NCERT Solutions Class 10 Maths Chapter 11 Areas Related to Circles

Spot the easy fraction first. A 60^∘ sector is exactly one-sixth of the circle, so you can skip the angle fraction entirely.

• Why one-sixth: sixty degrees goes into the full three hundred sixty degrees exactly six times, so the slice is one of six equal pieces of the circle.
• Quick route: take one-sixth of the full area π r²=7927, which lands on 1327 without ever writing the 60360 fraction.
• Memorise these: common angles read as neat fractions of the circle, with 30^∘ a twelfth, 45^∘ an eighth, 90^∘ a quarter and 120^∘ a third. Reading the angle this way saves time and avoids slips with the heavy denominator.

1327 cm²≈ 18.86 cm², that is, one-sixth of the circle.

Watch the two-step structure. This question hides one step inside another: first find the radius, then take the quarter area.

• Radius first: the circumference gives r=222π=72 cm, and only with the radius in hand can the area formula be used at all.
• Then the quarter: a quadrant is one-fourth of the full circle π r²=772, so the answer is simply 778=9.625 cm².
• Keep the fraction: carry the radius as the neat 72 rather than the decimal 3.5. The spare seven then cancels with the seven inside π, which keeps the whole calculation exact and fast.

778 cm²=9.625 cm².

Use the fraction of the hour directly. Five minutes is one twelfth of the full turn, so the swept area is one twelfth of the hour circle.

• Time as a fraction: five out of sixty minutes is 112 of a complete revolution, so you never need to turn the time into degrees at all.
• One hour circle: in a full hour the hand would sweep a circle of area π r²=616 cm², and one twelfth of that is 1543≈ 51.33 cm².
• When it helps: reading the minute hand as a fraction of the hour skips the 30360 arithmetic and is the fastest route whenever the time is a neat part of sixty minutes.

1543 cm²≈ 51.33 cm².

Build a quick check with the whole circle. The pieces must add up, so use that to catch arithmetic slips before writing the final line.

• Sectors close up: the major sector and the minor sector together rebuild the full circle of area 314 cm². Since 235.5+78.5=314, both sector answers are confirmed at once.
• Segment by subtraction: the minor segment is the minor sector minus the right triangle, that is 78.5-50=28.5 cm², and the major segment would be 314-28.5=285.5 cm² if asked.
• Why this works: whenever a circle is split into a sector and its triangle, the parts you compute should always reassemble the whole circle. That closure test is a fast, reliable check.

Minor segment 28.5 cm² and major sector 235.5 cm².

Keep the surd until the last line. The segment answer carries a square root, so handle that root with care to keep the answer accurate.

• Exact pieces: the arc and the sector are whole numbers, twenty two and two hundred thirty one, so neither of them needs any rounding and you can write both down with full confidence.
• Carry the root: keep the equilateral triangle area as the exact surd 44134 right up to the end, and only swap in 3≈ 1.732 at the very last moment to get about 190.96 cm².
• Subtract last: doing the subtraction 231-190.96 once gives 40.04 cm² cleanly. Rounding the root too early is the single biggest cause of a wrong segment area, so use the value of the root the question supplies, or leave it as a surd when no value is given.

Arc 22 cm, sector 231 cm², segment ≈ 40.04 cm².

Reuse the circle total. Both segments sit inside the same circle, so once you have the minor one the major one comes for free.

• Minor first: the small segment is sector minus triangle, which is 117.75-97.3125=20.4375 cm², and this is the only piece you actually have to compute from scratch.
• Major for free: the big segment is just the whole circle minus that small piece, 706.5-20.4375=686.0625 cm², so you avoid building a large sector and a second triangle.
• Built-in check: the two pieces add back to the full circle, since 20.4375+686.0625=706.5 cm². Reporting each one to two decimal places is enough for full marks here.

Minor segment 20.44 cm², major segment 686.06 cm².

Find the triangle without trigonometry. Many students would rather avoid the sine of an obtuse angle, and there is a neat way around it using only the basic right-angle ratios.

• Drop a perpendicular: let OM meet the chord at right angles. The half angle at the centre is then 60^∘, which gives OM=6 cm for the height and AM=63 cm for half the chord, both from familiar ratios.
• Area from base and height: the full chord is AB=123 cm, so the triangle is 12× AB× OM=363, the very same value the sine formula produces.
• Same answer: putting 3=1.73 turns that into 62.28 cm², and the segment is 150.72-62.28=88.44 cm². Pick whichever route you trust more; the perpendicular method needs only the simple thirty and sixty degree ratios.

88.44 cm².

Confirm the quarter fits inside the field. The plain quarter-circle formula is correct only while the rope is no longer than the side of the square, so always check that first.

• Short rope: with five metres the grazing patch is a quarter circle of 19.625 m², and since five is well under fifteen the whole patch sits inside the field.
• Longer rope: ten metres gives a quarter circle of 78.5 m², and as ten is still less than the fifteen metre side the arc never reaches a far edge.
• The increase: subtracting the two gives 78.5-19.625=58.875 m² of extra grass. Had the rope ever exceeded the side, the patch would spill over the boundary and you would have to subtract that overflow, so the check matters.

Grazing area 19.625 m²; increase 58.875 m².

Use the angle of each sector as a cross-check. Ten equal sectors mean each one carries the same central angle, and computing the area from that angle should match the one-tenth answer exactly.

• Find the angle: sharing the full turn among ten equal pieces gives thirty six degrees to each sector, which is the natural number to feed into the area formula here.
• Area by angle: the sector with that thirty six degree angle works out to 96.25 mm², and this lands on the very same figure as taking one-tenth of the whole circle.
• General rule: cutting any circle into equal pieces always produces equal angles and equal areas, so whether you reason by the angle or by the simple fraction you reach an identical number. Two routes agreeing is a quick, dependable way to confirm the sector area is right.

Wire 285 mm; each sector 96.25 mm².

One-eighth of the circle is faster. Rather than carry the angle fraction, take a straight one-eighth of the whole circle and the arithmetic stays much tidier.

• Whole circle: the full flat umbrella has area π r²=445507 cm², which is the single quantity every rib gap is measured against.
• One slice: eight equal ribs cut it into eight equal parts, so one gap is one-eighth of that, giving 2227528≈ 795.54 cm².
• Why it helps: reading the gap between two ribs as one eighth of the circle turns the worded problem into a single division and avoids slips with the heavy three sixty.

2227528 cm²≈ 795.54 cm².

Double first, then divide. Pulling the factor of two inside the sum early keeps every number on the page manageable.

• Fold in the two: since the wipers do not overlap, the total is twice one sector. Combining that two with the three sixty lets you work with 115180 of a single circle instead of a half circle doubled later.
• One multiplication: the top becomes 115× 22× 625 and the bottom 180× 7, so the whole job collapses into one fraction rather than two passes.
• Final figure: that fraction reduces to 158125126≈ 1254.96 cm², the same as the step method. Folding the doubling in at the start avoids handling a large single-sector decimal and then doubling it.

158125126 cm²≈ 1254.96 cm².

Reduce the fraction before multiplying. Cutting the angle fraction down to its lowest terms first keeps every number small.

• Simplify early: the angle fraction 80360 reduces to 29 once you divide top and bottom by forty, which is far easier to carry through.
• Scale the circle: multiplying the squared distance by pi gives 854.865, and a clean ninth of twice that is the sea area you want.
• Clean finish: the result is 189.97 km² to two decimals. Reducing the fraction before the heavy multiplication is a small habit that prevents large-number slips, and it matters most when the radius itself is a decimal like this one.

≈ 189.97 km².

Keep the cost step separate from the area step. Many students work the area out correctly and then forget the final step that turns it into money, so treat the two as separate jobs.

• One design: a single segment is the sector minus its equilateral triangle, which comes to 77.47 cm². This is the only design you actually have to compute by hand.
• All six: the table cover carries six identical designs, so the decorated region is six times that, or 464.8 cm² of worked silver thread in total.
• Convert to rupees: the question asks for cost, not area, so the last line multiplies the area by the rate to give 464.8× 0.35, that is about one hundred sixty two rupees. Stopping at the area would leave the answer incomplete and throw away the final mark, which is why this step is held apart.

Designs cover ≈ 464.8 cm² and cost ≈ Rs. 162.68.

Use a known case to test the options. Feed a full circle into each choice, since the right formula must hand back the whole circle area, and that screens the wrong ones out at a glance.

• Test value: the exact sector area is p360π R², and at the full turn it returns π R², exactly the whole circle as it should.
• Match option D: putting the full turn into option (D) also lands on π R², so that choice carries the very same value as the exact form.
• Read the key: options (A) and (C) contain a circumference rather than an area and fail at once. Among the printed keys the marked answer is (B), while the exact algebraic form lines up with (D). Substituting a value you already know is a reliable way to test any select-the-formula question.

Keyed answer (B); the exact area formula is p360×π R².