Area of Plane Figures 193
With the advancement of civilization on this planet earth, the human beings were faced with
the following types of things in their immediate environment : part of the earth on which they
live (plane), fruit and vegetables trees, birds and animals, stone crests and other solids or plane
objects. The desire to know more about them, to have a measure of sizes and capacity and
to acquire them was a natural phenomenon. With this came into being the science of
measurement of land and knowledge about rectilinear and solid figures, which was given the
name of mensuration.
Old civilization like Greek, Egyptians, Arabian and especially Indian contributed a lot to the
development of this science. In the old Indian scriptures, there are rules laid down to perform
special types of yajnas to fulfil specific desires in specially built altars (vedis) These vedis were
in the shape of squares, rectangles, and circles made of cuboidal, cubical and spherical bricks.
The whole lot of “Sulba Sutras” and Srauta Sutras contain the knowledge of methods of
construction of such vedis and the properties of the used shapes.
“Baudhayan Sulba” Apastamba Sulba and Katyayan Sulba which date back to 8th century B.C.
are famous among all these sutras. After that even the Indian mathematician like Aryabhatt
(476 AD), Brahmgupta (598 AD), Bhaskaracharya (1114 AD) and Varahmihir (1120 AD)
devoted much of their works in providing rules for finding perimeters and areas of plane figures
and volumes of solids. One of the oldest text “Surya Sidhant” deals with the shapes (mostly
spherical) and orbits (mainly elliptical) of heavenly bodies.
In this text, we shall study about the perimeter and area of plane figures and surface area, total
surface area and volumes of solids.
Area of Plane Figures
In our day-to-day life, we have to estimate the amount of wood required for a rectangular table
top or the area of cloth required for table cover, or to find the area of paths, parks, plots and
In this lesson, we shall study the methods of finding perimeters and areas of plane figures,
especially square, rectangle, triangle, special quadrilaterals and circle. We will also study
methods to find areas of rectangular or circular paths.
After studying this lesson the learner will be able to :
identify rectilinear and non-rectilinear plane figures.
explain the meaning of perimeter and area of closed figures.
find the perimeter and area of rectilinear figures, like a square, a rectangle, a triangle,
a quadrilateral, a trapezium, a parallelogram and a rhombus.
find the area of a triangle using Hero’s formula.
find the area of rectilinear paths of different types in a rectangle (or a rectangular
find the circumference and area of a circle.
find area and perimeter of a sector.
find the area of circular paths inside or outside a circle (or a circular enclosure).
Solve problems from day-to-day life situations based on the above concepts
23.3 EXPECTED BACKGROUND KNOWLEDGE
Measurement of line-segments
Knowledge and conversion of units of measurements
Area of Plane Figures 195
Four fundamental operations on numbers
Drawing plane geometric figures
Properties of special quadrilaterals
23.4 PERIMETER OF PLANE FIGURES
The distance covered to walk along the boundary of a plane figure is called its perimeter. In
connection with the circle, this distance covered is called circumference of the circle. The
perimeter is measured in terms of linear units.
23.5 AREA OF PLANE FIGURES
The measure of the planar region enclosed by a plane figure is called its area.
This is measured in terms of square units.
23.5.1 A Unit Square
A square of side 1 cm is called a unit cm square and its area is taken to be 1 cm2 and its perimeter
is taken to be 4 cm.
Similarly a square of side 1 m is called a unit meter square and its area is taken to be 1 m2
and its perimeter as 4 m.
In general, area of a square of side ‘a’ unit is a2 sq. unit and its perimeter is 4a
Example 23.1 : Find the area of a square whose perimeter is 80 meters.
Solution : Perimeter of square = 4a = 80 meters
∴ a= or 20 metres
∴ Area of square = a2 sq. m
= 20 × 20 sq. m or 400 sq m.
Example 23.2. The area of a square park is 625 sq m. Find the length of the wire required
to fence around the park.
Solution. Area of square park (a2) = 625 sq. m
∴ Side of square (a) = 625 m = 25 × 25 m or 25 m
Now, length of wire = Perimeter of square
= 4a = 4 × 25 m = 100 m
i.e. length of the wire required to fence the park is 100 m.
Example 23.3. Find the area of a square whose diagonal is 15 m.
Solution. Diagonal of the square ABCD = AC = AB2 + BC 2
= a 2 + a 2 or 2a m
∴ 2 a = 15 or a = 2
Area of the square ABCD = a2 sq m = or 112.5 sq m.
23.6. RECTILINEAR FIGURES
The figures whose boundaries are formed by line-segments are called rectilinear figures. For
example square, rectangle, triangle, trapezium, quadrilateral, parallelogram and rhombus are
rectilinear figures while a circle and a sector are non-rectilinear figures.
23.7 PERIMETER AND AREA OF A RECTANGLE
Take 4 centimetre squares and join them in a row as shown in Fig. 23.3(i). Add another row
of four centimetre square over the and above the first row to form the rectangle ABCD as shown
in Fig 23.3(ii).
Area of Plane Figures 197
Rectangle ABCD is formed by 8 small unit squares. Therefore the planar region enclosed by
it is 8 cm2.
You can see that 4 × 2 = 8.
If we take another rectangle PQRS with base
formed by 5 unit squares and has four such rows
(See Fig. 23.4), we see that it contains 20 or
(5 × 4) unit cm squares.
∴ Its area is 20 cm2
i.e. (5 × 4) cm2
or 20 cm2 Fig. 23.4
From here we can generalise that the area of a rectangle of length l cm and breadth b cm can
be written as (l × b) cm2 or lb cm2.
In case of Fig 23.3(ii), the distance covered to go around the rectangle ABCD is
(4 + 2 + 4 + 2) cm or 12 cm.
Similarly, the distance covered to go around the rectangle PQRS in Fig. 23.4 is
(5 + 4 + 5 + 4) cm or 18 cm.
Do you observe that in each case the distance covered is obtained by adding 2 × length and
2 × breadth?
i.e. Perimeter of a Rectangle = 2(l + b)
A square is a special rectangle in which length and breadth are equal.
Thus, the area of a
square = (side)2 square units
rectangle = (length × breadth) square units
And, the perimeter of a
square = 4 × side linear units
rectangle = 2 (length + breadth) linear units
Example 23.4 : The length and breadth of a rectangular field are 23.7 m and 14.5 m respecively.
(i) the area of the field
(ii) the length of wire required to put a fence around the boundary of the field.
Solution : (i) Area of the field = Length × breadth
= (23.7 × 14.5) sq m
= 343.65 sq m
(ii) Length of wire = Perimeter of the field
= 2(length + breadth)
= 2(23.7 + 14.5) m
= 2(38.2) or 76.4 m
Example 23.5 : A rectangular plot measures 400 m × 121 m. Find the side of a square plot
having area equal to the rectangular plot.
Solution : Area of rectangular plot = (400 × 121) sq m = (20)2 × 112 sq m
∴ Area of square plot = (20)2 × 112 sq m
∴ Side of square plot = 202 × 112 m
= 220 m.
23.8. AREA OF A PARALLELOGRAM
We know from our study of geometry that
parallelograms on the same base and between the
same parallels are equal in area.
∴ Area (||gm ABCD) = Area Rect. PBCQ
= (BC × PB) sq units.
∴ Area of a Parallelogram = Base × Altitude square units
Example 23.6 : Find the area of a parallelogram whose base and altitude are 12 cm and
8 cm respectively.
Solution : We know that Area of ||gm ABCD = Base × altitude
= (12 × 8) cm2
= 96 cm2.
23.9. AREA OF A TRIANGLE
We know that a diagonal of a parallelogram divides it into two triangles of equal area
∴ Area of ∆ DBC = Area of ||gm ABCD
= Base × Altitude
∴ Area of a triangle = Base × Altitude square units
Area of Plane Figures 199
Example 23.7 : The base of a triangular field is three times its altitude. If the cost of cultivating
the field at the rate of Rs 15 per square metre is Rs 20250, find the base and the altitude of
Solution : Area of the field = sq m
= 1350 sq m
Let the altitude of the triangular field be x m
∴ Base = 3x m
∴ Area of the field = .3x.x = x 2 sq m
As given in the question,
1350 × 2
or x2 = = 900
or x = 30
Thus, the base of the field is 90 m and its altitude is 30 m.
23.10 AREA OF A TRAPEZIUM
In Fig. 23.7, ABCD is a trapezium in which AB || DC.
∴ Area of trapezium ABCD
= Area of ∆ ABD + Area ∆ BCD
= .a.h + .b.h
= (a + b).h
2 Fig. 23.7
Thus, the area of a trapezium
= (sum of the parallel sides) × Distance between them square units
Example 23.8 : Find the area of a trapezium lengths of whose parallel sides are 20 cm and
12 cm and the distance between them is 5 cm.
Solution : We know that
Area of Trapezium = (Sum of the Parallel sides) × Distance between them
= [20 + 12] × 5 cm2
= × 32 × 5 cm2
= 80 cm2.
23.11 AREA OF A QUADRILATERAL
Area of quadrilateral ABCD
= Area ∆ ABC + Area ∆ ACD
(See Fig. 23.8)
1 AC. h + 1 AC. h
= 1 2 2
1 AC h + h
1 2 , g
where h1 and h2 are lengths of perpendiculars
from vertices B and D on diagonal AC Fig. 23.8
Example 23.9 : Find the area of a field in the shape of a quadrilateral, one of whose diagonals
is 40 m long and the perpendiculars from the other two vertices on this diagonal are 32 m
and 18 m.
Solution : Area of quadrilateral field ABCD
= Area of ∆ ABC + Area of ∆ ACD
= 2 × 40 × 32 + 2 × 40 × 18 sq m
= (640 + 360) sq m = 1000 sq m.
23.12 AREA OF RHOMBUS
In Fig. 23.10, ABCD is a rhombus whose diagonal AC and BD bisect each other at right angles.
Thus : Area of rhombus ABCD = Area ∆ ABD + Area ∆ BCD
1 BD. AP + 1 BD. PC
= BD (AP + PC) = (BD) (AC)
Thus, the area of rhombus
= × product of its diagonals square units.
Area of Plane Figures 201
Example 23.10. The diagonals of a rhombus are of length 12 cm and 8 cm. Find the area of
Solution. We know that
Area of a Rhombus = × Product of Diagonals
= (12 × 8) cm2
= 48 cm2
Example 23.11. Find the area of a rhombus, one of whose diagonals is 12 cm and the side
is 10 cm.
Solution : In Fig.23.11, let ABCD be the given rhombus with AB = BC = CD = AD = 10 cm
and the diagonal BD = 12 cm.
AC and BD bisect (at right angles) at O (See Fig. 23.11)
∴ OB = OD =6 cm
∴ OA = 102 − 62 cm = 8 cm
∴ Diagonal AC = 16 cm
Area of rhombus = (product of diagonals)
1 Fig. 23.11
= × 12 × 16 cm2
= 96 cm2
23.13 AREA OF A TRIANGLE USING HERO’S FORMULA
The area of triangle ABC whose sides are given by a, b and c is given by
∆ = b gb gb g
s s− a s− b s− c
where s =
This formula is known as Hero’s formula after the name of Greek Mathematician Heron of
Alexendria. The formula was also obtained by the Indian Mathematicians Brahmagupta and
Example 23.12 : The sides of a triangular field are 165 m, 143 m and 154 m. Find the area
of the field.
Solution : Here we use the Hero’s formula for finding the area of the field
∆ = b gb gb g
s s− a s− b s− c
where a, b, c are sides and s =
165 + 143 + 154 = 462
∴ s= or 231 m
∴ ∆ = b gb gb
231 231 − 165 231 − 143 231 − 154 sq mg
= 231b66gb88gb77g sq m
= 11 × 7 × 3 × 3 × 2 × 11 × 2 × 2 × 2 × 11 × 7 × 11 sq m
= 11 × 11 × 7 × 3 × 2 × 2 = 10164 sq m.
Thus area of the triangular field = 10164 sq m
Example 23.13 : Find the area of a field in the shape of a trapezium whose parallel sides are
of length 11 m and 25 m and the non-parallel sides are of length 15 m and 13 m.
Solution : Let ABCD be a field in the shape of a trapezium with AB || CD in which
AB = 25 m, CD = 11 m, AD = 13 m and BC = 15 m [Fig. 23.12]
Draw CE || DA and let CF be perpendicular to AB
Area of ∆ BCE = b gb gb
21 21 − 14 21 − 15 21 − 13 g
13 + 14 + 15
because s = or 21 m
= 21 × 7 × 6 × 8 or 84 sq m
Also, area of ∆ BCE = b g
1 14 .CF
= 7 CF
∴ 7 CF = 84 or CF = 12 m Fig. 23.12
∴ Area of trapezoidal field ABCD
= (sum of parallel sides) × distance between them
= (11 + 25) × 12 sq m
= 216 sq m
23.14 AREA OF RECTANGULAR PATHS
In Fig. 23.13, ABCD is a rectangular park of dimensions a × b and let there be a path of uniform
width c all around the park as shown
Area of Plane Figures 203
∴ Area of path = Area of Rectangle PQRS –
Area of rectangle ABCD
= (PQ × QR) – (AB × BC)
= [(a + 2c). (b + 2c)] – (a × b)
= ab + 2bc + 2ac + 4c2 – ab
= 4c2 + 2c(a + b) Fig. 23.13
In Fig. 23.14, ABCD is a rectangular park of dimensions a × b, in which PQRS and EFGH
are two perpendicular paths of width c, parallel to the sides of the park.
Area of paths = Area of path EFGH + Area of PQRS – Area of path QLMN
= a.c + b.c – c.c
= (a + b – c)c
Let us take some examples to illustrate :
Example 23.14 : A rectangular hall 15 m long and
12 m broad is surrounded by a verandah 2 m wide.
Find the area of verandah.
Solution : Dimension of rectangular enclosure PQRS
are (15 + 2 × 2) m and (12 + 2 × 2)m or 19 m and
From Fig. 23.15, we see that Fig. 23.15
Area of Verandah = Area of rectangle PQRS – Area of rectangle ABCD
= (19 × 16 – 15 × 12) sq m
= (304 – 180) sq m or 124 sq m
Example 23.15 : A rectangular piece of land
measures 100 m by 80 m. From the centre of
each side a path 5 m wide runs across up to the
opposite side. Find the area of the paths.
Solution : Area of paths = Area of rectangle
ABCD + Area of
rectangle PQRS –
Area of square
= [(100 × 5) +(80 × 5) – (5 × 5)] sq m
= [500 + 400 – 25] sq m
= 875 sq m
CHECK YOUR PROGRESS 23.1
1. The area of a square field is 225 square metres. Find its perimeter.
2. What will be the diagonal of a square whose perimeter is 60 m?
3. The length and breadth of a rectangular field are 22.5 m and 12.5 m. Find
(i) the area of the field, (ii) the length of the wire required to put a fence along the
boundary of the field.
4. The sides of a rectangular field of area 726 sq metres are in the ratio
3 : 2. Find the length of its diagonal correct up to one decimal place.
5. Find the area of a triangular field whose sides are 50 m, 78 m and 112 m. Also, find the
length of the perpendicular from the opposite vertex to the side measuring 112 m.
6. Find the area of a field in the form of a quadrilateral ABCD in which
AB = 165 m, BC = 143 m, CD = 85 m, AD = 85 m and the diagonal AC = 154 m.
7. Find the area of a parallelogram with base and altitude of length 20 cm and 12 cm
8. The parallel sides of a trapezium are 20 metres and 16 metres long respectively and the
distance between them is 12 m. Find its area.
9. The perimeter of a rhombus is 146 cm and one of its diagonals is 48 cm, find the other
diagonal and the area of the rhombus.
10. Find the area of a quadrilateral one of whose diagonals is 30 metres long and the lengths
of perpendiculars from the other two vertices to this diagonal are 10 m and 14 m
Area of Plane Figures 205
11. A rectangular courtyard 120 m long and 90 m broad has a path of uniform width 5 m
on the inside running round it. Find the area of the path.
12. A rectangular lawn 80 metres by 60 metres has two roads, each 2 metres wide running
in the middle of it, one parallel to the length and the other parallel to the breadth. Find
the area of the roads.
23.15 CIRCUMFERENCE AND AREA OF A CIRCLE
Recall that a circle is the locus of a point which moves in a plane in such a way that its distance
from a fixed point (in the same plane) always remains constant. The fixed point (O) is called
the centre of the circle and the constant distance is called the radius of the circle. Any straight
line drawn through the centre of circle, whose end point lie on the circle is called a diameter,
whose length is equal to twice the radius of the circle.
We state without any logical proof that the ratio of the circumference of a circle to its diameter
is always constant which is also equal to the ratio of the area enclosed by a circle to the square
of its radius (r).
i.e. = 2 = constant.
The value of this constant is nearly 3 and is usually
denoted by the Greek letter π (pie), which is an irrational
number but its value, correct to four places of decimals
is 3.1416, to avoid lengthy calculations the value of π
is often taken to be .
7 Fig. 23.17
∴ = π or C = 2πr
and = π or A = π r2 .
Thus, circumference of a circle = 2πr linear units
and, the area of the circle = πr2 square units.
Example 23.16 : Find the circumference and area of a circle of radius 3.5 cm
[Use π = 22/7]
Solution : Circumference of the circle = 2πr
22 × 35
= 2× . cm
= 22 cm
Also, area of the circle = π r 2
= × 3.5 × 3.5 cm2
= 38.5 cm2
Example 23.17 : The radius of a wheel is 42 cm. How many revolutions will it make in going
26.4 km ?
Solution : Distance travelled in one revolution = Circumference of the wheel
22 × 42
= 2πr = 2 × cm
= 264 cm
∴ No. of revolutions required to travel 26.4 km
26.4 × 1000 × 100
Example 23.18 : The difference between the circumference and the radius of a circle is
74 metres. Find the diameter of the circle.
Solution : We have
2 π r – r = 74
or r (2 π – 1) = 74
r 44 − 1 = 74
or r F 37 I = 74
74 × 7
or r= or 14 m
∴ The diameter of the circle = 2(14) m = 28 m.
23.16 PERIMETER AND AREA OF SECTOR OF A CIRCLE
Two radii OA and OB enclose a portion of the circular region making central angle θ . The
region is called a sector of the circle.
In Fig. 23.18, sector AOB is the sector with central angle
θ. Let ‘l’ be the length of arc AB. The,
[Corresponding arcs subtend proportional central
l θ θ
or = or l = 2 πr.
2πr 360° 360° Fig. 23.18
Area of Plane Figures 207
∴ Perimeter of the sector = OA + OB + AB
= 2r + 2 πr θ
Also area of sector is given by
Area of sector AOB θ
πr 2 360°
or Area of sector AOB = πr 2 .
23.17. AREA OF CIRCULAR PATHS
If we have a circular field of radius ‘r’, surrounded by a path of
uniform width (d) so that r + d = R (say)
then, the area of the circular path
= Area of the outer circle – Area of the inner circle
= ( π R2 – π r2) sq unit
= π (R2 – r2) sq unit
Let us take some examples based on the above formula :
Example 23.19 : A circular road runs round a circular garden. If the difference between the
circumference of outer circle and inner circle is 66 meters, find the width of the road.
Solution : Let the radius of inner circle be r
and that of outer circle be R
∴ Width of the road = (R – r) = d (say)
Now, we have 2 π R – 2 π r = 66
or 2 π (R – r) = 66
∴ d = (R – r) = 66 × 7 × 1 m = 10.5 m.
Thus width of the road is 10.5 m Fig. 23.20
Example 23.20 : A path of width 2 meters runs around a circular plot whose circumference
is 75 metres. Find
(i) the area of path.
(ii) the cost of gravelling the path at Rs 7 per square meter.
Solution : Here 2 π r = 75
i.e. 2 × 22 × r = 528
528 × 1 × 7
or r= = 12
7 2 22
i.e., the radius of the plot is 12 m.
(i) Area of path = π (R2 – r2) =
22 14 2 − 12 2
22 × 52 2 1144
= m = sq m
FG 1144 × 7IJ
(ii) Cost = Rs H 7 K
= Rs 1144.
Example 23.21. Find the area of sector of a circle whose radius
is 6 cm when
(a) the angle at the centre is 35°
(b) when the length of arc is 22 cm
Solution. (a) Area of sector = π r 2 .
22 × 6 × 6 × 35
= 11 sq cm
(b) Here length of arc l = 22 cm
∴ 2 πr θ = 22 cm
Area of sector = π r 2 .
1 r.2 πr θ
1 r.l 1
= = × 6 × 22 sq cm = 66 sq cm
CHECK YOUR PROGRESS 23.2
1. Find the areas and circumference of the circles with radius (i) 1.4 m (b) 4.9 m.
2. The difference between the circumference and diameter of a circle is 210 m. Find the radius
of the circle.
Area of Plane Figures 209
3. The difference between the area of a circle and the square of its radius is 105 sq m. Find
the circumference of circle.
4. The driving wheel of a locomotive engine, 1.4 m in radius, makes 70 revolutions in one
minute. Find the speed of the train in km/hr.
5. A circular park of radius 21 m has a road 7 m wide on its outside all around. Find the
area of the road.
6. Find the areas of sectors of a circle with radius 3.5 m having central angle
(i) 60° (ii) 90°.
7. The length of the minute hand of a clock is 7 cm. Find the area covered by the minute
hand in 6 minutes. Use π = 7 .
LET US SUM UP
Area of a square of side a = a2 sq units and perimeter = 4a units
Perimeter of rectangle = 2(length + breadth) units
Area of rectangle = (length × breadth) sq units
Area of a triangle = × base × height sq units
= b gb gb g
s s − a s − b s − c sq units
where a, b anc c are sides of the triangle and
Area of a parallelogram = Base × Altitude sq units
Area of a trapezium = (sum of parallel sides) × Distance between them sq units
Area of a rhombus = (Product of the diagonals) sq units
Area of rectangular path = Area of outer rectangular enclosure – Area of inner rectangular
Area of perpendicular paths in the middle of a rectangular field
= Area of path parallel to length + Area of path parallel to
breadth – Area of common square.
Circumference of a circle = 2 π r units
Area of a circle = π r2 sq units
Area of circular path = π (R2 – r2), where R > r
Area of a sector of a circle = π r 2 . sq units
Perimeter of a sector = 2 r + 2 πr 360°
1. The side of a square park is 37.5 m. Find its area.
2. The perimeter of a square is 480 m. What is its area ?
3. How long will a man take to walk round the boundary of a square field of area
40000 sq meters at the rate of 4 km an hour ?
4. The length of a room is three times its breadth. If its breadth is 4.5 m, find the area of
5. The perimeter of a rectangle is 980 meters; and the length is to breadth is 5 : 2. Find
6. Find the area of each of the following parallelograms :
(i) side 20 m and the corresponding altitude 12 m.
(ii) one side is 13m, second side is 14 m and the diagonal is 15 m.
7. The area of a rectangular field is 27000 sq meters and the ratio between its length and
breadth is 6 : 5. Find the cost of wire required to go four times round the field at the
rate of Rs 7 per 10 metres of length of wire.
8. Find the area of the following trapezia :
Lengths of parallel sides Perpendicular distance between them
(i) 30 m and 20 m 15 m
(ii) 15.5 m and 10.5 m 7.5 m
(iii) 17 m and 40 m 14.6 m
(iv) 40 m and 22 m 12 m
9. Find the area of a piece of land in the shape of a quadrilateral one of whose diagonals
is 20 m and lengths of perpendiculars from the other two vertices on the diagonal are
of length 12 m and 18 m respectively.
Area of Plane Figures 211
10. Find the area of a field in the form of a trapezium whose parallel sides measure 48 m
and 160 m and the non-parallel sides measure 50 m and 78 m respectively.
11. Find the area of a quadrilateral ABCD in which AB = 85 meters, BC = 143 metres,
CD = 165 metres, DA = 85 metres and DB = 154 metres.
12. Find the areas of the following triangles whose sides are :
(i) 25 m, 60 m, 65 m
(ii) 60 m, 111 m, 153 m
13. The sides of a triangle are 51 m, 52 m, and 53 m. Find the perpendicular from the opposite
side on the side of length 52 m. Also find the areas of the two triangles into which the
original triangle is divided.
14. Find the area of a rhombus, one of whose diagonals measures 8 m and the side is 5 m.
15. The difference between the two parallel sides of a trapezium is 8 metres, the perpendicular
between them is 24 metres and the area of the trapezium is 312 sq metres. Find the lengths
of the two parallel sides.
16. A rectangular piece of land measures 200 m by 150 m. From the centre of each side a
path of 10 m wide runs across up to the opposite side. Find the area of paths.
17. A rectangular plot of grass measures 65 m by 40 m. It has a path of uniform width
8 m all around inside it. Find the cost of spreading red sand stone on the path at the rate
of Rs 5.25 per sq m.
18. A rectangular plot of land measuring 30 m by 20 m has two paths each 2 m wide on both
the sides (inside and outside) of the boundary. Find the total area of the two paths.
19. A path 3 metres wide runs around a circular park whose radius is 9 meters. Find the area
of the path.
20. The difference between the circumference and diameter of a circle is 30 m. Find the radius
of the circle.
21. From the circular piece of cardboard with radius 1.47 m, a sector with central angle 60°
has been removed. Find the area of the portion removed.
22. A circular plot with a radius of 15 m has a road 2 m wide running all around inside it.
Find the area of the road.
Check Your Progress 23.1
1. 60 m 2. 15 2 m 3. 281.25 sq m, 70 m
4. 39.7 m 5. 1680 sq m, 30 m 6. 12936 sq m
7. 240 sq m 8. 216 sq m 9. 55 m, 1320 sq m
10. 360 sq m 11. 2000 sq m 12. 276 sq m
Check Your Progress 23.2
1. (i) 6.16 sq m, 8.8 m (ii) 75.46 sq m, 30.8 m
2. 49 m 3. 44 m 4. 36.96 km/hr
5. 1078 sq m 6. (i) 6.41 sq m (ii) 9.6 sq m
7. 15.4 cm.
1. 1406.25 sq m 2. 14400 sq m 3. 12 min
4. 60.75 sq m 5. 49000 sq m 6. (i) 240 sq m (ii) 168 sq m
7. Rs 1848
8. (i) 375 sq m (ii) 97.5 sq m (iii) 416.1 sq m (iv) 372 sq m
9. 300 sq m 10. 3120 sq m 11. 12936 sq m
12. (i) 750 sq m (ii) 2754 sq m 13. 45 m, 540 sq m, 630 sq m
14. 24 sq m 15. 17 m, 9 m 16. 3400 sq m
17. Rs 7476 18. 400 sq m 19. 198 sq m
20. 7 m 21. 1.1319 sq m 22. 176 sq m
Area of Plane Figures 213