Chapter 6 exercise answers: Measuring Space: Perimeter and Area

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1. Perimeter of a circle is its circumference: C = 2πr.
2. Put C = 44 and π = 22/7: 44 = 2 × 22/7 × r.
3. So 44 = 44/7 × r, which gives r = 44 × 7/44 = 7.

AnswerThe radius is 7 cm.

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(i) radius 7 cm

1. C = 2πr = 2 × 22/7 × 7 = 44 cm exactly.
2. To 3 significant figures this is written 44.0 cm.

Answer44.0 cm

(ii) radius 10 cm

1. C = 2πr = 2 × 22/7 × 10 = 440/7 = 62.857… cm.
2. The fourth figure is 5, so the third rounds up: 62.9 cm.
3. This uses π = 22/7, as the exercise set says. With π itself the circumference is 62.83… cm, which would round to 62.8 cm, so say which value of π you used.

Answer62.9 cm

(iii) radius 12 cm

1. C = 2πr = 2 × 22/7 × 12 = 528/7 = 75.428… cm.
2. The fourth figure is 2, so the third stays: 75.4 cm.

Answer75.4 cm

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(i) the radius is 3.5 cm … angle at the centre is 60°

1. Arc length = (θ/360) × 2πr.
2. = (60/360) × 2 × 22/7 × 3.5 = (1/6) × 22 = 11/3 cm.

Answer11/3 cm, about 3.67 cm

(ii) the radius is 6.3 m … angle at the centre is 120°

1. Arc length = (θ/360) × 2πr.
2. = (120/360) × 2 × 22/7 × 6.3 = (1/3) × 39.6 = 13.2 m.

Answer13.2 m

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1. Arc length = (75/360) × 2π × 14 = (75/360) × 88 = 55/3 cm.
2. Perimeter of the sector = arc + 2 radii = 55/3 + 2 × 14 = 55/3 + 28 = 139/3 cm.

Answer139/3 cm, about 46.33 cm.

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(i) 60 m, 80 m

1. The boundary is two straight sides of 80 m and two semicircular ends. Each end stands on the 60 m width, so its radius is 30 m.
2. The two semicircles together make one full circle: 2π × 30 = 60π.
3. Perimeter = 2 × 80 + 60π = 160 + 60 × 22/7 = 160 + 1320/7 = 2440/7 m.

Answer2440/7 m ≈ 348.57 m

(ii) 8 cm, 12 cm

1. The outer curve is a semicircle on 12 cm (radius 6 cm); the inner curve is a semicircle on 8 cm (radius 4 cm).
2. The two flat pieces at the bottom are each (12 − 8)/2 = 2 cm.
3. Perimeter = π × 6 + π × 4 + 2 + 2 = 10π + 4 = 220/7 + 28/7 = 248/7 cm.

Answer248/7 cm ≈ 35.43 cm

(iii) 10 cm

1. The dashed square is not part of the boundary. The boundary is 4 semicircles, one on each 10 cm side (radius 5 cm).
2. Each semicircle is π × 5 = 5π, so perimeter = 4 × 5π = 20π = 20 × 22/7 = 440/7 cm.

Answer440/7 cm ≈ 62.86 cm

(iv) 12 cm

1. The dashed triangle is equilateral with side 12 cm. The boundary is 3 semicircles, one on each side (radius 6 cm).
2. Perimeter = 3 × π × 6 = 18π = 18 × 22/7 = 396/7 cm.

Answer396/7 cm ≈ 56.57 cm

(v) 14 cm

1. The dashed lines make a plus shape of five squares, each of side 14 cm (all their sides are marked equal).
2. At the end of each of the 4 arms is a semicircle on a 14 cm side (radius 7 cm): each is π × 7 = 7π.
3. Between neighbouring arms is a quarter circle of radius 14 cm, centred at a corner of the middle square: each is (1/4) × 2π × 14 = 7π.
4. Perimeter = 4 × 7π + 4 × 7π = 56π = 56 × 22/7 = 176 cm.

Answer176 cm

(vi) 28 cm

1. The 28 cm base is marked into 4 equal parts of 7 cm. Over the whole base is one big semicircle of radius 14 cm: π × 14 = 14π.
2. On each 7 cm part is a small semicircle of radius 7/2 cm (two dip below the base, two rise above it); each is π × 7/2, so the four give 14π.
3. Perimeter = 14π + 14π = 28π = 28 × 22/7 = 88 cm.

Answer88 cm

(vii) 8 cm, 6 cm

1. The dashed triangle has a right angle between the 8 cm and 6 cm sides, so its third side is √(8² + 6²) = √100 = 10 cm.
2. The boundary is 3 semicircles, one on each side, with radii 4 cm, 3 cm and 5 cm.
3. Perimeter = π × 4 + π × 3 + π × 5 = 12π = 12 × 22/7 = 264/7 cm.

Answer264/7 cm ≈ 37.71 cm

(viii) 4 cm, 4 cm, 4 cm

1. The big semicircle stands on 4 + 4 + 4 = 12 cm, so its radius is 6 cm: length 6π.
2. The three small semicircles stand on 4 cm each (radius 2 cm): 3 × 2π = 6π.
3. Perimeter = 6π + 6π = 12π = 12 × 22/7 = 264/7 cm.

Answer264/7 cm ≈ 37.71 cm

(ix) 10 cm, 10 cm

1. The big curve is a semicircle on the whole base of 10 + 10 = 20 cm (radius 10 cm): length 10π.
2. The two small curves are semicircles on 10 cm each (radius 5 cm), one above the left half and one below the right half: 5π + 5π = 10π.
3. Perimeter = 10π + 10π = 20π = 20 × 22/7 = 440/7 cm.

Answer440/7 cm ≈ 62.86 cm

Watch this explained “Reading the figures”, 5:07 into Perimeter puzzles: composite curved boundaries reduce to arcs you already know · हिंदी में देखें

Open NCERT p. 130Checked by computer

(i) How far does the car … to complete one revolution?

1. One revolution of the tyre carries the car forward by a distance equal to the tyre's circumference.
2. Circumference = πd = 22/7 × 56 = 176 cm.

Answer176 cm

(ii) How many revolutions does the tyre make … travels 10 km?

1. 10 km = 10×1000×100 cm = 1,000,000 cm.
2. Number of revolutions = total distance ÷ circumference = 1,000,000 ÷ 176 = 62500/11.

Answer62500/11 revolutions, i.e. 5681 and 9/11 revolutions (about 5681.8).

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(i) The centres of the arcs are the midpoints of the sides …

1. Let M be the midpoint of the bottom side and O the centre of the square. M is 7 cm from both bottom corners and 7 cm from O, so the circle of radius 7 cm about M passes through all three.
2. Each petal edge runs from a corner to O. From M, the corner lies along the side and O lies straight up, so the angle at M is 90°: each edge is a quarter circle of radius 7 cm.
3. There are 4 petals with 2 edges each, so 8 quarter circles, each (1/4) × 2π × 7 = 7π/2.
4. Total = 8 × 7π/2 = 28π = 28 × 22/7 = 88 cm.

Answer88 cm

(ii) The centres of the arcs are the vertices of the hexagon

1. In a regular hexagon the centre O is as far from each vertex as the side length, 42 cm. So the circle of radius 42 cm about a vertex V passes through both neighbouring vertices and through O.
2. Each petal edge runs from a neighbouring vertex W to O. Triangle VWO has all three sides 42 cm, so it is equilateral and the angle at V is 60°: each edge is 60/360 = 1/6 of a circle of radius 42 cm.
3. There are 6 petals with 2 edges each, so 12 arcs, each (1/6) × 2π × 42 = 14π.
4. Total = 12 × 14π = 168π = 168 × 22/7 = 528 cm.

Answer528 cm

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1. Perimeter (circumference) of a circle is C = 2πr, which is directly proportional to r.
2. So the ratio of two circles' circumferences equals the ratio of their radii: C1:C2 = r1:r2.
3. Given C1:C2 = 5:4, the ratio of the radii is also 5:4.

Answer5:4

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1. ABCD is a rectangle with BC = 8 cm and DC = 10 cm, so AD = BC = 8 cm, and AD is parallel to BC.
2. E lies on BC, so triangle ADE has base AD = 8 cm, and its apex E lies on line BC, which is parallel to AD at a distance equal to the rectangle's width, 10 cm.
3. Area = (1/2) × base × height = (1/2) × 8 × 10 = 40 cm², whatever point of BC is chosen as E.

Answer40 cm²

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1. Drop perpendiculars from the ends of the shorter parallel side to the longer one; by symmetry each foot cuts off (40−20)/2 = 10 cm on either side.
2. Each slanted side is the hypotenuse of a right triangle with base 10 cm and the trapezium's height as the other leg: height = √(26²−10²) = √(676−100) = √576 = 24 cm.
3. Area = (1/2)×(sum of parallel sides)×height = (1/2)×(40+20)×24 = 720 cm².

Answer720 cm²

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1. The third side = perimeter − (8+11) = 32−19 = 13 cm.
2. Semi-perimeter s = 32/2 = 16 cm.
3. By Heron's formula, Area = √[s(s−a)(s−b)(s−c)] = √[16×8×5×3] = √1920 = 8√30 cm².

Answer8√30 cm², about 43.82 cm²

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1. Let the sides be 3x, 5x, 7x. Then 3x+5x+7x = 300, so 15x = 300 and x = 20.
2. The sides are 60 m, 100 m and 140 m; semi-perimeter s = 150 m.
3. By Heron's formula, Area = √[150×90×50×10] = √6,750,000 = 1500√3 m².

Answer1500√3 m², about 2598.08 m²

Watch this explained “Half the perimeter”, 2:18 into Heron's formula: area from the three sides alone · हिंदी में देखें

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1. Let the shorter diagonal be d, so the longer one is 2d.
2. Area of a rhombus = (1/2) × product of diagonals = (1/2)×d×2d = d².
3. d² = 128, so d = √128 = 8√2 cm.

Answer8√2 cm, about 11.31 cm

Open NCERT p. 142One way to think about it

1. Both triangles PCD and QCD have the same base CD.
2. P and Q lie on AB, which is parallel to CD (opposite sides of a parallelogram), so every point of AB is the same perpendicular distance from CD.
3. Equal base and equal height give equal area for the two triangles, whatever P and Q are on AB.

In shortarea(∆PCD) : area(∆QCD) = 1 : 1 — the two areas are always equal.

Watch this explained “Same base, same parallels”, 6:26 into From rectangle to parallelogram: area survives rearrangement · हिंदी में देखें

Open NCERT p. 142One way to think about it

1. Diagonal PR divides parallelogram PQRS into triangles PQR and PSR, which are congruent, so ar(∆PQR) = ar(∆PSR).
2. These two triangles share the base PR, so equal areas mean equal heights: the perpendicular distance from Q to line PR equals that from S to line PR.
3. For any point O on PR, ar(∆PQO) = (1/2)×PO×(height of Q from PR) and ar(∆PSO) = (1/2)×PO×(height of S from PR).
4. Since the two heights are equal, ar(∆PQO) = ar(∆PSO).

In shortar(∆PSO) = ar(∆PQO), proved.

Watch this explained “Same base, same parallels”, 6:26 into From rectangle to parallelogram: area survives rearrangement · हिंदी में देखें

Open NCERT p. 142One way to think about it

1. Let ABCD be the 4-gon (convex, as usually drawn) and P, Q, R, S the midpoints of AB, BC, CD, DA. The figure PQRS is what is left of ABCD after cutting off the four corner triangles APS, BPQ, CQR and DRS.
2. Corner B: in ∆ABC, CP is a median (P is the midpoint of AB), so ar(∆BPC) = (1/2) ar(∆ABC). In ∆BPC, PQ is a median (Q is the midpoint of BC), so ar(∆BPQ) = (1/2) ar(∆BPC) = (1/4) ar(∆ABC).
3. Corner D, in the same way: in ∆ACD, AR is a median, so ar(∆ADR) = (1/2) ar(∆ACD); in ∆ADR, RS is a median, so ar(∆DRS) = (1/4) ar(∆ACD).
4. Corners A and C, using the other diagonal BD: ar(∆APS) = (1/4) ar(∆ABD) and ar(∆CQR) = (1/4) ar(∆CBD).
5. Diagonal AC splits ABCD into ∆ABC and ∆ACD, and diagonal BD splits it into ∆ABD and ∆CBD. So the four corners add up to (1/4) ar(ABCD) + (1/4) ar(ABCD) = (1/2) ar(ABCD).
6. Therefore ar(PQRS) = ar(ABCD) − (1/2) ar(ABCD) = (1/2) ar(ABCD).

In shortar(PQRS) = (1/2) ar(ABCD): joining the midpoints of the sides in order gives a figure with half the area of the 4-gon. The proof needs only the fact that a median halves a triangle's area; why PQRS is a parallelogram is left, as the book says, to the chapter on quadrilaterals.

Open NCERT p. 143One way to think about it

1. Since D is the midpoint of BC, AD is a median, and a median splits a triangle into two equal areas: ar(∆ABD) = ar(∆ACD).
2. BD = DC, so ar(∆PBD) = ar(∆PCD) too, since these two triangles have equal bases BD, DC and the same height from P to line BC.
3. Subtracting, ar(∆ABD) − ar(∆PBD) = ar(∆ACD) − ar(∆PCD), i.e. ar(∆ABP) = ar(∆ACP), because P lies on AD between A and D.

In shortar(∆ABP) = ar(∆ACP), proved.

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1. Let the square have side a. Drop a perpendicular from P to side AB and another to side CD; these two distances add up to a, since AB and CD are opposite sides a apart.
2. ar(∆PAB) + ar(∆PCD) = (1/2)×a×(distance to AB) + (1/2)×a×(distance to CD) = (1/2)×a×a = a²/2, whatever point P is.
3. By the same argument using the other pair of opposite sides, ar(∆PBC) + ar(∆PDA) = a²/2 too.
4. So the red total and the green total are always both half the square's area, and their ratio is 1:1.

Answer1 : 1 — the red and green regions always have equal total area, whatever point P is.

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Open NCERT p. 143One way to think about it

1. Since D is the midpoint of AB, CD is a median of ∆ABC, so ar(∆BDC) = (1/2) ar(∆ABC).
2. P lies on BC, so DP splits ∆BDC into ∆BDP and ∆DPC: ar(∆BDC) = ar(∆BDP) + ar(∆DPC).
3. Since CQ || PD, the points C and Q lie on one line parallel to DP. So ∆DPC and ∆DPQ have the same base DP and the same height, and ar(∆DPC) = ar(∆DPQ).
4. The line CQ is parallel to PD and passes through C, which is farther from B than P is, so it meets BA farther from B than D (as Fig. 6.34 shows). So D lies between B and Q, and ∆BDP and ∆DPQ together make up ∆BPQ: ar(∆BDP) + ar(∆DPQ) = ar(∆BPQ).
5. Therefore ar(∆BPQ) = ar(∆BDP) + ar(∆DPC) = ar(∆BDC) = (1/2) ar(∆ABC).

In shortar(∆BPQ) = (1/2) ar(∆ABC), proved.

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1. A sector's area is its angle's share of the whole circle: (θ/360) × π × r².
2. Here θ = 60°, r = 7 cm, and π = 22/7 (the value this chapter uses).
3. Area = (60/360) × 22/7 × 7² = (1/6) × 22/7 × 49.
4. = (1/6) × 154 = 77/3 cm² (≈ 25.67 cm²).

Answer77/3 cm² (≈ 25.67 cm²)

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1. First recover the radius from the circumference: C = 2πr.
2. 44 = 2 × 22/7 × r, so r = 44 × 7 ÷ 44 = 7 cm.
3. A quadrant is a quarter of the circle, so its area is (1/4) × π × r².
4. Area = (1/4) × 22/7 × 49 = 77/2 cm² (= 38.5 cm²).

Answer77/2 cm² (= 38.5 cm²)

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1. In 60 minutes the minute hand turns a full 360°, so in 10 minutes it turns (10/60) × 360° = 60°.
2. The tip of the hand sweeps out a sector of radius 7 cm and angle 60°.
3. Area = (60/360) × 22/7 × 7² = 77/3 cm² (≈ 25.67 cm²).

Answer77/3 cm² (≈ 25.67 cm²)

Watch this explained “Sectors in use”, 7:52 into A sector's area is its angle's share of the whole · हिंदी में देखें

Open NCERT p. 148Checked by computer

(i) minor sector (that subtends 90° at the centre)

1. The minor sector's angle is 90°, given directly by the chord.
2. Area = (90/360) × π × r² with r = 10 cm and π ≈ 3.14 (as this question asks).
3. = (1/4) × 3.14 × 100 = 78.5 cm² = 157/2 cm².

Answer157/2 cm² (= 78.5 cm²)

(ii) major sector (that subtends 270° at the centre)

1. The major sector takes the rest of the circle, so its angle is 360° − 90° = 270°.
2. Area = (270/360) × 3.14 × 100 = 235.5 cm² = 471/2 cm².
3. Check: 78.5 + 235.5 = 314 cm², exactly π × 10² — the whole circle.

Answer471/2 cm² (= 235.5 cm²)

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1. Join the ends A and B of the chord to the centre O. OA = OB = 15 cm, so triangle OAB is isosceles and its two base angles are equal. They add up to 180° − 60° = 120°, so each is 60°: triangle OAB is equilateral, with every side 15 cm.
2. Sector OAB (60°) = (60/360) × π × r² = (1/6) × 3.14 × 225 = 117.75 cm².
3. Triangle OAB (equilateral, side 15 cm) = (√3/4) × 15² = (1.73/4) × 225 = 97.3125 cm².
4. Minor segment = sector − triangle = 117.75 − 97.3125 = 20.4375 cm².
5. Whole circle = π × r² = 3.14 × 225 = 706.5 cm².
6. Major segment = whole circle − minor segment = 706.5 − 20.4375 = 686.0625 cm².

AnswerMinor segment = 20.4375 cm² (= 327/16 cm²); major segment = 686.0625 cm² (= 10977/16 cm²).

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1. Each wiper blade sweeps a sector of radius 28 cm and angle 120°.
2. Area of one sector = (120/360) × 22/7 × 28² = (1/3) × 22/7 × 784 = 2464/3 cm² (≈ 821.33 cm²).
3. The two wipers' sweeps don't overlap, so the total cleaned area is just twice one blade's area.
4. Total = 2 × 2464/3 = 4928/3 cm² (≈ 1642.67 cm²).

Answer4928/3 cm² (≈ 1642.67 cm²)

Watch this explained “Sectors in use”, 7:52 into A sector's area is its angle's share of the whole · हिंदी में देखें

Open NCERT p. 148One way to think about it

1. Join the ends A and B of the chord to the centre O. OA = OB = r, so triangle OAB is isosceles and its base angles are equal. They add up to 180° − 60° = 120°, so each is 60°: triangle OAB is equilateral, with side r.
2. Sector OAB (60° out of 360°) = (60/360) × πr² = πr²/6.
3. The height of triangle OAB splits its base into r/2 and r/2, so by Pythagoras the height is √(r² − r²/4) = (√3/2)r. Triangle area = (1/2) × r × (√3/2)r = (√3/4)r².
4. Minor segment = sector − triangle = πr²/6 − (√3/4)r² = r²(π/6 − √3/4).
5. Notice that π multiplies only the first term. The expression printed in the question, πr²(1/6 − √3/4), multiplies the √3/4 by π as well — and it cannot be an area: 1/6 ≈ 0.167 is smaller than √3/4 ≈ 0.433, so it is negative (about −0.84r²). It is a misprint for r²(π/6 − √3/4).
6. Check with Question 5 (r = 15, π ≈ 3.14, √3 ≈ 1.73): 225 × (3.14/6 − 1.73/4) = 20.4375 cm², the same minor segment found there.

In shortMinor segment = r²(π/6 − √3/4) ≈ 0.0906r². The form printed in the book, πr²(1/6 − √3/4), is a misprint: it is negative (≈ −0.84r²), and an area cannot be negative.

Watch this explained “The answer that came out negative”, 6:58 into A sector's area is its angle's share of the whole · हिंदी में देखें

Open NCERT p. 148One way to think about it

1. Call the triangle ABC, its side a, and the centre of the circle O. O is the same distance r from B and from C, so it lies on the perpendicular bisector of BC. In an equilateral triangle that line is the height from A; let it meet BC at its midpoint D.
2. By Pythagoras in triangle ABD, the height AD = √(a² − (a/2)²) = (√3/2)a.
3. O lies on AD with OA = r, so OD = AD − r. In the right triangle OBD: r² = OD² + BD² = (AD − r)² + (a/2)².
4. Expanding: r² = AD² − 2·AD·r + r² + a²/4, so 2·AD·r = AD² + a²/4 = 3a²/4 + a²/4 = a². Hence r = a²/(2·AD) = a²/(√3a) = a/√3, that is, a = √3r.
5. Triangle area = (√3/4)a² = (√3/4) × 3r² = (3√3/4)r².
6. Circle area = πr², so the ratio = (3√3/4)r² ÷ πr² = 3√3/(4π) ≈ 5.196/12.566 ≈ 0.413. The r² cancels, so the ratio is the same for every circle.

In shortRatio = 3√3/(4π) ≈ 0.413, for a circle of any radius.

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Open NCERT p. 148One way to think about it

1. Call the square ABCD, with side a. Its diagonals AC and BD are equal and cut each other in half, so the point where they cross is the same distance from all four corners. That point is the centre of the circle, and each half-diagonal is a radius — so each diagonal is a diameter, 2r.
2. By Pythagoras in triangle ABC (right angle at B): AC² = a² + a² = 2a². So (2r)² = 2a², which gives a² = 2r².
3. Square area = a² = 2r²; circle area = πr².
4. Ratio = 2r² ÷ πr² = 2/π ≈ 2/3.1416 ≈ 0.637. The r² cancels, so the ratio is the same for every circle.

In shortRatio = 2/π ≈ 0.637, for a circle of any radius.

Watch this explained “A picture that looks like an answer”, 8:49 into A sector's area is its angle's share of the whole · हिंदी में देखें

Open NCERT p. 148One way to think about it

1. The hexagon here must be a regular one (all sides and angles equal); for any other hexagon the ratio is not fixed.
2. Call the corners A, B, C, D, E, F and join each to the centre O. The six triangles OAB, OBC, …, OFA have sides r, r and one equal hexagon side, so they are congruent (SSS), and their six angles at O are equal: each is 360° ÷ 6 = 60°.
3. In triangle OAB, OA = OB = r and the angle at O is 60°, so (as in Question 7) it is equilateral with side r, and its area is (√3/4)r².
4. Hexagon area = 6 × (√3/4)r² = (3√3/2)r². Ratio = (3√3/2)r² ÷ πr² = 3√3/(2π) ≈ 5.196/6.283 ≈ 0.827.
5. Why exactly twice Question 8: joining alternate corners A, C, E gives the equilateral triangle of Question 8 inside the same circle. It cuts three triangles ABC, CDE and EFA off the hexagon.
6. OABC has all four sides equal to r, so it is a rhombus, and its diagonal AC cuts it into two congruent triangles OAC and BAC. So triangle ABC has the same area as triangle OAC. In the same way CDE matches OCE, and EFA matches OEA.
7. Triangle ACE is made of OAC, OCE and OEA; the three cut-off triangles have exactly the same total area. So triangle ACE is exactly half the hexagon, and its ratio 3√3/(4π) is exactly half of 3√3/(2π).
8. Be careful: triangle ACE is not three alternate small triangles such as OAB, OCD, OEF. Those have the right total area but form a pinwheel meeting only at O — a different region.

In shortRatio = 3√3/(2π) ≈ 0.827 for a regular hexagon. It is exactly twice Question 8's 3√3/(4π) ≈ 0.413 because the inscribed equilateral triangle ACE is exactly half of the hexagon.

Watch this explained “A picture that looks like an answer”, 8:49 into A sector's area is its angle's share of the whole · हिंदी में देखें

Open NCERT p. 149One way to think about it

1. For (a + b)(a − b) = a² − b² (take b smaller than a): draw a square of side a, and mark a smaller square of side b in one corner.
2. Removing that small b × b square leaves an L-shaped region of area a² − b².
3. Cut the L-shape along the line that continues one edge of the small square. This gives two rectangles: one a by (a − b), and one b by (a − b).
4. Turn the b by (a − b) rectangle through a right angle and place it against the end of the other, matching the (a − b) sides. Together they form one rectangle, (a + b) long and (a − b) wide.
5. The pieces did not overlap and nothing was stretched, so a² − b² = (a + b)(a − b): the same region, with its area added up two ways.
6. For (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca: draw a square of side a + b + c, and split each side into lengths a, b, c in the same order.
7. Rule lines across at those marks: the square splits into a 3 × 3 grid of 9 pieces — three squares of areas a², b², c², and six rectangles, two each of area ab, bc and ca.
8. Adding the 9 pieces gives a² + b² + c² + 2ab + 2bc + 2ca, which is also the area of the whole square, (a + b + c)².

In shortBoth identities come from cutting a square into non-overlapping pieces and adding their areas two ways: once as the whole figure, once as its parts. One figure is shown for each; other cuttings work too.

Watch this explained “The pictures prove the algebra”, 9:10 into From rectangle to parallelogram: area survives rearrangement · हिंदी में देखें

Open NCERT p. 149Checked by computer

1. Base = perimeter − two equal sides = 40 − 15 − 15 = 10 cm.
2. All three sides are now known (15, 15, 10 cm), so use Heron's formula: s = perimeter/2 = 20 cm.
3. Area = √[s(s−a)(s−b)(s−c)] = √(20 × 5 × 5 × 10) = √5000.
4. √5000 = √(2500 × 2) = 50√2 cm² (≈ 70.71 cm²).

Answer50√2 cm² (≈ 70.71 cm²)

Watch this explained “Second check: two equal sides”, 5:48 into Heron's formula: area from the three sides alone · हिंदी में देखें

Open NCERT p. 149Checked by computer

1. Area = (1/2) × base × height, so 60 = (1/2) × 10 × height, giving height = 12 cm.
2. In an isosceles triangle, the height from the apex meets the base at its midpoint, splitting it into two halves of 5 cm each.
3. Each equal side is the hypotenuse of a right triangle with legs 12 cm (height) and 5 cm (half the base).
4. Equal side = √(12² + 5²) = √(144 + 25) = √169 = 13 cm.

Answer13 cm

Watch this explained “Which route is shorter?”, 9:28 into Heron's formula: area from the three sides alone · हिंदी में देखें

Open NCERT p. 149Checked by computer

1. For a right triangle, area = (1/2) × leg1 × leg2, so 54 = (1/2) × 12 × leg2, giving leg2 = 9 cm.
2. The hypotenuse is √(12² + 9²) = √(144 + 81) = √225 = 15 cm.
3. Perimeter = 12 + 9 + 15 = 36 cm.

Answer36 cm

Watch this explained “Which route is shorter?”, 9:28 into Heron's formula: area from the three sides alone · हिंदी में देखें

Open NCERT p. 149Checked by computer

1. Let the sides be 2x, 3x and 4x cm. Then 2x + 3x + 4x = 45, so 9x = 45 and x = 5.
2. The sides are 10 cm, 15 cm and 20 cm.
3. Semi-perimeter s = 45/2 cm. Then s − 10 = 25/2, s − 15 = 15/2 and s − 20 = 5/2.
4. Heron's formula: Area = √[(45/2) × (25/2) × (15/2) × (5/2)] = √(45 × 25 × 15 × 5) ÷ √16 = √84375 ÷ 4.
5. 84375 = 5625 × 15 and 5625 = 75², so √84375 = 75√15.
6. Area = 75√15/4 cm² ≈ 72.62 cm².

Answer75√15/4 cm² (≈ 72.62 cm²)

Watch this explained “Half the perimeter”, 2:18 into Heron's formula: area from the three sides alone · हिंदी में देखें

Open NCERT p. 149Checked by computer

1. First way — Heron's formula: s = (7 + 24 + 25)/2 = 28 cm.
2. Area = √[28 × (28−7) × (28−24) × (28−25)] = √(28 × 21 × 4 × 3) = √7056 = 84 cm².
3. Second way — notice 7² + 24² = 49 + 576 = 625 = 25², so this is actually a right triangle with legs 7 and 24.
4. Area = (1/2) × 7 × 24 = 84 cm² — the same answer both ways.

Answer84 cm²

Watch this explained “Third check: three, four, five”, 6:36 into Heron's formula: area from the three sides alone · हिंदी में देखें

Open NCERT p. 149Checked by computer

1. Each rotation carries the wheel forward by exactly its circumference: C = π × d = 22/7 × 60 = 1320/7 cm.
2. In 100 rotations, distance = 100 × 1320/7 = 132000/7 cm (≈ 18857.14 cm).

Answer132000/7 cm (≈ 18857.14 cm)

Watch this explained “Two ways to write one length”, 0:00 into Arc length as the central angle's share of the circumference · हिंदी में देखें

Open NCERT p. 150Checked by computer

1. From C = 2πr: 66 = 2 × 22/7 × r, so r = 66 × 7 ÷ 44 = 21/2 cm.
2. A quadrant's area is (1/4) × π × r².
3. Area = (1/4) × 22/7 × (21/2)² = (1/4) × 22/7 × 441/4 = 693/8 cm² (= 86.625 cm²).

Answer693/8 cm² (= 86.625 cm²)

Watch this explained “A quarter disc, by a quarter-turn”, 1:21 into A sector's area is its angle's share of the whole · हिंदी में देखें

Open NCERT p. 150Checked by computer

1. One complete turn covers the wheel's circumference: C = 2πr = 2 × 22/7 × 28 = 176 cm.
2. 1 km = 100000 cm, so the number of turns = 100000 ÷ 176 = 6250/11 (≈ 568.18 turns).

AnswerOne turn covers 176 cm; the wheel turns 6250/11 times (≈ 568.18 times) in a 1 km journey.

Watch this explained “Two ways to write one length”, 0:00 into Arc length as the central angle's share of the circumference · हिंदी में देखें

Open NCERT p. 150Checked by computer

1. Let the first rectangle have length l and breadth b, and the second have length L and breadth B, with l ≥ b and L ≥ B.
2. Same perimeter: 2(l + b) = 2(L + B), so l + b = L + B.
3. Same area: l × b = L × B.
4. Use the identity (l − b)² = (l + b)² − 4lb. The right-hand side is the same for both rectangles (equal sums, equal products), so (l − b)² = (L − B)².
5. Both differences are 0 or positive, so l − b = L − B.
6. Now the sum and the difference of the sides agree: l = ½[(l + b) + (l − b)] = ½[(L + B) + (L − B)] = L, and then b = B.
7. So the two rectangles have the same length and the same breadth, and one can be placed exactly on the other.
8. Example: a rectangle with perimeter 22 cm and area 30 cm² needs two sides adding to 11 and multiplying to 30; the only such pair is 5 and 6.

AnswerYes. Two rectangles with the same area and the same perimeter have the same length and breadth, so they are congruent.

Watch this explained “Same area and same perimeter”, 8:23 into From rectangle to parallelogram: area survives rearrangement · हिंदी में देखें

Open NCERT p. 150One way to think about it

1. In the trapezium, the top side a and the bottom side b are parallel, and h is the perpendicular distance between them.
2. Draw a line through the top-right corner, parallel to the left slanted side, meeting the bottom side at a new point (this is the construction shown in the figure). It cuts the trapezium into a parallelogram and a triangle.
3. The parallelogram has base a and height h, so its area = a × h (area of parallelogram = base × height).
4. The triangle left over has base (b − a) and the same height h. Its area is half of a parallelogram with that base and height, so its area = ½ × (b − a) × h.
5. Area of trapezium = area of parallelogram + area of triangle = a×h + ½(b−a)×h.
6. Simplify: a×h + ½(b−a)×h = ½×h×(2a + b − a) = ½×h×(a+b).

In shortArea of trapezium = ½(a + b)h, matching the given formula.

Watch this explained “A trapezium, three ways over”, 7:23 into From rectangle to parallelogram: area survives rearrangement · हिंदी में देखें

Open NCERT p. 150One way to think about it

1. Draw one diagonal of the trapezium. It splits the trapezium into two triangles.
2. Both triangles have the same height h — the perpendicular distance between the two parallel sides a and b.
3. One triangle stands on the side of length a, so its area = ½ × a × h.
4. The other triangle stands on the side of length b, so its area = ½ × b × h.
5. Area of trapezium = ½ah + ½bh = ½h(a + b).

In shortArea of trapezium = ½(a + b)h — the same formula as before.

Watch this explained “A trapezium, three ways over”, 7:23 into From rectangle to parallelogram: area survives rearrangement · हिंदी में देखें

Open NCERT p. 150One way to think about it

1. Take two identical (congruent) copies of the trapezium, each with parallel sides a and b and height h.
2. Turn the second copy through half a turn (upside down) and slide it up against the first, matching a slanted side of one to a slanted side of the other.
3. The two trapezia together now form a parallelogram: its base is a + b (side a of one trapezium lies end to end with side b of the other), and its height is still h.
4. Area of this parallelogram = base × height = (a + b) × h.
5. The parallelogram is made of two identical trapezia, so one trapezium is exactly half of it.

In shortArea of one trapezium = ½(a + b)h.

Watch this explained “A trapezium, three ways over”, 7:23 into From rectangle to parallelogram: area survives rearrangement · हिंदी में देखें

Open NCERT p. 150One way to think about it

(i) using algebra

1. Let the kite be ABCD with AB = AD and CB = CD. Its diagonals are AC = p and BD = q, and they cross at O.
2. Triangles ABC and ADC are congruent (SSS: AB = AD, CB = CD, AC common), so ∠BAO = ∠DAO.
3. Then triangles ABO and ADO are congruent (SAS: AB = AD, AO common, equal angles at A). So BO = OD = q/2, and ∠AOB = ∠AOD. These two angles add up to 180°, so each is 90°: AC ⊥ BD.
4. Diagonal AC splits the kite into triangles ABC and ADC, both on base AC = p. Their heights are BO = q/2 and OD = q/2.
5. Area of kite = ½ × p × (q/2) + ½ × p × (q/2) = pq/4 + pq/4 = ½ × p × q.

In shortArea of kite = ½ × p × q, half the product of its diagonals.

(ii) using geometry

1. From (i), the diagonals AC = p and BD = q cross at right angles at O.
2. Through A and C draw lines parallel to BD, and through B and D draw lines parallel to AC. They make a rectangle around the kite with sides p and q, so its area is p × q.
3. The two diagonals cut this rectangle into 4 smaller rectangles, each with one corner at O.
4. Each side of the kite (AB, BC, CD, DA) is a diagonal of one of these small rectangles. It cuts that small rectangle into two congruent right triangles, one inside the kite and one outside it.
5. So exactly half of each small rectangle lies inside the kite, which means the kite is exactly half of the big rectangle.
6. Area of kite = ½ × p × q.

In shortThe kite is exactly half of the p × q rectangle drawn around it, so its area = ½ × p × q.

Open NCERT p. 150One way to think about it

(i) Rectangle ABCD has sides a, b, and rectangle PQRS has sides 2a

1. Area of ABCD = a × b = ab.
2. Area of PQRS = 2a × 2b = 4ab = 4 × area of ABCD.
3. Check: join the midpoints of opposite sides of PQRS. This cuts it into 4 rectangles, each a by b, which is ABCD itself.

In shortYes. PQRS has area 4ab, 4 times the area ab of ABCD, and 4 copies of ABCD fill it exactly in a 2-by-2 arrangement.

(ii) ∆ABC has sides a, b, c, and ∆PQR has sides 2a

1. By Heron's formula, ∆ABC has semi-perimeter s = (a + b + c)/2 and area √(s(s − a)(s − b)(s − c)).
2. ∆PQR has sides 2a, 2b, 2c, so its semi-perimeter is 2s, and 2s − 2a = 2(s − a), 2s − 2b = 2(s − b), 2s − 2c = 2(s − c).
3. Area of ∆PQR = √(2s × 2(s − a) × 2(s − b) × 2(s − c)) = √16 × √(s(s − a)(s − b)(s − c)) = 4 × area of ∆ABC.
4. Check: let QR = 2a, RP = 2b, PQ = 2c. Mark the midpoints L of QR, M of RP and N of PQ, and join them. A segment joining the midpoints of two sides is half the third side, so NM = a, LN = b and LM = c.
5. This cuts ∆PQR into 4 triangles: PNM (sides c, b, a), QNL (c, b, a), RLM (a, b, c) and the middle one LMN (a, b, c). All four have sides a, b, c, so each is congruent to ∆ABC (SSS).

In shortYes. ∆PQR has 4 times the area of ∆ABC, and 4 copies of ∆ABC fit into it exactly. The lines joining the midpoints of its sides cut it into 4 such triangles, with the middle one turned half a turn.

(iii) ∆ABC has sides a, b, c, and ∆PQR has sides 3a

1. ∆PQR has sides 3a, 3b, 3c, so its semi-perimeter is 3s, and 3s − 3a = 3(s − a), and so on.
2. Area of ∆PQR = √(3s × 3(s − a) × 3(s − b) × 3(s − c)) = √81 × √(s(s − a)(s − b)(s − c)) = 9 × area of ∆ABC.
3. Check: divide each side of ∆PQR into three equal parts, and through the division points draw lines parallel to the sides. This makes a grid of small triangles, with 1, 3 and 5 triangles in the three rows from the top: 9 in all.
4. Every grid line is cut into equal pieces that are one-third of 3a, 3b or 3c long, so every small triangle has sides a, b, c. The 6 upright ones and the 3 upside-down ones are all congruent to ∆ABC.

In shortYes. ∆PQR has 9 times the area of ∆ABC, and 9 copies of ∆ABC fit into it exactly: 6 upright and 3 turned half a turn.

Watch this explained “Double the sides”, 9:07 into A triangle is half a parallelogram · हिंदी में देखें

Open NCERT p. 151Checked by computer

1. Fig. 6.43: call the triangle PQR, with P at the top, Q at the bottom left and R at the bottom right. The tick marks show that D is the midpoint of PQ (triple ticks on both halves), and that E and F cut PR into three equal parts, E being nearer P (double ticks on all three pieces).
2. The shaded region is the quadrilateral QDEF. It is the triangle PQF with the small triangle PDE taken away.
3. Two triangles with the same apex and their bases on one line have the same height, so their areas are in the ratio of their bases. Triangle PQF has apex Q and base PF = ⅔ PR, so area(PQF) = ⅔ × area(PQR).
4. In the same way, area(PQE) = ⅓ × area(PQR), because PE = ⅓ PR. Triangle PDE has apex E, like triangle PQE, and its base PD is ½ of PQ. So area(PDE) = ½ × ⅓ × area(PQR) = ⅙ × area(PQR).
5. Shaded area = ⅔ − ⅙ = 4/6 − 1/6 = 3/6 = ½ of the triangle.
6. Fig. 6.44: take the side of the square as 2 units, so its area is 4. Each corner is joined to the midpoint of a side it does not touch, all turning the same way. The figure looks the same after a quarter turn, so the lines form two pairs of parallel lines at right angles, and the shaded region is a square.
7. Take one pair: the line from the top-left corner to the midpoint of the bottom side, and the line from the midpoint of the top side to the bottom-right corner. Together with the top and bottom sides they make a parallelogram with base 1 (along the top) and height 2, so its area is 1 × 2 = 2.
8. The slanting side of this parallelogram goes 1 across and 2 down, so by Pythagoras its length is √(1² + 2²) = √5. Taken as the base, it gives area = √5 × d, where d is the gap between the two parallel lines. So √5 × d = 2, and d = 2/√5.
9. The shaded square has two opposite sides on these two lines, so its side is d and its area is d² = 4/5.
10. Fraction shaded = (4/5) ÷ 4 = 1/5.

AnswerFig. 6.43: 1/2 of the triangle is shaded. Fig. 6.44: 1/5 of the square is shaded.

Open NCERT p. 151Checked by computer

1. Fig. 6.45 has 3 identical circles in a row, each touching the top and bottom of the rectangle and touching its neighbours, so the rectangle's height equals one diameter (2r) and its width equals 3 diameters (6r).
2. Area of the 3 circles = 3 × πr². Area of the rectangle = 6r × 2r = 12r². Fraction covered = 3πr²/12r² = π/4.
3. Fig. 6.46 has 4 such circles instead of 3: the height is still one diameter (2r), and the width is now 4 diameters (8r).
4. Area of the 4 circles = 4 × πr². Area of the rectangle = 8r × 2r = 16r². Fraction covered = 4πr²/16r² = π/4 — the same fraction as before.

AnswerBoth Fig. 6.45 and Fig. 6.46 have exactly π/4 of the rectangle covered by circles.

Open NCERT p. 151One way to think about it

1. Conjecture: whatever the number of circles n, as long as they are arranged in a single touching row that exactly fits the rectangle's height, the circles always cover exactly π/4 of the rectangle.
2. Testing n = 10: the rectangle is 20r by 2r, area 40r². Circles' area = 10πr². Fraction = 10πr²/40r² = π/4. Matches.
3. Testing n = 20: the rectangle is 40r by 2r, area 80r². Circles' area = 20πr². Fraction = 20πr²/80r² = π/4. Matches.
4. Testing n = 50: the rectangle is 100r by 2r, area 200r². Circles' area = 50πr². Fraction = 50πr²/200r² = π/4. Matches.
5. General proof: for n circles of radius r in a row, the rectangle is (2nr) by (2r), so its area = 4nr², and the circles' total area is nπr². Fraction = nπr²/4nr² = π/4 for every value of n, since the n's cancel exactly.

In shortThe circles always cover exactly π/4 of the rectangle, whatever the number of circles in the row.

Open NCERT p. 151Checked by computer

1. In the figure, the top row has 4 identical small rectangles standing side by side, and the bottom row has 5 of the very same rectangles side by side (turned the other way), both rows spanning the same total width W.
2. Since the top row's 4 rectangles span the whole width W, each one's long side is L = W/4. Since the bottom row's 5 rectangles span the same width W, each one's short side is B = W/5.
3. So the two sides of one small rectangle are L = W/4 and B = W/5, which gives L = (5/4)B.
4. All 9 identical rectangles tile the big rectangle exactly, so their combined area is the given 72 cm²: 9 × L × B = 72, i.e. L × B = 8.
5. Substituting L = (5/4)B into LB = 8: (5/4)B² = 8, so B² = 32/5, giving B = 4√10/5 cm and L = √10 cm.
6. Perimeter of one small rectangle = 2(L + B) = 2(√10 + 4√10/5) = 2 × (9√10/5) = 18√10/5 cm.

AnswerPerimeter of each small rectangle = 18√10/5 cm ≈ 11.38 cm.

Open NCERT p. 152One way to think about it

1. Call the top vertex P and the bottom side QR, which is cut into three equal parts at X and Y (in order Q, X, Y, R). The blue triangle is PQX and the red triangle is PYR.
2. Equal areas: QX = YR, since each is one-third of QR. Both triangles have their apex at P, so both have the same height h, the perpendicular distance from P to the line QR. So area(PQX) = ½ × QX × h = ½ × YR × h = area(PYR).
3. Cutting, step 1: in the blue triangle, mark the midpoints of PQ and PX and cut along the line joining them. Turn the small top triangle half a turn about the midpoint of PX, so its tip lands on X. It fits beside the lower piece, and together they make a parallelogram on the base QX with half the height.
4. Step 2: slide this parallelogram along the base line until its base sits exactly on YR. The two bases are equal, so they match.
5. Step 3: cut the parallelogram along its diagonal from the bottom-right corner (at R) to the top-left corner. Slide the right-hand triangle to the left by the length of the base, so that it sits against the other end. The parallelogram still stands on YR with the same height, but now its slanting sides are parallel to PY, and its top-left corner is the midpoint of PY.
6. Step 4: this is exactly the parallelogram that step 1 would make from the red triangle. So cut along the line from R to the midpoint of PR, and turn the piece on the right half a turn about that midpoint. It swings up to become the top corner of the red triangle, with its tip at P.
7. Every move was a straight cut, a slide or a half turn, and none of these changes any area. So the pieces now cover the red triangle exactly, with no gaps and no overlaps. Traced back, the blue triangle has been cut into 5 pieces.

In shortThe blue and red triangles have equal area: their bases are equal (each one-third of the bottom side) and they share the same height from the common top vertex. Cutting the blue triangle into 5 pieces as above and moving them covers the red triangle exactly.

Watch this explained “The median”, 4:40 into A triangle is half a parallelogram · हिंदी में देखें

Open NCERT p. 152One way to think about it

1. Let the square have side s. The quarter circle has its centre at the bottom-left corner O and radius s. The two semicircles stand on the two sides that meet at O (the left side and the bottom side), each with diameter s.
2. Each semicircle lies inside the quarter circle, because two points on a circle are never more than a diameter apart, so every point of each semicircle is within s of O.
3. Region A is the part inside both semicircles. Region B is the part of the quarter circle outside both semicircles. The rest of the quarter circle is two white pieces: W₁, inside the left semicircle only, and W₂, inside the bottom semicircle only.
4. Quarter circle = ¼ × πs² = ¼πs². Each semicircle = ½ × π(s/2)² = ⅛πs², so the two semicircles together = ¼πs², the same as the quarter circle.
5. Quarter circle = A + W₁ + W₂ + B. Left semicircle = A + W₁ and bottom semicircle = A + W₂, so the two semicircles together = 2A + W₁ + W₂. A is counted twice because it lies in both.
6. The totals are equal, so A + W₁ + W₂ + B = 2A + W₁ + W₂. Taking away A + W₁ + W₂ from both sides leaves B = A.

In shortArea(A) = Area(B). The two semicircles together have the same area as the quarter circle (¼πs² each). In the semicircles A is counted twice and B is not covered at all, so the two regions must have equal area.

Open NCERT p. 152Checked by computer

1. Each semicircle has a side of the square (2 units) as its diameter, so its radius is 1 and its centre is the midpoint of that side. Each one bulges into the square.
2. Look at the petal in the bottom-left corner. It lies between the semicircle on the bottom side (centre M, the midpoint of the bottom side) and the semicircle on the left side (centre N, the midpoint of the left side).
3. The corner and the centre of the square are both 1 unit from M, and both 1 unit from N. So both semicircles pass through the corner and through the centre of the square, and the petal runs from the corner to the centre.
4. Seen from M, the corner is straight along the bottom side and the square's centre is straight up, so the angle between them is 90°. The petal's edge from this semicircle is therefore a quarter circle of radius 1, with length ¼ × 2π × 1 = π/2. The same is true of its other edge, from the semicircle centred at N.
5. Perimeter: 4 petals × 2 edges × π/2 = 4π units (about 12.57 units).
6. Area of one petal: draw the straight line from the corner to the centre of the square. It cuts the petal into two equal pieces. Each piece is a quarter circle of radius 1 (area π/4) minus a right triangle with legs 1 and 1 (area ½), so it has area π/4 − ½.
7. One petal = 2 × (π/4 − ½) = π/2 − 1. Four petals = 4 × (π/2 − 1) = 2π − 4 square units (about 2.28 square units).

AnswerPerimeter of the flower = 4π units (about 12.57 units). Area of the flower = (2π − 4) square units (about 2.28 square units).

Watch this explained “The square flower”, 8:05 into Perimeter puzzles: composite curved boundaries reduce to arcs you already know · हिंदी में देखें

Open NCERT p. 152One way to think about it

1. Let R be the radius of the larger circle and r the radius of the smaller circle, both centred at O.
2. Since BC is tangent to the smaller circle at A, the radius OA is perpendicular to BC, and OA = r.
3. OA is then a perpendicular dropped from the centre O onto the chord BC of the larger circle, so OA bisects BC: BA = AC = l/2.
4. In right triangle OAB (right angle at A), by Pythagoras: OB² = OA² + AB², i.e. R² = r² + (l/2)².
5. So R² − r² = l²/4.
6. Area of the green ring between the two circles = πR² − πr² = π(R² − r²) = π × l²/4 = ¼πl².

In shortArea of the green region = ¼πl², without needing to know R or r separately.

Open NCERT p. 153One way to think about it

1. Read the figure: the semicircles on the two shorter sides are drawn outwards, and the semicircle on the hypotenuse is drawn inwards, over the triangle. A and B are the green crescents, and C is the right-angled triangle itself.
2. Let the two shorter sides be p and q and the hypotenuse h. By Pythagoras, p² + q² = h².
3. The right-angle corner lies on the semicircle on the hypotenuse. To see this, complete the triangle to a rectangle: its diagonals are equal and bisect each other, so the midpoint of the hypotenuse is h/2 from the right-angle corner too. So the inner arc runs from one end of the hypotenuse, through the right-angle corner, to the other end.
4. So the semicircle on the hypotenuse = triangle C + the two white pieces X (beside side p) and Y (beside side q). Its arc has the larger radius h/2, so it is flatter than the semicircles on p and q, and X and Y lie inside them. So the semicircle on p = X + A, and the semicircle on q = Y + B.
5. A semicircle on a side of length d has area ½ × π(d/2)² = πd²/8. Multiplying p² + q² = h² by π/8 gives πp²/8 + πq²/8 = πh²/8, that is, (X + A) + (Y + B) = C + X + Y.
6. Take away X and Y from both sides: A + B = C.

In shortArea(A) + Area(B) = Area(C). The two crescents together have exactly the same area as the right-angled triangle.

Open NCERT p. 153Checked by computer

1. Let the centres be A and B and the common radius r. Each circle passes through the other's centre, so AB = r. Let the circles cross at C and D.
2. AC = BC = r (radii) and AB = r, so triangle ABC is equilateral and ∠CAB = 60°. In the same way ∠DAB = 60°, so ∠CAD = 120°.
3. C and D are each r away from both A and B, so both lie on the perpendicular bisector of AB. The chord CD therefore cuts AB at right angles at its midpoint M, with AM = r/2, and it splits the red region into two equal halves.
4. The half on B's side is the part of circle A beyond the chord CD: the sector ACD minus the triangle ACD.
5. Sector ACD has angle 120°, which is one-third of the circle, so its area is (120/360) × πr² = πr²/3.
6. In the right triangle AMC, MC² = r² − (r/2)² = 3r²/4, so MC = (√3/2)r and CD = √3 r. Triangle ACD has base CD = √3 r and height AM = r/2, so its area is ½ × √3 r × r/2 = (√3/4)r².
7. Half of the region = πr²/3 − (√3/4)r². The whole region = 2 × [πr²/3 − (√3/4)r²] = (2π/3 − √3/2)r².

AnswerArea of the region enclosed by the two circles = r²(2π/3 − √3/2), about 1.23r² square units.

Watch this explained “The triangle nobody drew”, 1:07 into Perimeter puzzles: composite curved boundaries reduce to arcs you already know · हिंदी में देखें

Open NCERT p. 153One way to think about it

1. Let the rectangle have width W and height H, with one corner (call it the origin) at the vertex all three triangles touch. Let M be the marked point on the top side and R the marked point on the right side.
2. Drop a line straight down from M and a line straight across from R; they meet at the interior point N, so MN is vertical and NR is horizontal, at right angles — N is the fourth marked point.
3. Let p be the horizontal distance from the origin to M along the top side, and q be the vertical distance from the bottom up to R on the right side. Then MN = H − q and NR = W − p.
4. Triangle C (= M, N, R) is a right triangle with legs MN and NR, so C = ½(H−q)(W−p).
5. Triangle A (= origin, M, N) stands on base MN = (H−q), at horizontal distance p from that base's line, so A = ½(H−q)(p).
6. Triangle B (= origin, N, R) stands on base NR = (W−p), at vertical distance q from that base's line, so B = ½(W−p)(q).
7. Adding: A + C = ½(H−q)(p) + ½(H−q)(W−p) = ½(H−q)×W, and B + C = ½(W−p)(q) + ½(W−p)(H−q) = ½(W−p)×H.
8. So 2(A+C)(B+C)/C = 2 × [½W(H−q)] × [½H(W−p)] / [½(H−q)(W−p)] = WH, exactly the area of the rectangle.

In shortArea of rectangle = WH = 2(A+C)(B+C)/C, verified by writing all three triangle areas in terms of the two right-triangle legs meeting at the interior point.

Open NCERT p. 153One way to think about it

1. Let O be the centre of the big semicircle on diameter AC, with radius r = OA = OB = OC, where B is the point on the big semicircle directly above O, so angle AOB = 90°.
2. By Pythagoras in right triangle AOB: AB² = OA² + OB² = r² + r² = 2r².
3. Area of the quarter-circle sector OAB (radii OA, OB and the arc between them) = ¼ × πr² = πr²/4.
4. Area of the semicircle drawn on AB as diameter = ½ × π × (AB/2)² = π×AB²/8 = π×(2r²)/8 = πr²/4 — the same as the sector's area.
5. The sector OAB splits into triangle OAB plus the small segment S cut off between chord AB and the big circle's own arc: Area(sector) = Area(triangle OAB) + Area(S).
6. The semicircle on AB splits into that very same segment S (the part it shares with the big circle's disc) plus the crescent-shaped lune E that sticks out beyond the big circle's arc: Area(semicircle on AB) = Area(S) + Area(E).
7. Since Area(sector OAB) = Area(semicircle on AB) [both πr²/4], and both equal Area(S) plus one more piece, that extra piece must match: Area(triangle OAB) = Area(E).

In shortThe lune (one shaded region) and the triangle OAB (the other shaded region) have equal area, because the quarter-circle sector and the semicircle on AB have exactly the same area (πr²/4), and subtracting the shared segment S from each leaves equal remainders.

Watch this explained “A segment is a sector, less a triangle”, 6:03 into A sector's area is its angle's share of the whole · हिंदी में देखें