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Class 9 SELINA Solutions Maths Chapter 20 - Area and Perimeter of Plane Figures

Area and Perimeter of Plane Figures Exercise Ex. 20(A)

Solution 1

Since the sides of the triangle are 18cm,24cm and 30cm respectively.

 

Hence area of the triangle is

 

Again

 

Hence

 

Solution 2

Let the sides of the triangle are

a=3x

b=4x

c=5x

Given that the perimeter is 144 cm.

hence

3 x plus 4 x plus 5 x equals 144
rightwards double arrow 12 x equals 144
rightwards double arrow x equals 144 over 12
rightwards double arrow x equals 12

s equals fraction numerator a plus b plus c over denominator 2 end fraction equals fraction numerator 12 x over denominator 2 end fraction equals 6 x equals 72

The sides are a=36 cm, b=48 cm and c=60 cm

Area of the triangle is

A equals square root of s open parentheses s minus a close parentheses open parentheses s minus b close parentheses open parentheses s minus c close parentheses end root
space space space equals square root of 72 open parentheses 72 minus 36 close parentheses open parentheses 72 minus 48 close parentheses open parentheses 72 minus 60 close parentheses end root
space space space equals square root of 72 cross times 36 cross times 24 cross times 12 end root
space space space equals square root of 746496
space space space equals 864 space c m squared

 

Solution 3

(i)

Area of the triangle is given by

 

(ii)

Again area of the triangle

 

Solution 4

Solution 5

Since the perimeter of the isosceles triangle is 36cm and base is 16cm. hence the length of each of equal sides are

Here

It is given that

Let 'h' be the altitude of the isosceles triangle.

Since the altitude from the vertex bisects the base perpendicularly, we can apply Pythagoras Theorem.

Thus we have,

h equals square root of a squared minus open parentheses b over 2 close parentheses squared end root equals 1 half square root of 4 a squared minus b squared end root
W e space k n o w space t h a t space
A r e a space o f space t h e space t r i a n g l e equals 1 half cross times b a s e cross times a l t i t u d e

 

Solution 6

It is given that

A r e a equals 192 space s q. c m
b a s e equals 24 space c m

K n o w i n g space t h e space l e n g t h space o f space e q u a l space s i d e comma space a comma space a n d space b a s e comma space b comma space space o f space a n space i s o s c e l e s space t r i a n g l e comma
t h e space a r e a space c a n space b e space c a l c u l a t e d space u sin g space t h e space f o r m u l a comma
A equals 1 fourth cross times b cross times square root of 4 a squared minus b squared end root

Let 'a' be the length of an equal side.

 

Hence perimeter=

Solution 7

From  ,

 

Area of

 

 

Area of

Now

Solution 8

Given , AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and  DBC = 90o

 

Hence perimeter=8+10+13+5=36cm

Area of

 

Area of

 

Now

Solution 9

Area space of space the space rectangular space field equals fraction numerator 49572 over denominator 36.72 end fraction equals 135000
Let space the space height space of space the space triangle space be space straight x
135000 equals 1 half cross times straight x cross times 3 straight x
rightwards double arrow straight x squared equals 90000
rightwards double arrow straight x equals 300
Height space equals 300 space straight m
Base space equals 900 space straight m

Solution 10

(i)

 

(ii)

Consider the following figure.

 

 

  

 

 

 

 

(iii)

Solution 11

Let the height of the triangle be x cm.

Equal sides are (x+4) cm.

According to Pythagoras theorem,

 

Hence perimeter=

Area of the isosceles triangle is given by

Here a=20cm

b=24cm

hence

 

Solution 12

Each side of the triangle is

Hence the area of the equilateral triangle is given by

The height h of the triangle is given by

Solution 13

The area of the triangle is given as 150sq.cm

 

Hence AB=15cm,AC=20cm and

 

Solution 14

Let the two sides be x cm and (x-3) cm.

Now

 

Hence the sides are 12cm, 9cm and

The required perimeter is 12+9+15=36cm.

Solution 15

 

 

Since AB=AC and

 

Now

Area and Perimeter of Plane Figures Exercise Ex. 20(B)

Solution 1

Solution 2

Solution 3

Consider the figure:

From the right triangle ABD we have

The area of right triangle ABD will be:

Again from the equilateral triangle BCD we have

Therefore the area of the triangle BCD will be:

Hence the area of the quadrilateral will be:

Solution 4

The figure can be drawn as follows:

Here ABD is a right triangle. So the area will be:

Again

Now BCD is an isosceles triangle and BP is perpendicular to BD, therefore

From the right triangle DPC we have

So

Hence the area of the quadrilateral will be:

Solution 5

Let the width be x and length 2x km.

Hence

Hence the width is 100m and length is 200m

The required area is given by

Solution 6

Length of the laid with grass=85-5-5=75m

Width of the laid with grass=60-5-5=50m

Hence area of the laid with grass is given by

Solution 7

Area of the rectangle is given by

 

Let h be the height of the triangle ,then

Solution 8

Consider the following figure.

Thus the required area = area shaded in blue + area shaded in red

= Area ABPQ +  Area TUDC +  Area A'PUD' +  Area QB'C'T

= 2Area ABPQ + 2Area QB'C'T

=2(Area ABPQ +Area QB'C'T)

A r e a space o f space t h e space f o o t p a t h space i s space g i v e n space b y
A equals 2 cross times open parentheses 25 plus 25 plus 17 plus 17 close parentheses
space space space space equals 168 space s q. space m
space space space space equals 1680000 space s q. c m
H e n c e space n u m b e r space o f space t i l e s space r e q u i r e d equals 1680000 over 400 equals 4200

Solution 9

Perimeter of the garden

Again, length of the garden is given to be 120 m. hence breadth of the garden b is given by

Hence area of the field

Solution 10

Length of the rectangle=x

Width of the rectangle=

Hence its perimeter is given by

Again it is given that the perimeter is 4400cm.

Hence

 Length of the rectangle=1400 cm = 14 m

Solution 11

(i)

Breadth of the verandah=x

Length of the verandah=x+3

According to the question

 

(ii)

From the above equation

Hence breadth=3m

Length =3+3=6m

Solution 12

Consider the following figure.

Thus, the area of the shaded portion

=Area(ABCD) + Area(EFGH) - Area(IJKL)…(1)

 

 

Solution 13

First we have to calculate the area of the hall.

We need to find the cost of carpeting of 80 cm = 0.8 m wide carpet, if the rate of carpeting is Rs. 25. Per metre.

Then

C o s t equals fraction numerator 25 over denominator 0.8 end fraction cross times 1440
space space space space space space space equals R s.45 comma 000

Solution 14

Let a be the length of each side of the square.

Hence

Hence

 

And

Solution 15

Consider the following figure.

(i)

The length of the lawn = 30 - 2 - 2 = 26 m

The breadth of the lawn = 12 - 2 = 10 m

(ii)

The orange shaded area in the figure is the required area.

Area of the flower bed is calculated as follows:

Solution 16

Number of tiles required

Fraction of floor uncovered=

Solution 17

Since

Hence the distance between the shorter sides is 16m.

Solution 18

At first we have to calculate the area of the triangle having sides, 28cm, 26cm and 30cm.

Let the area be S.

By Heron's Formula,

Area of a Parallelogram = 2 × Area of a triangle

 = 2 × 336

 = 672 cm2

We know that,

Area of a parallelogram = Height × Base

672 = Height × 26

Height = 25.84 cm

the distance between its shorter sides is 25.84 cm.

Solution 19

(i)

We know that

Here A=216sq.cm

AC=24cm

BD=?

Now

 

(ii)

Let a be the length of each side of the rhombus.

 

(iii)

Perimeter of the rhombus=4a=60cm.

Solution 21

Let a be the length of each side of the rhombus.

 

We know that,

Solution 22

The diagram is redrawn as follows:

Here

AF=1.2m,EF=0.3m,DC=0.6m,BK=1.8-0.6-0.3=0.9m

Hence

Solution 23

Here we found two geometrical figure, one is a triangle and other is the trapezium.

Now

hence area of the whole figure=150+240=630sq.m

Solution 24

We can divide the field into three triangles and one trapezium.

Let A,B,C be the three triangular region and D be the trapezoidal region.

Now

 

Area of the figure=Area of A+ Area of B+ Area of C+ Area of D

=3900+1250+312.5+2062.5

=7525sq.m

Solution 25

Let x be the width of the footpath.

Then

Again it is given that area of the footpath is 360sq.m.

Hence

Hence width of the footpath is 3m.

Solution 26

Area of the square is 484.

Let a be the length of each side of the square.

Now

Hence length of the wire is=4x22=88m.

(i)

Now this 88m wire is bent in the form of an equilateral triangle.

 

(ii)

Let x be the breadth of the rectangle.

Now

Hence area=16x28=448m2

Solution 27

(i)

 

A r e a space o f space triangle E B C equals 1 fourth cross times 8 cross times square root of 4 cross times 10 squared minus 8 squared end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 fourth cross times 8 cross times 18.3
space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 36.6 space c m squared

Again

(ii)

(iii)

For the triangle EBC,

S=19cm

Let h be the height.

 

A r e a space o f space triangle E B C equals 1 half cross times 12 cross times h
rightwards double arrow 59.9 equals 6 h
rightwards double arrow h equals fraction numerator 59.9 over denominator 6 end fraction equals 9.98 space c m
A r e a space o f space A B C D equals 1 half cross times open parentheses 20 plus 32 close parentheses cross times 9.98
space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half cross times 52 cross times 9.98
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 259.48 space c m squared


(iv)

In the given figure, we can observe that the non-parallel sides are equal and hence it is an isosceles trapezium.

 

Therefore, let us draw DE and CF perpendiculars to AB.

Thus, the area of the parallelogram is given by

Since space AB space equals space AE space plus space EF space plus space FB space and space CD space equals space EF space equals space 18 space cm comma space we space have
30 space equals space AE space plus space 18 space plus space FB
rightwards double arrow space 30 space equals space AE space plus space 18 space plus space AE
rightwards double arrow 2 AE space plus space 18 space equals space 30
rightwards double arrow space 2 AE space equals space 30 space minus space 18
rightwards double arrow space 2 AE space equals space 12
rightwards double arrow space AE space equals space 6 space cm
Now comma space consider space the space right space angled space triangle space ADE.
AD squared space equals space AE squared space plus space DE squared
rightwards double arrow space 12 squared space equals space 6 squared space plus space DE squared
rightwards double arrow 144 space equals space 36 space plus space DE squared
rightwards double arrow space DE squared space equals space 144 space minus space 36
rightwards double arrow space DE squared space equals space 108
rightwards double arrow space DE space equals space square root of 36 cross times 3 end root
rightwards double arrow space DE space equals space 6 square root of 3
Area space open parentheses parallelogram ABCD close parentheses space equals space Area open parentheses triangle ADE close parentheses space plus space Area open parentheses parallelogram DEFC close parentheses space plus space Area open parentheses triangle CFB close parentheses
rightwards double arrow Area open parentheses parallelogram ABCD close parentheses space equals space 1 half cross times 6 cross times 6 square root of 3 plus 18 cross times 6 square root of 3 plus 1 half cross times 6 cross times 6 square root of 3
rightwards double arrow Area open parentheses parallelogram ABCD close parentheses space equals space 6 cross times 6 square root of 3 plus 18 cross times 6 square root of 3
rightwards double arrow Area open parentheses parallelogram ABCD close parentheses space equals space 144 square root of 3 equals 249.41 cm squared


Solution 28

Let b be the breadth of rectangle. then its perimeter

2 open parentheses x plus b close parentheses equals 70
x plus b equals 35
b equals 35 minus x

 

 Again

x cross times b equals 300
x open parentheses 35 minus x close parentheses equals 300
x squared minus 35 x plus 300 equals 0
open parentheses x minus 15 close parentheses open parentheses x minus 20 close parentheses equals 0
x equals 15 comma 20

 

Hence its length is 20cm and width is 15cm.

Solution 29

Let b be the width of the rectangle.

Again perimeter of the rectangle is 104m.

Hence

Hence

length=32m

width=20m.

Solution 30

Let a be the length of the sides of the square.

According to the question

Hence sides of the square are 12cm each and

Length of the rectangle =2a=24cm

Width of the rectangle=a+6=18cm.

Solution 31

Solution 32

Length of the wall=45+2=47m

Breath of the wall=30+2=32m

Hence area of the inner surface of the wall is given by

A equals open parentheses 47 cross times 2 cross times 2.4 close parentheses plus open parentheses 32 cross times 2 cross times 2.4 close parentheses
equals 225.6 plus 153.6
equals 379.2 space m squared

Solution 33

Let a be the length of each side.

Hence length of the wire=96cm

(i)

For the equilateral triangle,

(ii)

Let the adjacent side of the rectangle be x and y cm.

Since the perimeter is 96 cm, we have,

2 open parentheses x plus y close parentheses equals 96

Hence

Hence area of the rectangle is = 26 x 22 = 572 sq.cm

Solution 34

Let 'y' and 'h' be the area and the height of the first parallelogram respectively.

Let 'height' be the height of the second parallelogram

base of the first parallelogram= cm

base of second parallelogram= cm

Solution 35

Solution 36

L e t space a space a n d space b space b e space t h e space s i d e s space o f space t h e space r e c tan g l e
S i n c e space t h e space p e r i m e t e r space i s space 92 space m comma space w e space h a v e comma
2 open parentheses a plus b close parentheses equals 92
rightwards double arrow a plus b equals 46 space m.... left parenthesis 1 right parenthesis
A l s o space g i v e n space t h a t space d i a g o n a l space o f space a space t r a p e z i u m space i s space 34 space m.
rightwards double arrow a squared plus b squared equals 34 squared.... left parenthesis 2 right parenthesis
W e space k n o w space t h a t
open parentheses a plus b close parentheses squared minus a squared minus b squared equals 2 a b
F r o m space e q u a t i o n s space left parenthesis 1 right parenthesis space a n d space left parenthesis 2 right parenthesis comma space w e space h a v e comma
46 squared minus 34 squared equals 2 a b
rightwards double arrow 2 a b equals 960
rightwards double arrow a b equals 960 over 2
rightwards double arrow a b equals 480 space m squared

Solution 20

Let a be the length of each side of the rhombus.

(i)

It is given that,

AC=24cm

We have to find BD.

Now

 

Hence the other diagonal is 10cm.

(ii)

begin mathsize 11px style Area thin space of thin space the space rhombus space equals space 1 half cross times AC cross times BD
thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space space space space space space space space space space space space space space space space equals space 1 half cross times 24 cross times 10
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 120 space sq. space cm end style

Area and Perimeter of Plane Figures Exercise Ex. 20(C)

Solution 1

Let r be the radius of the circle.

(i)

                              = 88 cm

(ii)

         = 616 cm2

Solution 2

L e t space r space b e space t h e space r a d i u s space o f space t h e space c i r c u l a r space f i e l d
left parenthesis i right parenthesis
2 πr equals 308
rightwards double arrow straight r equals fraction numerator 308 over denominator 2 straight pi end fraction
rightwards double arrow straight r equals 308 over 2 cross times 7 over 22
rightwards double arrow straight r equals 49 space straight m
open parentheses ii close parentheses
area equals πr squared
space space space space space space space space equals 22 over 7 cross times open parentheses 49 close parentheses squared
space space space space space space space space equals 7546 space straight m squared

Solution 3

Let r be the radius of the circle.

Solution 4

Circumference of the first circle

Circumference of the second circle

Let r be the radius of the resulting circle.

Solution 5

Circumference of the first circle

Circumference of the second circle

Let r be the radius of the resulting circle.

Hence area of the circle

Solution 6

Let the area of the resulting circle be r.

Hence the radius of the resulting circle is 20cm.

Solution 8

Area of the rectangle is given by

For the largest circle, the radius of the circle will be half of the sorter side of the rectangle.

Hence r=21cm.

Hence

Solution 9

Area of the square is given by

Since there are four identical circles inside the square.

Hence radius of each circle is one fourth of the side of the square.

Area remaining in the cardboard is

Solution 10

Let the radius of the two circles be 3r and 8r respectively.

 

According to the question

 

Hence radius of the smaller circle is

Area of the smaller circle is given by

A equals πr squared equals 22 over 7 cross times 21 squared equals 1386 space cm squared

Solution 11

Let the diameter of the three circles be 3d, 5d and 6d respectively.

Now

Solution 12

Solution 13

Let r be the radius of the circular park.

a r e a space o f space t h e space p a r k equals straight pi cross times open parentheses 8.75 close parentheses squared equals 240.625 space straight m squared

Radius of the outer circular region including the path is given by

R equals 8.75 plus 3.5
space space space space equals 12.25 space m

Area of that circular region is

A equals straight pi cross times open parentheses 12.25 close parentheses squared equals 471.625 space straight m squared

Hence area of the path is given by

A r e a space o f space t h e space p a t h equals 471.625 minus 240.625 equals 231 space m squared

Solution 14

L e t space r space b e space t h e space r a d i u s space o f space t h e space c i r c u l a r space g a r d e n space A.
S i n c e space t h e space c i r c u m f e r e n c e space o f space t h e space g a r d e n space A space i s space 1.760 space K m equals 1760 m comma space w e space h a v e comma
2 πr equals 1760 space straight m
rightwards double arrow straight r equals fraction numerator 1760 cross times 7 over denominator 2 cross times 22 end fraction equals 280 space straight m
Area space of space garden space straight A equals πr squared equals 22 over 7 cross times 280 squared space straight m squared
Let space straight R space be space the space radius space of space the space circular space garden space straight B.
Since space the space area space of space garden space straight B space is space 25 space times space the space area space of space garden space straight A comma space we space have comma
πR squared equals 25 cross times πr squared
rightwards double arrow πR squared equals 25 cross times straight pi cross times 280 squared space
rightwards double arrow straight R squared equals 1960000
rightwards double arrow straight R equals 1400 space straight m
Thus space circumference space of space garden space straight B space equals 2 πR equals 2 cross times 22 over 7 cross times 1400 equals 8800 space straight m space equals 8.8 space Km

 

 

 

Solution 15

 D i a m e t e r space o f space t h e space w h e e l equals 84 space c m
T h u s comma space r a d i u s space o f space t h e space w h e e l equals 42 space c m
C i r c u m f e r e n c e space o f space t h e space w h e e l equals 2 cross times 22 over 7 cross times 42 equals 264 space c m
I n space 264 space c m comma space w h e e l space i s space c o v e r i n g space o n e space r e v o l u t i o n.
T h u s comma space i n space 3.168 space K m equals 3.168 cross times 100000 space c m comma space n u m b e r space o f space r e v o l u t i o n s
c o v e r e d space b y space t h e space w h e e l equals fraction numerator 3.168 over denominator 264 end fraction cross times 100000 equals 1200

Solution 16

C i r c u m f e r e n c e space o f space t h e space w h e e l equals d i s tan c e space c o v e r e d space b y space t h e space w h e e l space i n space o n e space r e v o l u t i o n
T h u s comma space w e space h a v e comma
C i r c u m f e r e n c e equals 2 cross times 22 over 7 cross times 80 over 2 equals 251.43 space c m
T h u s comma space t h e space n u m b e r space o f space r e v o l u t i o n s space c o v e r e d space
b y space t h e space w h e e l space i n space 1100000 space c m equals fraction numerator 1100000 over denominator 251.43 end fraction almost equal to 4375

Solution 17

Solution 18

Time interval is 9.05 space minus 8.30 equals 35 space m i n u t e s

Area covered in one 60 minutes=

Hence area swept in 35 minutes is given by

A equals 201 over 60 cross times 35 equals 117 1 third space c m squared

 

Solution 19

Let R and r be the radius of the big and small circles respectively.

G i v e n space t h a t space t h e space c i r c u m f e r e n c e space o f space t h e space b i g g e r space c i r c l e space i s space 396 space c m
T h u s comma space w e space h a v e comma
2 πR equals 396 space cm
rightwards double arrow straight R equals fraction numerator 396 cross times 7 over denominator 2 cross times 22 end fraction
rightwards double arrow straight R equals 63 space cm
Thus comma space area space of space the space bigger space circle equals πR squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 22 over 7 cross times 63 squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 12474 space cm squared
Also space given space that space the space circumference space of space the space smaller space circle space is space 374 space cm
rightwards double arrow 2 πr equals 374
rightwards double arrow straight r equals fraction numerator 374 cross times 7 over denominator 2 cross times 22 end fraction
rightwards double arrow straight r equals 59.5 space cm
Thus comma space the space area space of space the space smaller space circle equals πr squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 22 over 7 cross times 59.5 squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 11126.5 space cm squared
Thus space the space area space of space the space shaded space portion space equals 12474 minus 11126.5 equals 1347.5 space cm squared

Solution 20

From the given data, we can calculate the area of the outer circle and then the area of inner circle and hence the width of the shaded portion.

G i v e n space t h a t space t h e space c i r c u m f e r e n c e space o f space t h e space o u t e r space c i r c l e space i s space 132 space c m
T h u s comma space w e space h a v e comma space 2 πR equals 132 space cm
rightwards double arrow straight R equals fraction numerator 132 cross times 7 over denominator 2 cross times 22 end fraction
rightwards double arrow straight R equals 21 space cm
Area space of space the space bigger space circle equals πR squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 22 over 7 cross times 21 squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1386 space cm squared
Also space given space the space area space of space the space shaded space portion.
Thus space the space area space of space the space inner space circle equals Area space of space the space outer space circle minus Area space of space the space shaded space portion
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1386 minus 770
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 616 space cm squared
rightwards double arrow πr squared equals 616
rightwards double arrow straight r squared equals fraction numerator 616 cross times 7 over denominator 22 end fraction
rightwards double arrow straight r squared equals 196
rightwards double arrow straight r equals 14 space cm
Thus comma space the space width space of space the space shaded space portion equals 21 minus 14 equals 7 space cm
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space

 

Solution 21

Let the radius of the field is r meter.

Therefore circumference of the field will be:  meter.

Now the cost of fencing the circular field is 52,800 at rate 240 per meter.

Therefore

Thus the radius of the field is 35 meter.

Now the area of the field will be:

Thus the cost of ploughing the field will be:

Solution 22

Let r and R be the radius of the two circles.

Putting the value of r in (2)

Hence the radius of the two circles is 3cm and 7cm respectively.

Solution 23

Solution 24

Since the diameter of the circle is the diagonal of the square inscribed in the circle.

Let a be the length of the sides of the square.

Hence

 

Hence the area of the square is 98sq.cm.

Solution 25

Let space apostrophe straight a apostrophe space be space the space length space of space each space side space of space an space equilateral space triangle space formed.
Now comma space area space of space equilateral space triangle space formed equals 484 square root of 3 space cm squared
rightwards double arrow fraction numerator square root of 3 over denominator 4 end fraction straight a squared equals 484 square root of 3
rightwards double arrow straight a squared equals 4 cross times 484
rightwards double arrow straight a equals 2 cross times space 22 equals 44 space cm
Then comma space perimeter space of space equilateral space triangle equals 3 straight a equals 3 cross times 44 equals 132 space cm
Now comma space length space of space wire space equals space perimeter space of space equilateral space triangle equals circumference space of space circle
rightwards double arrow circumference space of space circle equals 132 space cm
rightwards double arrow 2 πr equals 132 space space space space space space left parenthesis straight r space is space radius space of space circle right parenthesis
rightwards double arrow straight r equals fraction numerator 132 cross times 7 over denominator 2 cross times 22 end fraction equals 21 space cm
therefore space Area space of space circle equals πr squared equals 22 over 7 cross times 21 cross times 21 equals 1386 space cm squared

Solution 7

Area of the circle having radius 85m is

begin mathsize 11px style straight A equals straight pi cross times open parentheses 5 close parentheses squared
space space space equals 7225 space πm squared end style

Let r be the radius of the circle whose area is 49times of the given circle.

 

Hence circumference of the circle

Area and Perimeter of Plane Figures Exercise Ex. 20(D)

Solution 1

Let the sides of the triangle be

a = 12x

b = 5x

c = 13x

Given that the perimeter = 450 cm

12x + 5x + 13x = 450

30x = 450

x = 15

Hence, the sides of a triangle are

a = 12x = 12(15) = 180 cm

b = 5x = 5(15) = 75 cm

c = 13x = 13(15) = 195 cm

Now,

Solution 2

Let the sides of the triangle be

a = 26 cm, b = 28 cm and c = 30 cm

Now,

Solution 3

Construction: Draw CM AB

  

In right-angled triangle CMB,

BM2 = BC2 - CM2 = (15)2 - (9)2 = 225 - 81 = 144

BM = 12 m

Now, AB = AM + BM = 23 + 12 = 35 m

Solution 4

Let the sides of two squares be a and b respectively.

Then, area of one square, S1 = a2

And, area of second square, S2 = b2

Given, S1 + S2 = 400 cm2

a2 + b2 = 400 cm2 ..(1)

Also, difference in perimeter = 16 cm

4a - 4b = 16 cm

a - b = 4

a = (4 + b)

Substituting the value of 'a' in (1), we get

(4 + b)2 + b2 = 400

16 + 8b + b2 + b2 = 400

2b2 + 8b - 384 = 0

b2 + 4b - 192 = 0

b2 + 16b - 12b - 192 = 0

b(b + 16) - 12(b + 16) = 0

(b +16)(b - 12) = 0

b + 16 = 0 or b - 12 = 0

b = -16 or b = 12

Since, the side of a square cannot be negative, we reject -16.

Thus, b = 12

a = 4 + b = 4 + 12 = 16

Hence, the sides of a square are 16 cm and 12 cm respectively.

Solution 5

Diagonal of a square = 24 cm

Solution 6

Solution 7

Solution 8

Let the radii of two circles be r1 and r2 respectively.

Sum of the areas of two circles = 130π sq. cm

πr12 + πr22 = 130π

r12 + r22 = 130 ….(i)

Also, distance between two radii = 14 cm

r1 + r2 = 14

r1 = (14 - r2)

Substituting the value of r1 in (i), we get

(14 - r2)2 + r22 = 130

196 - 28r2 + r22 + r22 = 130

2r22 - 28r2 + 66 = 0

r22 - 14r2 + 33 = 0

r22 - 11r2 - 3r2 + 33 = 0

r2 (r2 - 11) - 3 (r2 - 11) = 0

(r2 - 11) (r2 - 3) = 0

r2 = 11 or r2 = 3

r1 = 14 - 11 = 3 or r1 = 14 - 3 = 11

Thus, the radii of two circles are 11 cm and 3 cm respectively.

Solution 9

 

Solution 10

 

Solution 11

  

 

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