DAV Class 8 Maths Chapter 14 Brain Teasers Solutions

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 cm³ 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 = l³
= (1)³ cm³
= 1 cm³
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 V₁.
Then, V₁ = πr²h ……(i)
If the height of the cylinder becomes its half i.e., \(\frac{h}{2}\)
then its volume, say V₂ = πr² \(\frac{h}{2}\) = \(\frac{\pi r^2 h}{2}\) ……(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 cm²
(b) 44 cm²
(c) 46 cm²
(d) 48 cm²
Solution:
Area of Trapezium = \(\frac{1}{2}\) × height × sum of parallel sides
= \(\frac{1}{2}\) × 4 × (10 + 12)
= \(\frac{1}{2}\) × 4 × 22
= 44 cm²
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 V₁.
Then, V₁ = l³
Now, the side of a new cube be 2l and its volume be V₂.
Then, V₂ = (2l)³ = 8l³
∴ \(\frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{8 l^3}{l^3}=\frac{8}{1}\)
∴ V₂ : V₁ = 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 m³.
Solution:
Let the side of the cube be l.
Then, the volume of the cube = l³
⇒ 729 = l³
⇒ 9 × 9 × 9 = l³
⇒ l³ = 9³
⇒ l = 9 m
Surface area of the cube = 6l²
= 6 × 9²
= 6 × 81
= 486 m²

(iii) The volume of a cylinder is 2,376 cm³. 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 = πr²h
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 cm²

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 cm²

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 m², 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,
CE² = EF² + CF² [From Pythagoras Theorem]
⇒ (15)² = (12)² + CF²
⇒ 225 = 144 + CF²
⇒ CF² = 225 – 144
⇒ CF² = 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 cm²

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 m²

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 r₁ = 3.5 cm
thickness = 0.7 cm
∴ outer radius r₂ = 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
= 2πr₁h + 2πr₂h + \(\pi\left(r_2^2+r_1^2\right)\)
= 2πh(r₁ + r₂) + \(\pi\left(r_2^2+r_1^2\right)\)
= 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 cm²

Question 7.
A cylinder with a curved surface area of 1250 m² 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
⇒ 2x² = 1250
⇒ x² = \(\frac{1250}{2}\) = 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 m²?
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 cm²
Area of the playground = 1584 m² = 15840000 cm²
∴ 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 m³, find the number of hours it will take to fill the reservoir.
Solution:
Volume of the reservoir = 108 m³ = 108000 litres [∵ 1 m³ = 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/cm³, find the weight of the pipe.
Solution:
Here, exterior radius r₂ = \(\frac{8}{2}\) = 4 cm
Thickness = 1 cm
∴ Inner radius r₁ = 4 – 1 = 3 cm
∴ Volume of the iron pipe = \(\pi\left[r_2^2-r_1^2\right] \times h\)
= \(\frac{22}{7}\) × [4² – 3²] × 2100
= \(\frac{22}{7}\) × 7 × 2100
= 46200 cm³
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 = πr²h
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20\)
= 770 m³
Let the height of the platform be h m.
∴ Volume of the platform = 22 × 14 × h m³
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 cm³ and its height is 7 cm. Find the lateral surface area and total surface area.
Solution:
Volume of cylinder = πr²h
⇒ 448π = π × r² × 72
⇒ r² = \(\frac{448 \pi}{7 \times \pi}\) = 64
⇒ r = 8 cm
Lateral surface area = 2πrh
= 2 × \(\frac{22}{7}\) × 8 × 7
= 352 cm²
Total surface area = 2πrh + 2πr²
= 352 + 2 × \(\frac{22}{7}\) × 7 × 7
= 352 + 308
= 660 cm²

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)
⇒ πr² = 154
⇒ \(\frac{22}{7}\) × r² = 154
⇒ r² = \(\frac{154 \times 7}{22}\) = 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 m² is ₹ 924 and the cost of painting the walls at the rate of ₹ 5.50 per m² 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 m²
⇒ \(\frac{3x}{2}\) × x = 24
⇒ 3x² = 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 m²
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 cm²
Hence, the required area = 350 cm²

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,
AB² + AD² = DB² [By Pythagoras Theorem]
⇒ AB² + (24)² = (26)²
⇒ AB² + 576 = 676
⇒ AB² = 676 – 576
⇒ AB² = 100
⇒ AB = 10
Area of DABCD = area of ΔDAB + area of ΔDCB
= \(\frac{1}{2}\) × AB × AD + \(\frac{\sqrt{3}}{4}\) × DB²
= \(\frac{1}{2}\) × 10 × 24 + \(\frac{\sqrt{3}}{4}\) × (26)²
= 120 + 169√3
= 120 + 169 × 1.732
= 120 + 292.37
= 412.37 cm²

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,
AE² + ED² = AD² [By Pythagoras Theorem]
⇒ (8)² + h² = (17)²
⇒ 64 + h² = 289
⇒ h² = 289 – 64
⇒ h² = 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 m²
Hence, the required area = 750 m²

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 m²
Hence, the required area = 1087.5 m²

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 m³
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 m²
Lateral surface area = 2[l + b] × h
= 2[8.5 + 6.4] × \(\frac{1}{2}\)
= 14.9 m²
Hence, the required volume = 27.2 m³
Surface area = 123.70 m²
Lateral surface area = 14.9 m²

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 cm³ 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 = 225x²
⇒ x² = 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
⇒ l² + b² + h² = 121
(l + b + h)² = (19)²
⇒ l² + b² + h² + 2lb +2bh + 2lh = 361
⇒ 121 + 2[lb + bh + 2lh] = 361
⇒ 2[lb + bh + lh] = 361 – 121 = 240 m²
Hence, the total surface area = 240 m²

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 cm²
Hence, the required surface area = 504 cm²

Question 10.
A cylindrical tank has a capacity of 6160 m³. Find its depth if its radius is 14 m. Also, find the cost of painting on a curved surface at ₹ 3 per m².
Solution:
Volume = πr²h
⇒ 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 m²
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 m², find its volume.
Solution:
Given that (r + h) = 37 m
and total surface area = 1628 m²
∴ 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 = πr²h
= \(\frac{22}{7}\) × 7 × 7 × 30
= 4620 cm³
Hence, the required volume = 4620 m³

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 m³
The volume of water issuing from the pipe in 1 sec = πr²h
= \(\frac{22}{7} \times \frac{35}{2 \times 100} \times \frac{35}{2 \times 100}\) × 6 m³
∴ 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 m³

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