NCERT Solutions for Chapter 11 Mensuration Class 8 Maths

Answer

Given: The side of a square = 60 m and the length of rectangular field = 80 m

According to question,

Perimeter of rectangular file = Perimeter of square field

⇒ 2(l+b) = 4 × Side

⇒ 2(80 + b) = 4 × 60 

⇒ (80 + b) = 240/2 

⇒ (80 + b) = 120

⇒ b = 120 - 80

⇒ b = 40 m

Hence, the breadth of the rectangular field is 40 m.

Now, Area of Square field= (Side)2 
= (60)2 sq.m = 3600 sq.m

Area of Rectangular field = (length × breadth)

= 80 × 40 sq. m = 3200 sq. m

Hence, area of square field is larger.

Answer

Side of a square plot = 25 m

∴ Area of square plot = (Side)2 = (25)2 = 625 m2

Length and Breadth of the house is 20 m and 15 m respectively

∴ Area of the house = (length x breadth )

= 20 × 15 = 300 m2

Area of garden = Area of square plot – Area of house

= (625 – 300) = 325 m2

∵ Cost of developing the garden around the house is Rs.55

∴ Total Cost of developing the garden of area 325 sq. m = Rs.(55 × 325)

= Rs.17,875

Answer

Given: Total length of the diagram = 20 m

Diameter of semi circle on both the ends = 7 m

∴  Radius of semi circle = diameter/ 2 = 7/2 = 3.5 m

Length of rectangular field = [Total length - (radius of semicircle on both side)]

={20 – (3.5 + 3.5)}

= 20 – 7 = 13 m

Breadth of the rectangular field = 7 m

∴ Area of rectangular field = ( l x b)

= (13 × 7) ⇒ 91 m2

Area of two semi circles = 2 × 1/2 πr2 

= 2 × 1/2 × 22/7 × 3.5 × 3.5  = 38.5 m2

Total Area of garden = (91 + 38.5)⇒129.5 m2

Perimeter of two semi circles = 2 × πr = 2 × 22/7 × 3.5 

= 22 m

Hence, Perimeter of garden = (22 + 13 + 13)m = 48 m

Answer

Base of flooring tile = 24 cm 
⇒ 0.24 m

height of a flooring tile = 10 cm
⇒0.10 m [1cm = 1/100 m]

Now, Area of flooring tile= Base × Altitude

= 0.24 × 0.10 sq. m

= 0.024 m2

∴ Number of tiles required to cover the floor = Area of floor/area of one tile 

= 1080/0.024 

= 45000 tiles

Hence 45000 tiles are required to cover the floor.

Answer

Answer

Answer

Let the length of the other parallel side be = b cm

Length of one parallel side = 10 am and height (h) = 4 cm

Area of trapezium = 1/2 (sum of parallel sides) Height 

= 34 = 1/2 ( a + b)h 

= 34 = 1/2 (10 + b) × 4

= 34 = ( 10 + b) × 2 

= 34 = 20 + 2b 

= 34 - 20 = 2b 

= 14 = 2b 

= 7 = b 

= b = 7 

Hence another required parallel side is 7 cm.

Answer

Given: BC = 48 m, CD = 17 m,

AD = 40 m and perimeter = 120 m

∵ Perimeter of trapezium ABCD = Sum of all sides

120 =(AB + BC + CD + DA)

120 = AB + 48 + 17 + 40

120 = AB + 105

(120 – 105) = AB

AB = 15 m

Now Area of the field = 1/2 × ( Sum of parallel sides) × Height 

= 1/2 × (BC + AD) × AB

= 1/2 × (48 + 40) × 15 m2

= 1/2 ×(88) × 15  m2

= 1/2 × (1320)  m2

= 660  m2

Hence area of the field ABCD is 660  m2

Answer

Here h1 = 13 m, h2 = 8 m and AC = 24 m

Area of quadrilateral ABCD = Area of △ ABC + Area of △ADC

= 1/2 b × h1+ 1/2 b × h2 

= 1/2 b ( h1 + h2) 

= 1/2 × 24 ( 13 + 8) m2

= 1/2 × 24 (21) m2

= 12 × 21 12 x 21 m2

= 252  m2

Hence required area of the field is 252  m2

Answer

Given: d1 =7.5 cm and d2 = 12 cm

Area of rhombus = >12x(Product of digonals)

= 12 x (d1 x d2 )

= >12x (7.5 x12) cm2

= 45  cm2

Hence area of rhombus is 45  cm2

Answer

Rhombus is also a kind of Parallelogram.

∴ Area of rhombus= Base × Altitude

= (6 × 4)  cm2

= 24 cm2

Also Area of rhombus = 12× (d1 x d2 )

24 = 1/2 ×( 8 × d2 )
24 = 4 d2
244 cm = d2
d2 = 6 cm

Hence, the length of the other diagonal is 6 cm.

Answer

Here, d1 = 45 cm and d2 = 30 cm

∵ Area of one tile = >12x (d1 x d2 )

= >12 x(45 x 30 )

=>12(1350)

= 675 cm2

So, the area of one tile is 675 cm2

Area of 3000 tiles = 675 × 3000 cm2

= 2025000 cm2

= >2025000100∗100 m2
[1 cm = >1100 m, Here cm2 = Cm x cm = 1100 x 1100 m2 ]

= 202.50 m2

∵  Cost of polishing the floor per sq. meter = Rs. 4

∴ Cost of polishing the floor per 202.50 sq. meter =Rs. 4 × 202.50 = Rs. 810

Hence the total cost of polishing the floor is Rs. 810.

Answer

Given: Perpendicular distance (h) AM = 100 m

Area of the trapezium shaped field = 10500 m2

Let side along the road AB= x m

side along the river CD = 2x m

∴Area of the trapezium field = 12× (AB + CD) x AM

10500 = 1/2 ( x + 2x) × 100

10500 = 3x × 50

3x = 10500/50 

x = 10500/(50 × 3)

x = 70 m

Hence the side along the river = 2m = (2 ×70) = 140 m.

Answer

Given: Octagon having eight equal sides, each 5 m.

Construction: Join HC and GD It will divide the octagon into two equal trapezium.

And AM is perpendicular on HC and EN is perpendicular on GD

Area of trap. ABCD = Area of trap. GDFE ....................................(1)

Area of two trapeziums = (area of trap. ABCH + area of trap. GDFE)

= (area of trap. ABCH + area of trap. ABCH) (by statement 1).
= (2 x area of trap. ABCH )

= (2 x 12x (sum of parallel sides ) x height)

= (2 x 12x(AB + CH) x AM)

=(11 + 5 ) x 4 m2

=(16) x 4

= 64 m2

And Area of rectangle (HCDG ) = length × breadth

= HC x HG = 11 ×  5 = 55 m2

∴ Total area of octagon = Area of 2 Trapezium + Area of Rectangle

= 64 m2 + 55 m2 = 119 m2

Answer

First way: By Jyoti’s diagram,

Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP

= 1/2 (AP + BC) x CP + 1/2 (ED + AP) ×  DP

= 1/2  (30 + 15 ) x CP + 1/2  (15 + 30) × DP

= 1/2 (30 + 15) (CP + DP)

= 1/2 × 45 × CD

= 1/2 × 45 × 15 

=337.5 m2

Second way: By Kavita’s diagram

Here, a perpendicular AM drawn to BE. AM = 30 – 15 = 15 m

Area of pentagon = Area of △ ABE + Area of square BCDE

={12×15×15}+(15 × 15) m2

= (112.5 + 225.0) m2

= 337.5 m2

Hence total area of pentagon shaped park = 337.5 m2

Answer

Here two of given figures (I) and (II) are similar in dimensions. And also figures (III) and (IV) are similar in dimensions.

 Area of figure (I) = Area of trapezium

= 1/2 (a+b) × h = 1/2 (28 + 20) × 4 

= 1/2 ×48 × 4  = 96 cm2

Also Area of figure (II) = 96 cm2

Now Area of figure (III)

Area of trapezium = 1/2 (a + b) × h 

= 1/2 (24 + 16) × 4

= 1/2 × 40 × 4 

= 80 cm2

Also Area of figure (IV) = 80 cm2

Answer

 (a) Length of cuboidal box (l)= 60 cm

Breadth of cuboidal box (b) = 40 cm

Height of cuboidal box (h) = 50 cm

∴ Total surface area of cuboidal box = 2(lb+bh+hl)

= 2 (60 × 40 + 40 × 50 + 50 × 60) cm2

= 2 (2400 + 2000 + 3000) cm2

= 2 × 7400 cm2

= 14800 cm2

(b) Length of the cube is 50 cm

∴ Total surface area of cuboidal box =6(side)2

=6 ( 50)2 cm2

= 6 (2500) cm2

= 15000 cm2

Thus, the cuboidal box (a) requires the lesser amount of materal.

Answer

Given: Length of suitcase box (l) = 80 cm, Breadth of suitcase box (b) = 48 cm

And Height of cuboidal box (h) = 24 cm

∴ Total surface area of suitcase box= 2(lb+bh+hl)

= 2 (80 × 48 + 48 × 24 + 24 × 80) cm2

= 2 (3840 + 1152 + 1920)

= 2 × 6912 = 13824 cm2

Area of Tarpaulin cloth = Surface area of suitcase

⇒ l × b= 13824

⇒ l × 96 = 13824

⇒ l = 13824/96 

= 144 cm

Required tarpaulin for 100 suitcases = (144 × 100) cm

= 14400 cm

= 144 m [ 1cm = 1100m]

Thus, 144 m tarpaulin cloth required to cover 100 suitcases.

Answer

Answer

Length of cabinet (l) = 2 m

Breadth of cabinet (b) = 1 m

Height of cabinet (h) = 1.5 m

 Surface area of cabinet = (Area of Base of cabinet (Cuboid) + Area of four walls)

=lb+2(l+b)h
={2 × 1 + 2 (1 + 2) 1.5 } m2

= 2 + 2 (3) 1.5 m2
=2+6 (1.5) m2
= (2 + 9.0) m2
= 11 m2

Hence required surface area of cabinet is 11 m2

Answer

Length of wall (l) = 15 m
Breadth of wall (b) = 10 m
Height of wall (h) = 7 m

∴ Total Surface area of classroom=(Area of Base of ceiling (Cuboid) + Area of four walls)

= lb+2(l+b)h
= (15 × 10 + 2 (10 +15) (7)) m2

= (150 + 2 (25) (7)) m2
= (150 + 350) m2
= 500 m2

Area of one can is 100 m2

Now Required number of cans = Area of hall/ Area of one can = 500/100 = 5 cans

Hence 5 cans are required to paint the room.

Answer

Diameter of cylinder = 7 cm
∴ Radius of cylinder (r) = 7/2  cm
Height of cylinder (h)  = 7 cm
Lateral surface area of cylinder = 2πrh 

= 2 × 22/7 × 7/2 × 7 

= 154 cm2

Now lateral surface area of cube = 4(Side)2=4(7)2  cm2
= (4 × 49) cm2

= 196 cm2
Hence the cube has larger lateral surface area.

Answer

Radius of cylindrical tank (r) = 7 m
Height of cylindrical tank (h) = 3 m

Total surface area of cylindrical tank = (Curved surface area + Area of upper end (circle)+ Area of Lower (circle) end)

= (2πrh+πr2+πr2)

= (2πrh+2πr2)

=2πr(h+r)

= 2 × 22/7 × 7(3 + 7)  m2

= 44 × 10  m2

= 440  m2

Hence 440  m2 metal sheet is required.

Answer

Lateral surface area of hollow cylinder = 4224 cm2

Height of hollow cylinder = 33 cm

Curved surface area of hollow cylinder = 2πrh 

⇒ 4224 = 2 × 22/7 × r × 33 

⇒ r = (4224 × 7)/(2 × 22 × 33) 

= (64 × 7)/22  cm

Now Length of rectangular sheet = 2πr

l = 2 × 22/7 × (64 × 7)/22

= 128 cm

Perimeter of rectangular sheet =  2(l + b)

= 2 (128 + 33)
= 2 x 161
= 322 cm

Hence perimeter of rectangular sheet is 322 cm.

Answer

Diameter of road roller = 84 cm
∴ Radius of road roller (r) = d/2 = 84/2 
= 42 cm

Length of road roller (h)  = 1 m = 100 cm

Curved surface area of road roller = 2πrh 

= 2 × 22/7 × 42 × 100 

= 26400 cm2

∴ Area covered by road roller in 750 revolutions = 26400 × 750 cm2

= 1,98,00,000 cm2
= 1980 m2 [∵ 1 m2= 10,000 cm2]

Thus, the area of the road is 1980 m2.

Answer

Diameter of cylindrical container = 14 cm

∴ Radius of cylindrical container (r) = d/2 = 14/2 = 7 cm

Height of cylindrical container = 20 cm

Height of the label (h) = (20 – 2 – 2)
= 16 cm

Curved surface area of label = 2πrh 

= 2×22/7×7×16
= 704 cm2

Hence the area of the label of 704 cm2.

Answer

(a) Volume (it is measure of the amount of space inside of a solild figures)

(b) Surface area (the outside part or uppermost layer of the soild figures)

(c) Volume

Answer

Yes, we can say that volume of cylinder B is greater, Because radius of cylinder B is greater than that of cylinder A.

Diameter of cylinder A = 7 cm

⇒ Radius(r) of cylinder A = 7/2 cm and Height(h) of cylinder A = 14 cm

∴ Volume of cylinder A = πr2h 

= 22/7 × 7/2 ×7/2 × 14 

= 539 cm3

Now Diameter of cylinder B = 14 cm

⇒ Radius of cylinder B = 14/2  = 7 cm and Height of cylinder B = 7 cm

∴ Volume of cylinder A = πr2h 

= 22/7 × 7 ×7 ×7 cm3

= 1078 cm3

Since the cylinder A and cylinder B is open from upper end then it will exclude from the Total surface area

Total surface area of cylinder A = ( Area of lower end circle + curved surface area of cylinder)

=πr2 +  2πrh)

=πr(r+2h)

=227 x 
>72+2x14)

= 11 72+28)

= 11(31.5) cm2 = 346.5 cm2

Total surface area of cylinder B = πr(2h + r) 

= 227x 7(2x7+7)

= 22 × (14 + 7)

= 22 × 21 = 462 cm2

Yes, cylinder with greater volume also has greater surface area.

Answer

Let the Length, breadth and height of the cuboid be l, b, h.

Base of the cuboid is form a Recatangle so,that the Base(Reactangle) Area is (Length x Breadth)

Base area of cuboid = 180 cm2

L x B = 180 cm2 ......(1)

Volume of cuboid = l ×b × h 

Volume of cuboid = 900 cm3 

(lb)h = 900 (From eq. 1)

(180) h = 900

h = 900/180

= 5 m

Hence the height of cuboid is 5 m.

Answer

Given: Length of cuboid (l) = 60 cm,

Breadth of cuboid (b) = 54 cm and

Height of cuboid (h) = 30 cm

We know that, Volume of cuboid = l × b ×h

= (60 × 54 × 30) cm3 

And Volume of cube = (Side)3

= 6 × 6 × 6 cm3 

∴ Number of small cubes = (Volume of cuboid)/( Volume of cube) = (60 × 54 × 30)/( 6 × 6× 6) 

= 450

Hence required number of small cubes are 450.

Answer

Given: Volume of cylinder = 1.54 m3  and Diameter of cylinder = 140 cm

∴ Radius (r) = d/2 = 140/2  = 70 cm

=70100m = 0.7m [$1cm=1/100\mathrm{m]}$

Volume of cylinder = πr2h 

⇒ 1.54 = 22/7 × 0.7 × 0.7 × h 

⇒ h = (1.54 × 7)/(22 × 0.7 × 0.7)

⇒ h = ( 154 × 7 × 10 × 10)/( 22 × 7 × 7 × 100)

= 1 m

Hence height of the cylinder is 1 m.

Answer

Given: Radius of cylindrical tank (r) = 1.5 m

Height of cylindrical tank (h)  = 7 m

Volume of cylindrical tank = πr2h 

= 22/7 × 1.5 × 1.5 × 7 

= 49.5 m3

= 49.5 × 1000 liters [∵ 1 m3 = 1000 liters]

= 49500 liters

Hence required quantity of milk is 49500 liters that can be stored in the tank.

Answer

(ii)

Answer

Volume of reservoir = 108m3 

= 108 x 1000 litres[1m3=1000l]

=108000 litres

Since water is pouring into reservoir @ 60 litres per minute and in

Time taken to fill the reservoir = 10800060x 160hours

= 30 hours

Hence, 30 hours it will take to fill the reservoir.