Math Ncert Exemplar 2019 Solutions for Class 8 Maths Chapter 11 Mensuration are provided here with simple step-by-step explanations. These solutions for Mensuration are extremely popular among class 8 students for Maths Mensuration Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Math Ncert Exemplar 2019 Book of class 8 Maths Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Math Ncert Exemplar 2019 Solutions. All Math Ncert Exemplar 2019 Solutions for class 8 Maths are prepared by experts and are 100% accurate.
Page No 344:
Question 1:
In questions âthere are four options out of which one is correct.
A cube of side 5 cm is painted on all its faces. If it is sliced into 1 cubic centimetre cubes, how many 1 cubic centimetre cubes will have exactly one of their faces painted?
(a) 27
(b) 42
(c) 54
(d) 142
Answer:
The side of the cube measures 5 cm. It is sliced into five equal parts of side 1 cm each.
In one face of the cube, there are a total of 9 cubes painted with one side.
Since there are 6 faces in a cube.
Therefore, total number of painted faces = 9 × 6 = 54
Hence, the correct answer is option (c).
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Question 2:
In questions âthere are four options out of which one is correct.
A cube of side 4 cm is cut into 1 cm cubes. What is the ratio of the surface areas of the original cube and cut-out cubes?
(a) 1 : 2
(b) 1 : 3
(c) 1 : 4
(d) 1 : 6
Answer:
The side of a cube is 4 cm. It is cut into cubes of side 1 cm each.
Now, volume of the original cube with side 4 cm = (side)3 = (4)3 = 64 cm3
Also, the volume of the cut-out cubes of side 1 cm = (side)3 = (1)3 = 1 cm3
Now, surface area of 64 cut-out cubes = 64 × 6 × (side)2 = 64 × 6 × (1)2
And
surface area of original cubes = 6 × (side)2 = 6 × (4)2
Thus, the required ratio is 1 : 4.
Hence, the correct answer is option (c).
Page No 345:
Question 3:
In questions âthere are four options out of which one is correct.
A circle of maximum possible size is cut from a square sheet of board. Subsequently, a square of maximum possible size is cut from the resultant circle. What will be the area of the final square?
Answer:
Let the initial square have side x.
Therefore,
Area of the initial square = x × x = x2
The maximum possible size of the circle that can be obtained from this square will have the diameter x.
Now, the diagonal of the final square will be equal to the diameter of the circle. Thus, the diagonal of the final square is x.
Therefore,
Area of the final square = Area of rhombus with diagonals x (âµ square is a rhombus with equal diagonals)
Hence, the correct answer is option (b).
Page No 345:
Question 4:
In questions âthere are four options out of which one is correct.
What is the area of the largest triangle that can be fitted into a rectangle of length l units and width w units?
(a) lw/2
(b) lw/3
(c) lw/6
(d) lw/4
Answer:
Consider a rectangle of length l units and width w units.
Area of triangle =
The area of the largest triangle that can be fitted inside this rectangle =
=
Hence, the correct answer is option (a).
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Question 5:
In questions âthere are four options out of which one is correct.
If the height of a cylinder becomes of the original height and the radius is doubled, then which of the following will be true?
(a) Volume of the cylinder will be doubled.
(b) Volume of the cylinder will remain unchanged.
(c) Volume of the cylinder will be halved.
(d) Volume of the cylinder will be of the original volume.
Answer:
Consider a cylinder with radius r and height h.
So,
Volume of cylinder =
If the height of a cylinder becomes of the original height and the radius is doubled, then
H =
and R = 2r
Hence, the correct answer is option (b).
Page No 346:
Question 6:
In questions âthere are four options out of which one is correct.
If the height of a cylinder becomes of the original height and the radius is doubled, then which of the following will be true?
(a) Curved surface area of the cylinder will be doubled.
(b) Curved surface area of the cylinder will remain unchanged.
(c) Curved surface area of the cylinder will be halved.
(d) Curved surface area will be of the original curved surface.
Answer:
Consider a cylinder with radius r and height h.
So,
Curved surface area of cylinder =
If the height of a cylinder becomes of the original height and the radius is doubled, then
H =
and R = 2r
Thus, the curved surface area of the cylinder will be halved.
Hence, the correct answer is option (c).
Page No 346:
Question 7:
In questions âthere are four options out of which one is correct.
If the height of a cylinder becomes of the original height and the radius is doubled, then which of the following will be true?
(a) Total surface area of the cylinder will be doubled.
(b) Total surface area of the cylinder will remain unchanged.
(c) Total surface of the cylinder will be halved.
(d) None of the above.
Answer:
Consider a cylinder with radius r and height h.
So,
Total surface area of cylinder =
If the height of a cylinder becomes of the original height and the radius is doubled, then
H =
and R = 2r
Hence, the correct answer is option (d).
Page No 346:
Question 8:
In questions âthere are four options out of which one is correct.
The surface area of the three coterminous faces of a cuboid are 6, 15 and 10 cm2 respectively. The volume of the cuboid is
(a) 30 cm3
(b) 40 cm3
(c) 20 cm3
(d) 35 cm3
Answer:
The three coterminous faces of a cuboid are 6, 15 and 10 cm2 respectively.
So, let
l × h = 6 cm2
l × b = 15 cm2
b × h = 10 cm2
Now, volume of a cuboid = lbh
Here, lbh = 30 cm3.
Hence, the correct answer is option (a).
Page No 346:
Question 9:
In questions âthere are four options out of which one is correct.
A regular hexagon is inscribed in a circle of radius r. The perimeter of the regular hexagon is
(a) 3r
(b) 6r
(c) 9r
(d) 12r
Answer:
Consider a regular hexagon inscribed in a circle of radius r.
Now, a regular hexagon is made up of six identical triangles. Thus, the angle made by each triangle at the centre is .
Consider AOB.
OA = OB = r
Since, angles opposite to equal sides are equal.
∴ ∠OBA = ∠OAB
⇒ ∠OBA = ∠OAB = 60°
Thus, a regular hexagon is made up of six equilateral triangles with side r.
Therefore,
Perimeter of the regular hexagon = 6 × side
= 6 × r
= 6r
Hence, the correct answer is option (b).
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Question 10:
In questions âthere are four options out of which one is correct.
The dimensions of a godown are 40 m, 25 m and 10 m. If it is filled with cuboidal boxes each of dimensions 2 m × 1.25 m × 1 m, then the number of boxes will be
(a) 1800
(b) 2000
(c) 4000
(d) 8000
Answer:
Volume of a cuboid = lbh
Here,
Volume of the godown = 40 m × 25 m × 10 m
Volume of the cuboidal boxes = 2 m × 1.25 m × 1 m
Hence, the correct answer is option (c).
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Question 11:
In questions âthere are four options out of which one is correct.
The volume of a cube is 64 cm3. Its surface area is
(a) 16 cm2
(b) 64 cm2
(c) 96 cm2
(d) 128 cm2
Answer:
The volume of a cube with side a = (a)3
Here,
(a)3 = 64
⇒ a = 4 cm
Now, surface area of a cube with side a = 6a2
Here,
Surface area of the cube = 6 × (4)2
= 96 cm2
Hence, the correct answer is option (c).
Page No 347:
Question 12:
In questions âthere are four options out of which one is correct.
If the radius of a cylinder is tripled but its curved surface area is unchanged, then its height will be
(a) tripled
(b) constant
(c) one sixth
(d) one third
Answer:
Consider a cylinder with radius r and height h.
So,
Curved surface area of cylinder =
If the radius of a cylinder is tripled, then R = 3r. Let the new height be H.
Since, its curved surface area is unchanged.
Thus, its height will be one third.
Hence, the correct answer is option (d).
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Question 13:
In questions âthere are four options out of which one is correct.
How many small cubes with edge of 20 cm each can be just accommodated in a cubical box of 2 m edge?
(a) 10
(b) 100
(c) 1000
(d) 10000
Answer:
Volume of cube = (side)3
Volume of a cuboid = lbh
Here,
Volume of the cubical box = 2 m × 2 m × 2 m
Volume of the smaller cubes = 20 cm × 20 cm × 20 cm
= 0.2 m × 0.2 m × 0.2 m
Hence, the correct answer is option (c).
Page No 347:
Question 14:
In questions âthere are four options out of which one is correct.
The volume of a cylinder whose radius r is equal to its height is
Answer:
Consider a cylinder with radius and height r.
So,
Hence, the correct answer is option (c).
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Question 15:
In questions âthere are four options out of which one is correct.
The volume of a cube whose edge is 3x is
(a) 27x3
(b) 9x3
(c) 6x3
(d) 3x3
Answer:
Volume of cube = (side)3
Here, edge of the cube is 3x.
So,
Volume of the cube with side 3x = (3x)3
= 27x3
Hence, the correct answer is option (a).
Page No 347:
Question 16:
In questions âthere are four options out of which one is correct.
The figure ABCD is a quadrilateral in which AB = CD and BC = AD. Its area is
(a) 72 cm2
(b) 36 cm2
(c) 24 cm2
(d) 18 cm2
Answer:
In the given quadrilateral, there are two triangles, ABC and ADC.
Consider ABC and ADC.
AB = CD (given)
BC = AD (given)
AC = AC (common)
by SSS congruence.
This means that both the triangles will have the same area.
Hence, the correct answer is option (b).
Page No 347:
Question 17:
In questions âthere are four options out of which one is correct.
What is the area of the rhombus ABCD below if AC = 6 cm, and BE = 4cm?
(a) 36 cm2
(b) 16 cm2
(c) 24 cm2
(d) 13 cm2
Answer:
In the given rhombus, there are two triangles, ABC and ADC.
Consider ABC and ADC.
AB = AD (sides of a rhombus are equal)
BC = DC (sides of a rhombus are equal)
AC = AC (common)
by SSS congruence.
This means that both the triangles will have the same area.
Hence, the correct answer is option (c).
Page No 348:
Question 18:
In questions âthere are four options out of which one is correct.
The area of a parallelogram is 60 cm2 and one of its altitude is 5 cm. The length of its corresponding side is
(a) 12 cm
(b) 6 cm
(c) 4 cm
(d) 2 cm
Answer:
Given that, area of a parallelogram is 60 cm2 and one of its altitude is 5 cm.
Now, area of a parallelogram = b × h
Here,
Hence, the correct answer is option (a).
Page No 348:
Question 19:
In questions âthere are four options out of which one is correct.
The perimeter of a trapezium is 52 cm and its each non-parallel side is equal to 10 cm with its height 8 cm. Its area is
(a) 124 cm2
(b) 118 cm2
(c) 128 cm2
(d) 112 cm2
Answer:
Given that, the perimeter of the trapezium is 52 cm and each of its non-parallel sides is 10 cm with its height 8 cm.
Now,
Perimeter of trapezium = Sum of parallel sides + sum of non parallel sides
= Sum of parallel sides + (10 + 10)
= Sum of parallel sides + 20
So,
52 = Sum of parallel sides + 20
⇒ Sum of parallel sides = 32 cm
Now,
Hence, the correct answer is option (c).
Page No 348:
Question 20:
In questions âthere are four options out of which one is correct.
Area of a quadrilateral ABCD is 20 cm2 and perpendiculars on BD from opposite vertices are 1 cm and 1.5 cm. The length of BD is
(a) 4 cm
(b) 15 cm
(c) 16 cm
(d) 18 cm
Answer:
Given that, the area of a quadrilateral ABCD is 20 cm2 and the perpendiculars on BD from opposite vertices are 1 cm and 1.5 cm.
Now,
Area of a quadrilateral ABCD = ar(ADB) + ar(BCD)
Hence, the correct answer is option (c).
Page No 348:
Question 21:
In questions âthere are four options out of which one is correct.
A metal sheet 27 cm long, 8 cm broad and 1 cm thick is melted into a cube. The side of the cube is
(a) 6 cm
(b) 8 cm
(c) 12 cm
(d) 24 cm
Answer:
Volume of the metal sheet = 27 cm × 8 cm × 1 cm
= 216 cm2
Since, the metal sheet is melted and recast into a cube.
Therefore, their volume remains the same.
∴ volume of the cube = 216 cm2
Now, volume of a cube of side a = (a)3
So,
216 = (side)3
⇒ side of the cube = 6 cm
Hence, the correct answer is option (a).
Page No 348:
Question 22:
In questions âthere are four options out of which one is correct.
Three cubes of metal whose edges are 6 cm, 8 cm and 10 cm respectively are melted to form a single cube. The edge of the new cube is
(a) 12 cm
(b) 24 cm
(c) 18 cm
(d) 20 cm
Answer:
When three cubes are melted to form a new cube, the sum of the volumes of the three cubes is equal to the volume of the single cube formed later.
Now, volume of a cube = (side)3
Here,
Thus, volume of the new cube = 1728 cm3 = (side)3
⇒ side = 12 cm
Hence, the correct answer is option (a).
Page No 348:
Question 23:
In questions âthere are four options out of which one is correct.
A covered wooden box has the inner measures as 115 cm, 75 cm and 35 cm and thickness of wood as 2.5 cm. The volume of the wood is
(a) 85,000 cm3
(b) 80,000 cm3
(c) 82,125 cm3
(d) 84,000 cm3
Answer:
Given that, a covered wooden box has the inner measures as 115 cm, 75 cm and 35 cm and thickness of wood as 2.5 cm.
Now,
Volume of a cuboid = lbh
Thus,
Volume of the inner box = 115 × 75 × 35
= 301875 cm3
Since, the thickness of the box is 2.5 cm.
Therefore, the dimensions of the outer box becomes:
l = 115 cm + 5 cm = 120 cm
b = 75 cm + 5 cm = 80 cm
h = 35 cm + 5 cm = 40 cm
∴ The volume of the outer box = 120 × 80 × 40
= 384000 cm3
Therefore,
Volume of the wood = 384000 − 301875 = 82125 cm3
Hence, the correct answer is option (c).
Page No 348:
Question 24:
In questions âthere are four options out of which one is correct.
The ratio of radii of two cylinders is 1 : 2 and heights are in the ratio 2 : 3. The ratio of their volumes is
(a) 1 : 6
(b) 1 : 9
(c) 1 : 3
(d) 2 : 9
Answer:
Given that, the ratio of radii of two cylinders is 1 : 2 and heights are in the ratio 2 : 3.
So, let the radius and height of the first cylinder be x and 2y and that of the second cylinder be 2x and 3y.
Consider a cylinder with radius r and height h.
So,
Volume of cylinder =
Here,
Hence, the correct answer is option (a).
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Question 25:
In questions âthere are four options out of which one is correct.
Two cubes have volumes in the ratio 1 : 64. The ratio of the area of a face of first cube to that of the other is
(a) 1 : 4
(b) 1 : 8
(c) 1 : 16
(d) 1 : 32
Answer:
Given that, the two cubes have volumes in the ratio 1 : 64.
Now,
Volume of a cube = (side)3
Here,
Therefore, let the edge of the first and the second cube be x and 4x respectively.
Now,
Hence, the correct answer is option (c).
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Question 26:
In questions âthere are four options out of which one is correct.
The surface areas of the six faces of a rectangular solid are 16, 16, 32, 32, 72 and 72 square centimetres. The volume of the solid, in cubic centimetres, is
(a) 192
(b) 384
(c) 480
(d) 2592
Answer:
Given that, the surface areas of the six faces of a rectangular solid are 16, 16, 32, 32, 72 and 72 cm2.
So, let
l × b = 16
b × h = 32
h × l = 72
Multiplying (1), (2) and (3),
Now, the volume of the solid = lbh
∴ Volume of the solid = 192 cm3
Hence, the correct answer is option (a).
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Question 27:
In questions âthere are four options out of which one is correct.
Ramesh has three containers.
(a) Cylindrical container A having radius r and height h,
(b) Cylindrical container B having radius 2r and height h, and
(c) Cuboidal container C having dimensions r × r × h
The arrangement of the containers in the increasing order of their volumes is
(a) A, B, C
(b) B, C, A
(c) C, A, B
(d) cannot be arranged
Answer:
The volume of a cylinder with radius r and height h =
(a) Container A has radius r and height h.
(b) Container B has radius 2r and height h.
(c) Container C has dimensions r × r × h.
Now,
Therefore, the arrangement of the containers in the increasing order of their volumes is C, A, B.
Hence, the correct answer is option (c).
Page No 349:
Question 28:
In questions âthere are four options out of which one is correct.
If R is the radius of the base of the hat, then the total outer surface area of the hat is
(a) πr (2h + R)
(b) 2πr (h + R)
(c) 2πrh + πR2
(d) None of these
Answer:
Given that, R is the radius of the base of the hat, height of the hat is h and radius of the cylindrical part of the hat is r.
Now, the TSA of the cylindrical part of the hat does not include the surface area of the base as it is open.
∴ TSA of the cylindrical part of the hat =
Also, the TSA of the hat should include the surface area of the base of the hat. Again, this does not include the surface area of the inner circular part.
∴ TSA of the base of the hat =
Thus,
TSA of the hat = +
Hence, the correct answer is option (c).
Page No 349:
Question 29:
Fill in the blanks to make the statements true.
A cube of side 4 cm is painted on all its sides. If it is sliced in 1 cubic cm cubes, then number of such cubes that will have exactly two of their faces painted is __________.
Answer:
Given that, a cube of side 4 cm is painted on all its sides and it is sliced in 1 cubic cm cubes.
Now, the number of cubes with exactly two of their faces painted will be the ones present on the edges. But the 8 smaller cubes on the corners of the cube will have three faces painted.
Since, a cube has 12 edges with the two middle cubes having two faces painted.
Therefore,
the number of cubes with exactly two painted faces = 12 × 2
= 24
Hence, a cube of side 4 cm is painted on all its sides. If it is sliced in 1 cubic cm cubes, then number of such cubes that will have exactly two of their faces painted is 24 .
Page No 349:
Question 30:
Fill in the blanks to make the statements true.
A cube of side 5 cm is cut into 1 cm cubes. The percentage increase in volume after such cutting is __________.
Answer:
Since, a cube of side 5 cm is cut into 1 cm cubes.
Therefore, the volume of the bigger cube and the smaller cubes remains the same.
Hence, the percentage increase in volume after such cutting is 0% .
Page No 349:
Question 31:
Fill in the blanks to make the statements true.
The surface area of a cuboid formed by joining two cubes of side a face to face is __________.
Answer:
When two cubes of side a are joined face to face, the length of the resulting cuboid becomes 2a.
Now,
Total surface area of a cuboid = 2(lb + bh + hl)
Therefore,
Surface area of the new cuboid = 2 × (2a × a + a × a + a × 2a)
= 2 × (2a2 + a2 + 2a2)
= 2 × (5a2)
= 10a2
Hence, the surface area of a cuboid formed by joining two cubes of side a face to face is 10a2.
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Question 32:
Fill in the blanks to make the statements true.
If the diagonals of a rhombus get doubled, then the area of the rhombus becomes __________ its original area.
Answer:
Area of a rhombus with diagonals d1 and d2 =
Here, the diagonals of a rhombus get doubled.
Hence, if the diagonals of a rhombus get doubled, then the area of the rhombus becomes four times its original area.
Page No 350:
Question 33:
Fill in the blanks to make the statements true.
If a cube fits exactly in a cylinder with height h, then the volume of the cube is __________ and surface area of the cube is __________.
Answer:
If a cube fits exactly in a cylinder with height h, then the edge of the cube is h.
Now,
Volume of cube = (edge)3
and
Surface area of the cube = 6(edge)2
Here,
Volume of the cube = h3
and
Surface area of the cube = 6h2
Hence, if a cube fits exactly in a cylinder with height h, then the volume of the cube is h3 and surface area of the cube is 6h2.
Page No 350:
Question 34:
Fill in the blanks to make the statements true.
The volume of a cylinder becomes __________ the original volume if its radius becomes half of the original radius.
Answer:
Consider a cylinder with radius r and height h.
So,
Volume of cylinder =
Now, if its radius becomes half of the original radius, then
Hence, the volume of a cylinder becomes one-fourth the original volume if its radius becomes half of the original radius.
Page No 350:
Question 35:
Fill in the blanks to make the statements true.
The curved surface area of a cylinder is reduced by ____________ per cent if the height is half of the original height.
Answer:
Consider a cylinder with radius r and height h.
So,
Curved surface area of cylinder =
If the height is half of the original height,
Now,
Hence, the curved surface area of a cylinder is reduced by 50 per cent if the height is half of the original height.
Page No 350:
Question 36:
Fill in the blanks to make the statements true.
The volume of a cylinder which exactly fits in a cube of side a is __________.
Answer:
If a cylinder fits exactly in a cube of side a, then the height of the cylinder is a and its radius is .
Now,
Volume of a cylinder with radius r and height h =
Here,
Hence, the volume of a cylinder which exactly fits in a cube of side a is .
Page No 350:
Question 37:
Fill in the blanks to make the statements true.
The surface area of a cylinder which exactly fits in a cube of side b is __________.
Answer:
If a cylinder fits exactly in a cube of side b, then the height of the cylinder is b and its radius is .
Now,
TSA of a cylinder with radius r and height h =
Here,
Hence, the surface area of a cylinder which exactly fits in a cube of side b is .
Page No 350:
Question 38:
Fill in the blanks to make the statements true.
If the diagonal d of a quadrilateral is doubled and the heights h1 and h2 falling on d are halved, then the area of quadrilateral is __________.
Answer:
Given that, the diagonal d of a quadrilateral is doubled and the two triangles into which the quadrilateral is divided with the heights h1 and h2 falling on d are halved.
Now,
new diagonal = d' = 2d
new heights, h1' and h2' = and respectively.
Therefore,
Hence, if the diagonal d of a quadrilateral is doubled and the heights h1 and h2 falling on d are halved, then the area of quadrilateral is .
Page No 350:
Question 39:
Fill in the blanks to make the statements true.
The perimeter of a rectangle becomes __________ times its original perimeter, if its length and breadth are doubled.
Answer:
Perimeter of a rectangle with length l and breadth b = 2(l + b)
Here, the length and breadth are doubled.
So,
new length = 2l
new breadth = 2b
∴ Perimeter of the rectangle = 2(2l + 2b)
= 2 × 2(l + b)
= 2 × original perimeter
Hence, the perimeter of a rectangle becomes two times its original perimeter, if its length and breadth are doubled.
Page No 350:
Question 40:
Fill in the blanks to make the statements true.
A trapezium with 3 equal sides and one side double the equal side can be divided into __________ equilateral triangles of _______ area.
Answer:
Let three equal sides of a trapezium be x and fourth side be 2x.
Now, an equilateral triangle has three equal sides.
Therefore,
Hence, a trapezium with 3 equal sides and one side double the equal side can be divided into three equilateral triangles of equal area.
Page No 350:
Question 41:
Fill in the blanks to make the statements true.
All six faces of a cuboid are __________ in shape and of ______ area.
Answer:
All six faces of a cuboid are rectangular in shape and of unequal area.
Page No 350:
Question 42:
Fill in the blanks to make the statements true.
Opposite faces of a cuboid are _________ in area.
Answer:
Opposite faces of a cuboid are equal in area.
Page No 350:
Question 43:
Fill in the blanks to make the statements true.
Curved surface area of a cylinder of radius h and height r is _______.
Answer:
Curved surface area of a cylinder of radius r and height h is .
Here,
Curved surface area of a cylinder of radius h and height r is .
Hence, the curved surface area of a cylinder of radius h and height r is .
Page No 350:
Question 44:
Fill in the blanks to make the statements true.
Total surface area of a cylinder of radius h and height r is _________.
Answer:
Total surface area of a cylinder of radius r and height h =
Here,
Total surface area of a cylinder of radius h and height r =
Hence, the total surface area of a cylinder of radius h and height r is .
Page No 350:
Question 45:
Fill in the blanks to make the statements true.
Volume of a cylinder with radius h and height r is __________.
Answer:
The volume of a cylinder with radius r and height h =
Here,
Volume of a cylinder with radius h and height r =
Hence, the volume of a cylinder with radius h and height r is .
Page No 351:
Question 46:
Fill in the blanks to make the statements true.
Area of a rhombus = product of _________.
Answer:
Let d1 and d2 be two diagonals of a rhombus.
Area of a rhombus =
Hence, area of a rhombus = product of diagonals.
Page No 351:
Question 47:
Fill in the blanks to make the statements true.
Two cylinders A and B are formed by folding a rectangular sheet of dimensions 20 cm × 10 cm along its length and also along its breadth respectively. Then volume of A is ________ of volume of B.
Answer:
Volume of a cylinder with radius r and height h =
Cylinder A is formed by folding a rectangular sheet of dimensions 20 cm × 10 cm along its length. Therefore, the circumference of the base of the cylinder is 20 cm and height of the cylinder is 10 cm.
and h1 = 10
Similarly, cylinder B is formed by folding a rectangular sheet of dimensions 20 cm × 10 cm along its breadth. Therefore, the circumference of the base of the cylinder is 10 cm and height of the cylinder is 20 cm.
and h2 = 20
Now,
Hence, volume of A is twice of volume of B.
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Question 48:
Fill in the blanks to make the statements true.
In the above question, curved surface area of A is ________ curved surface area of B.
Answer:
CSA of a cylinder with radius r and height h =
Cylinder A is formed by folding a rectangular sheet of dimensions 20 cm × 10 cm along its length. Therefore, the circumference of the base of the cylinder is 20 cm and height of the cylinder is 10 cm.
and h1 = 10
Similarly, cylinder B is formed by folding a rectangular sheet of dimensions 20 cm × 10 cm along its breadth. Therefore, the circumference of the base of the cylinder is 10 cm and height of the cylinder is 20 cm.
and h2 = 20
Now,
Hence, the curved surface area of A is equal to the curved surface area of B.
Page No 351:
Question 49:
Fill in the blanks to make the statements true
__________ of a solid is the measurement of the space occupied by it.
Answer:
Volume of a solid is the measurement of the space occupied by it.
Page No 351:
Question 50:
Fill in the blanks to make the statements true
__________ surface area of room = area of 4 walls.
Answer:
Lateral surface area of room = area of 4 walls.
Page No 351:
Question 51:
Fill in the blanks to make the statements true
Two cylinders of equal volume have heights in the ratio 1 : 9. The ratio of their radii is __________.
Answer:
Given that, two cylinders of equal volume have heights in the ratio 1 : 9.
Now,
Volume of a cylinder with radius r and height h =
Let r1 and r2 be the radii of the two cylinders.
Hence, the ratio of their radii is 3 : 1.
Page No 351:
Question 52:
Fill in the blanks to make the statements true
Two cylinders of same volume have their radii in the ratio 1 : 6, then ratio of their heights is __________.
Answer:
Given that, two cylinders of same volume have their radii in the ratio 1 : 6.
Now,
Volume of a cylinder with radius r and height h =
Let the radii of the two cylinders be x and 6x respectively. Let h1 and h2 be the heights of the two cylinders.
Since, they have the same volume.
Hence, the ratio of their heights is 36 : 1.
Page No 351:
Question 53:
State whether the statements are true (T) or false (F).
The areas of any two faces of a cube are equal.
Answer:
True
Consider a cube of side a. Now, area of each face of the cube is a square of side a.
Thus,
Area of each face of a cube = a2
Hence, the statement is true.
Page No 351:
Question 54:
State whether the statements are true (T) or false (F).
The areas of any two faces of a cuboid are equal.
Answer:
False
Let l, b and h be the length, breadth and height of the cuboid respectively.
Now, area of rectangle = l × b
Here,
ar(ABCD) = lb
ar(EFGH) = lb
ar(ABFE) = bh
ar(CDHG) = bh
ar(BCGF) = lh
ar(ADHE) = lh
Thus, the area of opposite faces of a cuboid are equal.
Hence, the statement is false.
Page No 351:
Question 55:
State whether the statements are true (T) or false (F).
The surface area of a cuboid formed by joining face to face 3 cubes of side x is 3 times the surface area of a cube of side x.
Answer:
The surface area of a cube of side x = 6x2
The surface area of a cuboid = 2(lb + bh + hl)
Here,
Surface area of the cube with side x = 6x2
The cuboid formed by joining three cubes will have:
Length = 3x
Breadth = x
Height = x
∴ Surface area of a cuboid formed = 2 × (3x × x + x × x + 3x × x)
= 2 × (3x2 + x2 + 3x2)
= 14x2
And 14x2 ≠ 3(6x2).
Hence, the statement is false.
Page No 351:
Question 56:
State whether the statements are true (T) or false (F).
Two cuboids with equal volumes will always have equal surface areas.
Answer:
False
Consider two cuboids such that:
Dimensions of the first cuboid = 5 cm, 4 cm and 6 cm
Dimensions of the second cuboid = 4 cm, 3 cm and 10 cm
Here, the volume of both the cuboids is the same, i.e., 120 cm3. (âµ Volume of cuboid = lbh)
Surface area of the first cuboid = 2(5 × 4 + 4 × 6 + 6 × 5) [âµ SA of cuboid = 2(lb + bh + hl)]
= 2(20 + 24 + 30)
= 148 cm2
And
Surface area of the second cuboid = 2(4 × 3 + 3 × 10 + 10 × 4) [âµ SA of cuboid = 2(lb + bh + hl)]
= 2(12 + 30 + 40)
= 164 cm2
Hence, the statement is false.
Page No 351:
Question 57:
State whether the statements are true (T) or false (F).
The area of a trapezium become 4 times if its height gets doubled.
Answer:
False
Original area of a trapezium =
Here, the height gets doubled.
Hence, the statement is false.
Page No 351:
Question 58:
State whether the statements are true (T) or false (F).
A cube of side 3 cm painted on all its faces, when sliced into 1 cubic centimetre cubes, will have exactly 1 cube with none of its faces painted.
Answer:
True
Given that, a cube of side 3 cm is painted on all its faces and sliced into 1 cubic centimetre cubes.
Now,
volume of the cube of side 3 cm = 3 × 3 × 3 = 27 cubic cm
Number of smaller cubes of volume 1 cubic cm = = 27
So, the total number of smaller cubes obtained in this process = 27
Out of these, the eight cubes at the corners are painted on three sides. The middle cube on all the edges are painted on two sides. The middle cube of all the faces is painted on one side. Thus, only one cube is left that has no painted side.
Hence, the statement is true.
Page No 351:
Question 59:
State whether the statements are true (T) or false (F).
Two cylinders with equal volume will always have equal surface areas.
Answer:
False
Volume of a cylinder =
Let the radius of the two cylinders be 2 cm and 3 cm respectively. Also, let their heights be 9 cm and 4 cm respectively.
Volume =
The volume of both the cylinders is equal.
Now,
and
Now, . Thus, their surface areas are not equal.
Hence, the statement is false.
Page No 351:
Question 60:
State whether the statements are true (T) or false (F).
The surface area of a cube formed by cutting a cuboid of dimensions 2 × 1 × 1 in 2 equal parts is 2 sq. units.
Answer:
False
Given that, a cuboid of dimensions 2 cm × 1 cm × 1 cm is cut into two equal parts. So, the side of the new cube formed will be 1 cm.
Thus,
Surface area of the cube formed = 6 × (side)2
= 6 × (1)2
= 6 cm2
And
6 cm2 ≠ 2 cm2.
Hence, the given statement is false.
Page No 351:
Question 61:
State whether the statements are true (T) or false (F).
Ratio of area of a circle to the area of a square whose side equals radius of circle is 1 : π.
Answer:
False
Given that, the side of the square equals the radius of circle.
Now,
Hence, the given statement is false.
Page No 352:
Question 62:
Solve the following:
The area of a rectangular field is 48 m2 and one of its sides is 6 m. How long will a lady take to cross the field diagonally at the rate of 20 m/minute?
Answer:
Given that, the area of a rectangular field is 48 m2 and one of its sides is 6 m.
Let b = 6 m.
Now,
Area of rectangular field = l × b
⇒ 48 = l × b
⇒ 48 = l × 6
⇒ l = 8 cm
The lady has to cross the field diagonally. This means that she will cross the rectangular field along its diagonal.
Applying Pythagoras theorem in ABC,
Therefore, the lady has to cross 10 m with a speed of 20 m/minute.
Since, she can cross 20 m in one minute.
Therefore, she can cover 10 m in half the time.
Hence, she will take half a minute or 30 seconds to cover the field diagonally.
Page No 352:
Question 63:
Solve the following:
The circumference of the front wheel of a cart is 3 m long and that of the back wheel is 4 m long. What is the distance travelled by the cart, when the front wheel makes five more revolutions than the rear wheel?
Answer:
Given that, the circumference of the front wheel of a cart is 3 m long and that of the back wheel is 4 m long.
Let the number of revolutions made by the front wheel be (x + 5).
Thus, the rear wheel makes x revolutions.
The distance moved by the front wheel in one revolution is 3 m while that moved by the back wheel is 4 m. Now, the distance travelled by both the wheel is the same for the cart to move properly.
Therefore,
the distance covered by the cart = 4 × 15
= 60 m
Hence, the distance covered by the cart is 60 m.
Page No 352:
Question 64:
Solve the following:
Four horses are tethered with equal ropes at 4 corners of a square field of side 70 metres so that they just can reach one another. Find the area left ungrazed by the horses.
Answer:
Given that, four horses are tethered with equal ropes at 4 corners of a square field of side 70 m so that they just can reach one another.
So, on one side of the square field, two horses are tethered at the corners.
⇒ length of the rope used for each horse =
Now, the four horses will graze areas in shape of quadrants of a circle of radius 35 m.
Subtracting (1) from (2),
Ungrazed area = 4900 m2 − 3850 m2
= 1050 m2
Hence, the area left ungrazed by the four horses is 1050 m2
Page No 352:
Question 65:
Solve the following:
The walls and ceiling of a room are to be plastered. The length, breadth and height of the room are 4.5 m, 3 m, and 350 cm respectively. Find the cost of plastering at the rate of Rs 8 per m2.
Answer:
Given that, the length, breadth and height of the room are 4.5 m, 3 m and 350 cm (= 3.5 m) respectively.
The walls and ceiling of a room are to be plastered.
Now, the cost of plastering at the rate of â¹8 per m2.
Therefore,
Total cost of plastering = 66 × 8
= â¹528
Page No 352:
Question 66:
Solve the following:
Most of the sailboats have two sails, the jib and the mainsail. Assume that the sails are triangles. Find the total area of each sail of the sail boats to the nearest tenth.
Answer:
(i) Area of the sails = ar(ADC) + ar(CEB) + ar(ABE)
(ii) Area of the sails = ar(ABC) + ar(CDE)
(iii) Since, ADB is the sail in the sailboat while GEF is the flag.
Therefore, area of the sail will not include the area of GEF.
Area of the sails = ar(ADC) + ar(ABC)
Page No 352:
Question 67:
Solve the following:
The area of a trapezium with equal non-parallel sides is 168 m2. If the lengths of the parallel sides are 36 m and 20 m, find the length of the non-parallel sides.
Answer:
Given that, area of a trapezium with equal non-parallel sides is 168 m2 and the lengths of the parallel sides are 36 m and 20 m.
Now,
Area of trapezium =
Applying Pythagoras theorem in AED,
Since, the non parallel sides are equal.
Therefore, AD = BC = 10 m.
Page No 352:
Question 68:
Solve the following:
Mukesh walks around a circular track of radius 14 m with a speed of 4 km/hr. If he takes 20 rounds of the track, for how long does he walk?
Answer:
Given that, Mukesh walks around a circular track of radius 14 m with a speed of 4 km/hr.
This means that in 1 hour, he covers 4 km = 4000 m.
Now,
Distance covered by him in 1 round = Circumference of the circular track of radius 14 m
Since, he covers 4000 m in 60 mins.
Therefore,
Page No 353:
Question 69:
Solve the following:
The areas of two circles are in the ratio 49 : 64. Find the ratio of their circumferences.
Answer:
Given that, the areas of two circles are in the ratio 49 : 64.
Let C1 represent the first circle and C2 represent the second one. Let their respective radii be r1 and r2.
Now, ratio of their circumference is given as:
Hence, the ratio of their circumferences is 7 : 8.
Page No 353:
Question 70:
Solve the following:
There is a circular pond and a footpath runs along its boundary. A person walks around it, exactly once keeping close to the edge. If his step is 66 cm long and he takes exactly 400 steps to go around the pond, find the diameter of the pond.
Answer:
Given that, his step is 66 cm long and he takes exactly 400 steps to go around the pond.
So,
Circumference of the pond = 66 × 400
= 26400 cm
= 264 m
Let the radius of the pond be r.
Hence, the diameter of the pond is 84 m.
Page No 353:
Question 71:
Solve the following:
A running track has 2 semicircular ends of radius 63 m and two straight lengths. The perimeter of the track is 1000 m. Find each straight length.
Answer:
Given that, a running track has two semicircular ends of radius 63 m and two straight lengths.
Since, there are two semi circular ends of radius 63 m.
Therefore,
Now, the perimeter of the track is 1000 m.
Since, the two straight paths are identical.
Therefore,
Length of each straight path = = 302 m
Page No 353:
Question 72:
Solve the following:
Find the perimeter of the given figure.
Answer:
In the figure, there are two quadrants (because angle made is a right angle) of a circle with radius 6.3 cm.
So,
Perimeter of the figure = 4 × radius of the circle + 2 × circumference of two quadrants of the circle
= 4 × 6.3 + 2 ×
Hence, the perimeter of the figure is 45 cm.
Page No 353:
Question 73:
Solve the following:
A bicycle wheel makes 500 revolutions in moving 1 km. Find the diameter of the wheel.
Answer:
Given that, a bicycle wheel makes 500 revolutions in moving 1 km.
Thus, 500 revolutions of the wheel cover 1000 m.
∴ In one revolution, are covered.
Now,
Page No 353:
Question 74:
Solve the following:
A boy is cycling such that the wheels of the cycle are making 140 revolutions per hour. If the diameter of the wheel is 60 cm, calculate the speed in km/h with which the boy is cycling.
Answer:
Given that, the wheels of the cycle are making 140 revolutions per hour. Also, the diameter of the wheel is 60 cm.
Now,
In 140 revolutions made by the wheels in an hour,
Page No 353:
Question 75:
Solve the following:
Find the length of the largest pole that can be placed in a room of dimensions 12 m × 4 m × 3 m.
Answer:
Given that, the largest pole is to be placed in a room of dimensions 12 m × 4 m × 3 m.
The largest pole in the given figure is side AF.
Applying Pythagoras theorem in AFC,
⇒ AF = 13 m (âµ length cannot be negative)
Hence, the length of the largest pole that can be placed in a room of dimensions 12 m × 4 m × 3 m is 13 m.
Page No 354:
Question 76:
Find the area of the following fields. All dimensions are in metres.
Answer:
Area of a rectangle = l × b
Area of triangle =
The figure can be shown as:
Total area of the figure = ar(FEH) + ar(EDCI) + ar(CLB) + ar(ILBK) + ar(ABK) + ar(AJG) + ar(FHJM) + ar(FMG)
Hence, the area of the given figure is 53000 m2.
Page No 354:
Question 77:
Find the area of the following fields. All dimensions are in metres.
Answer:
Area of triangle =
Area of trapezium =
Area of the figure = ar(FCD) + ar(FCBH) + ar(BHA) + ar(AGE) + ar(EGD)
Hence, the area of the figure is 30100 m2.
Page No 354:
Question 78:
Find the area of the shaded portion in the following figures.
Answer:
The shaded portion is a triangle with base PQ = 36 m and height RQ = 24 m.
Therefore,
Hence, the area of the shaded region is 432 m2.
Page No 354:
Question 79:
Find the area of the shaded portion in the following figures.
Answer:
The given figure is a triangle with base = AC = 40 m and height 16 m. Also, a rectangle of length 10 m and breadth 8 m is unshaded. Thus, its area is to be subtracted from the area of the triangle to obtain the area of the shaded region.
Now,
Area of triangle =
and
Area of rectangle = l × b
Therefore,
Hence, area of the shaded region is 240 m2.
Page No 354:
Question 80:
Find the area of the shaded portion in the following figures.
Answer:
In the given figure,
Area of the shaded region = Area of the Parallelogram ABCD − Area of the triangle AEB
Now,
Area of a parallelogram = Base × Corresponding Height
and
Area of triangle =
Therefore,
Page No 354:
Question 81:
Find the area of the shaded portion in the following figures.
Answer:
The given figure contains a trapezium ABCD, a rectangle with sides 40 cm and 20 cm and a circle of radius 7 cm.
Now,
Area of trapezium =
Area of triangle =
Area of circle =
Therefore,
Area of the shaded figure = Area of the trapezium ABCD − Area of rectangle − Area of circle
Page No 355:
Question 82:
Find the area of the shaded portion in the following figures.
Answer:
The given figure has:
(i) a trapezium of height 4 cm and parallel sides as 6 cm and 9 cm (= 12 − 3).
(ii) a trapezium of height 3 cm and parallel sides as 8 cm and 12 cm (= 16 − 4).
(iii) a rectangle of sides 4 cm and 3 cm.
Now,
Area of rectangle = l × b
Area of trapezium =
Therefore,
Page No 355:
Question 83:
Find the area of the shaded portion in the following figures.
Answer:
The shaded region has a circle of diameter 21 cm, four triangles of base 7 cm and height 7 cm and a square of side 7 cm.
Now,
Area of circle =
Area of a square = s × s
Area of triangle =
Thus,
Area of the shaded region = Area of the circle − Area of four triangles − Area of a square
Page No 355:
Question 84:
Find the area of the shaded portion in the following figures.
Answer:
The given figure contains a square of side 12 cm, a semi circle of diameter 12 cm and a quadrant of a circle of radius 6 cm.
Now,
Area of semi circle =
Area of a square = s × s
Area of a quadrant =
Page No 355:
Question 85:
Find the area of the shaded portion in the following figures.
Answer:
The given figure has:
(i) two triangles of base 6 cm and sides 5 cm
(ii) a square of side 6 cm
(iii) two semi circles of radius 3 cm
The height of the triangle is given as:
Now,
Area of triangle =
Area of a square = s × s
Area of semi circle =
Therefore,
Page No 355:
Question 86:
Find the volume of each of the given figure if volume = base area × height.
Answer:
(a) Volume of the figure = base area × height
(b) Volume of the cuboid = base area × height
(c) Volume of the figure = base area × height
Page No 355:
Question 87:
A cube of side 5 cm is cut into as many 1 cm cubes as possible. What is the ratio of the surface area of the original cube to that of the sum of the surface areas of the smaller cubes?
Answer:
Given that, a cube of side 5 cm is cut into as many 1 cm cubes as possible.
Now,
Volume of the larger cube = (side)3
= (5)3
= 125 cm3
Volume of a smaller cube = (side)3
= (1)3
= 1 cm3
Now,
SA of the larger cube = 6(side)2
= 6(5)2
= 150 cm2
SA of 125 smaller cubes = 125 × 6(side)2
= 125 × 6(1)2
= 750 cm2
Hence, the ratio of the surface area of the original cube to that of the sum of the surface areas of the smaller cubes is 1 : 5.
Page No 356:
Question 88:
A square sheet of paper is converted into a cylinder by rolling it along its side. What is the ratio of the base radius to the side of the square?
Answer:
Consider a square sheet of side s.
Now, when the square sheet of paper is converted into a cylinder by rolling it along its side, the height of the cylinder becomes s and the circumference of its base is also s.
Since, the circumference of its base is s.
Therefore,
Hence, the ratio of the base radius to the side of the square is .
Page No 356:
Question 89:
How many cubic metres of earth must be dug to construct a well 7 m deep and of diameter 2.8 m?
Answer:
Given that, a well 7 m deep and of diameter 2.8 m is to be constructed by digging the earth.
Now, the amount of earth to be dug for its construction will be equal to the volume of the well.
So,
Volume of the cylinder =
Here,
r = 1.4 m
h = 7 m
Page No 356:
Question 90:
The radius and height of a cylinder are in the ratio 3 : 2 and its volume is 19,404 cm3. Find its radius and height.
Answer:
Given that, the radius and height of a cylinder are in the ratio 3 : 2 and its volume is 19,404 cm3.
Let its radius be 3x and its height be 2x.
So,
Volume of the cylinder =
Therefore,
Radius = 3x = 3 × 7 = 21 cm
Height = 2x = 2 × 7 = 14 cm
Page No 356:
Question 91:
The thickness of a hollow metallic cylinder is 2 cm. It is 70 cm long with outer radius of 14 cm. Find the volume of the metal used in making the cylinder, assuming that it is open at both the ends. Also find its weight if the metal weighs 8 g per cm3.
Answer:
Given that, a hollow cylinder has height 70 cm, outer radius 14 cm and thickness is 2 cm.
Thus, the inner radius of the cylinder = 14 − 2 = 12 cm.
Now, for a hollow cylinder with height h, outer radius R and inner radius r,
Volume of the cylinder =
So,
Now, the metal weighs 8 g per cm3.
Page No 356:
Question 92:
Radius of a cylinder is r and the height is h. Find the change in the volume if the
(a) height is doubled.
(b) height is doubled and the radius is halved.
(c) height remains same and the radius is halved.
Answer:
Consider a cylinder of radius r and height h. Then,
Volume of the cylinder =
(i) Let the new height be 2h.
Hence, the volume of the cylinder becomes double the initial volume.
(ii) Let the new height be 2h and the new radius be .
Hence, the volume of the cylinder becomes half the initial volume.
(iii) Let the new radius be .
Hence, the volume of the cylinder becomes one-fourth the initial volume.
Page No 356:
Question 93:
If the length of each edge of a cube is tripled, what will be the change in its volume?
Answer:
Let the initial edge of the cube be x.
So,
Initial volume of the cube =
Now, if the length of each edge of a cube is tripled, then the new length becomes 3x.
So,
Hence, the new volume is 27 time the original volume of the cube.
Page No 356:
Question 94:
A carpenter makes a box which has a volume of 13,400 cm3. The base has an area of 670 cm2. What is the height of the box?
Answer:
Given that, a box has a volume of 13,400 cm3 and base has an area of 670 cm2.
Thus,
lb = 670
Now, volume of a cuboid = lbh
So,
Hence, the height of the box is 20 cm.
Page No 356:
Question 95:
A cuboidal tin box opened at the top has dimensions 20 cm × 16 cm × 14 cm. What is the total area of metal sheet required to make 10 such boxes?
Answer:
Given that, a cuboidal tin box opened at the top has dimensions 20 cm × 16 cm × 14 cm.
Hence, the total area of metal sheet required to make 10 such boxes is 13280 cm2.
Page No 356:
Question 96:
Find the capacity of water tank, in litres, whose dimensions are 4.2 m, 3 m and 1.8 m?
Answer:
Volume of cuboidal tank = lbh
Hence, the capacity of water tank is 22680 L.
Page No 356:
Question 97:
How many cubes each of side 0.5 cm are required to build a cube of volume 8 cm3?
Answer:
Given that, the volume of a cube is 8 cm3.
Now,
Volume of a smaller cube = (side)3
= 0.5 × 0.5 × 0.5
= 0.125 cm3
Hence, 64 cubes each of side 0.5 cm are required to build a cube of volume 8 cm3.
Page No 356:
Question 98:
A wooden box (including the lid) has external dimensions 40 cm by 34 cm by 30 cm. If the wood is 1 cm thick, how many cm3 of wood is used in it?
Answer:
Given that, a wooden box (including the lid) has external dimensions 40 cm × 34 cm × 30 cm. Also, the thickness of the wood is 1 cm.
Thus, the dimensions of the inner box is 38 cm × 32 cm × 28 cm.
Now,
Page No 356:
Question 99:
A river 2 m deep and 45 m wide is flowing at the rate of 3 km per hour. Find the amount of water in cubic metres that runs into the sea per minute.
Answer:
Given that, a river is 2 m deep and 45 m wide and it flows at the rate of 3 km per hour.
The speed of the river is 3 km per hour.
⇒ In 60 minutes, 3000 m are covered by water.
⇒ In 1 minute, 50 m are covered by water.
So, the length of the water flowing into the sea in one minute be 50 m. And the river is 2 m deep and 45 m wide.
Page No 357:
Question 100:
Find the area to be painted in the following block with a cylindrical hole. Given that length is 15 cm, width 12 cm, height 20 cm and radius of the hole 2.8 cm.
Answer:
Given that, a block with a cylindrical hole has length 15 cm, width 12 cm, height 20 cm and radius of the hole 2.8 cm.
Now,
Area to be painted = TSA of cuboid − 2 × Area of circle
Page No 357:
Question 101:
A truck carrying 7.8 m3 concrete arrives at a job site. A platform of width 5 m and height 2 m is being constructed at the site. Find the length of the platform, constructed from the amount of concrete on the truck.
Answer:
Given that, a truck carries 7.8 m3 concrete. A platform of width 5 m and height 2 m is being constructed at the site.
Let the length of the platform be l.
Now, the volume of the platform to be constructed must be equal to the volume of the concrete carried by the truck.
Hence, the length of the platform, constructed from the amount of concrete on the truck is 78 cm.
Page No 357:
Question 102:
A hollow garden roller of 42 cm diameter and length 152 cm is made of cast iron 2 cm thick. Find the volume of iron used in the roller.
Answer:
Given that, a hollow garden roller has 42 cm diameter and length 152 cm is made of cast iron 2 cm thick.
Thus, radius of the roller is 21 cm.
Since, the thickness is 2 cm.
Therefore,
inner radius of the roller = 21 − 2 = 19 cm
Now, the volume of iron used for making a 2 cm thick roller will be the difference between the volumes of the outer and the inner cylinders.
So,
Page No 357:
Question 103:
Three cubes each of side 10 cm are joined end to end. Find the surface area of the resultant figure.
Answer:
Given that, three cubes each of side 10 cm are joined end to end.
Thus, the length of the resultant cuboid is 30 cm (= 10 × 3) and the breadth and height remain 10 cm.
Therefore,
Surface area of the cuboid formed = 2(lb + bh + hl)
= 2(30 × 10 + 10 × 10 + 10 × 30)
= 2(300 + 100 + 300)
= 1400 cm2
Page No 357:
Question 104:
Below are the drawings of cross sections of two different pipes used to fill swimming pools. Figure A is a combination of 2 pipes each having a radius of 8 cm. Figure B is a pipe having a radius of 15 cm. If the force of the flow of water coming out of the pipes is the same in both the cases, which will fill the swimming pool faster?
Answer:
Given that, figure A is a combination of 2 pipes each having a radius of 8 cm and figure B is a pipe having a radius of 15 cm.
Now, for comparing, consider that the water in the pipes flows for a constant height h.
Let r = 8 cm and R = 15 cm.
So,
and
Now,
Hence, the pipe with radius 15 cm will fill the pool faster.
Page No 357:
Question 105:
A swimming pool is 200 m by 50 m and has an average depth of 2 m. By the end of a summer day, the water level drops by 2 cm. How many cubic metres of water is lost on the day?
Answer:
Given that, a swimming pool is 200 m by 50 m and the water level drops by 2 cm.
For the lost water, depth is 2 cm (= 0.02 m), length is 200 m and breadth is 50 m.
∴ Volume of water lost = lbh
= 200 × 50 × 0.02
= 200 m3
Hence, the water is lost on the day is 200 m3.
Page No 358:
Question 106:
A housing society consisting of 5,500 people needs 100 L of water per person per day. The cylindrical supply tank is 7 m high and has a diameter 10 m. For how many days will the water in the tank last for the society?
Answer:
Given that, a housing society consisting of 5,500 people needs 100 L of water per person per day.
Thus,
Daily requirement of water in the society = 5500 × 100 = 550000 L
Now,
Therefore,
Hence, the water in the tank will last for one day.
Page No 358:
Question 107:
Metallic discs of radius 0.75 cm and thickness 0.2 cm are melted to obtain 508.68 cm3 of metal. Find the number of discs melted (use π = 3.14).
Answer:
Given that, some metallic discs of radius 0.75 cm and thickness 0.2 cm are melted to obtain 508.68 cm3 of metal.
Now,
Volume of a metallic disc =
Thus,
Hence, 1440 metallic discs were melted.
Page No 358:
Question 108:
The ratio of the radius and height of a cylinder is 2 : 3. If its volume is 12,936 cm3, find the total surface area of the cylinder.
Answer:
Given that, the volume of a cylinder is 12,936 cm3 and the ratio of the radius and height of a cylinder is 2 : 3.
Let radius of the cylinder be 2x and its height be 3x.
Now,
Therefore,
Page No 358:
Question 109:
External dimensions of a closed wooden box are in the ratio 5 : 4 : 3. If the cost of painting its outer surface at the rate of Rs 5 per dm2 is Rs 11,750, find the dimensions of the box.
Answer:
Given that, the external dimensions of a closed wooden box are in the ratio 5 : 4 : 3.
So, let the length of the box be 5x, the breadth be 4x and the height be 3x.
Now,
TSA of the box = 2(lb + bh + hl)
= 2(5x × 4x + 4x × 3x + 3x × 5x)
= 2(20x2 + 12x2 + 15x2)
= 94x2 sq units .....(1)
Now, the cost of painting its outer surface at the rate of â¹5 per dm2 is â¹11,750.
Equating (1) and (2),
Therefore,
length of the box be 5x = 5 × 5 = 25 dm,
the breadth be 4x = 4 × 5 = 20 dm and
the height be 3x = 3 × 5 = 15 dm.
Page No 358:
Question 110:
The capacity of a closed cylindrical vessel of height 1 m is 15.4 L. How many square metres of metal sheet would be needed to make it?
Answer:
Given that, the volume of a closed cylindrical vessel of height 1 m is 15.4 L.
As,
1 m3 = 1000 L
⇒ 15.4 L =
So,
Now, a metal sheet would be needed to make this vessel.
Page No 358:
Question 111:
What will happen to the volume of the cube, if its edge is:
(a) tripled
(b) reduced to one-fourth?
Answer:
Let the initial edge of the cube be a.
So,
Initial volume of the cube = a3
(a) If its edge is tripled, the new edge becomes 3a.
Therefore,
Volume of the cube = (3a)3
= 27a3
Hence, its volume becomes 27 times the original volume.
(b) If its edge is reduced to one-fourth, the new edge becomes .
Therefore,
Hence, its volume becomes times the original volume.
Page No 358:
Question 112:
A rectangular sheet of dimensions 25 cm × 7 cm is rotated about its longer side. Find the volume and the whole surface area of the solid thus generated.
Answer:
Given that, a rectangular sheet of dimensions 25 cm × 7 cm is rotated about its longer side.
Thus, height of the cylinder so formed is 7 cm and the circumference of the base of the cylinder is 25 cm.
So,
Also,
Page No 358:
Question 113:
From a pipe of inner radius 0.75 cm, water flows at the rate of 7 m per second. Find the volume in litres of water delivered by the pipe in 1 hour.
Answer:
Given that, a pipe has inner radius 0.75 cm (= 0.0075 m) and water flows through it at the rate of 7 m per second.
Now,
Volume of a cylinder =
In one second, 7 m of water flows through the pipe.
So, in one hour, m of water flows through the pipe. Thus, this is the height of the water flowing through the pipe in one hour.
Page No 358:
Question 114:
Four times the area of the curved surface of a cylinder is equal to 6 times the sum of the areas of its bases. If its height is 12 cm, find its curved surface area.
Answer:
Given that, four times the area of the curved surface of a cylinder is equal to 6 times the sum of the areas of its bases. Also, the height of the cylinder is 12 cm.
Therefore,
Now,
Page No 358:
Question 115:
A cylindrical tank has a radius of 154 cm. It is filled with water to a height of 3 m. If water to a height of 4.5 m is poured into it, what will be the increase in the volume of water in kl?
Answer:
Given that, a cylindrical tank has a radius of 154 cm and its height increases from 3 m to 4.5 m when water is poured into it.
So, the increase in the height of water = 4.5 − 3 = 1.5 cm.
The increase in the volume of water is equal to the volume of water added afterwards. For the water that is added afterwards, the height is 1.5 m and the radius is 154 cm (= 1.54 m).
Page No 358:
Question 116:
The length, breadth and height of a cuboidal reservoir is 7 m, 6 m and 15 m respectively. 8400 L of water is pumped out from the reservoir. Find the fall in the water level in the reservoir.
Answer:
Given that, the length, breadth and height of a cuboidal reservoir is 7 m, 6 m and 15 m respectively.
So,
Initial level of the water in the reservoir = 7 × 6 × 15
= 630 m3
= 630000 L (âµ 1 m3 = 1000 L)
Now, 8400 L of water is pumped out from the reservoir.
Therefore,
The fall in the level of water = 630000 L − 8400 L
= 621600 L
Page No 358:
Question 117:
How many bricks of size 22 cm × 10 cm × 7 cm are required to construct a wall 11 m long, 3.5 m high and 40 cm thick, if the cement and sand used in the construction occupy part of the wall?
Answer:
Given that, a wall 11 m (= 1100 cm) long, 3.5 m (= 350 cm) high and 40 cm thick is to be constructed and cement and sand used in the construction occupy part of the wall.
Now,
Also,
Volume of a brick = 22 × 10 × 7
Therefore,
Hence, 9000 bricks will be required.
Page No 359:
Question 118:
A rectangular examination hall having seats for 500 candidates has to be built so as to allow 4 cubic metres of air and 0.5 square metres of floor area per candidate. If the length of hall be 25 m, find the height and breadth of the hall.
Answer:
Given that, the hall has to have space for 500 candidates keeping in mind that each gets 4 cubic metres of air and 0.5 square metres of floor area.
Now,
Total area required for 500 candidates = 500 × 0.5 = 250 m2
And
Total volume of the hall = 500 × 4 = 2000 m3
Now, the length of hall be 25 m.
So,
l × b = 250 (âµ Area of the floor = lb)
⇒ b = 10 m
Also,
2000 = lbh
⇒ 2000 = 25 × 10 × h
Hence, the height and breadth of the hall are 8 m and 10 m respectively.
Page No 359:
Question 119:
The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the ratio between the height and radius of the cylinder.
Answer:
Given that, .
Now,
Hence, h : r = 1 : 1.
Page No 359:
Question 120:
A birthday cake has two tiers as shown in the figure below. Find the volume of the cake.
Answer:
The volume of the cake is the sum of the volumes of the two cuboids of which it is made up of.
Now,
Volume of the larger cuboid = 20 × 20 × 15 = 6000 cm3
Volume of the smaller cuboid = 10 × 10 × 5 = 500 cm3
∴ Total volume of the cake = 6000 + 500 = 6500 cm3
Page No 359:
Question 121:
Work out the surface area of following shapes in questions (use π = 3.14).
Answer:
TSA of the horizontal part = TSA of horizontal cuboid − SA of square of side 1 cm
= 2(lb + bh + hl) − s × s
= 2(3 × 1 + 1 × 1 + 1 × 3) − 1 × 1
= 2(3 + 1 + 3) − 1
= 13 cm2
TSA of the vertical part = TSA of vertical cuboid − SA of square of side 1 cm
= 2(lb + bh + hl) − s × s
= 2(1 × 1 + 1 × 4 + 4 × 1) − 1 × 1
= 2(1 + 4 + 4) − 1
= 17 cm2
Therefore,
SA of the figure = 13 + 17 = 30 cm2
Disclaimer: The answer given in the NCERT Exemplar is incorrect. The correct answer is as per the given calculations.
Page No 359:
Question 122:
Work out the surface area of following shapes in questions (use π = 3.14).
Answer:
TSA of the cuboidal part = 2(lb + bh + hl) − Area of square of side 12 cm
= 2(12 × 12 + 12 × 24 + 24 × 12) − 12 × 12
= 1296 cm2
TSA of the cubical part = 5 × s × s
= 5 × 12 × 12
= 720 cm2
Hence, total surface area of the figure = 1296 + 720 = 2016 cm2.
Page No 359:
Question 123:
Work out the surface area of following shapes in questions (use π = 3.14).
Answer:
The dimensions of the solid are as follows:
So,
Area of the upper surface = Area of the lower surface
= 18 × 16 (âµ Area of rectangle = lb)
= 288 cm2
Area of the back of the solid = 18 × (3 + 2 + 18) (âµ Area of rectangle = lb)
= 18 × 23
= 414 cm2
Area of the front of the solid = 18 × 18
= 324 cm2
Area of front of solid with height 3 cm and length 18 cm = 18 × 3
= 54 cm2
Area of front of solid with height 2 cm and length 18 cm = 18 × 2
= 36 cm2
Area of the sides of the solid = 2[3 × 3 + 2 × (8 + 3) + 16 × 18]
= 2[9 + 22 + 288]
= 638 cm2
∴ TSA of the solid = 2 × 288 + 414 + 324 + 54 + 36 + 638
= 2042 cm2
Page No 359:
Question 124:
Work out the surface area of following shapes in questions (use π = 3.14).
Answer:
The TSA of the following figure will be the sum of the TSA of the cube and the CSA of the cylinder.
TSA of the cube = 6(side)2
= 6 × 5 × 5
= 150 cm2
∴ Total SA of the solid = 150 + 251.2
= 401.2 cm2
Page No 359:
Question 125:
Water flows from a tank with a rectangular base measuring 80 cm by 70 cm into another tank with a square base of side 60 cm. If the water in the first tank is 45 cm deep, how deep will it be in the second tank?
Answer:
Volume of the water in the first tank = lbh
= 80 × 70 × 45
= 252000 cm3
Now, the volume of both the tanks should be the same as water flows from one to another.
Volume of the second tank = (side)2 × h
= 60 × 60 × h
Hence, the depth of the water in the second tank is 70 cm.
Page No 360:
Question 126:
A rectangular sheet of paper is rolled in two different ways to form two different cylinders. Find the volume of cylinders in each case if the sheet measures 44 cm × 33 cm.
Answer:
Given that, the sheet measures 44 cm × 33 cm.
Now, when it is rolled along side 44 cm, its height becomes 44 cm and the circumference of its base becomes 33 cm.
So,
Hence, volume of cylinder is 3811.5 cm3.
Similarly, when it is rolled along side 33 cm, its height becomes 33 cm and the circumference of its base becomes 44 cm.
So,
Hence, volume of cylinder is 5082 cm3.
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