NCERT Exemplar Class 10 Maths Chapter 11 Exercise 11.3 Solutions

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.