ROUTERA


Chapter 15 Perimeter and Area of Plane Figures

Class 10th R. S. Aggarwal Maths Solution
CBSE Class 10 Maths
R. S. Aggarwal Solution


Perimeter and Area of Plane Figures Exercise Ex. 15A

Solution 1

Solution 2

Let a = 42 cm, b = 34 cm and c = 20 cm

(i)Area of triangle =

       

(ii)Let base = 42 cm and corresponding height = h cm

          Then area of triangle =

          Hence, the height corresponding to the longest side = 16 cm

Solution 3

Let a = 18 cm, b = 24 cm, c = 30 cm

Then,2s = (18 + 24 + 30) cm = 72 cm 

           s = 36 cm

(s a) = 18cm, (s b) = 12 cm and (s c) = 6 cm

(i)Area of triangle =

(ii)Let base = 18 cm and altitude = x cm

          Then, area of triangle =

           Hence, altitude corresponding to the smallest side = 24 cm

Solution 4

On dividing 150 m in the ratio 5 : 12 : 13, we get

Length of one side =

Length of the second side = 

Length of third side =

Let a = 25 m, b = 60 m, c = 65 m

(s a) = 50 cm, (s b) = 15 cm, and (s c) = 10 cm

Hence, area of the triangle = 750 m2

Solution 5

On dividing 540 m in the ratio 25 : 17 : 12, we get

Then,

 

Therefore, area of triangular field

 

Cost of ploughing 100 m2 field = Rs. 40

Then, cost of ploughing 9000 m2 field

Solution 6

Let the length of one side be x cm

Then the length of other side = {40 (17 + x)} cm = (23 - x) cm

Hypotenuse = 17 cm

Applying Pythagoras theorem, we get

Hence, area of the triangle = 60 cm2

Solution 7

Let the sides containing the right - angle be x cm and (x - 7) cm

One side = 15 cm and other = (15 - 7) cm = 8 cm

perimeter of triangle (15 + 8 + 17) cm = 40 cm

Solution 8

Let the sides containing the right angle be x and (x-2) cm

One side = 8 cm, and other (8-2) cm = 6 cm

                         = 10 cm

Therefore, perimeter of the triangle = 8 + 6 + 10 = 24 cm

Solution 9

  

Solution 10

Let each side of an equilateral triangle =   cm

Then,

 

Therefore, area of an equilateral triangle

 

Solution 11

Let each side of the equilateral triangle be a cm

Perimeter of equilateral triangle = 3a = (3x12) cm = 36 cm

Solution 12

Let each side of the equilateral triangle be a cm

area of equilateral triangle =

Height of equilateral triangle

Solution 13

Base of right angled triangle = 48 cm

Height of the right angled triangle =

Solution 14

Let the hypotenuse of right - angle triangle = 65cm 

Base = 60 m 

Perpendicular space equals space square root of left parenthesis hypotenuse right parenthesis squared minus open parentheses base close parentheses squared end root
equals square root of 65 squared minus 60 squared end root
equals square root of 625
equals 25 space cm

Area space of space triangle space equals space 1 half cross times base cross times height
equals 1 half cross times 60 cross times 25
equals 750 space cm squared

Hence, perpendicular = 25 cm and area of the triangle =750 cm2

Solution 15

The circumcentre of a right - triangle is the midpoint of the hypotenuse

Hypotenuse = 2 × (radius of circumcircle)

= (2 × 8) cm = 16 cm

Base = 16 cm, height = 6 cm

Area of right angled triangle

Hence, area of the triangle= 48 cm2

Solution 16

Let each equal side be a cm in length.

Then,

Hence, hypotenuse = 28.28 cm and perimeter = 68.28 cm

Solution 17

Let each equal side be a cm and base = 80 cm

perimeter of triangle = (2a + b) cm

= (2 41 + 80) cm

= (82 + 80) cm = 162 cm

Hence, perimeter of the triangle = 162 cm

Solution 18

Let the height be h cm, then a= (h + 2) cm and b = 12 cm

Squaring both sides,

Therefore, a = h + 2 = (8 + 2)cm = 10 cm

Hence, area of the triangle = 48 cm2.

Solution 19

Let ABC is a isosceles triangle. Let AC, BC be the equal sides

Then AC = BC = 10cm. Let AB be the base of ABC right angle at C.

Area of right isosceles triangle ABC

Hence, area = 50 cm2 and perimeter = 34.14 cm

Solution 20

Area of shaded region = Area of ABC – Area of DBC

First we find area of ABC

Second we find area of DBC which is right angled

Area of shaded region = Area of ABC – Area of DBC

= (43.30 - 24) = 19. 30

Area of shaded region = 19.3

Perimeter and Area of Plane Figures Exercise Ex. 15B

Solution 1

Let the length of a rectangular plot = L m

Given, perimeter = 80 m and breadth = 16 m

Now,

Perimeter of a rectangular plot = 2(Length + Breadth)

80 = 2(L + 16)

40 = L + 16

L = 24 m

 

Therefore, area of the plot = Length × Breadth

= 24 m × 16 m

= 384 m2

Solution 2

  

Solution 3

Let the other side of a rectangle =   cm

Then, by Pythagoras theorem,

And, area of a rectangle = 35 cm × 12 cm = 420 cm2

Solution 4

Let the breadth of a rectangular plot =   m

Given, area of the rectangular plot = 462 m2

⇒ Length × Breadth = 462

⇒ 28 ×   = 462 

 = 16.5 cm

Now, perimeter of a rectangular plot = 2(Length + Breadth)

= 2(28 + 16.5)

= 2(44.5)

= 89 m

Solution 5

  

Solution 6

  

Solution 7

Area of floor = Length Breadth

                 

Area of carpet = Length Breadth

                    =

Number of carpets =

                                                          = 216

Hence the number of carpet pieces required = 216

Solution 8

Area of verandah = (36 × 15) = 540

Area of stone = (0.6 × 0.5) [10 dm = 1 m]

Number of stones required = 

Hence, 1800 stones are required to pave the verandah.

Solution 9

Perimeter of rectangle = 2(l + b)

2(l + b) = 56 Þ l + b = 28 cm

b = (28 l) cm

Area of rectangle = 192

l (28 l) = 192

28l - = 192

- 28l + 192 = 0

- 16l 12l + 192 = 0

l(l 16) 12(l 16) = 0

(l 16) (l 12) = 0

l = 16 or l = 12

Therefore, length = 16 cm and breadth = 12 cm

Solution 10

Length of the park = 35 m

Breadth of the park = 18 m

Area of the park = (35 18) = 630

Length of the park with grass =(35 5) = 30 m

Breadth of the park with grass = (18- 5) m = 13 m

Area of park with grass = (30 13) = 390

Area of path without grass = Area of the whole park area of park with grass

                                     = 630 - 390 = 240

Hence, area of the park to be laid with grass = 240 m2

Solution 11

Length of the plot = 125 m

Breadth of the plot = 78 m

Area of plot ABCD = (125 78) = 9750

Length of the plot including the path= (125 + 3 + 3) m = 131 m

Breadth of the plot including the path = (78 + 3 + 3) m = 84 m

Area of plot PQRS including the path

                              = (131 84) = 11004

Area of path = Area of plot PQRS Area of plot ABCD

                  = (11004 9750)

                  = 1254

Cost of gravelling = Rs 75 per m2

Cost of gravelling the whole path = Rs. (1254 75) = Rs. 94050

Hence, cost of gravelling the path = Rs 94050

Solution 12(i)

Area of rectangular field including the foot path = (54 35)

Let the width of the path be x m

Then, area of rectangle plot excluding the path = (54 2x) (35 2x)

Area of path = (54 35) (54 2x) (35 2x)

                      (54 35) (54 2x) (35 2x) = 420

                      1890 1890 + 108x + 70x - 4 = 420

                      178x - 4 = 420

                      4 - 178x + 420 = 0

                      2 - 89x + 210 = 0

                      2 - 84x 5x + 210 = 0

                      2x(x 42) 5(x 42) = 0

                      (x 42) (2x 5) = 0

Solution 12(ii)

Let the width of the border = b m

Then, the length of a carpet = (8 - b - b) = (8 - 2b) m

And, the breadth of a carpet = (5 - b - b) = (5 - 2b) m

Area of a carpet = (8 - 2b) × (5 - 2b) = (40 - 26b + 4b2) m2

 

Length of a room = 8 m

Breadth of a room = 5 m

Area of a room = 8 × 5 = 40 m2

 

Given, area of border = 12 m2

 

Now,

Area of a carpet = Area of a room - Area of a border

⇒ (40 - 26b + 4b2) = 40 - 12

⇒ 4b2 - 26b + 12 = 0

⇒ 2b2 - 13b + 6 = 0

⇒ 2b2 - 12b - b + 6 = 0

⇒ 2b(b - 6) - (b - 6) = 0

⇒ (b - 6)(2b - 1) = 0

⇒ b - 6 = 0 or 2b - 1 = 0

⇒ b = 6 or b = ½

b ≠ 6 as width of the border cannot be greater than width of the room.

⇒ b = ½ = 0.5 m = 50 cm

 

Hence, the width of the carpet is 50 cm.

Solution 13

Let the length and breadth of a rectangular garden be 9x and 5x.

Then, area of garden = (9x 5x)m = 45

Length of park excluding the path = (9x 7) m

Breadth of the park excluding the path = (5x 7) m

Area of the park excluding the path = (9x 7)(5x 7)

Area of the path =

                           

(98x 49) = 1911

     98x = 1911 + 49

Length = 9x = 9 20 = 180 m

Breadth = 5x = 5 20 = 100 m

Hence, length = 180 m and breadth = 100 m

Solution 14

  

Solution 15

Let the width of the grass strip = b m

Then, the length of the pond = (50 - b - b) = (50 - 2b) m

And, the breadth of the pond = (40 - b - b) = (40 - 2b) m

Area of the pond = (50 - 2b) × (40 - 2b) = (2000 - 180b + 4b2) m2

 

Length of a park = 50 m

Breadth of a park = 40 m

Area of a park = 50 × 40 = 2000 m2

 

Given, area of the grass strip = 1184 m2

 

Now,

Area of the pond = Area of a park - Area of the grass strip

⇒ (2000 - 180b + 4b2) = 2000 - 1184

⇒ 4b2 - 180b + 1184 = 0

⇒ b2 - 45b + 296 = 0

⇒ b2 - 37b - 8b + 296 = 0

b(b - 37) - 8(b - 37) = 0

⇒ (b - 37)(b - 8) = 0

⇒ b - 37 = 0 or b - 8 = 0

⇒ b = 37 or b = 8

 

b ≠ 37 as width of the grass strip at both ends is exceeding 50 m.

⇒ b = 8 m

 

⇒ Length of the pond = 50 - 2(8) = 34 m and breadth of the pond = 40 - 2(8) = 24 m.

Solution 16

  

 

  

Solution 17

  

Solution 18

  

Solution 19

  

Solution 20

Area of the square =

Let diagonal of square be x

Length of diagonal = 16 cm

Side of square =

Perimeter of square = [4 side] sq. units

=[ 4 11.31] cm = 45.24 cm

Solution 21

Let d meter be the length of diagonal

Area of square field =

                                

Time taken to cross the field along the diagonal

                                          

Hence, man will take 6 min to cross the field diagonally.

Hence, man will take 6 min to cross the field diagonally.

Solution 22

  

Solution 23

Rs. 14 is the cost of fencing a length = 1m

Rs. 28000 is the cost of fencing the length=

Perimeter = 4 side = 2000

side = 500 m

Area of a square =

= 250000

Cost of mowing the lawn =

Solution 24

  

Solution 25

ABCD be the given quadrilateral in which AD = 24 cm, BD = 26 cm, DC = 26 cm and BC = 26 cm

By Pythagoras theorem

For area of equilateral DBC, we have

              a = 26 cm

Area of quad. ABCD = Area of ABD + Area of DBC

                            = (120 + 292.37) = 412.37

Perimeter ABCD = AD + AB + BC + CD

                       = 24 cm + 10 cm + 26 cm + 26 cm

                       = 86 cm

Solution 26

Area of quad. ABCD = Area of ABC + Area of ACD

Now, we find area of a ACD

Area of quad. ABCD = Area of ABC + Area of ACD

                           

Perimeter of quad. ABCD = AB + BC + CD + AD

                                    =(17 + 8 + 12 + 9) cm

                                    = 46 cm

Perimeter of quad. ABCD = 46 cm

Solution 27

Area of quad. ABCD = Area of ABD + Area ofDBC

For area of ABD

Let a = 42 cm, b = 34 cm, and c = 20 cm

For area of DBC

a = 29 cm, b = 21 cm, c = 20 cm

Solution 28

Area of the ||gm = (base height) sq. unit

= (25 16.8)

Solution 29

Longer side = 32 cm, shorter side = 24 cm

Distance between longer sides = 17.4 cm

Let the distance between the shorter sides be x cm

Area of ||gm = (longer side distance between longer sides)

= (shorter side distance between the short sides)

distance between the shorter side = 23.2 cm

Solution 30

Solution 31

Area ofparallelogram = 2 area of DABC

Opposite sides of parallelogram are equal

AD = BC = 20 cm

And AB = DC = 34 cm

In ABC we have

a = AC = 42 cm

b = AB = 34 cm

c = BC = 20 cm

Then, (s a) = 6 cm, (s b) = 14 cm and (s c) = 28 cm

Solution 32

 

 

                       

We know that the diagonals of a rhombus, bisect each other at right angles

OA = OC = 15 cm,

And OB = OD = 8 cm

And AOB = 90

By Pythagoras theorem, we have

Solution 33

(i)Perimeter of rhombus = 4 side

4 side = 60 cm

 

By Pythagoras theorem

OB = 12 cm

OB = OD = 12 cm

BD = OB + OD = 12 cm + 12 cm = 24 cm

 

                       Length of second diagonal is 24 cm

             

             (ii) Area of rhombus =

Solution 34

(i)Area of rhombus = 480

One of its diagonals = 48 cm

Let the second diagonal =x cm

 

 

Hence the length of second diagonal 20 cm

 

(ii)We know that the diagonals of a rhombus bisect each other at right angles

AC = 48, BD = 20 cm

OA = OC = 24 cm and OB = OD = 10 cm

By Pythagoras theorem , we have

 

(iii)Perimeter of the rhombus = (4 26) cm = 104 cm

Solution 35

  

Solution 36

Areaof cross section =

Solution 37

Let ABCD be a given trapezium in which

AB = 25, CD = 11

BC = 15, AD = 13

Draw CE || AD

In ||gm ADCE, AD || CE and AE || CD

AE = CD = 11 cm,

And BE = AB BE

           = 25 11 = 14 cm

In BEC,

Area of BEC =

Let height of BEC is h

Area of BEC = 

From (1) and (2), we get

7h = 84 h = 12 m

Area of trapezium ABCD

Perimeter and Area of Plane Figures Exercise MCQ

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Perimeter and Area of Plane Figures Exercise Test Yourself

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Solution 15

Let the side of a square lawn = S metre

Then,

4S × 14 = 2800

4S = 200

S = 50 m

Area of a square lawn = (50 m)2 = 2500 m2

 

Now,

cost of mowing 100 m2 lawn = Rs. 54

Then, cost of mowing 2500 m2 lawn   

Therefore, the cost of mowing the lawn is Rs. 1350.

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