The DAV Maths Book Class 8 Solutions Pdf and **DAV Class 8 Maths Chapter 14 Brain Teasers** Solutions of Mensuration offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 14 Brain Teasers Solutions

Question 1A.

Tick (✓) the correct option.

(i) The volume of a cube of side 0.01 m in cm^{3} is

(a) 0.001

(b) 1

(c) 0.0001

(d) 0.000001

Solution:

Side of the cube, l = 0.01 m = 1 cm

The volume of the cube = l^{3}

= (1)^{3} cm^{3}

= 1 cm^{3}

Thus, option (b) is the correct answer.

(ii) If the height of a cylinder is halved, its volume will be

(a) \(\frac{1}{2}\) times

(b) \(\frac{1}{3}\) times

(c) 2 times

(d) 3 times

Solution:

Let the radius of the cylinder be r, height be h and volume be V_{1}.

Then, V_{1} = πr^{2}h ……(i)

If the height of the cylinder becomes its half i.e., \(\frac{h}{2}\)

then its volume, say V_{2} = πr^{2} \(\frac{h}{2}\) = \(\frac{\pi r^2 h}{2}\) ……(ii)

From (i) and (ii), it is clear that volume becomes half when the height of the cylinder is halved.

Thus, option (a) is the correct answer.

(iii) The area of a trapezium having two parallel sides as 10 cm and 12 cm and height as 4 cm is

(a) 42 cm^{2}

(b) 44 cm^{2}

(c) 46 cm^{2}

(d) 48 cm^{2}

Solution:

Area of Trapezium = \(\frac{1}{2}\) × height × sum of parallel sides

= \(\frac{1}{2}\) × 4 × (10 + 12)

= \(\frac{1}{2}\) × 4 × 22

= 44 cm^{2}

Thus, option (b) is the correct answer.

(iv) The number of faces a tetrahedron has

(a) 14

(b) 12

(c) 6

(d) 4

Solution:

A tetrahedron has 4 faces. Thus, option (d) is the correct answer.

(v) If the length of the side of a cube is doubled, then the ratio of volumes of the new cube and the original cube is

(a) 1 : 2

(b) 2 : 1

(c) 4 : 1

(d) 8 : 1

Solution:

Let the side of a cube be l and its volume be V_{1}.

Then, V_{1} = l^{3}

Now, the side of a new cube be 2l and its volume be V_{2}.

Then, V_{2} = (2l)^{3} = 8l^{3}

∴ \(\frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{8 l^3}{l^3}=\frac{8}{1}\)

∴ V_{2} : V_{1} = 8 : 1

Thus, option (d) is the correct answer.

Question 1B.

Answer the following questions.

(i) Find the number of edges of a polyhedron having 20 faces and 12 vertices.

Solution:

Here, F = 20, V = 12 and E = ?

Using Euler’s formula, we have

F + V – E = 2

⇒ 20 + 12 – E = 2

⇒ 32 – 2 = E

⇒ E = 30

(ii) Find the surface area of a cube whose volume is 729 m^{3}.

Solution:

Let the side of the cube be l.

Then, the volume of the cube = l^{3}

⇒ 729 = l^{3}

⇒ 9 × 9 × 9 = l^{3}

⇒ l^{3} = 9^{3}

⇒ l = 9 m

Surface area of the cube = 6l^{2}

= 6 × 9^{2}

= 6 × 81

= 486 m^{2}

(iii) The volume of a cylinder is 2,376 cm^{3}. If the diameter of its base is 12 cm, then find its height.

Solution:

The base diameter of the cylinder = 12 cm

∴ The base radius of the cylinder = 6 cm

Let height of the cylinder be h cm.

Then, the volume of the cylinder = πr^{2}h

2376 = \(\frac{22}{7}\) × 6 × 6 × h (∵ π = \(\frac{22}{7}\))

⇒ h = \(\frac{2376 \times 7}{22 \times 6 \times 6}\)

⇒ h = 21 cm

(iv) Draw the solid represented by the views given below:

Solution:

(v) Find the total surface area of a cuboid of length 10 cm, breadth 8 cm, and height 6 cm.

Solution:

Given: l = 10 cm, b = 8 cm and h = 6 cm

Total surface area of a cuboid = 2(lb + bh + hl)

= 2{(10 × 8) + (8 × 6) + (6 × 10)}

= 2(80 + 48 + 60)

= 2 × 188

= 376 cm^{2}

Question 2.

Find the area of the given quadrilateral.

Solution:

Area of the given quadrilateral = \(\frac{1}{2}\) × AC × (BE + FD)

= \(\frac{1}{2}\) × 6 × (3.8 + 7.2)

= \(\frac{1}{2}\) × 6 × 11

= 33 cm^{2}

Question 3.

A field is in the shape of a trapezium. One of its parallel sides is twice the other parallel side and distance between them is 100 m. If the area of the field is 10,500 m^{2}, find the length of the parallel sides.

Solution:

Let one of the parallel sides be x m

∴ other parallel side = 2x m

Area of the trapezium = \(\frac{1}{2}\) × (sum of parallel sides) × height

⇒ 10500 = \(\frac{1}{2}\) × (x + 2x) × 100

⇒ 10500 = 3x × 50

⇒ x = \(\frac{10500}{3 \times 50}\)

⇒ x = 70 m

Hence, the two parallel sides are 70 m and 70 × 2 = 140 m

Question 4.

Two parallel sides DC and AB of a trapezium are 12 cm and 36 cm respectively. If nonparallel sides are each equal to 15 cm, find the area of the trapezium.

Solution:

Draw CE || DA.

∴ ADCE is a parallelogram.

∴ DA = CE = 15 cm

and EB = 36 – 12 = 24 cm

∴ EF = \(\frac{24}{2}\) = 12 cm

∴ In right triangle CFE,

CE^{2} = EF^{2} + CF^{2} [From Pythagoras Theorem]

⇒ (15)^{2} = (12)^{2} + CF^{2}

⇒ 225 = 144 + CF^{2}

⇒ CF^{2} = 225 – 144

⇒ CF^{2} = 81

⇒ CF = 9

Area of trapezium ABCD = \(\frac{1}{2}\) × (AB + DC) × CF

= \(\frac{1}{2}\) × (36 + 12) × 9

= \(\frac{1}{2}\) × 48 × 9

= 216 cm^{2}

Question 5.

The length, breadth, and height of a cuboidal box are 2 m 10 cm, 1 m, and 80 cm respectively. Find the area of canvas required to cover this box.

Solution:

Here, l = 2 m 10 cm = 2.1 m

b = 1 m

h = 80 cm = 0.8 m

∴ Surface area of the cuboidal box = 2[lb + bh + lh]

= 2[2.1 × 1 + 1 × 0.8 + 2.1 × 0.8]

= 2[2.1 + 0.8 + 1.68]

= 2[4.58]

= 9.16 m^{2}

Question 6.

A metallic pipe is 0.7 cm thick. Inner radius of the pipe is 3.5 cm and length is 5 dm. Find its total surface area.

[Hint: Total surface area = Inner surface area + Outer surface area + Area of two rims]

Solution:

Here, inner radius r_{1} = 3.5 cm

thickness = 0.7 cm

∴ outer radius r_{2} = 3.5 + 0.7 = 4.2 cm

height h = 5 dm = 50 cm

∴ Total surface area = Inner surface area + Outer surface area + Area of two rims

= 2πr_{1}h + 2πr_{2}h + \(\pi\left(r_2^2+r_1^2\right)\)

= 2πh(r_{1} + r_{2}) + \(\pi\left(r_2^2+r_1^2\right)\)

= 2 × \(\frac{22}{7}\) × 50(3.5 + 4.2) + \(\frac{22}{7}\) × (3.52 + 4.22)

= 2 × \(\frac{22}{7}\) × 50 × 7.7 + \(\frac{22}{7}\) × (12.25 + 17.64)

= 2 × \(\frac{22}{7}\) × 50 × 7.7 + \(\frac{22}{7}\) × 29.89

= \(\frac{22}{7}\) × (2 × 50 × 7.7 + 29.89)

= \(\frac{22}{7}\) × (770 + 29.89)

= \(\frac{22}{7}\) × 799.89

= 2453.88 cm^{2}

Question 7.

A cylinder with a curved surface area of 1250 m^{2} is formed from a rectangular metallic sheet. Find the dimensions of the rectangular sheet if its length is double its breadth.

Solution:

Let the breadth be x m.

∴ Length = 2x m

Area of the rectangular sheet = Curved surface of the sheet

⇒ 2x × x = 1250

⇒ 2x^{2} = 1250

⇒ x^{2} = \(\frac{1250}{2}\) = 625

⇒ x = 25

Hence, length = 2 × 25 = 50 m and breadth = 25 m.

Question 8.

The diameter and length of a roller are 84 cm and 120 cm respectively. In how many revolutions, can the roller level the playground of area 1584 m^{2}?

Solution:

Here, r = \(\frac{84}{2}\) = 42 cm, h = 120 cm

Curved surface area = 2πrh

= 2 × \(\frac{22}{7}\) × 42 × 120

= 44 × 6 × 120

= 31680 cm^{2}

Area of the playground = 1584 m^{2} = 15840000 cm^{2}

∴ Number of revolutions = \(\frac{15840000}{31680}\) = 500

Hence, the number of revolutions required = 500

Question 9.

Water is poured into the reservoir at the rate of 60 liters per minute. If the volume of the reservoir is 108 m^{3}, find the number of hours it will take to fill the reservoir.

Solution:

Volume of the reservoir = 108 m^{3} = 108000 litres [∵ 1 m^{3} = 1000 l]

Rate of flow of water = 60 liters per minute

= 60 × 60

= 3600 litres per hour

∴ Number of hours taken by water to fill the reservoir = \(\frac{108000}{3600}\) = 30 hours.

Hence, the required number of hours = 30.

Question 10.

An iron pipe is 21 m long and its exterior diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weight is 8 g/cm^{3}, find the weight of the pipe.

Solution:

Here, exterior radius r_{2} = \(\frac{8}{2}\) = 4 cm

Thickness = 1 cm

∴ Inner radius r_{1} = 4 – 1 = 3 cm

∴ Volume of the iron pipe = \(\pi\left[r_2^2-r_1^2\right] \times h\)

= \(\frac{22}{7}\) × [4^{2} – 3^{2}] × 2100

= \(\frac{22}{7}\) × 7 × 2100

= 46200 cm^{3}

Weight of the pipe = 8 × 46200 g = 369600 g = 369.6 kg

Question 11.

A well is dug 20 m deep and it has a diameter of 7 m. The earth that is so dug out is spread evenly on a rectangular plot 22 m long and 14 m broad. What is the height of the platform formed?

Solution:

The volume of the earth dug out = πr^{2}h

= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20\)

= 770 m^{3}

Let the height of the platform be h m.

∴ Volume of the platform = 22 × 14 × h m^{3}

Volume of platform = Volume of the earth dug out

⇒ 22 × 14 × h = 770

⇒ h = \(\frac{770}{22 \times 14}\) = 2.5 m

Hence the required height = 2.5 m.

Question 12.

The volume of a right circular cylinder is 448π cm^{3} and its height is 7 cm. Find the lateral surface area and total surface area.

Solution:

Volume of cylinder = πr^{2}h

⇒ 448π = π × r^{2} × 72

⇒ r^{2} = \(\frac{448 \pi}{7 \times \pi}\) = 64

⇒ r = 8 cm

Lateral surface area = 2πrh

= 2 × \(\frac{22}{7}\) × 8 × 7

= 352 cm^{2}

Total surface area = 2πrh + 2πr^{2}

= 352 + 2 × \(\frac{22}{7}\) × 7 × 7

= 352 + 308

= 660 cm^{2}

Question 13.

Verify Euler’s formula for

(i) Square pyramid

(ii) Triangular prism

(iii) Rectangular prism.

Solution:

(i) For square pyramid,

F = 5, V = 5, E = 8

From Euler’s formula,

F + V – E = 2

⇒ 5 + 5 – 8 = 2

⇒ 2 = 2

Hence verified.

(ii) For triangular prism,

F = 5, V = 6, E = 9

From Euler’s formula,

F + V – E = 2

⇒ 5 + 6 – 9 = 2

⇒ 2 = 2

Hence verified.

(iii) For rectangular prism,

F = 6, V = 8, E = 12

From Euler’s formula,

F + V – E = 2

⇒ 6 + 8 – 12 = 2

⇒ 2 = 2

Hence verified.

### DAV Class 8 Maths Chapter 14 HOTS

Question 1.

In a cylindrical pipe (see figure), the area of the surfaces is given. Find the radius and height of the cylinder.

Solution:

The surface area of the circular face of the cylinder = 154 sq. units (Given)

⇒ πr^{2} = 154

⇒ \(\frac{22}{7}\) × r^{2} = 154

⇒ r^{2} = \(\frac{154 \times 7}{22}\) = 49

⇒ r = 7 units

The lateral surface area of the cylinder = 880 sq. units

⇒ 2πrh = 880

⇒ 2 × \(\frac{22}{7}\) × 7 × h = 880

⇒ h = \(\frac{880}{2 \times 22}\)

⇒ h = 20 units

Thus, the radius of the cylinder = 7 units, and the height = 20 units.

Question 2.

The length of a room is 50 percent more than its breadth. The cost of carpeting the room at the rate of ₹ 38.50 per m^{2} is ₹ 924 and the cost of painting the walls at the rate of ₹ 5.50 per m^{2} is ₹ 1,320. Find the dimensions of the room.

Solution:

Let the breadth of the room, i.e., b be x m.

Then, the length of the room, i.e., l = x + x × \(\frac{50}{100}\) = \(\frac{3x}{2}\) m

Area of the floor of the room = \(\frac{924}{38.50}\)

⇒ l × b = 24 m^{2}

⇒ \(\frac{3x}{2}\) × x = 24

⇒ 3x^{2} = 48

⇒ x = 4 m

∴ Length of the room = 6 m

and breadth of the room = 4 m

Area of the room to be painted = \(\frac{1320}{5.50}\) = 240 m^{2}

Area of four walls = 240

⇒ 2(l + b) × h = 240

⇒ 2(6 + 4) × h = 240

⇒ 2 × 10 × h = 240

⇒ h = 12 m

Thus, length = 6 m, breadth = 4 m and height = 12 m

### DAV Class 8 Maths Chapter 1 Enrichment Questions

Question 1.

This is a net of a triangular prism.

Can you tell me the 3D shape that you will get using the given net?

Draw the net of-

(i) a triangular pyramid.

(ii) a square prism.

(iii) a cone.

Solution:

We will get the right circular cylinder using the net shown alongside.

(i) Net of a triangular pyramid

(ii) Net of a square prism

(iii) Net of a cone

Additional Questions:

Question 1.

Find the area of a quadrilateral one of whose diagonal is 25 cm long and the lengths of perpendiculars from the other two vertices are 16.4 cm and 11.6 cm respectively.

Solution:

Area of ABCD = \(\frac{1}{2}\) × AC × (DE + BF)

= \(\frac{1}{2}\) × 25 × (11.6 + 16.4)

= \(\frac{1}{2}\) × 25 × 28

= 350 cm^{2}

Hence, the required area = 350 cm^{2}

Question 2.

Calculate the area of quadrilateral ABCD in which ABCD is an equilateral triangle with each side equal to 26 cm, ∠BAD = 90°, and AD = 24 cm.

Solution:

In ΔABD,

AB^{2} + AD^{2} = DB^{2} [By Pythagoras Theorem]

⇒ AB^{2} + (24)^{2} = (26)^{2}

⇒ AB^{2} + 576 = 676

⇒ AB^{2 }= 676 – 576

⇒ AB^{2} = 100

⇒ AB = 10

Area of DABCD = area of ΔDAB + area of ΔDCB

= \(\frac{1}{2}\) × AB × AD + \(\frac{\sqrt{3}}{4}\) × DB^{2}

= \(\frac{1}{2}\) × 10 × 24 + \(\frac{\sqrt{3}}{4}\) × (26)^{2}

= 120 + 169√3

= 120 + 169 × 1.732

= 120 + 292.37

= 412.37 cm^{2}

Question 3.

The two parallel sides of a trapezium are 58 m and 42 m long. The other two sides being equal are 17 m, find its area.

Solution:

Draw DE ⊥ AB and CF ⊥ AB.

∴ DC = EF = 42 m

and AE = FB = \(\frac{58-42}{2}\) = 8 m

In right ΔAED,

AE^{2} + ED^{2} = AD^{2} [By Pythagoras Theorem]

⇒ (8)^{2} + h^{2} = (17)^{2}

⇒ 64 + h^{2} = 289

⇒ h^{2} = 289 – 64

⇒ h^{2} = 225

⇒ h = 15 m

∴ Area of the trapezium ABCD = \(\frac{1}{2}\) × (AB + DC) × h

= \(\frac{1}{2}\) × (42 + 58) × 15

= \(\frac{1}{2}\) × 100 × 15

= 750 m^{2}

Hence, the required area = 750 m^{2}

Question 4.

The adjoining figure shows a field with the measurement given in meters. Find the area of the field.

Solution:

Area of the given field = ar(ΔAZB) + ar(Trap. ZBCX) + ar(ΔDXC) + ar(ΔDAE)

= [\(\frac{1}{2}\) × 10 × 25] + [\(\frac{1}{2}\) × (30 + 25) × 15] + [\(\frac{1}{2}\) × (30 × 12] + [\(\frac{1}{2}\) × (12 + 15 + 10) × 20]

= 125 + 412.5 + 180 + 370

= 1087.5 m^{2}

Hence, the required area = 1087.5 m^{2}

Question 5.

Find the volume, surface area, and lateral surface area of a cuboid whose length is 8.5 m, breadth is 6.4 m, and height is 50 cm.

Solution:

Here, l = 8.5 m, b = 6.4 m, h = 50 cm = \(\frac{1}{2}\) m

Volume = l × b × h

= 8.5 × 6.4 × \(\frac{1}{2}\)

= 27.2 m^{3}

Surface area = 2[lb + bh + lh]

= 2[8.5 × 6.4 + 6.4 × \(\frac{1}{2}\) + \(\frac{1}{2}\) × 8.5]

= 2[54.4 + 3.2 + 4.25]

= 2 × 61.5

= 123.70 m^{2}

Lateral surface area = 2[l + b] × h

= 2[8.5 + 6.4] × \(\frac{1}{2}\)

= 14.9 m^{2}

Hence, the required volume = 27.2 m^{3}

Surface area = 123.70 m^{2}

Lateral surface area = 14.9 m^{2}

Question 6.

Find the length of the longest rod that can be placed in a room measuring 12 m × 9 m × 8 m.

Solution:

Length of the longest rod = Length of the diagonal

= \(\sqrt{l^2+b^2+h^2}\)

= \(\sqrt{(12)^2+(9)^2+(8)^2}\)

= \(\sqrt{144+81+64}\)

= \(\sqrt{289}\)

= 17 m

Hence, the required length = 17 m

Question 7.

The volume of a cuboid is 14400 cm^{3} and its height is 15 cm. The cross-section of the cuboid is a rectangle having its sides in the ratio 5 : 3. Find the perimeter of the base.

Solution:

Let the length be 5x m and breadth be 3x m.

∴ Volume of cuboid = (Area of cross-section) × height

⇒ 14400 = (5x × 3x) × 15

⇒ 14400 = 225x^{2}

⇒ x^{2} = 64

⇒ x = 8 cm

Length = 5 × 8 = 40 cm

Breadth = 3 × 8 = 24 cm

∴ Perimeter of base = 2[l + b]

= 2[40 + 24]

= 2 × 64

= 128 cm

Hence, the required perimeter = 128 cm.

Question 8.

The sum of the length, breadth, and depth of a cuboid is 19 cm and the length of its diagonal is 11 cm. Find the surface area of the cuboid.

Solution:

l + b + h = 19 and \(\sqrt{l^2+b^2+h^2}\) = 11

⇒ l^{2} + b^{2} + h^{2} = 121

(l + b + h)^{2} = (19)^{2}

⇒ l^{2} + b^{2} + h^{2} + 2lb +2bh + 2lh = 361

⇒ 121 + 2[lb + bh + 2lh] = 361

⇒ 2[lb + bh + lh] = 361 – 121 = 240 m^{2}

Hence, the total surface area = 240 m^{2}

Question 9.

Three cubes each of side 6 cm, are joined end to end. Find the surface area of the resulting cuboid.

Solution:

Length of the resulting cuboid = 6 × 3 = 18 cm

breadth = 6 cm and height = 6 cm

∴ Surface area = 2[lb + bh + lh]

= 2[18 × 6 + 6 × 6 + 6 × 18]

= 2[108 + 36 + 108]

= 2 × 252

= 504 cm^{2}

Hence, the required surface area = 504 cm^{2}

Question 10.

A cylindrical tank has a capacity of 6160 m^{3}. Find its depth if its radius is 14 m. Also, find the cost of painting on a curved surface at ₹ 3 per m^{2}.

Solution:

Volume = πr^{2}h

⇒ 6160 = \(\frac{22}{7}\) × 14 × 14 × h

⇒ h = \(\frac{6160 \times 7}{22 \times 14 \times 14}\)

⇒ h = 10 m

Curved surface area = 2πrh

= 2 × \(\frac{22}{7}\) × 14 × 10

= 880 m^{2}

Cost of painting of its curved surface = 880 × ₹ 3 = ₹ 2640

Hence, the required depth is 10 m, and the required cost = ₹ 2640

Question 11.

The sum of the radius of the base and the height of a cylinder is 37 m. If the total surface area of the cylinder is 1628 m^{2}, find its volume.

Solution:

Given that (r + h) = 37 m

and total surface area = 1628 m^{2}

∴ T.S.A. = 2πr(h + r)

⇒ 1628 = 2 × \(\frac{22}{7}\) × r × 37

⇒ r = \(\frac{1628 \times 7}{2 \times 22 \times 37}\)

⇒ r = 7 m

∵ r + h = 37

⇒ 7 + h = 37

⇒ h = 37 – 7 = 30 cm

∴ Volume = πr^{2}h

= \(\frac{22}{7}\) × 7 × 7 × 30

= 4620 cm^{3}

Hence, the required volume = 4620 m^{3}

Question 12.

A swimming pool 70 m long, 44 m wide, and 3 m deep is filled by water issuing from a pipe of diameter 35 cm, at 6 m per second. How many hours does it take to fill the pool?

Solution:

The volume of the swimming pool = l × b × h

= 70 × 44 × 3

= 210 × 44

= 9240 m^{3}

The volume of water issuing from the pipe in 1 sec = πr^{2}h

= \(\frac{22}{7} \times \frac{35}{2 \times 100} \times \frac{35}{2 \times 100}\) × 6 m^{3}

∴ Volume of water flowing into swimming pool = \(\frac{22}{7} \times \frac{35}{2 \times 100} \times \frac{35}{2 \times 100}\) × 6 × 60 × 60 m^{3}

Hence, the required time = \(4 \frac{4}{9}\) hours.