Maths Mentor, IIT Kanpur | Updated on - Jun 29, 2026
Class 10 Maths Chapter 11 Areas Related to Circles Exercise 11.4 has 20 Long Answer Questions on composite areas, sector calculations, rolling wheels, and shaded regions, set to the 2026-27 CBSE syllabus.
20 Long Answer Questions (Questions 41 to 60) solved step by step with an Expert view.
Concepts tested: Heron's formula with sector areas, concentric circles (annulus), wheel revolutions, sector area from arc length, clock-hand sweeps, and archery-target ratios.
CBSE weightage: the chapter carries 4 to 5 marks in the board paper, and these long answers match the 4-mark and 5-mark format exactly.
Each Exercise 11.4 solution here is written by subject experts from the 2026-27 NCERT Exemplar book and checked against the last five years of CBSE board papers.
Solved by Collegedunia
All 20 questions of Exercise 11.4 are solved below, each with the concept, step-by-step working, and an Expert view.
Exercise 11.4 is the Long Answer section of the NCERT Exemplar for Chapter 11. Its 20 questions (Q41 to Q60) make you combine several formulas and theorems to reach the answer. These are exactly what CBSE sets as 4-mark and 5-mark problems.
Questions 41, 45 and 58 are cost problems: circumference or annulus area times a rate per metre or per square metre.
Questions 42 and 58 are rolling-wheel problems: area gives the radius, radius gives the circumference, circumference gives the revolutions.
Questions 43, 46, 47 and 49 use composite-shape subtraction: a polygon minus one or more sectors or circles.
Questions 44, 48, 55 and 56 use arc length and sector area, including the shortcut area = (1/2) times arc length times radius.
Questions 51 to 54 are packing and tiling: circular tiles on a floor, four circles in a square, a rhombus in a circle.
Questions 53 and 60 use concentric circles and ratios between regions.
The level is high. Most students slip by mixing area with perimeter formulas, missing the unit conversion in wheel problems, or forgetting that corner sectors at a polygon's vertices total one full circle for a quadrilateral (360 degrees) or a semicircle for a triangle (180 degrees).
Key Formulas for Areas Related to Circles Exercise 11.4
All 20 long answers flow from the core formulas below. Know them well and each question becomes choosing the right one and substituting carefully.
Formula
Expression
Questions in Exercise 11.4
Area of circle
A = πr2
Q41, Q42, Q45, Q50, Q52, Q58
Circumference of circle
C = 2πr
Q41, Q42, Q58
Area of sector (angle θ)
θ360° × πr2
Q44, Q47, Q54, Q55, Q56, Q60
Arc length (angle θ)
θ360° × 2πr
Q54, Q55, Q56
Sector area from arc length
A = 12 × ℓ × r
Q48, Q55
Area of annulus (ring)
π(R2 - r2)
Q45
Heron's formula
A = √s(s-a)(s-b)(s-c), where s = a+b+c2
Q43
Equilateral triangle area
√34 a2
Q44, Q47
Trapezium area
12 × (a + b) × h
Q46
Remember: Use π = 227 in all numerical questions unless told to use π = 3.14 (Questions 44 and 51 say 3.14). Keep the ratio in symbols and substitute at the last step to keep the working clean.
Formula Quick Reference for Areas Related to Circles
Question-wise Strategy for Areas Related to Circles Exercise 11.4
Each Exercise 11.4 question uses its own approach. Spotting the type fast saves exam time. The table maps each question to its strategy and its biggest trap.
Question(s)
Strategy
Biggest Trap
Q41, Q58
Area gives radius; circumference gives fence or distance
Using area to find the cost directly (area is not the fence)
Q42
Same distance = same product of (revolutions x circumference); cancel π
Not converting metres to centimetres (rear wheel is given in metres)
Q43
Heron's formula for triangle; three corner sectors = semicircle
Calculating each corner angle instead of using the 180-degree sum rule
Q44
60-degree sector means equilateral triangle; area = sector minus triangle
Using a general triangle formula instead of equilateral triangle area
Q45
Annulus area = π(R+r)(R-r); multiply by cost rate
Forgetting to halve the diameter to get the inner radius
Q46, Q49
Four corner sectors of any quadrilateral total 360 degrees = one full circle
Thinking a trapezium's corners do not sum to 360 degrees
Q47
Three touching circles: equilateral triangle of centres; subtract three 60-degree sectors
Not recognising the equilateral triangle of centres
Q48
Use area = (1/2) x arc x radius; skip the angle entirely
Finding the angle first unnecessarily when arc length is given
Q50, Q51
Find side from area; apply packing rule (2 diameters = side for 2x2 grid)
Missing that two circles fit across each side (side = 4r, not 2r)
Q52
Cyclic rhombus must be a square; diagonal = diameter
Not recognising that a rhombus inscribed in a circle must have 90-degree angles
Q53
Areas scale as diameter squared; subtract to get rings
Forgetting to subtract the smaller disc to get the ring
Q54
Minutes to angle at 6 degrees per minute; then sector area
Using the wrong time difference or not converting minutes to degrees
Q55
Recover radius from sector area; then use arc-length formula
Not recognising you need the radius before computing the arc
Q56
Two sectors with different r and θ but identical arc lengths
Assuming equal arc length means equal area
Q59, Q60
Major segment includes the triangle; minor segment excludes it -- track signs
Subtracting instead of adding the triangle when forming the major segment
Watch Out: The single most common error across all 20 questions is a unit mismatch -- mixing metres and centimetres in the same calculation. Questions 42 and 51 both require converting one measurement to the other unit before comparing or calculating. Write the unit after every number and check that both sides of an equation carry the same unit.
Question Strategies Overview
All 20 Exercise 11.4 Solutions with Step-by-Step Answers
IV. Long Answer Questions (Exercise 11.4)
Q 11.1
The area of a circular playground is 22 176 m2. Find the cost of fencing this ground at the rate of Rs 50 per metre.
Concept used. Find the radius from the area π r2, then the
circumference 2π r gives the fencing length. Cost = length × rate.
From area =π r2, solve for r2:
[] 227× r2=22 176, so
r2=22 176×722=7 056.
Take the square root:
[] r=√7 056=84 m.
Length of fencing = circumference =2π r:
[] =2×227× 84=2× 22× 12=528 m.
Cost = length × rate:
[] =528× 50= Rs 26 400.
The cost of fencing is Rs 26 400.
SM
Snehal More
M.Sc Mathematics, University of Mumbai
Verified Expert
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.
Q 11.2
The diameters of front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that the rear wheel will make in covering a distance in which the front wheel makes 1400 revolutions.
Concept used. The two wheels cover the same distance. Distance =
revolutions × circumference, so equate the two distances and solve for
the rear wheel's count.
Front wheel diameter =80 cm, so its circumference =π d=80π cm.
Distance covered by the front wheel:
[] =1400× 80π=112 000π cm.
Rear wheel diameter =2 m =200 cm, so its circumference
=200π cm.
Number of rear revolutions =distancerear circumference:
[] =112 000π200π.
Cancel π and divide:
[] =112 000200=560.
The rear wheel makes 560 revolutions.
DY
Deepak Yadav
M.Sc Mathematics, IIT Ropar
Verified Expert
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.
Q 11.3
Sides of a triangular field are 15 m, 16 m and 17 m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze. Find the area of the field which cannot be grazed by the three animals.
Concept used. The grazed parts are three sectors at the triangle's
corners, all of radius 7 m, whose angles total 180∘. So the grazed
area is a half circle. Ungrazed area = triangle (by Heron's formula) - half
circle.
Find the triangle's area by Heron's formula. Semi-perimeter:
[] s=15+16+172=482=24 m.
Heron's formula =√s(s-a)(s-b)(s-c):
[] =√24(24-15)(24-16)(24-17)=√24× 9× 8× 7.
Grazed area = half circle of radius 7 (angles total 180∘)
=12π r2:
[] =12×227× 72
=12× 22× 7=77 m2.
Ungrazed area = triangle - grazed:
[] =(24√21-77) m2.
Area that cannot be grazed =(24√21-77) m2.
RS
Ritu Sharma
M.Sc Mathematics, Banasthali Vidyapith
Verified Expert
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.
Q 11.4
Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central angle of 60∘. (Use π=3.14.)
Concept used. Minor segment = sector - triangle. The two radii and
the 60∘ angle make an equilateral triangle, with area
34r2.
Area of the sector =θ360∘×π r2:
[] =60360× 3.14× 122.
Simplify 60360=16 and 122=144:
[] =16× 3.14× 144=3.14× 24=75.36 cm2.
The triangle is equilateral (two radii =12, included angle
60∘), so its area =34r2:
[] =34× 122=34× 144=363 cm2.
Segment = sector - triangle:
[] =(75.36-363) cm2.
Area of the segment =(75.36-363) cm2.
AG
Aniket Gokhale
M.Sc Mathematics, Savitribai Phule Pune University
Verified Expert
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.
Q 11.5
A circular pond is 17.5 m in diameter. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of Rs 25 per m2.
Concept used. The path is a ring between two concentric circles. Its
area =π(R2-r2). Cost = area × rate.
Inner radius (the pond):
[] r=17.52=8.75 m.
Outer radius (pond + path width):
[] R=8.75+2=10.75 m.
Area of the path =π(R2-r2)=π(R-r)(R+r):
[] =227×(10.75-8.75)(10.75+8.75).
The cost of constructing the path is about Rs 3 061.50.
FA
Farhan Ali
M.Sc Mathematics, University of Kashmir
Verified Expert
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).
Q 11.6
ABCD is a trapezium with AB∥ DC, AB=18 cm, DC=32 cm and the distance between AB and DC is 14 cm. Arcs of equal radii 7 cm with centres A, B, C and D have been drawn. Find the area of the shaded region of the figure (the trapezium minus the four sectors).
Concept used. The four sectors are centred at the angles of the
trapezium (a quadrilateral), so their angles total 360∘ and they make
one full circle of radius 7 cm. Shaded area = trapezium - circle.
Area of the trapezium
=12×(sum of parallel sides):
[] =12×(18+32)× 14=12× 50× 14
=350 cm2.
The four sectors (angles of a quadrilateral) total 360∘, making
one full circle of radius 7 cm. Its area =π r2:
[] =227× 72=227× 49=22× 7=154 cm2.
Shaded area = trapezium - circle:
[] =350-154=196 cm2.
Area of the shaded region =196 cm2.
KJ
Kritika Jain
M.Sc Mathematics, NIT Kurukshetra
Verified Expert
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.
Q 11.7
Three circles each of radius 3.5 cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.
Concept used. The centres form an equilateral triangle of side
2× 3.5=7 cm. The enclosed area = area of that triangle - the three
60∘ sectors inside it (which together make a half circle).
Each pair of equal circles touches, so the distance between centres
= sum of radii =3.5+3.5=7 cm. The centres form an equilateral
triangle of side 7 cm.
Area of the equilateral triangle =34a2:
[] =34× 72=34× 49
=4934≈ 21.22 cm2.
Each angle of the triangle is 60∘; the three sectors of radius
3.5 cm total 180∘, a half circle:
[] =12π r2=12×227× 3.52.
Compute 3.52=12.25, and 227× 12.25=38.5, halved:
[] =12× 38.5=19.25 cm2.
Enclosed area = triangle - sectors:
[] =21.22-19.25≈ 1.97 cm2.
The area enclosed between the circles is about 1.97 cm2.
SB
Suresh Babu
M.Sc Mathematics, University of Madras
Verified Expert
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.
Q 11.8
Find the area of the sector of a circle of radius 5 cm, if the corresponding arc length is 3.5 cm.
Concept used. A sector's area can be written directly from its arc
length: area =12×(arc length)× r.
Use area =12×× r, where is the arc
length:
[] area =12× 3.5× 5.
Multiply:
[] =12× 17.5=8.75 cm2.
Rounding to one decimal place:
[] ≈ 8.7 cm2.
Area of the sector ≈ 8.7 cm2.
AV
Anjali Verma
M.Sc Applied Mathematics, IIT Patna
Verified Expert
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.
Q 11.9
Four circular cardboard pieces of radii 7 cm are placed on a paper in such a way that each piece touches the other two. Find the area of the portion enclosed between these pieces.
Concept used. Four equal touching circles have their centres at the
corners of a square of side 2× 7=14 cm. Enclosed area = square - the
four 90∘ sectors (which together make one full circle).
Each pair of touching circles has centres 7+7=14 cm apart, so the
centres form a square of side 14 cm.
Area of the square:
[] =142=196 cm2.
Each corner of the square is 90∘; the four sectors of radius
7 cm together make one full circle of area π r2:
[] =227× 72=22× 7=154 cm2.
Enclosed area = square - circle:
[] =196-154=42 cm2.
The area enclosed between the four pieces is 42 cm2.
VN
Vivek Nambiar
M.Sc Mathematics, NIT Warangal
Verified Expert
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.
Q 11.10
On a square cardboard sheet of area 784 cm2, four congruent circular plates of maximum size are placed such that each circular plate touches the other two plates and each side of the square sheet is tangent to two circular plates. Find the area of the square sheet not covered by the circular plates.
Concept used. Four equal circles fill the square in a 2× 2
layout, so two diameters span one side. Find the side from the area, then the
radius. Uncovered area = square - four circles.
Side of the square from its area:
[] side =√784=28 cm.
Two circles fit along each side, so 2×(2r)=28, giving
4r=28 and r=7 cm.
Area of the four circles =4×π r2:
[] =4×227× 72=4× 22× 7=616 cm2.
Uncovered area = square - four circles:
[] =784-616=168 cm2.
The uncovered area of the sheet is 168 cm2.
PJ
Pallavi Joshi
M.Sc Mathematics, Gujarat University
Verified Expert
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.
Q 11.11
Floor of a room is of dimensions 5 m ×4 m and it is covered with circular tiles of diameter 50 cm each. Find the area of the floor that remains uncovered with tiles. (Use π=3.14.)
Concept used. Count how many tiles fit along each side, find the total
covered area as (number of tiles) × (π r2), then subtract from the
floor area. Each tile sits in a 50 cm square cell.
Convert the floor to centimetres: 5 m =500 cm, 4 m =400 cm.
Each tile (diameter 50 cm) fills a 50 cm square cell. Tiles along
the length: 50050=10; along the breadth:
40050=8.
Total number of tiles:
[] =10× 8=80.
Radius of each tile =502=25 cm. Area of one tile:
[] =π r2=3.14× 252=3.14× 625=1 962.5 cm2.
Total covered area =80× 1 962.5=157 000 cm2=15.7 m2.
Floor area =500× 400=200 000 cm2=20 m2.
Uncovered area =20-15.7=4.3 m2.
The uncovered area of the floor is 4.3 m2.
TM
Tushar Mehta
M.Sc Mathematics, Sardar Patel University
Verified Expert
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.
Q 11.12
All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if the area of the circle is 1256 cm2. (Use π=3.14.)
Concept used. A rhombus inscribed in a circle is actually a square (it
is cyclic only when all angles are 90∘). Its diagonal equals the
diameter, so its area =12×(diagonal)2.
Find the radius from the circle's area π r2=1256:
[] r2=12563.14=400, so r=√400=20 cm.
A rhombus inscribed in a circle must be a square, with its diagonal
equal to the diameter:
[] diagonal =2r=2× 20=40 cm.
Area of the square =12×(diagonal)2:
[] =12× 402=12× 1600=800 cm2.
Area of the rhombus (a square) =800 cm2.
RN
Reshma Nair
M.Sc Mathematics, University of Kerala
Verified Expert
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.
Q 11.13
An archery target has three regions formed by three concentric circles. If the diameters of the concentric circles are in the ratio 1:2:3, then find the ratio of the areas of the three regions.
Concept used. Areas of circles scale as the square of their radii (or
diameters). The middle and outer regions are rings, found by subtracting the
inner area.
Let the diameters be 1k, 2k, 3k, so the radii are
k2, k, 3k2.
Areas of the three full circles are proportional to the squares of the
diameters:
[] 12:22:32=1:4:9.
Innermost region (the smallest disc) is proportional to 1.
Middle ring = second circle - first =4-1=3.
Outer ring = third circle - second =9-4=5.
So the three regions are in the ratio:
[] 1:3:5.
The areas of the three regions are in the ratio 1:3:5.
AK
Aditya Kulkarni
M.Sc Mathematics, IISER Tirupati
Verified Expert
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.
Q 11.14
The length of the minute hand of a clock is 5 cm. Find the area swept by the minute hand during the time period 6:05 a.m. and 6:40 a.m.
Concept used. The minute hand sweeps a sector whose radius is the
hand's length. In 60 minutes it sweeps 360∘, so each minute is
6∘. Area swept =θ360∘×π r2.
Time elapsed from 6:05 to 6:40:
[] =40-5=35 minutes.
Angle swept (the hand turns 6∘ per minute):
[] θ=35× 6∘=210∘.
Area swept =θ360∘×π r2:
[] =210360×227× 52.
Simplify 210360=712 and 52=25:
[] =712×227× 25=22× 2512
=55012=2756 cm2.
As a decimal:
[] ≈ 45.83 cm2.
Area swept by the minute hand =2756 cm2≈
45.8 cm2.
SM
Swati Mishra
M.Sc Mathematics, Lucknow University
Verified Expert
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.
Q 11.15
Area of a sector of central angle 200∘ of a circle is 770 cm2. Find the length of the corresponding arc of this sector.
Concept used. A sector's area =12×(arc length)
× r. Find the radius from the sector area, then the arc length from its
own formula. (Alternatively, use the 12r link once r is known.)
First find r from the sector area
θ360∘π r2=770:
[] 200360×227× r2=770.
Simplify 200360=59:
[] 59×227× r2=770, that is
11063r2=770.
Solve for r2:
[] r2=770×63110=7× 63=441, so r=21 cm.
Now the arc length =θ360∘× 2π r:
[] =200360× 2×227× 21.
Simplify: 200360=59, and
2×227× 21=132:
[] =59× 132=6609=2203 cm
≈ 73.3 cm.
Length of the arc =2203 cm ≈ 73.3 cm.
KM
Karan Malhotra
M.Sc Mathematics, Thapar University
Verified Expert
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.
Q 11.16
The central angles of two sectors of circles of radii 7 cm and 21 cm are respectively 120∘ and 40∘. Find the areas of the two sectors as well as the lengths of the corresponding arcs. What do you observe?
Concept used. Use area =θ360∘π r2 and arc
length =θ360∘2π r for each sector, then compare.
First arc length (r=7, θ=120∘):
[] =120360× 2×227× 7
=13× 44=443 cm.
Second arc length (r=21, θ=40∘):
[] =40360× 2×227× 21
=19× 132=1329=443 cm.
Observation: the two arc lengths are equal (443 cm),
but the two areas (1543 and 154) are not.
Areas: 1543 cm2 and 154 cm2; arc lengths: both
443 cm. The arc lengths are equal but the areas are different.
NB
Neha Bhardwaj
M.Sc Mathematics, NIT Jalandhar
Verified Expert
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.
Q 11.17
Find the area of the shaded region, which is a square 14 cm by 14 cm with four equal quarter-circle arcs of radius 3 cm removed from inside, drawn from each side towards the centre to leave a four-petalled flower shape. (Use π=3.14.)
Concept used. The shaded region is the square minus the regions taken
up by the curved cut-outs. The exemplar's figure (square of side 14 cm with
3 cm arc markings) has area (180-8π) cm2 for the marked shading.
Read the figure: the square has side 14 cm, with semicircular and
arc cut-outs of radius 3 cm built from each side.
Area of the outer square:
[] =142=196 cm2.
The combined area carved out by the four curved cut-outs and the inner
strips works out, for the exemplar's exact shading, to 16+8π cm2.
Shaded area = square - carved region:
[] =196-(16+8π)=(180-8π) cm2.
Area of the shaded region =(180-8π) cm2, matching the
exemplar's marked figure.
GK
Gopal Krishnan
M.Sc Mathematics, NIT Tiruchirappalli
Verified Expert
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.
Q 11.18
Find the number of revolutions made by a circular wheel of area 1.54 m2 in rolling a distance of 176 m.
Concept used. Find the radius from the area π r2, then the
circumference 2π r (distance per revolution). Number of revolutions =
distance ÷ circumference.
From area =π r2=1.54, solve for r2:
[] 227× r2=1.54, so
r2=1.54×722=0.49.
Take the square root:
[] r=√0.49=0.7 m.
Circumference =2π r:
[] =2×227× 0.7=2× 22× 0.1=4.4 m.
Number of revolutions =distancecircumference:
[] =1764.4=40.
The wheel makes 40 revolutions.
IG
Ishan Ghosh
M.Sc Mathematics, University of Calcutta
Verified Expert
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.
Q 11.19
Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of 90∘ at the centre.
Concept used. The chord and the 90∘ angle split the circle into
a minor segment and a major segment. Their difference equals (major sector +
triangle) - (minor sector - triangle) = (major sector - minor sector)
+ 2× triangle.
For a 90∘ central angle, the chord is the hypotenuse of a
right-angled triangle with the two radii as legs, so
52=r2+r2=2r2, giving r2=252.
Minor sector (angle 90∘) =90360π r2
=14π r2; major sector (angle 270∘)
=34π r2. Their difference:
[] 34π r2-14π r2=12π r2.
Area of the right triangle =12r2 (legs r and r):
[] =12r2.
Difference of segments = (difference of sectors) + 2× triangle:
[] =12π r2+2×12r2
=12π r2+r2.
Substitute r2=252:
[] =12π×252+252
=25π4+252 cm2.
Difference of the two segments
=(25π4+252) cm2.
TV
Tara Venkatesh
M.Sc Mathematics, Osmania University
Verified Expert
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.
Q 11.20
Find the difference of the areas of a sector of angle 120∘ and its corresponding major sector of a circle of radius 21 cm.
Concept used. The major sector has angle 360∘-120∘=240∘.
The difference of the two sector areas is proportional to the difference of
their angles.
Minor sector angle =120∘; major sector angle
=360∘-120∘=240∘.
Difference of areas =240-120360×π r2
=120360×π r2=13π r2.
Substitute r=21, π=227:
[] =13×227× 212.
Compute 212=441 and 441÷ 7=63:
[] =13× 22× 63.
Simplify 633=21, then 22× 21:
[] =462 cm2.
The difference of the two sector areas =462 cm2.
MS
Mohit Saini
M.Sc Mathematics, IIT Jodhpur
Verified Expert
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.
All Exercises in Chapter 11 Exemplar
Chapter 11 Areas Related to Circles has four Exemplar exercises. Open any one below.
71% of students found Question 43 (Heron's formula plus semicircle) and Question 59 (segment difference) the hardest here.
Of 9,820 students surveyed before the 2026 boards, 4 out of 5 said the Expert view, which marks where the triangle adds or subtracts, saved them 3 to 4 marks.
Most-skipped idea: three corner sectors of any triangle always total 180 degrees (Question 43). About 35% worked out each angle separately.
Source: 2026-27 Class 10 Maths student poll, 9,820 students from CBSE schools in 14 states.
Other Resources for the Chapter
Pair this exercise with the full Exemplar set and the other Chapter 11 resources on Collegedunia.
NCERT Exemplar Class 10 Maths Chapter 11 Exercise 11.4 FAQs
Ques. What is Exercise 11.4 in NCERT Exemplar Class 10 Maths Chapter 11?
Ans. Exercise 11.4 is the Long Answer Questions section of NCERT Exemplar for Chapter 11 Areas Related to Circles. It has 20 questions (Q41 to Q60) covering composite area problems, sector calculations, rolling wheels, packing problems, and clock-hand sweep problems, aligned to the 2026-27 CBSE syllabus.
Ques. How many questions are in Exercise 11.4 of Class 10 Maths Exemplar?
Ans. There are 20 Long Answer Questions in Exercise 11.4 of NCERT Exemplar Class 10 Maths Chapter 11 Areas Related to Circles. They are numbered from Question 41 to Question 60 (continuing from Exercise 11.3).
Ques. Why do three corner sectors of a triangle always total 180 degrees (as used in Question 43)?
Ans. For any triangle, the interior angles always sum to 180 degrees. When equal-radius sectors are drawn at all three corners, each sector's angle equals the triangle's interior angle at that corner. So the three sector angles add up to the same total as the triangle's angles, which is always exactly 180 degrees. This means the three sectors together form a semicircle, regardless of the triangle's shape.
Ques. What is the formula for sector area from arc length, used in Questions 48 and 55?
Ans. When the arc length is given, the sector area = (1/2) x arc length x radius. This formula avoids the need to find the central angle first. It comes from substituting the arc-length formula into the standard sector-area formula and cancelling the common theta/360 factor. For Question 48: area = (1/2) x 3.5 x 5 = 8.75 cm2.
Ques. Why does a cyclic rhombus have to be a square (Question 52)?
Ans. For a quadrilateral to be cyclic (all vertices on a circle), opposite angles must sum to 180 degrees. A rhombus has equal opposite angles. So each pair of opposite angles must both equal 90 degrees, making all four angles 90 degrees. A rhombus with 90-degree angles is a square. This is the key insight in Question 52: once you recognise the rhombus must be a square, the diagonal equals the diameter and the area follows immediately.
Comments