NCERT Solutions Class 9 Maths Ch 6: Perimeter and Area

The radius is 7 cm.

1. Use C = 2πr44 = 2 × 227 × r ⇒ r = 44 × 744 = 7 cm

(i) 44.0 cm (ii) 62.8 cm (iii) 75.4 cm.

1. Use C = 2πr(i) 2π × 7 = 43.98… ≈ 44.0 cm(ii) 2π × 10 = 62.83… ≈ 62.8 cm(iii) 2π × 12 = 75.39… ≈ 75.4 cm

(i) 113 ≈ 3.67 cm. (ii) 13.2 m.

1. Arc length = θ360 × 2πr
2. (i)60360 × 2 × 227 × 3.5 = 16 × 22 = 113 ≈ 3.67 cm
3. (ii)120360 × 2 × 227 × 6.3 = 13 × 44 × 0.9 = 13.2 m

The perimeter is about 46.33 cm.

1. Curved part (arc)l = 75360 × 2 × 227 × 14 = 524 × 88 = 553 ≈ 18.33 cm
2. Straight partsThe two straight sides are radii: 14 + 14 = 28 cm.
3. Perimeter18.33 + 28 ≈ 46.33 cm

The perimeter is 348.4 m.

1. Straight edgesTwo straight edges of 80 m each: 160 m.
2. Curved endsThe two semicircular ends make one circle of diameter 60 m.π × 60 = 3.14 × 60 = 188.4 m
3. Perimeter160 + 188.4 = 348.4 m

The perimeter is 35.4 cm.

1. Outer arc12 × π × 12 = 6π = 18.84 cm
2. Inner arc12 × π × 8 = 4π = 12.56 cm
3. Straight base piecesThe base is 12 cm and the gap is 8 cm, so the two straight pieces total 12 − 8 = 4 cm (2 cm each).
4. Perimeter18.84 + 12.56 + 4 = 35.4 cm

The perimeter is 62.8 cm.

1. Only the arcsThe dashed square is inside the shape, so only the four semicircles (diameter 10 cm) form the boundary.
2. Four semicircles = two circles2 × π × 10 = 2 × 3.14 × 10 = 62.8 cm

The perimeter is 56.52 cm.

1. One semicircleDiameter 12 cm: 12 × π × 12 = 6π = 18.84 cm
2. Three semicircles3 × 18.84 = 56.52 cm

The perimeter is 176 cm.

1. Read the figureThe dotted grid has 3 × 3 equal small squares, and the marked 14 cm is the side s of one small square. The boundary has no straight parts: it is made of 4 semicircles (diameter s, on the middle square of each side) and 4 quarter circles (radius s, one at each corner, centred at the inner grid corners).
2. 4 semicircles4 × 12 × π × 14 = 28π
3. 4 quarter circles4 × 14 × 2π × 14 = 28π
4. Perimeter28π + 28π = 56π = 56 × 227 = 176 cm

The perimeter is 88 cm.

1. PiecesThe 28 cm base is cut into four equal parts of 7 cm. The top is a large semicircle of diameter 28 cm. The bottom has four small semicircles of diameter 7 cm (two curving up, two curving down).
2. Top arc12 × π × 28 = 14π
3. Four small arcs4 × 12 × π × 7 = 14π
4. Perimeter14π + 14π = 28π = 28 × 227 = 88 cm

The perimeter is 12π = 37.68 cm.

1. Find the hypotenuse√(6² + 8²) = √100 = 10 cm
2. Three semicirclesThe boundary is made of semicircles on the sides 6, 8 and 10 cm.12π(6) + 12π(8) + 12π(10) = 3π + 4π + 5π = 12π
3. Value12 × 3.14 = 37.68 cm

The perimeter is 12π = 37.68 cm.

1. Large arcDiameter 4 + 4 + 4 = 12 cm: 12 × π × 12 = 6π
2. Three small arcs3 × 12 × π × 4 = 6π
3. Perimeter6π + 6π = 12π = 37.68 cm

The perimeter is 20π = 62.8 cm.

1. Large arcDiameter 10 + 10 = 20 cm: 12 × π × 20 = 10π
2. Two small arcsTwo semicircles of diameter 10 cm make one full circle: π × 10 = 10π
3. Perimeter10π + 10π = 20π = 62.8 cm

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

1. (i) One revolution = circumferenceπ × 56 = 227 × 56 = 176 cm (1.76 m)
2. (ii) 10 km in centimetres10 km = 10 × 1000 × 100 = 10,00,000 cm
3. Number of revolutions10,00,000176 ≈ 5681.8 ≈ 5682

(i) 88 cm. (ii) 528 cm.

1. (i) The square flowerThe arcs have their centres at the midpoints of the sides, so each has radius 7 cm. Each petal runs from a corner of the square to the centre, and is bounded by two quarter circles (each makes a 90° angle at its centre). There are 4 petals, so 8 quarter circles = 2 full circles.2 × 2π × 7 = 28π = 28 × 227 = 88 cm
2. (ii) The hexagon flowerThe arcs have their centres at the vertices of the hexagon and radius 42 cm. Each petal runs from a vertex to the centre, and its two arcs each subtend 60° at their centres (the triangle formed by a side and two radii is equilateral). There are 6 petals, so 12 arcs of 60°, which is 12 ÷ 6 = 2 full circles.2 × 2π × 42 = 168π = 168 × 227 = 528 cm

The ratio of the radii is also 5 : 4.

1. Perimeter is proportional to the radiusC₁C₂ = 2πr₁2πr₂ = r₁r₂
2. ConclusionThe common factor 2π cancels, so r₁ : r₂ = C₁ : C₂ = 5 : 4.

The area is 40 cm².

1. Base and heightTake AD (8 cm) as the base. E is on the opposite side BC, so its distance from AD is the length of the rectangle, 10 cm.
2. Area12 × 8 × 10 = 40 cm²

The area is 720 cm².

1. Find the heightDrop perpendiculars from the ends of the shorter side. The bottom side (40 cm) is split into 20 cm in the middle and two equal parts of (40 − 20) ÷ 2 = 10 cm.
2. Heighth² + 10² = 26² ⇒ h² = 676 − 100 = 576 ⇒ h = 24 cm
3. Area12 × (40 + 20) × 24 = 30 × 24 = 720 cm²

The area is √1920 = 8√30 ≈ 43.82 cm².

1. Third side8 + 11 + c = 32 ⇒ c = 13 cm
2. Semi-perimeters = 32 ÷ 2 = 16 cm
3. Heron’s formulaA = √[16(16 − 8)(16 − 11)(16 − 13)] = √(16 × 8 × 5 × 3) = √1920 ≈ 43.82 cm²

The area is 1500√3 ≈ 2598.08 m².

1. Sides3x + 5x + 7x = 300 ⇒ x = 20, so the sides are 60 m, 100 m, 140 m
2. Semi-perimeters = 150 m
3. Heron’s formulaA = √[150 × 90 × 50 × 10] = √6,750,000 = 1500√3 ≈ 2598.08 m²

The shorter diagonal is 8√2 ≈ 11.31 cm.

1. Area of a rhombusArea = ½ × d₁ × d₂. Let the diagonals be d and 2d.
2. Solve128 = 12 × d × 2d = d² ⇒ d = √128 = 8√2 ≈ 11.31 cm

The ratio is 1 : 1: the two areas are equal.

1. Same baseBoth triangles have the base CD.
2. Same heightP and Q lie on AB, and AB ∥ CD, so the distance of P (and of Q) from CD is the distance between the parallel lines, the height of the parallelogram.
3. Comparearea(△PCD) = 12 × CD × h = area(△QCD)

Area(△PSO) = area(△PQO).

1. Same baseBoth triangles have the base PO, which lies on the diagonal PR.
2. Equal heightsThe diagonal PR of a parallelogram makes triangles PQR and PSR that are congruent, so Q and S are at the same distance from the line PR. So the heights from Q and S to PO are equal.
3. ConclusionSame base and equal heights, so area(△PSO) = area(△PQO).

The figure formed by the midpoints is a parallelogram, and its area is half of the area of the 4-gon.

1. It is a parallelogramLet ABCD be the 4-gon with midpoints E, F, G, H of AB, BC, CD, DA. In △ABC, EF ∥ AC and EF = ½ AC (midpoint theorem). In △ADC, HG ∥ AC and HG = ½ AC. So EF ∥ HG and EF = HG: EFGH is a parallelogram.
2. The corner trianglesTriangle EBF is similar to △ABC with sides half as long, so its area is ¼ of area(△ABC). In the same way, △GDH is ¼ of area(△ADC). So △EBF + △GDH = ¼ of the 4-gon.
3. The other two cornersSimilarly, using the diagonal BD: △FCG + △HAE = ¼ of the 4-gon.
4. SubtractThe four corner triangles total ½ of the 4-gon, so the parallelogram EFGH has area 1 − ½ = ½ of the area of the 4-gon.

Area(△ABP) = area(△ACP).

1. Triangle ABCThe median AD divides △ABC into two triangles of equal area (equal bases BD = DC, same height from A): area(△ABD) = area(△ACD).
2. Triangle PBCIn the same way, PD is a median of △PBC, so area(△PBD) = area(△PCD).
3. Subtractarea(△ABD) − area(△PBD) = area(△ACD) − area(△PCD) ⇒ area(△ABP) = area(△ACP)

The ratio is 1 : 1.

1. Red regionLet the side be s. △PAB has base AB = s and height h₁ (the distance of P from AB). △PCD has base CD = s and height h₂ = s − h₁.Red = 12 s h₁ + 12 s (s − h₁) = s²2
2. Green regionIn the same way, with the sides BC and DA the heights add up to s:Green = s²2
3. RatioRed : green = 1 : 1. Each region is half of the square wherever P is.

Area(△BPQ) = ½ area(△ABC).

1. Triangles on the same base△PDQ and △PDC have the same base PD, and Q and C lie on the line CQ parallel to PD. So they are between the same parallels and area(△PDQ) = area(△PDC).
2. Add △BPDarea(△BPQ) = area(△BPD) + area(△PDQ) = area(△BPD) + area(△PDC) = area(△BDC)
3. MedianCD is a median of △ABC (D is the midpoint of AB), so area(△BDC) = ½ area(△ABC).
4. ConclusionArea(△BPQ) = ½ area(△ABC).

The area is 77/3 ≈ 25.67 cm².

1. Area of a sector = θ360 × πr²60360 × 227 × 7 × 7 = 16 × 154 = 773 ≈ 25.67 cm²

The area is 38.5 cm².

1. Radius2π r = 44 ⇒ r = 7 cm
2. Quadrant = quarter of the circle14 × 227 × 7 × 7 = 14 × 154 = 38.5 cm²

The area swept is 77/3 ≈ 25.67 cm².

1. Angle sweptThe hand turns 360° in 60 minutes, so in 10 minutes it turns 60°.
2. Sector of radius 7 cm and angle 60°60360 × 227 × 49 = 773 ≈ 25.67 cm²

(i) 78.5 cm². (ii) 235.5 cm².

1. (i) Minor sector90360 × 3.14 × 10 × 10 = 0.25 × 314 = 78.5 cm²
2. (ii) Major sector270360 × 3.14 × 100 = 0.75 × 314 = 235.5 cm²

Check: 78.5 + 235.5 = 314 = π × 10², the whole circle.

Minor segment ≈ 20.44 cm². Major segment ≈ 686.06 cm².

1. Sector60360 × 3.14 × 225 = 117.75 cm²
2. TriangleWith the two radii and the chord, the triangle has two sides 15 cm and an angle of 60°, so it is equilateral with side 15 cm.√34 × 15² = 0.433 × 225 ≈ 97.31 cm²
3. Minor segment = sector − triangle117.75 − 97.31 = 20.44 cm²
4. Major segment = circle − minor segment3.14 × 225 − 20.44 = 706.5 − 20.44 = 686.06 cm²

The total area is about 1642.67 cm².

1. One blade (a sector of radius 28 cm, angle 120°)120360 × 227 × 28 × 28 = 24643 ≈ 821.33 cm²
2. Two blades2 × 821.33 = 1642.67 cm²

Area of the minor segment = πr²(1/6 − √3/(4π)).

1. Sector60360 πr² = πr²6
2. TriangleThe triangle with two radii and the chord is equilateral with side r:area = √34 r²
3. Segment = sector − triangleπr²6 − √34 r² = πr²(16 − √34π)

The ratio is 3√3 ÷ 4π ≈ 0.413.

1. Split the triangleJoin the centre to the three vertices: this gives 3 triangles, each with two sides r and an angle of 120° between them. Each has area ½ r² sin 120° = √34 r².
2. Area of the triangle3 × √34 r² = 3√34 r²
3. Ratio3√3 r² ÷ 4πr² = 3√34π ≈ 0.413

The ratio is 2 ÷ π ≈ 0.637.

1. Side of the squareThe diagonal of the square is the diameter 2r, so the side s satisfies s√2 = 2r, which gives s = r√2.
2. Area of the squares² = (r√2)² = 2r²
3. Ratio2r²πr² = 2π ≈ 0.637

The ratio is 3√3 ÷ 2π ≈ 0.827, exactly twice the ratio for the triangle.

1. Split the hexagonA regular hexagon splits into 6 equilateral triangles of side r.
2. Area of the hexagon6 × √34 r² = 3√32 r²
3. Ratio3√3 r² ÷ 2πr² = 3√32π ≈ 0.827
4. Why twiceThe equilateral triangle inscribed in the same circle has area 3√34 r², which is exactly half of the hexagon’s area 3√32 r². So the hexagon-to-circle ratio is twice the triangle-to-circle ratio.

(a + b)² is a square of side a + b cut into a², b² and two rectangles ab. The other two identities are drawn below.

1. How the figure shows (a + b)²The big square has side a + b. It is cut into a square a², a square b² and two rectangles of area ab. Adding the four parts gives a² + 2ab + b².
2. (a + b)(a − b) = a² − b²Start with a square of side a (area a²) and remove a small square of side b from one corner. The L-shaped region that is left has area a² − b². Cut it into two rectangles, a × (a − b) and b × (a − b), and put them side by side: they form one rectangle of length a + b and width a − b, with area (a + b)(a − b).

3. (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2caDraw a square of side a + b + c and divide each side into parts a, b and c. This gives nine regions: three squares a², b², c² along the diagonal, and six rectangles: two of ab, two of bc and two of ca. Their total is the identity.

The area is 50√2 ≈ 70.71 cm².

1. Base40 − (15 + 15) = 10 cm
2. Heron’s formulas = 20 cm.A = √[20 × 5 × 5 × 10] = √5000 = 50√2 ≈ 70.71 cm²

The equal sides are 13 cm each.

1. Height60 = 12 × 10 × h ⇒ h = 12 cm
2. Equal sideThe height bisects the base, giving a right triangle with legs 5 cm and 12 cm.L = √(12² + 5²) = √169 = 13 cm

The perimeter is 36 cm.

1. Other leg54 = 12 × 12 × b ⇒ b = 9 cm
2. Hypotenusec = √(12² + 9²) = √225 = 15 cm
3. Perimeter12 + 9 + 15 = 36 cm

The area is 75√15 ÷ 4 ≈ 72.62 cm².

1. Sides2x + 3x + 4x = 45 ⇒ x = 5, so the sides are 10, 15, 20 cm
2. Heron’s formulas = 22.5 cm.A = √[22.5 × 12.5 × 7.5 × 2.5] = √5273.4375 ≈ 72.62 cm²

The area is 84 cm² by both methods.

1. Method 1: Heron’s formulas = (7 + 24 + 25) ÷ 2 = 28 cm.A = √[28 × 21 × 4 × 3] = √7056 = 84 cm²
2. Method 2: right triangle7² + 24² = 49 + 576 = 625 = 25², so the triangle is right-angled with legs 7 and 24.A = 12 × 7 × 24 = 84 cm²

The cyclist travels 60π m ≈ 188.5 m.

1. One rotationcircumference = π × 60 = 60π cm
2. 100 rotations100 × 60π = 6000π cm = 60π m ≈ 188.5 m

The area is 86.625 cm².

1. Radius2πr = 66 ⇒ r = 33π = 10.5 cm (π = 227)
2. Quadrant14 × 227 × 10.5² = 86.625 cm²

The car travels 176 cm per turn, and the wheel turns about 568 times in 1 km.

1. One turn2πr = 2 × 227 × 28 = 176 cm
2. 1 km = 1,00,000 cm1,00,000176 ≈ 568.18 turns

Yes, they are congruent.

1. The sum and the productIf the sides of a rectangle are l and w, the perimeter gives the sum l + w, and the area gives the product lw.
2. They fix l and wThe numbers l and w are the two roots of x² − (l + w)x + lw = 0. A quadratic equation has only two roots, so the pair {l, w} is the same for both rectangles.
3. ConclusionBoth rectangles have the same two side lengths, so they are congruent.

Area = ½(a + b)h.

1. Cut the trapeziumDraw a line from the end of the shorter side a, parallel to the slanted side. This splits the trapezium into a parallelogram (base a, height h) and a triangle (base b − a, height h).
2. Add the areasah + 12(b − a)h = h(a + b2 − a2) = h(a2 + b2) = 12(a + b)h

Area = ½(a + b)h.

1. Draw a diagonalA diagonal divides the trapezium (parallel sides a and b, height h) into two triangles.
2. First triangleIts base is a (the side parallel to b) and its height is h: area = ½ah.
3. Second triangleIts base is b and its height is also h (the parallel sides are always h apart): area = ½bh.
4. AddArea = 12ah + 12bh = 12(a + b)h

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

1. Make the parallelogramTake a trapezium with parallel sides a and b and height h. Turn an identical copy upside down (half a turn) and join it to the first along a slanted side. The new figure is a parallelogram.
2. Its base and heightThe base is a + b (the two parallel sides together) and the height is h.
3. Area of the parallelogram(a + b) × h
4. Area of one trapeziumThe parallelogram is made of two identical trapeziums, so one trapezium has area 12(a + b)h.

Area of a kite = ½ × d₁ × d₂.

1. (i) AlgebraThe diagonals AC = d₁ and BD = d₂ meet at O at a right angle. The four right triangles have areas ½·AO·BO, ½·OC·BO, ½·OC·DO and ½·AO·DO.Area = 12(AO·BO + OC·BO + OC·DO + AO·DO) = 12(AO + OC)(BO + DO) = 12 d₁ d₂
2. (ii) GeometryDraw a rectangle round the kite, with sides through its four vertices parallel to the diagonals. The rectangle is d₁ by d₂ and the kite covers exactly half of it (each of the four corner triangles is matched by a triangle of the kite). So the area of the kite is ½ d₁ d₂.

Doubling the sides gives 4 times the area, and tripling the sides gives 9 times the area. Yes, the copies fit in each case.

1. (i) RectanglesArea of ABCD = ab. Area of PQRS = (2a)(2b) = 4ab, which is 4 times. Yes, 4 copies fit: put two along the length and two along the width.
2. (ii) Triangles, sides doubledBy Heron’s formula, doubling every side multiplies s and each (s − a) by 2, so the area is multiplied by 2² = 4. Yes, 4 copies fit: join the midpoints of the sides of △PQR to cut it into 4 triangles, each with sides a, b, c.
3. (iii) Triangles, sides tripledTripling the sides multiplies the area by 3² = 9. Yes, 9 copies fit: divide each side into three equal parts and draw lines parallel to the sides through these points. This cuts △PQR into 9 triangles, each congruent to △ABC.

Exactly ½ of the triangle is shaded.

1. Name the pointsLet the triangle be PQR with P at the top and Q at the bottom left. The left side PQ is cut into two equal parts at M. The right side PR is cut into three equal parts at N₁ and N₂ (N₁ nearer to P). The shaded region is the quadrilateral Q M N₁ N₂.
2. Triangle QN₁N₂Its base N₁N₂ is ⅓ of PR and its height (from Q to PR) is the same as that of the whole triangle. So its area is ⅓ of the area of △PQR.
3. Triangle QMN₁M is the midpoint of PQ, so △QMN₁ and △PMN₁ have equal bases and the same height from N₁: each is half of △PQN₁. And △PQN₁ is ⅓ of △PQR (base PN₁ = ⅓ PR). So △QMN₁ = ½ × ⅓ = ⅙.
4. Total⅓ + ⅙ = 12

Exactly ⅕ of the square is shaded.

1. Set upTake a square of side 2 with vertices (0, 0), (2, 0), (2, 2), (0, 2). Join each vertex to the midpoint of a side, as in the figure: the four lines are x − 2y = 0, 2x + y = 4, x − 2y = −2 and 2x + y = 2.
2. Side of the inner squareTwo of the lines, x − 2y = 0 and x − 2y = −2, are parallel and the distance between them is 2√5. The other two lines are perpendicular to them, so the shaded figure is a square of side 2√5.
3. AreaInner area = 45,   whole square = 4 ⇒ fraction = 4/54 = 15

Another way to see it: the square can be cut and the four triangles moved so that the figure is made of 5 equal squares, one of which is the shaded one.

The circles cover π ÷ 4 ≈ 0.785 (78.5%) of the rectangle.

1. DimensionsLet each circle have radius r. The rectangle is as tall as one circle: 2r, and as long as three circles: 3 × 2r = 6r.
2. AreasCircles: 3πr²   and   rectangle: 6r × 2r = 12r²
3. Fraction3πr²12r² = π4 ≈ 0.785

The circles cover π ÷ 4 ≈ 0.785 (78.5%) of the rectangle.

1. DimensionsThe rectangle is 2r tall and 4 × 2r = 8r long.
2. AreasCircles: 4πr²   and   rectangle: 8r × 2r = 16r²
3. Fraction4πr²16r² = π4 ≈ 0.785

This is the same as for three circles, because adding a circle adds a square of side 2r to the rectangle and a circle of radius r to the circles.

Conjecture: the circles always cover π ÷ 4 of the rectangle, however many circles there are.

1. Test the cases

   Number of circles	Circle area	Rectangle area	Fraction
   10	10πr²	(10 × 2r)(2r) = 40r²	10π40 = π4
   20	20πr²	(20 × 2r)(2r) = 80r²	20π80 = π4
   50	50πr²	(50 × 2r)(2r) = 200r²	50π200 = π4

2. Proof for n circlesCircles: nπr²   and   rectangle: (n × 2r)(2r) = 4nr²Fraction = nπr²4nr² = π4. The n and the r² cancel, so the fraction does not depend on n.

The perimeter of each small rectangle is 18√10 ÷ 5 ≈ 11.38 cm.

1. Relation between the sidesLet each small rectangle be l by w. The top row has 4 rectangles lying flat (width l each) and the bottom row has 5 rectangles standing (width w each). Both rows fill the width of the big rectangle: 4l = 5w, so l = 1.25w.
2. Use the areaOne small rectangle has area 72 ÷ 9 = 8 cm².l × w = 1.25w² = 8 ⇒ w² = 6.4 ⇒ w = √6.4 ≈ 2.53 cml = 1.25w = √10 ≈ 3.16 cm
3. Perimeter2(l + w) = 2(3.16 + 2.53) ≈ 11.38 cm (exactly 18√105)

The two triangles have equal bases and the same height, so their areas are equal. Both can be cut into a rectangle of the same size.

1. Equal areasThe lines are drawn from one vertex to the trisection points of the opposite side. So the bases of the blue and red triangles are equal (each is ⅓ of the side) and they have the same height (the distance of the vertex from that side).Area = 12 × b × h for both
2. Cut and rearrangeAny triangle with base b and height h can be cut into a rectangle b × (h ÷ 2) as drawn below. Do this to the blue triangle to get the rectangle. Then do the cuts for the red triangle backwards, starting from the same rectangle. The blue triangle has been rearranged into the red one (using at most 3 + 3 = 6 pieces).

Area A = area B.

1. Set upLet the side of the square be s. The quarter circle has radius s, so its area is ¼πs². Each semicircle has diameter s, so each has area ½π(s ÷ 2)² = ⅛πs².
2. The regionsA is the overlap of the two semicircles. B is the part of the quarter circle that is outside both semicircles. Each semicircle lies inside the quarter circle.
3. Area of the union of the semicirclesTogether the semicircles have area ⅛πs² + ⅛πs² = ¼πs². This counts the overlap A twice, so the area covered by the two semicircles is ¼πs² − A.
4. Find BB = quarter circle − union of the semicircles = 14πs² − (14πs² − A) = A

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

1. PerimeterEach semicircle has radius 1. The 4 petals are bounded by 8 quarter circles of radius 1, which make 2 full circles.2 × 2π × 1 = 4π ≈ 12.57 units
2. AreaThe four semicircles have total area 4 × ½π(1)² = 2π. They cover the whole square (area 4), and the petals are the parts covered twice.Area of the flower = 2π − 4 ≈ 2.28 square units

The area of the ring is ¼πl².

1. RadiiLet the larger radius be R and the smaller one r. BC touches the smaller circle at A, so OA ⊥ BC, and OA = r. The perpendicular from the centre bisects the chord: AB = l ÷ 2.
2. Pythagoras in △OABR² = r² + (l2)² ⇒ R² − r² = l²4
3. Area of the ringπR² − πr² = π(R² − r²) = 14πl²

Area(A) + area(B) = area(C): the two crescents together are equal to the triangle.

1. Name the partsA and B are the two crescents (the parts of the semicircles on the legs that lie outside the semicircle on the hypotenuse). C is the triangle. Let the legs be a and b, and the hypotenuse c, so a² + b² = c². Let S₁ and S₂ be the two circular segments between the triangle and the hypotenuse semicircle, on the sides of the legs a and b.
2. Semicircle on the hypotenuseThe semicircle on the hypotenuse passes through the right-angle vertex, and is made of the triangle C and the two segments: ½π(c ÷ 2)² = C + S₁ + S₂.
3. Semicircles on the legsThe semicircle on a is A + S₁, and the semicircle on b is B + S₂.
4. Use a² + b² = c²π8a² + π8b² = π8c²So (A + S₁) + (B + S₂) = C + S₁ + S₂.
5. ConclusionCancel S₁ and S₂: A + B = C.

The common (red) region has area (2π ÷ 3 − √3 ÷ 2) r² ≈ 1.23r². The whole area covered by both circles is (4π ÷ 3 + √3 ÷ 2) r².

1. The trianglesThe centres A and B are r apart, and the circles meet at C and D. So AC = BC = AB = r and △ABC is equilateral; so is △ABD. Hence ∠CAD = 60° + 60° = 120°, and in the same way ∠CBD = 120°.
2. One circular segmentThe segment cut off by the chord CD in the circle with centre A is a sector of 120° minus △ACD:Sector = 120360πr² = πr²3△ACD = 12 r² sin 120° = √34 r²
3. The common region (red)It is made of two such segments (one from each circle):Area = 2(πr²3 − √34 r²) = (2π3 − √32) r² ≈ 1.23r²
4. The region covered by both circlesAdd the two circles and subtract the common region:2πr² − (2π3 − √32) r² = (4π3 + √32) r²

Area of the rectangle = 2(A + C)(B + C) ÷ C.

1. Name the lengthsLet the rectangle be w wide and h high. The top vertex of the triangles is at distance p from the left side. Let the vertical segment from it have length u, and the horizontal segment to the right side have length v. So w = p + v, and the lower triangle has height q = h − u.
2. The three areasA = 12 u p,   B = 12 v q,   C = 12 u v
3. Express w and hFrom A: p = 2A ÷ u. From B: q = 2B ÷ v.w = p + v = 2Au + v,   h = q + u = 2Bv + u
4. Multiplywh = (2Au + v)(2Bv + u) = 4ABuv + 2A + 2B + uvNow uv = 2C, sowh = 4AB2C + 2A + 2B + 2C = 2AB + 2AC + 2BC + 2C²C = 2(A + C)(B + C)C

The shaded crescent has the same area as the shaded triangle AOB.

1. Set upLet the radius of the circle with centre O be R, so OA = OB = R and ∠AOB = 90°. The shaded regions are the crescent (between the arc AFB and the semicircle on AB) and the triangle AOB.
2. Triangle AOBarea = 12 R²
3. Semicircle on ABAB² = R² + R² = 2R², so the radius of this semicircle is AB ÷ 2 and its area is12π(AB2)² = 12π × 2R²4 = πR²4
4. Segment between the chord AB and arc AFBQuarter circle − triangle = πR²4 − R²2
5. Crescent = semicircle on AB − segmentπR²4 − (πR²4 − R²2) = R²2
6. ConclusionThe crescent has area ½R², the same as triangle AOB.

The radius of the second lane is one lane width more than that of the first. For a lane width of 1.22 m the stagger is 2π × 1.22 ≈ 7.67 m. Yes, the same stagger is needed between the third and second lanes.

1. Difference in radiusEach lane is one lane width w wide (the standard width is 1.22 m), so r₂ − r₁ = w = 1.22 m.
2. StaggerIn one lap the runner goes round two semicircular bends, which together make one full circle. The extra distance for the outer lane is2πr₂ − 2πr₁ = 2π(r₂ − r₁) = 2π × 1.22 ≈ 7.67 mSo the runner in the second lane must start about 7.67 m ahead.
3. Third and second lanesYes. The extra distance is 2π times the difference of the radii, and that difference is the lane width w for any two neighbouring lanes. So the stagger is the same, 2πw, between every pair of adjacent lanes.

We can still show that the area is base × height, by adding and subtracting congruent triangles instead of cutting and moving a piece.

1. The problemWhen the parallelogram is very slanted, the perpendicular from C to the line AD meets the line outside the side AD (on its extension), so we cannot cut off a triangle and slide it across.
2. The fixExtend the line AD. Draw BF and CE perpendicular to it, with F and E on this line. Then FBCE is a rectangle with base BC and height h (the distance between the parallel lines).
3. Compare the areasLook at the trapezium FBCD. It is the parallelogram ABCD plus the triangle FBA. It is also the rectangle FBCE plus the triangle ECD. The triangles FBA and ECD are congruent (FB = EC = h, BA = CD, and each has a right angle), so they have the same area.
4. ConclusionTherefore area of parallelogram ABCD = area of rectangle FBCE = BC × h. The formula base × height holds however thin the parallelogram is.

No. The two side lengths do not fix the area, because the area also depends on the height (that is, on the angle between the sides).

1. RectangleThe angle between the sides is always 90°, so the area is length × breadth.
2. ParallelogramThe area is base × height, and the height depends on how much the figure is slanted. Take a parallelogram with sides 6 cm and 4 cm. If it is a rectangle, the height is 4 cm and the area is 24 cm². If it is pushed over so that the height is 2 cm, the sides are still 6 and 4 but the area is 12 cm². If it is flattened, the area gets closer and closer to 0.
3. What we needBesides the sides we also need the height, or an angle (or a diagonal). For example, a hinged frame with fixed sides can be pushed into parallelograms of many different areas.

Yes, it is possible.

1. The triangles△ABD and △ACD have equal areas. When they have equal bases on a line and the same height (as in Q59), we can cut each into a rectangle of the same size.
2. The planCut △ABD into a rectangle b × (h ÷ 2) (cut along the line joining the midpoints of the slanted sides, turn the small triangle to make a parallelogram, then move an end triangle to make a rectangle). Then cut △ACD in the same way and use those cuts in reverse: the rectangle turns into △ACD.
3. In generalThe same idea works for any two polygons of equal area: each can be cut into pieces that make the same rectangle (the Bolyai–Gerwien theorem). So equal area always means that one figure can be cut into pieces and rearranged to make the other.

There are both practical and non-practical reasons. This is an open question, and a few ideas are given below.

1. Practical reasons
   • The wheel: every point of a circle is the same distance from its centre, so a wheel rolls smoothly. Wheels are used in carts, pottery, clocks and machines.
   • Less material: a circle encloses the largest area for a given perimeter, so round pots, tanks, pipes and tents use little material.
   • Strength: round arches, domes and pipes spread forces evenly.
2. Other reasons
   • Nature: the sun, the full moon, the pupil of an eye and ripples in water are round, so the circle became a symbol of the sky and of life.
   • Symbols and art: a circle has no beginning and no end, so it stands for unity and the cycle of time. Mandalas, rangoli, rose windows and chakras use it.
   • Fairness: in a circle everyone sits at the same distance from the centre, as at a round table or in a circle of dancers.
3. Other usesThe circle is used to measure angles (360°), time (clock faces), directions (compass), and in gears, pulleys and turbines.