NCERT Exemplar Solutions Class 10 Maths Chapter 11

Cancel the common π and read off the relation.

• Why π goes: every circle here carries the same factor π, so the moment you write area as π r² that factor divides out of the whole equation and only the squared radii are left.
• The clue: this is why the answer is the Pythagoras-looking R₁²+R₂²=R² and not the tempting linear R₁+R₂=R.
• The trap: adding radii directly would be correct for perimeters, not for areas, so option (A) catches the unwary while (B) is the clean truth.

Option (B), R₁²+R₂²=R².

The 2π cancels, leaving a plain sum.

• Linear in radius: because circumference is linear in the radius, dividing through by the shared 2π turns the length equation into R₁+R₂=R exactly.
• No ambiguity: there is no inequality and nothing undecided here, so options (B), (C) and (D) are all wrong.
• The pairing: setting this beside Question 1 is deliberate, as squares add for areas while radii add for circumferences.

Option (A), R₁+R₂=R.

Same fence, more land for the circle.

• The picture: think of P as a length of fencing; bent into a circle it traps area P²4π, but folded into a square it traps only P²16.
• The winner: since 4π≈ 12.57 is smaller than 16, the circle encloses more, which is why round pipes, tanks and manhole covers are so common.
• Without numbers: start with the square and round off its corners at fixed perimeter, and every rounding pushes the boundary outward into new area until the limit is the circle, so it must out-enclose the square.

Option (B), the circle has the larger area.

Fix the base, push the top up.

• Natural base: the diameter 2r is the longest chord available, so it is the natural base to build the triangle on.
• Tallest apex: once the base is locked in, only the height can grow, and the tallest the apex can reach inside a semicircle of radius r is a height of r.
• The maximum: plugging into 12 bh gives 12(2r)(r)=r², and any other inscribed triangle is shorter or narrower, so r² is genuinely the largest possible.

Option (A), r² square units.

The whole ratio collapses to 4/π.

• Radius cancels: because the radius drops out, the answer is independent of size, so any circle and square with equal perimeters have areas in the ratio 4:π.
• Numerical form: converting π to 227 turns this into 14:11, which also confirms Question 3 because 14>11 tells us the circle's area is the larger one.
• A fact to store: keep 4:π≈ 1.27 in mind, since for equal perimeters the circle always beats the square by about 27%, and the same 14:11 holds for a tiny coin or a huge field alike.

Option (B), 14:11.

Diameters first, then squares.

• The trap: the tempting mistake here is to add the diameters or the radii straight off, which is wrong for areas.
• Squares add: areas combine through R²=r₁²+r₂², so R²=64+36=100 and the new radius is R=10 m.
• The signal: the numbers were chosen as the 6,8,10 Pythagorean set, a nice cue that you have applied the squares-add rule rather than a linear shortcut.

Option (A), 10 m.

The side is the diameter.

• Full width: an inscribed circle just fits between two opposite sides, so its full width 2r is the side 6, giving r=3 and area 9π.
• Keep π: leaving the answer in terms of π avoids any rounding error in an objective question.
• Sanity check: the circle's 9π≈ 28.3 cm² is less than the square's 36 cm², exactly as expected for a shape tucked inside.

Option (D), 9π cm².

Diagonal is the diameter; halve its square.

• Corners on circle: the corners of an inscribed square touch the circle, so the diagonal stretches right across as a diameter of 16 cm.
• Faster route: using area =12 d²=12(256)=128 cm² is quicker than first finding the side.
• Cross-check: the side would be 162=82, and (82)²=128 confirms the same answer.

Option (B), 128 cm².

Lengths add, so radii add.

• Cancel 2π: each circumference is 2π r and the shared 2π cancels, so R=r₁+r₂=18+10=28 cm.
• The trap: the wrong option 56 cm is exactly the sum of the diameters, which is what you get if you forget to halve.
• The pairing: this is the linear partner of Question 10, where it is the areas that combine instead of the lengths.

Option (C), 28 cm.

Squares add, then read the question again.

• Squares add: because areas combine, R²=576+49=625 and the radius is R=25 cm.
• The catch: the word ``diameter'' means the answer they want is actually 2R=50 cm, not the radius.
• The triple: the 7,24,25 set makes the square root instant, and the deliberate option 25 cm punishes anyone who forgets to double.

Option (D), 50 cm.

Halve the side before squaring.

• Squeezed inside: since the circle is squeezed inside the square, its diameter is the side a and its radius is a2.
• The 14 factor: squaring that radius brings in a factor of 14, so the true area is π a²4.
• The error: the claim is off by a factor of 4, which is exactly what using the side as the radius produces.

False; the correct area is π a²4 cm².

Side equals the diameter here.

• Touching condition: because the circle of radius a fits exactly between opposite sides of the square, each side is the diameter 2a.
• The perimeter: four such sides give 8a, so the statement is correct.
• Mirror of Q11: there the square was inside and the side became the diameter, and the two ideas are mirror images of the same touching condition.

True; perimeter =8a cm.

Doubling the area needs a 2 scale, not 2.

• Side ratio: the outer square's side is d while the inner square's side is d2, so the sides are in the ratio 2, not 2.
• Areas square it: areas scale as the square of lengths, so the true area ratio is (2)²=2, which is why the outer square is only twice the inner one.
• Why not four: the ``four times'' guess would need the sides in ratio 2, and that happens only if the outer side were the inner side doubled rather than the inner diagonal.

False; the outer square is twice, not four times, the inner square.

The word ``segment'' hides two cases.

• Minor case: for the minor segment you carve the triangle away from the sector, so it shrinks below the sector.
• Major case: for the major segment the same triangle is added on, pushing it above its sector; since the claim does not say which segment, one counterexample is enough to mark it false.
• The picture: draw any chord and you get a small minor cap and a big major piece, where the cap is clearly less than its sector but the big piece visibly bulges past its own sector.

False; the claim only holds for the minor segment.

Pick one form of the circumference.

• One per turn: a rolling wheel lays down exactly one circumference for every full turn it makes.
• Two forms: in radius form that distance is 2π r, and in diameter form it is π d, and the two are the same length.
• The error: writing 2π d mixes the two forms, effectively using r=d, so the correct one-revolution distance is π d cm.

False; one revolution is π d cm.

Counting turns is just division.

• One step: each revolution advances the wheel by its circumference 2π r, so the whole journey s contains s2π r of those steps.
• Units check: the formula is dimensionally clean, since metres divided by metres gives a pure count of turns.
• Verdict: the statement is therefore exactly right with no correction needed.

True; the wheel makes s2π r revolutions.

The crossover is at r=2.

• Turning point: solving π r²=2π r gives r=2 as the single radius where the two numbers match exactly.
• Which side wins: past that radius the squared growth of area wins, and below it the linear circumference is bigger; a check at r=1 (area 3.14 versus circumference 6.28) settles it.
• Units note: the units do not even match, since area is in square units and circumference in plain units, so the comparison is purely numerical and hinges entirely on the size of r.

False; true only for r>2.

Equal arcs trade radius for angle.

• Key product: arc length is proportional to the product rθ, so for equal arcs that product stays fixed.
• Angle halves: if the radius doubles from r to 2r, the angle must halve to keep the product steady, so the smaller circle's angle is twice the larger one's.
• Wording trap: the double-negative phrasing ``is this statement false?'' is the only tricky part, since the statement itself is correct.

True statement; ₁=2₂.

Same arc, bigger radius, bigger slice.

• The formula: using area =12r, two sectors with the same arc length still differ if their radii differ, because a larger radius stretches the slice outward and adds area.
• Verdict: the claim is false in general and becomes true only for the special case of equal radii, which is effectively the same circle.
• From a figure: lay the two arcs as equal bases, and the sector with the longer radius is the taller ``triangle'' with the greater area, since equal bases with unequal heights give unequal areas.

False; areas are equal only if the radii are equal.

One equation, two unknowns.

• Single condition: equal areas give just one condition ₁ r₁=₂ r₂, which two different pairs can satisfy.
• Counterexample: the worked pair (4× 3=6× 2) shows two sectors of equal area with unequal arcs, so equal arc length is certainly not necessary; this is the reverse of Question 19.
• The takeaway: fixing the area fixes only the product r, while fixing the arc fixes only , so neither condition alone pins both quantities and the ``must be equal'' claim breaks down for circles of different sizes.

False; only the product r must match.

The short side caps the circle.

• Short side rules: with a>b, the circle is stopped by the breadth b, so its diameter is b and its radius is b2.
• The 14 factor: squaring that radius brings the factor 14, giving the true area π b²4.
• Why a drops out: the length a never enters the answer, because there is always room to spare along the longer direction.

False; the area is π b²4 cm².

One number controls the whole circle.

• One parameter: the radius is the only parameter a circle has, so everything about the circle is decided once it is known.
• Radius then area: equal circumferences pin the radius, and once the radius is fixed the area π r² has no freedom left, so the areas must match.
• Unlike sectors: where a sector lets arc and angle vary independently, a full circle leaves no room for two different sizes to share a circumference.

True; equal circumference forces equal radius and hence equal area.

Run Question 22 backwards.

• Areas to radius: equal areas give r₁²=r₂², and since radii are positive lengths, this forces r₁=r₂.
• Radius to circumference: equal radii then force equal circumferences, completing the chain in the reverse direction.
• The big idea: the two questions together show that for a full circle every standard measure is locked to the radius, so matching one matches all of them.

True; equal area forces equal radius and hence equal circumference.

Use the diagonal shortcut.

• Diagonal spans: the corners of the square sit on the circle, so the diagonal stretches across as the diameter p.
• Half its square: then area =12 p²=p²2, which is exactly half of the stated p².
• The slip: the mistake in the statement is reading p as the side of the square rather than as its diagonal.

False; the area is p²2 cm².

Add the radii, skip the π.

• Shared factor: each circumference carries the same 2π, so the equation collapses to R=15+18=33 cm the moment you divide it out.
• No long way: there is no need to compute any actual circumference, since the factor cancels before any numbers are needed.
• Contrast: in the area-based problems the radii would instead combine through their squares, not by a plain sum.

R=33 cm.

Circle minus square, using the diagonal shortcut.

• Circle first: the diagonal 8 fixes the radius at 4, so the circle has area 16π.
• Square next: the square sits inside, and its area from the diagonal is 12(8²)=32.
• The crescents: subtracting leaves the four crescent-shaped pieces outside the square, totalling (16π-32) cm²≈ 18.3 cm², and reading the diagonal directly as the diameter is what makes this quick.

(16π-32) cm².

Cancel before you multiply.

• Reduce angle: the angle fraction reduces to 18, which keeps the later multiplication small.
• Share the 7: since 28²=784 shares the factor 7 with 227, we get 784÷ 7=112, leaving 18× 22× 112=308 cm².
• Why r=28: choosing a radius that is a multiple of 7 is what makes the 227 arithmetic come out to a whole number.

308 cm².

One turn =220 cm, then just divide.

• Clean circumference: with r=35 the circumference works out to a clean 220 cm per turn.
• Distance per minute: the bike covers 66 km in an hour, which is 110 000 cm in a minute, so it needs 110000220=500 turns each minute.
• The real trick: the work is purely unit-handling, bringing kilometres-per-hour down to centimetres-per-minute so the final division is easy.

500 revolutions per minute.

A quarter circle, nothing more.

• Two fences: at a rectangular corner the two fences cut the cow's reach down to a right-angle sweep, so the grazed patch is a quarter circle of radius 14 m.
• No spillover: since 14<16<20, that quarter circle never spills past a side, and the value is 14π(14)²=154 m².
• If the rope grew: had the rope been longer than 16 m, the quarter circle would spill over the shorter wall and you would subtract the overshoot, so checking 14<16 first justifies the clean 14π r² with no correction.

154 m².

Rectangle plus one whole circle.

• Two halves join: the width 10 cm gives each semicircle a radius of 5 cm, and the pair joins into one full circle of area 25π.
• Add them up: adding that to the rectangle's 380 cm² gives a total of (380+25π) cm².
• Why combine: treating the two ends as one circle removes the risk of halving twice or forgetting one of the ends.

(380+25π) cm².

Semicircle angle does the heavy lifting.

• Right angle at C: because AB is a diameter, C sees it at 90^∘, so the 6–8–10 triple gives AB=10 and r=5.
• Two areas: the circle is 78.5 cm² and the right triangle is 24 cm², which leaves 54.5 cm² shaded.
• Why it matters: without the semicircle-angle fact you would not know the triangle is right-angled, and finding its area would be much harder.

54.5 cm².

Straight addition of two pieces.

• Two pieces: the field is a rectangle of 32 m² with one half-disc tacked on the side.
• The half-disc: it has radius 2 m, giving 12π(2)²=2π m², so the sum is (32+2π) m²≈ 38.3 m².
• Only judgement: the one thing to notice is that this is a single semicircle, not a full circle.

(32+2π) m².

Composite areas are addition and subtraction.

• The marked answer: the exemplar shades the rectangle minus the rounded cut-outs, and its key records (248-4π) m² for the exact regions marked.
• Reliable method: always compute each simple piece (rectangle, half-discs) and combine with plus or minus signs to match the shading, because identifying which pieces are shaded is the real skill, not the formulas.
• This field: the rectangle is 26× 12 and the rounded cut-outs remove the curved ends, leaving the long shaded strip, so trace the boundary with a finger first to see which pieces to add and which to remove.

(248-4π) m², matching the exemplar's marked shading.

Sector minus equilateral triangle.

• The sector: it is 16 of the circle, namely 3083 cm².
• The triangle: the chord-and-two-radii triangle has a 60^∘ apex and two equal sides, so it is equilateral with area 493 cm², and recognising this spares you the trigonometry.
• Close in size: the sector 3083≈ 102.7 and the triangle 493≈ 84.9 are close, so the minor segment is a slim ≈ 17.8 cm²; leaving it as 3083-493 is exact and avoids rounding two awkward numbers.

(3083-493) cm².

Square minus a hidden full circle.

• Four slices: each corner contributes a 90^∘ slice of radius 6, and the four slices together total 360^∘, so they form exactly one full circle of area 113.04 cm².
• The subtraction: taking that circle out of the 144 cm² square leaves the curved region of 30.96 cm² shaded in the middle.
• Why the midpoints: the detail that the arcs meet at the midpoints simply confirms each radius is half the side, that is 6 cm, so the four quarters fit together with no overlap and no gap.
• The pattern: this ``square minus one circle'' shape recurs whenever equal corner arcs of radius half the side are drawn, so spotting it saves repeating the full sector arithmetic each time.

30.96 cm².

Three sectors become a semicircle.

• Angle-sum engine: because any triangle's angles add to 180^∘, the three equal-radius sectors merge into a half circle of area 12π(5)²=39.25 cm².
• The sliver: the equilateral triangle of side 10 is 253≈ 43.3 cm², so subtracting leaves the small curved-sided sliver of about 4.5 cm².
• No overlap: the radius 5 cm is exactly half the side, so the arcs reach only to the midpoints and never overlap, which lets you add the three 60^∘ pieces cleanly into one semicircle without double-counting.

(253-39.25) cm²≈ 4.5 cm².

Skip the angles, take half the circle.

• Half a circle: since ∠ P+∠ Q+∠ R=180^∘ for any triangle, the three sectors add up to a semicircle of radius 14, that is 12π(14)²=308 cm².
• Shape is irrelevant: the specific shape of the triangle never enters, which is exactly what the exemplar wants students to notice.
• The invariance: because only the radius and the fixed 180^∘ total matter, a needle-thin triangle or a near-equilateral one would still give 308 cm², and that invariance is the real lesson behind the 14 cm radius.

308 cm².

A ring is the gap between two discs.

• The gap: the road occupies the space between the 105 m park and the 126 m outer edge, so its area is π(R²-r²).
• Factor it: factoring as (R-r)(R+r)=21× 231 and using 227 gives the answer 15 246 m².
• Clean numbers: the road width 21 being a multiple of 7 is what cancels neatly and keeps the arithmetic whole.

15 246 m².

Four corners, one full circle.

• Angles total 360: a quadrilateral's interior angles always add to 360^∘, so the four equal-radius sectors assemble into one complete circle of radius 21.
• Read the area: its area 227(21)²=1 386 cm² is the entire shaded region, with no extra pieces to add or take away.
• Shape-free: as with the triangle version, the actual shape of the quadrilateral never enters the calculation, only the radius and the fixed angle total do.
• Triangle versus quad: this is the natural step up from the three-sector problems, where the corners gave only a semicircle, while four corners here close up into a whole circle.

1 386 cm².

One equation, solve for r.

• Wire is the arc: the 20 cm wire is precisely the arc, so the arc-length equation is 60360(2π r)=20.
• Reduce and solve: the angle fraction is 16, leaving π r3=20 and hence r=60π cm.
• Keep π below: keeping π in the denominator is fine, as the exemplar's answer is exactly this form, 60π cm ≈ 19.1 cm.

r=60π cm.

Area to radius to circumference to cost.

• Find the radius: dividing the area by 227 gives r²=7 056, a perfect square, so r=84 m.
• Then the cost: the fence is the circumference 528 m, and at Rs 50 a metre that comes to Rs 26 400.
• Common slip: the chain of steps is what trips students up, since many try to ``fence the area'', which has no meaning.

Rs 26 400.

Same road, fewer big turns.

• Only the ratio: both wheels roll the same distance, so the π cancels and only the ratio of the two diameters matters.
• Do the division: the rear wheel (200 cm) is 2.5 times the front (80 cm), so it needs 1400÷ 2.5=560 turns.
• Intuition check: bigger wheels make fewer revolutions for the same journey, which matches everyday experience.

560 revolutions.

Heron for the field, semicircle for the grazing.

• Field by Heron: Heron's formula gives the triangle's area as 24√21 m² from the three sides directly.
• Grazed half circle: the three ropes carve out sectors whose angles add to 180^∘, so together they graze a half circle of radius 7, worth 77 m², leaving (24√21-77)≈ 32 m² ungrazed.
• Looks can mislead: the 15–16–17 sides are close to equal so the field looks almost equilateral, but the method never used that, since Heron handled the exact area and the angle-sum gave the grazed semicircle.

(24√21-77) m².

Same recipe as the 14 cm case.

• The sector: it is 16 of the circle, namely 75.36 cm² when you take π=3.14.
• The triangle: the 60^∘ triangle is equilateral with area 363 cm², so their difference (75.36-363)≈ 13 cm² is the minor segment.
• What changed: the only difference from Question 34 is the radius and the chosen value of π, while the method is identical.

(75.36-363) cm².

Ring area times the rate.

• The annulus: the path is the gap between the 8.75 m pond and the 10.75 m outer edge, an annulus of area 227(2)(19.5)≈ 122.57 m².
• Then the cost: at Rs 25 per square metre this comes to about Rs 3 061.50.
• Factor first: factoring the squares as (R-r)(R+r) keeps the awkward decimals manageable throughout.

Rs 3 061.50 (about).

Trapezium minus one circle.

• The trapezium: the parallel-side formula gives its area as 350 cm² from the two parallel sides and the height.
• Corners make a circle: the four corner sectors, being angles of a quadrilateral, sum to 360^∘ and form a circle of radius 7, area 154 cm², so removing it leaves 196 cm².
• Slants do not matter: the trapezium's slanted sides do not change the fact that its four angles still total 360^∘.

196 cm².

The little curved triangle in the middle.

• Triangle of centres: the centres make an equilateral triangle of side 7 cm, area 4934≈ 21.22 cm².
• Subtract the sectors: inside it the three 60^∘ sectors join into a semicircle of 19.25 cm², so the tiny region the circles fence off is the difference, about 1.97 cm².
• Why so tiny: the semicircle almost fills the triangle, so the discs pack so snugly that only a pinhole survives, which is why close-packed circles waste so little room in pipe-bundling problems.

≈ 1.97 cm².

The arc-length shortcut.

• Skip the angle: rather than finding the angle from the arc, use area =12r=12(3.5)(5)=8.75 cm² directly.
• Where it comes from: this formula drops out by cancelling the θ360 that appears in both the arc-length and sector-area expressions.
• Final rounding: the exemplar rounds the result to 8.7 cm² for its recorded answer.

≈ 8.7 cm².

Square minus a circle, again.

• Square of centres: the centres form a 14 cm square of area 196 cm², and the four quarter-circle corners total one circle of 154 cm², leaving the central curved square of 42 cm².
• Familiar structure: it is the same square-minus-circle shape as the corner-arcs problem, just arising from touching discs instead of drawn arcs.
• See it for real: the radius 7 cm fixes the centre distance at 14 cm, the square's side, so the four right-angle sectors tile up to one circle, and the leftover is the curved square you see when four bottle caps are pressed together.

42 cm².

Side gives the radius; subtract four discs.

• Side then radius: the 784 cm² sheet has side 28 cm, and a 2× 2 arrangement makes two diameters fit across, so r=7.
• Subtract four discs: four such circles cover 616 cm², leaving 168 cm² of bare cardboard, with ``side =4r'' the key packing insight.
• Why 2× 2: the wording ``each side tangent to two plates'' forces the square grid rather than a slanted packing, fixing r at exactly 7 cm, after which the bare area is just the four corner gaps.

168 cm².

Tiles cover most, corners stay bare.

• Count the tiles: with 50 cm tiles the room holds 10× 8=80 of them in a grid.
• Covered area: each circle covers 1 962.5 cm², so all 80 cover 15.7 m² of the 20 m² floor, and the 4.3 m² difference is the total of all the little corner gaps.
• Sanity check: each tile fills about π4≈ 78.5% of its square cell, so roughly 21.5% of the floor should stay bare, and indeed 4.320=21.5% confirms the count and answer agree.

4.3 m².

The rhombus is secretly a square.

• Must be a square: for all four vertices to lie on a circle the rhombus must have right angles, which forces it to be a square.
• Diagonal is diameter: its diagonal is the diameter 40 cm, so its area is 12(40)²=800 cm².
• Circle's role: the circle's area only serves to deliver r=20, so spotting that the rhombus is a square is the whole problem.

800 cm².

Difference the squares.

• Discs first: full discs go as 1:4:9, the squares of the diameters, so start from there.
• Difference the bands: the bull's-eye is the disc 1, and the two outer bands are the differences 4-1=3 and 9-4=5, giving 1:3:5, where the odd numbers are exactly the gaps between consecutive squares.
• For an archer: the outer ring, though it looks thin, actually has the largest area (5 parts versus 3 and 1), which is why scoring rewards the small, hard-to-hit central disc most.

1:3:5.

Minutes become an angle, the hand is the radius.

• Minutes to angle: across 35 minutes the hand sweeps 35× 6^∘=210^∘, that is 712 of a full turn.
• Sweep the sector: with radius 5 cm the swept sector is 712π(5)²=2756≈ 45.8 cm², and the two clock times matter only through their 35-minute difference.
• Forward only: the hand sweeps forward, so from 6:05 to 6:40 it covers 210^∘, a major sweep past the straight-down mark, not the smaller angle the eye might guess.

2756 cm²≈ 45.8 cm².

Back out the radius, then the arc.

• Radius from area: the only unknown blocking the arc is the radius, so first rearrange the sector-area equation 59·227r²=770 to get r²=441 and hence r=21 cm.
• Arc from radius: with the radius now known, the same 200^∘ angle yields an arc of length 59× 2π(21)=2203≈ 73.3 cm.
• A second route: you can also skip the radius and use area =12r, which rearranges to =2× 77021=2203 in one step.
• Why two paths help: when a problem hands you both the area and the angle, reaching the same arc length by two different formulas is a strong check that the answer is right and no arithmetic slipped.

2203 cm ≈ 73.3 cm.

The arcs match, the areas do not.

• Equal arcs: both sectors come out to an arc of 443 cm, yet the larger-radius sector has the bigger area (154 versus 1543 cm²).
• The mechanism: using area =12r explains it at once, since equal but a tripled r triples the area, making the ``equal arcs need not mean equal areas'' rule tangible.
• Closes the loop: two real sectors, r=7,θ=120^∘ and r=21,θ=40^∘, share identical arcs but have areas in the ratio 1:3, settling the earlier reasoning questions with concrete numbers.

Areas 1543 and 154 cm²; arcs both 443 cm; equal arcs but unequal areas.

Square minus the curved cut-outs.

• The subtraction: taking the 196 cm² square and removing the marked arc-and-strip region of 16+8π cm² leaves (180-8π) cm², the exemplar's recorded answer.
• Reliable approach: peel any such figure into rectangles, full or partial circles, and combine them with the correct signs to reproduce the shading exactly.
• The value: here the marked region is (180-8π)≈ 154.9 cm², a little under the full square, and matching signs to the diagram is what separates a correct answer from a plausible wrong one.

(180-8π) cm².

Two short conversions, then divide.

• Area to radius: dividing the area by 227 gives r²=0.49, so r=0.7 m and the circumference is 4.4 m.
• Then divide: the journey 176 m then contains 1764.4=40 turns of the wheel.
• Why so clean: the tidy r=0.7 comes from the area being a neat 1.54 m², which is exactly 227(0.7)².

40 revolutions.

Track the triangle's sign carefully.

• Signs reinforce: the major segment is its sector plus the central triangle and the minor segment is its sector minus the same triangle, so subtracting the minor from the major leaves (major sector - minor sector) and a + 2× triangle that does not cancel.
• Carry r²: the chord of 5 cm subtends 90^∘, so the two radii are legs of a right isosceles triangle with the chord as hypotenuse, giving 5²=r²+r² and r²=252, and every area here depends on r² alone, so no messy square root is needed.
• Assemble it: the sector difference is 12π r² and twice the triangle is r², so substituting r²=252 gives 25π4+252≈ 32.1 cm².
• Common slip: students often take only the difference of the two sectors and forget that the chord's triangle belongs to both segments with opposite signs.

(25π4+252) cm².

Only the angle gap matters.

• The gap: the major sector (240^∘) minus the minor (120^∘) spans a 120^∘ difference, which is 13 of the circle.
• Read it off: with r=21 that difference is 13×227× 441=462 cm², which is faster than computing each sector separately and subtracting.
• Long-way check: the major sector is 240360π(21)²=924 cm² and the minor is 462 cm², so 924-462=462 cm², and the difference matching the minor sector is a quirk of the 120^∘ choice.

462 cm².