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**Maths Ace Class 8 Solutions Chapter 14 Mensuration**

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Maths Ace Prime Class 8 Math Book, Chapter 14, Mensuration. Here students can easily find step by step solutions of all the problems for Mensuration, Exercise 14.1, 14.2, 14.3, 14.4 and 14.5 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 14 solutions.

**Mensuration Exercise 14.1 Solution :**

**Question no – (1) **

**Solution : **

If we add a circle right ,of park A then it equals of park ’B’

**∴** Park ‘B’ has greater area

So, the area of circle,

= 32/7 × 15 × 15

= 706.5 m^{2}

Therefore, Park B has greater area.

**Question no – (2) **

**Solution : **

Let, side of one square x cm and the side of another square (x + 5) cm

**∴** (x + 5)^{2} – (x^{2}) = 325

= x^{2} + 10x + 25 – x^{2} = 325

= 10x = 325 – 25

= x = 300/10

= 30

**∴** Side of one square 30 cm and another side square,

= (30 + 5) cm

= 35 cm.

**Question no – (3) **

**Solution : **

According to the question,

**∴** Area of floor,

= (8.4 × 5.4) cm^{2}

= 45.36 m^{2}

= 453600 cm^{2}

**∴** Area of marble pieces –

= 1/2 × 40 × 30

= 600 cm^{2}

**∴** No of marble piece required,

= 453600/600

= 750 marble piece

**Question no – (4) **

**Solution : **

Area of rectangular lawn,

= (80 × 60) m^{2}

= 4800 m^{2}

**∴** Total cost,

= (4800 × 7.5) Rs

= 36000 Rs

Therefore, the cost of levelling the lawn is 36000 Rs.

**Question no – (5) **

**Solution :**

Area of square,

= (35)^{2}

= 1225 m^{2}

Area of rectangle

= (51 × 25) m^{2}

= 1275 m^{2}

**∴** Total area,

= (1225 + 1275) m^{2}

= 2500 m^{2}

Area of new square field = 2500 m^{2}

**∴** Let side, ‘S’

**∴** S × S = 2500

= s2 = 2500

= s = 50 m

Therefore, Side of the square field = 50 m.

**Question no – (6) **

**Solution :**

Area of path on length wise,

= (125 – 3) × 5

= 610 m^{2}

And, area of path on breadth wise,

= (125 – 5) × 3

= 360 m^{2}

**∴** Area of path of intersection,

= 5 × 3

= 15 m^{2}

**∴** Area of two pathways,

= (610 + 360 + 15)

= 985 m^{2}

**Question no – (7) **

**Solution :**

As per the given question,

The area of a parallelogram is 780 sq. cm.

If the length of one side is 26 cm

**∴** The corresponding height,

= 780/26

= 30 cm

Hence, the corresponding height will be 30 cm.

**Question no – (8) **

**Solution :**

In the give question,

One side of a parallelogram is = 15 cm,

The area is = 375 sq. cm

Let, length of side x cm,

**∴** x × 15 = 375

= x = 375/15

= 25 cm

Therefore, The Length of attitude will be 25 cm.

**Question no – (9) **

**Solution :**

In the question we get,

The base and height of a triangle are in the ratio 4 : 5.

The area is 640 sq. cm,

Let, base is 4x and height 5x

**∴** 1/2 × 4x × 5x = 640

= 10x^{2} = 640

= x^{2} = 64

= x = 8

**∴ **Base = (4 × 8) = 32 cm

**∴** height = (5 × 8) = 40 cm

Therefore, the base is 32 cm and height is 40 cm.

**Question no – (10) **

**Solution :**

As we know, Area = 1/2 × base × height

**∴** 3136 = 1/2 × B × 64

= 3136 = 32 B

= B = 3136/32

= 98 cm

Therefore, the Base is 98 cm.

**Question no – (11) **

**Solution :**

In the given question,

The base of a parallelogram is 12 cm and its height is 18 cm

**∴** Area of parallelogram,

= (12 × 18) cm^{2}

= 216 cm^{2}

Therefore, the area of parallelogram is 216 cm^{2}

**Question no – (12) **

**Solution :**

According to the question,

Area of ABCD,

= (10 × 12 + 20 × 12)

= 120 + 240

= 360 cm2 – 120 cm^{2}

= 240 cm^{2}

Therefore, the area of parallelogram ABCD is 240 cm^{2}

**Mensuration Exercise 14.2 Solution :**

**Question no – (1)**

**Solution : **

As per the question we get,

The lengths of parallel sides of a trapezium are = 20 cm and 14 cm.

The distance between them is = 10 cm

**∴** Area of trapezium,

= 1/2 × (20 + 14) × 10

= 1/2 × 34 × 10

= 170 cm^{2}

Thus, the area of the trapezium will be 170 cm^{2}

**Question no – (2) **

**Solution : **

According to the given question,

The area of a rhombus is = 225 sq. cm.

The length of one of its diagonals is = 15 cm

**∴** Area = 1/2 × d_{1} × d_{2}

= 225 = 1/2 × 15 × d_{2}

= d_{2} = 225 × 2/15

= d_{2} = 30 cm

Thus, the length of the other diagonal will be 30 cm.

**Question no – (3) **

**Solution : **

As per the question,

One side is equal to = 10 cm

The length of one of the diagonals is = 12 cm

**∴** The Area rhombus,

= 1/2 p √4a^{2} – p^{2}

= 1/2 × 12 √4 × 10^{2} – 12^{2}

= 6 √400 – 144

= 6 × √256

= 6 × 16

= 96 cm

Therefore, the area of rhombus will be 96 cm.

**Question no – (4) **

**Solution : **

According to the given question,

Diagonals of a rhombus are 16 cm and 30 cm

**∴** Area of rhombus,

= (1/2 × 16 × 30)

= 240 cm^{2}

**∴** Length of side,

= √d_{1}^{2} + d_{2}^{2}/2

= 16^{2} + 30^{2}/2

= √256 + 900/2

= √1156/2

= 17 cm

Hence, the area will be 240 cm^{2 }and the length of the side will be 17 cm.

**Question no – (5) **

**Solution : **

Let, sides are 4x and 5x

**∴ **1/2 × (4x + 5x) × 12 = 54

= 1/2 × 9x × 12 = 594

= x = 594 × 2/12 × 9

= x = 11 cm

**∴** Parallel sides are,

= (4 × 11) = 44 cm

= (5 × 11) = 55 cm

Therefore, the lengths of the parallel sides are 44 cm and 55 cm.

**Question no – (6) **

**Solution :**

Let, sides x and 2x

**∴ **1/2 × (x + 2x) × 24 = 900

= 1/2 × 3x × 24 = 900

= 36x = 900

= x = 900/36

= 25 cm

**∴** Two parallel sides are,

= 25 cm and

= (25 × 2)

= 50 cm

Therefore, the length of the two parallel sides are 25 cm and 50 cm.

**Question no – (7) **

**Solution : **

As per the question we know,

The area of a rhombus is = 966 sq. cm.

length of one of its diagonals is = 46 cm,

**∴** Area of rhombus,

= 1/2 × (d_{1} × d_{2})

= 966 = 1/2 × (46 × d_{2})

= 46 d_{2} = 966 × 2

= d_{2} = 966 × 2/46

= 42 cm

Therefore, the length of the other diagonal is 42 cm

**Question no – (8) **

**Solution : **

**1st,** Area of triangle,

= (1/2 × 48 × 16) m^{2}

= 384 m^{2}

**2nd,** Another area,

= (1/2 × 48 × 26) m^{2}

= 624 m^{2}

Now, the total area,

= (624 + 384) m^{2}

= 1008 m^{2}

Therefore, area of the field will be 1008 m^{2}

**Question no – (9) **

**Solution :**

Here, Area = 23.36 sq.cm

Let AC = x

OB = 3.2 cm

By question 1/2 × b × h = 23.36/2

= 1/2 × 3.2x = 11.68

= 1.6x = 11.68

= x = 11.68/1.6

= 7.3 cm

**∴** AC = 7.3 cm.

**Question no – (10) **

**Solution : **

In the given question we get,

lengths of the two diagonals of the hall are = 12 m and 16 m

The rate of painting = Rs 6.50 per sq. m

**∴** Area of rhombus,

= (1/2 × 12 × 16) m^{2}

= 96 m^{2}

**∴** Total cost of painting the floor of the rate 6.50 Rs,

= (6.50 × 96) Rs

= 624 Rs

Therefore, the cost of painting the floor will be 624 Rs,

**Mensuration Exercise 14.3 Solution : **

**Question no – (1) **

**Solution : **

**From figure – (a)**

Area of ABCDE = Area of △ABC × Area of △ACD × Area of △AED

In right angle, △ADC

AC^{2} = CD^{2} + AD^{2}

= 12^{2} + 5^{2}

= 144 + 25

= 169

**∴** AC = 13 cm

**∴** Area of △ABC

= (1/2 × 13 × 5.5)

= 35.75 cm^{2}

**∴** Area of △ADC

= 1/2 × 12 × 5

= 30 cm^{2}

Area of △AED

= (1/2 × 12 × 4.5) cm^{2}

= 27 cm^{2}

Total area,

= (35.75 + 30 + 27) cm^{2}

= 92.75 cm^{2}

**From figure – (b) **

Area of ABCDEFGH

= Area of ABG + Area of parallelogram CDEF + Area of trapezium BCFG

**∴** Area of △ABG = (1/2 × 10 × 3)

= 15 cm^{2}

Area of trapezium BCFG,

= 1/2 × 5 × (10 + 6)

= 1/2 × 5 × 16

= 40 cm^{2}

Area of parallelogram CDEF,

= (6 × 4.5) cm^{2}

= 27 cm^{2}

**∴** Total area,

= (15 + 40 + 27) cm^{2}

= 82 cm^{2}

**Question no – (2) **

**Solution : **

First, Area of floor,

= (25 × 20) m^{2}

= 500 m^{2}

= 5000000 cm^{2}

Now, Area of rhombus,

= (25 × 10) cm^{2}

= 250 cm^{2}

**∴** No of tiles,

= 5000000/250

= 20000 tiles

Therefore, the number of tiles required will be 20000 tiles.

**Question no – (3) **

**Solution : **

As per the given figure,

Area of grass portion,

= (22 × 10) m^{2}

= 220 m^{2 }

Area of flower bed = Area of trapezium ABGH + Area of trapezium DEFG

Now, Area of trapezium,

= 1/2 × 8 × (10 + 22) + 1/2 × (8) (10 + 22)

= (128 + 128) m^{2}

= 256 m^{2}

**∴** Total area,

= (220 + 256) m^{2}

= 476 m^{2}

Therefore, the area of the trapezium will be 476 m^{2}

**Mensuration Exercise 14.4 Solution : **

**Question no – (1) **

**Solution :**

**(a)** As per the question we know,

length = 14 cm,

breadth = 10 cm

height = 9 cm.

As we know, Total Surface area = 2 (l×b + bh + l×h)

**∴** Total surface area,

= 2 (lb + bh + hd)

= 2 (14 × 10 + 10 × 9 + × 14) cm^{2}

= 2 (140 + 90 + 126) cm2

= 2 × 356

= 712 cm^{2}

Therefore, the surface area of the cuboid will be 712 cm^{2}

**(b)** According to question we know,

Length = 16 cm,

Breadth= 12 cm

Height = 7 cm

As we know, Total Surface area = 2 (l × b + b × h + l × h)

**∴** Total surface is,

= 2 (16 × 12 + 12 × 7 + 7 × 16) cm^{2}

= 2 (192 + 84 + 112) cm2

= (2 × 388)cm^{2}

= 776cm^{2}

Hence, the surface area of the cuboid wise776cm^{2}

**(c)** In the given question,

Length = 24 cm,

Breadth = 20 cm

Height = 15 cm

As we know, Total Surface area = 2 (l × b + b × h + l × h)

**∴** Total Surface area,

= 2 (24 × 20 + 20 × 15 × 24)

= 2 (480 + 300 + 360)

= (2 × 1140)

= 2280cm^{2}

Thus, the surface area of the cuboid is 2280cm^{2}

**Question no – (2) **

**Solution : **

As per the question we know,

length = 400 cm = 4 m

breadth = 220 cm = 2 m 20 cm

Height = 120 cm = 1.20 m

As we know, Total Surface area = 2 (l × b + b × h + l × h)

**∴** Surface area,

= 2 (4 × 2.20 + 2. 20 × 1.20 + 1.20 × 4)

= 2 × 98.80 + 2.64 + 4.80)

= 32.48 m^{2}

Now, Cost of polishing,

= (32.48 × 24) Rs

= 779.52 Rs

Therefore, the cost of polishing the box will be 779.52 Rs.

**Question no – (3) **

**Solution : **

Given, Lateral surface area = 64m^{2}

= 4x^{2} = 64

= x^{2} = 16

= x = 4

**∴** Total surface area,

= 6 × 4^{2}

= 6 × 16

= 96 m^{2}

Therefore, the total surface area of the cube is 96 m^{2}

**Question no – (4) **

**Solution : **

First, Area of the room,

= 2 (8 + 6) × 4

= 2 × 14 × 4

= 112 sq.m

**∴ **Cost of point,

= (112 × 80)

= 8960 Rs.

Therefore, the cost of paint required will be 8960 Rs.

**Question no – (5) **

**Solution : **

Length = breadth …..(according to the question)

**∴** Total surface area = 120

= 2 (lb + bh + hl) = 120

= (lb + bh + hl) = 60

= (x^{2} + 2x + 2x) = 60

= x^{2} + 4x = 60

= x^{2} + 4x – 60 = 0

= – 10 + 6

**∴** x = 6

= L b = 6

Therefore, the Length is 8 m breadth 6 m.

**Question no – (6) **

**Solution : **

First, Area of four walls,

= 2 (8 + 4) ×

= 2 × 12 × 2

= 48 m^{2}

**∴** Number of blue tiles required,

= 48/1.5

= 32 tiles

**∴** Number of Area of floor,

= (8 × 4) m^{2}

= 32 m^{2}

**∴** Number of white tiles required,

= 32/2

= 16 tiles

**Mensuration Exercise 14.5 Solution : **

**Question no – (1) **

**Solution : **

**(a)** As per the question we know,

Length = 13 m,

Breadth = 10 m

Height = 5 m

As we know, Volume of cuboid = (l × b × h)

**∴** Volume of cuboid,

= (13 × 10 × 5) m^{3}

= 650 m^{3}

Therefore, the volume of a cuboid will be 650 m^{3}

**(b) **According to the question,

length = 0.5 m,

breadth = 0.3 m

depth = 0.02 m

As we know, Volume of cuboid = (l × b × h)

**∴** Volume of cuboid,

= (0.5 × 0.3 × 0.02)m^{3}

= 0.003 m^{3}

Therefore, the volume of a cuboid will be 0.003 m^{3}

**(c) **Given in the question,

length = 8 m,

breadth = 6 m

depth = 3 m.

As we know, Volume of cuboid = (l × b × h)

**∴** Volume of the cuboid,

= (8 × 6 × 3) m^{3}

= 144m^{3}

Therefore, the volume of a cuboid will be 144m^{3}

**Question no – (2) **

**Solution :**

Let, breadth x m

**∴** (8 × 6 × x) = 200 (….according to the question)

= x = 200/48

= 4.17 m

Therefore, the Breadth will be 4.17 m.

**Question no – (3) **

**Solution : **

As we know, Volume of cube of edge = (x × x × x) = x^{3}

**∴** Volume of cube,

= (10 10 × 10) cm^{3}

= 100 cm^{3}

= 1L

**Question no – (4) **

**Solution : **

According to question we know,

Cartons each having length 2.4 m, breadth 1.4 m and height 0.8 m.

Room of dimensions = 6 m × 7m × 8 m

**∴** Number of cartoons,

= 6 × 7 × 8/1.4 × 1.4 × 0.8

= 125 cartoons

Therefore, the number of cartons will be 125.

**Question no – (5) **

**Solution : **

Let, the sides are, 2x 3x, 5x

**∴** 2.x × 3x × 5x = 21870

= 30x^{3} = 21870

= x^{3} = 21870/30

= 729

= x = 9

**∴** Sides are,

= (2 × 9) = m = 18m

= (3 × 9) = 8T m = 27

= (5 × 9) m. 45m

**∴** Total surface area,

= 2 (18 × 27 + 27 × 45 + 45 × 18) m^{2}

= 5022 m^{2}

Therefore, total surface area of the cuboid is 5022 m^{2}

**Question no – (6) **

**Solution : **

Given in the question,

radius = 1.4 m

height = 10 m

**∴** Volume of well = πr^{2}h

= 22/7 × 1.4 × 1.4 × 10

= 61.6m^{3}

Therefore, the quantity of soil that had been taken out 61.6m^{3}

**Question no – (7) **

**Solution : **

As per the question we know,

Radius = 13/2 m = 6.5 m

Depth = 35 m

**∴** Volume of drum = πr^{2}h

= 22/7 × 6.5 × 6.5 × 35

= 4647.5 m^{3}

Now, half of the volume of drum,

**∴** Volume of water,

= 4647.5/2m^{3}

= 2323.75 m^{3}

Therefore, the volume of the water inside the drum will be 2323.75 m^{3}

**Question no – (8) **

**Solution : **

According to the given question,

Inner and outer walls of a 30 cm long pipe have a diameter of 15 cm and 17 cm,

**∴** Volume of the material used in pipe,

= π× h × (R^{2} – r^{2})

= 22/7 × 30 ((17/2)^{2} – 915/2)^{2})

= 22/7 × 30 (8.52 – 7.52)

= 22/7 × 30 × (8.5 + 7.5) (8.6 – 7.5)

= 22/7 × 30 × 16 × 1

= 1508.5 m^{3}

Therefore, the volume of the materials will be 1508.5 m^{3}

**Question no – (9) **

**Solution : **

As per the question,

Diameter = 8 ft

radius = 4 ft

**∴** Cost of painting,

= (20 × 2 × 22/7 × 4 × 14/2.5)

= 2816 Rs.

Therefore, the cost of painting the inner wall of a well 2816 Rs.

**Question no – (10) **

**Solution : **

Height = 3m = 300 cm

Circumference = 252

∴ 2πr = 252

= 2 × 22/7 × r = 252

= r = 252 × 7/44

= 40.09 cm

Now, lateral surface area,

= 2πrh = 2 × 22/7 40.09 × 300

= 75600 cm^{2}

Therefore, the lateral surface area will be 75600 cm^{2}

**Question no – (11) **

**Solution : **

According to the question we know,

r = 0.5/2 m

**∴ **Total curved surface area,

= 2 × 22/7 × 0.5/2 × 1.5

= 33/14 sqm

**∴ **Number of revolutions,

= 264/33/14

= 264 × 14/33

= 112 revolutions

Therefore, it will take 112 revolutions.

**Question no – (12) **

**Solution : **

In the given question,

Rectangular piece of paper having length 4.4 cm and breadth 1.5 cm

First, the height,

= 2πr = 4.4

= r = 4.4 × 7/2 × 22

= 0.7 cm

Now, the Volume,

= (22/7 × 0.7 × 0.7 × 1.5)

= 2.310cm^{3}

Therefore, the height will be 0.7 cm and the volume will be 2.310cm^{3}

**Next Chapter Solution : **

👉 Chapter 15 👈