute the similarity ratio to obtain BD/BA = 1/2.

The described Q lies on AB only for positions of P permitting that intersection; at P = B the construction is undefined.

Sector area = (60/360)π×7² = (1/6)×154.

44 = 2πr gives r = 7.

Quadrant area = π×49/4.

10 minutes is 10/60 of a full turn, or 60°.

Area = (1/6)π×7².

Use the specified π ≈ 3.14, with r² = 100.

Quarter-circle area = 314/4; three-quarter-circle area = 3×314/4.

Minor sector = (1/6)×3.14×225 = 117.75 cm².

Its centre-endpoint triangle is equilateral with side 15; area = (1.73/4)×225 = 97.3125 cm².

Minor segment = 117.75 − 97.3125; major segment = 3.14×225 minus the minor segment.

Under the stated blade-radius model, each sector has area (120/360)π×28².

Multiply by 2 because the cleaned regions do not overlap.

Sector area is πr²/6. The triangle is equilateral with side r and area √3 r²/4.

Subtract the triangle from the sector.

Its side is √3r, so triangle area = (√3/4)(3r²) = 3√3r²/4.

Divide by circle area πr².

Square diagonal = 2r; side² = diagonal²/2 = 2r².

Divide square area by πr².

The stated ratio requires a regular hexagon, as in the diagram context. Its six centre-side triangles are equilateral with side r.

Area = 6(√3/4)r² = 3√3r²/2. Divide by πr².

For a > b > 0, remove a b-by-b corner from an a-by-a square. Cut and rearrange the remainder into a rectangle of sides a + b and a − b.

For the three-term square, partition each side into a, b, c. The nine cells have areas a², b², c² and two of each ab, bc, ca.

Base = 40 − 2×15 = 10. Height = √(225 − 25) = 10√2.

60 = ½×10×h gives h = 12.

Each equal side is √(12² + 5²).

54 = ½×12×b gives b = 9.

Hypotenuse = √(144 + 81) = 15.

Scale = 45/(2 + 3 + 4) = 5. Sides 10, 15, 20; s = 22.5.

Area = √(22.5×12.5×7.5×2.5).

7² + 24² = 25², so it is right-angled: area = ½×7×24 = 84.

Heron gives s = 28, area = √(28×21×4×3) = 84.

Circumference = (22/7)60 cm. Multiply by 100 and convert cm to metres.

Radius = 66/(2×22/7) = 10.5 cm.

Quadrant area = (22/7)(10.5)²/4.

One turn covers 2(22/7)28 = 176 cm = 1.76 m.

For 1 km, divide 1000 by 1.76.

Let sides be x,y and u,v. Equal perimeters give x + y = u + v = s; equal areas give xy = uv = k.

Both pairs are roots of t² − st + k = 0, so the unordered pairs of side lengths coincide.

Extend the shorter base to create a parallelogram of base b and height h. The added triangle has base b − a and height h.

Subtract its area: bh − ½(b − a)h.

The two triangles have bases a and b and the same perpendicular height h.

Add ½ah + ½bh.

Rotate one copy through 180° and place the slanted sides together.

The resulting parallelogram has base a + b and height h. Divide its area by 2.

Let the perpendicular diagonals have lengths d₁ and d₂; one bisects the other.

Algebra: adding the four right-triangle areas gives ½d₁d₂.

Geometry: translate the four triangles into two rectangles of combined area ½d₁d₂, or pair the triangles along one diagonal and add their common-base areas.

(i) Doubling both rectangle sides gives area 4ab; bisect each side to form a 2-by-2 grid of four original rectangles.

(ii) Doubling all triangle sides doubles its height and quadruples area. Join side midpoints to make four congruent small triangles.

(iii) Tripling the sides multiplies area by 9. Draw parallel lines through side-trisection points; the grid has nine congruent triangles.

Triangle: the triangle up to the second trisection point is 2/3 of the whole. The unshaded apex triangle has side ratios 1/2 and 1/3, hence area 1/6. Difference = 1/2.

Square: choose unit-square coordinates. The four bounding lines meet at (1/5, 2/5), (2/5, 4/5), (4/5, 3/5), (3/5, 1/5).

Adjacent vectors (1/5, 2/5) and (2/5, −1/5) give determinant magnitude 1/5.

For n identical circles of radius r in one touching row, total circle area is nπr².

The tight rectangle measures 2nr by 2r and has area 4nr².

The quotient nπr²/(4nr²) = π/4 does not depend on n or r.

For 10, 20 and 50 circles the same factors cancel.

Each small rectangle has area 72/9 = 8 cm².

Four long sides span the top row; five short sides span the bottom row. Thus 4l = 5w and l = 5w/4.

lw = 8 gives w = 4√10/5 and l = √10.

Both outer shaded triangles have base one third of the same side and the same height, so their areas agree.

Cut each triangle along the line halfway up its height. Rotate the small top triangle 180° around a midpoint of a slanted side; this makes a parallelogram with base b and height h/2.

Cut off and translate an end triangle of the parallelogram to obtain a b-by-h/2 rectangle. If the slant spans more than one base width, repeat cuts at successive base-width intervals.

Do this for both shaded triangles. Overlay the cuts in the common rectangle, then reverse the red triangle’s rearrangement to transfer the subdivided blue pieces.

For square side s, the quarter-circle area is πs²/4.

The two semicircles each have area πs²/8, so their areas sum to the quarter-circle area.

Their overlap A is counted twice in that sum; their union is counted once. The remainder B of the quarter-circle equals the overlap A.

Each petal is bounded by two quarter-circle arcs of radius 1. Eight such arcs give perimeter 4π.

A petal comprises two 90° sectors minus two right triangles: area π/2 − 1.

Four petals give total area 2π − 4.

The inner radius is the perpendicular distance to BC. Since the perpendicular bisects BC, Pythagoras gives R² − r² = (l/2)².

Annulus area = π(R² − r²).

Let the right triangle legs be a,b and hypotenuse c. Pythagoras gives a² + b² = c².

Semicircle area on diameter d is πd²/8, so the two leg semicircles together equal the hypotenuse semicircle.

The hypotenuse semicircle is the triangle C plus the two circular segments shared with the leg semicircles. Subtract the shared segments from both sides.

The two centres and either intersection form an equilateral triangle of side r. Each relevant sector has central angle 120°.

The lens consists of two 120° sectors minus two equilateral triangles.

Area = 2(πr²/3 − √3r²/4).

Let rectangle width W, height H. Let the top internal point be at x = u, and the right internal point at height v.

C = ½(W − u)(H − v); A + C = ½W(H − v); B + C = ½H(W − u).

Hence 2(A + C)(B + C)/C = WH, assuming C > 0.

Let OA = OB = r; then AB = r√2 and the semicircle on AB has area π(r√2)²/8 = πr²/4.

The quadrant AOB also has area πr²/4.

The two equal-area regions share the circular segment between arc AFB and chord AB. Removing that shared segment leaves the shaded lune and triangle AOB.

Outer lanes have longer curved sections; starting ahead compensates for this extra distance.

For a complete lap with lane-spacing w, the difference is 2πw, independent of track radius.

For the same 400 m race on a 200 m track, the runners negotiate two laps: the total extra curved distance is 4πw, assuming they remain in their lanes. Actual relay break-line rules can alter the required stagger.

Using the textbook’s equal running-line offset model, adjacent radii differ by 1.22 m.

A full lap’s curved-distance difference is 2π×1.22 ≈ 7.6655 m with π ≈ 3.1416.

Semicircle length is πd/2.

The smaller diameters sum to the large diameter PQ, so their arc lengths sum to the same total.

Cut a triangular end piece and translate it along the base, reducing the horizontal shear while retaining the height and area.

Repeat until the perpendicular foot is within the base, then use the usual rectangle rearrangement.

With adjacent sides a,b and included angle θ, area is ab sin θ.

Keeping a,b fixed while changing θ changes the height and area.

For an obtuse triangle, extend the base to the perpendicular foot. Subtract the two right-triangle areas, giving ½h times the original base.

Alternatively, rotate a congruent triangle 180° about the midpoint of a side to form a parallelogram; parallel opposite edges follow from equal alternate angles.

Convert a triangle to a parallelogram by a half-height cut and a 180° rotation of its top piece; convert the parallelogram to a rectangle by translating triangular end pieces.

Equal-area rectangles can be transformed into each other by further straight cuts; applying these conversions establishes a finite dissection for equal-area polygons.

The minimum number depends on the particular shapes; no universal minimum follows from area alone.

Let sides be x and 20 − x with 0 < x < 20. There are infinitely many choices.

Area = x(20 − x) = 100 − (x − 10)²; the maximum is 100 at x = 10.

Areas approach 0 as x approaches 0 or 20, but a nondegenerate rectangle never has area 0.

For base b and height h, first make a rectangle of sides b and h/2.

On a line mark adjacent segments of lengths b and h/2, draw a semicircle on their combined diameter and erect a perpendicular at their join.

The perpendicular length is √(bh/2), by the intersecting-chord/geometric-mean construction. Construct a square on this length.

Possible practical examples: wheels, pulleys, round pots, towers and wells. A circle has constant radius and rotational symmetry.

Aesthetic and symbolic preferences can also influence designs; this reflection invites examples rather than one numerical answer.