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Question 1:

In questions â€‹there 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 â€‹there 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
Number of cutout cubes=Volume of original cubeVolume of smaller cube=641=64
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

Required ratio=6×4264×6=14

Thus, the required ratio is 1 : 4.
Hence, the correct answer is option (c).
 

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Question 3:

In questions â€‹there 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?

a 34 of original square b 12 of original square c 14 of original square d 23 of original 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         (∵ square is a rhombus with equal diagonals)
Area of final square=12×d1×d2=12×x×x=x22

Area of final square=Area of initial square2

Hence, the correct answer is option (b).

 

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Question 4:

In questions â€‹there 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 units and width w units.

Area of triangle = 12×Base×Height

The area of the largest triangle that can be fitted inside this rectangle = 12×l×w

                                                                                                               = lw2
Hence, the correct answer is option (a).

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Question 5:

In questions â€‹there are four options out of which one is correct.

If the height of a cylinder becomes 14 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 14 of the original volume.

Answer:

Consider a cylinder with radius r and height h.
So,
Volume of cylinder = πr2h

If the height of a cylinder becomes 14 of the original height and the radius is doubled, then
H14h
and R = 2r

Volume of new cylinder=πR2H=π×2r2×14h=π×4r2×14h=πr2h=Volume of the original cylinder
Hence, the correct answer is option (b).
 



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Question 6:

In questions â€‹there are four options out of which one is correct.

If the height of a cylinder becomes 14 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 14 of the original curved surface.

Answer:

Consider a cylinder with radius r and height h.
So,
Curved surface area of cylinder = 2πrh

If the height of a cylinder becomes 14 of the original height and the radius is doubled, then
H = 14h
and R = 2r

Curved surface area of new cylinder=2πRH=2π×2r×14h=2π×2r×14h=122πrh=12Curved surface area of the original cylinder

Thus, the curved surface area of the cylinder will be halved.
Hence, the correct answer is option (c).

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Question 7:

In questions â€‹there are four options out of which one is correct.

If the height of a cylinder becomes 14 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 = 2πrr+h

If the height of a cylinder becomes 14 of the original height and the radius is doubled, then
H = 14h
and R = 2r

Total surface area of new cylinder=2πRR+H=2π×2r×2r+14h=2π×r×8r+h2

Hence, the correct answer is option (d).

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Question 8:

In questions â€‹there 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

l×h×l×b×b×h=6×15×10l2b2h2=900lbh2=900lbh=30          length cannot be negative

Now, volume of a cuboid = lbh

Here, lbh = 30 cm3.
Hence, the correct answer is option (a).


 

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Question 9:

In questions â€‹there 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 360°6=60°.

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 â€‹there 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

Number of boxes that can be fitted in the godown=Volume of the godownVolume of the cuboidal boxes=40×25×102×1.25×1=4000

Hence, the correct answer is option (c).



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Question 11:

In questions â€‹there 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).

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Question 12:

In questions â€‹there 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 = 2πrh

If the radius of a cylinder is tripled, then R = 3r. Let the new height be H.

Curved surface area of new cylinder=2πRH=2π×3r×H=32π×r×H

Since, its curved surface area is unchanged.
2πrh=32πrHh=3HH=h3

Thus, its height will be one third.
Hence, the correct answer is option (d).

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Question 13:

In questions â€‹there 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

Number of small cubes that can be fitted in the cubical box=Volume of the cubical boxVolume of the smaller cubes=2×2×20.2×0.2×0.2=1000

Hence, the correct answer is option (c).

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Question 14:

In questions â€‹there are four options out of which one is correct.

The volume of a cylinder whose radius r is equal to its height is

a 14πr3b πr332c πr3d r38

Answer:

Consider a cylinder with radius and height r.
So,
Volume of cylinder=πr2r=πr3

Hence, the correct answer is option (c).

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Question 15:

In questions â€‹there 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).

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Question 16:

In questions â€‹there 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)
ABCADC by SSS congruence.

This means that both the triangles will have the same area.
Area of quadrilateral ABCD=arABC+arADC=12×12×3+12×12×3            area of triangle=12×b×h=36 cm2
Hence, the correct answer is option (b).
 

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Question 17:

In questions â€‹there 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)
ABCADC by SSS congruence.

This means that both the triangles will have the same area.
Area of rhombus ABCD=2×arABC=2×12×AC×BE            area of triangle=12×b×h=2×12×6×4
Area of rhombus ABCD=24 cm2

Hence, the correct answer is option (c).



Page No 348:

Question 18:

In questions â€‹there 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,
60=b×5b=12 cm

Hence, the correct answer is option (a).

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Question 19:

In questions â€‹there 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,
Area of a trapezium=12×Sum of parallel sides×Height=12×32×8=128 cm2
Hence, the correct answer is option (c).

 

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Question 20:

In questions â€‹there 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)
20=12×BD×1.5+12×BD×1              area of triangle=12×b×h=12×BD×1.5+1=12×BD×2.5

BD=20×22.5=16 cm

Hence, the correct answer is option (c).

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Question 21:

In questions â€‹there 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).

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Question 22:

In questions â€‹there 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,
Sum of the volumes of the three cubes=63+83+103=216+512+1000=1728

Thus, volume of the new cube = 1728 cm3 = (side)3
⇒ side = 12 cm

Hence, the correct answer is option (a).

 

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Question 23:

In questions â€‹there 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).


 

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Question 24:

In questions â€‹there 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 = πr2h

Here,
Ratio of their volumes=π×x2×2yπ×2x2×3y=π×x2×2yπ×4x2×3y=π×x2×2yπ×4x2×3y=16
Hence, the correct answer is option (a).

 



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Question 25:

In questions â€‹there 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,
Ratio of their volumes=s3S3164=s3S3143=sS3sS=14
Therefore, let the edge of the first and the second cube be x and 4x respectively.

Now,
Ratio of the area of a face of first cube to that of the other =x×x4x×4x=116
Hence, the correct answer is option (c).

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Question 26:

In questions â€‹there 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),
l×b×b×h×h×l=16×32×72l2b2h2=36864lbh2=1922lbh=192             length cannot be negative

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 â€‹there 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 12h, 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 = πr2h

(a) Container A has radius r and height h.
Volume of the container A=πr2h

(b) Container B has radius 2r and height 12h.
Volume of the container B=π2r212h=4πr212h=2πr2h

(c) Container C has dimensions r × r × h.
Volume of container C=r×r×h=r2h

Now,
r2h<πr2h<2πr2h
Therefore, the arrangement of the containers in the increasing order of their volumes is C, A, B.
Hence, the correct answer is option (c).
 

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Question 28:

In questions â€‹there 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 = 2πrh+r-πr2

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 = πR2-r2

Thus,
TSA of the hat = 2πrh+r-πr2 + πR2-r2
                        =2πrh+2πr2-πr2+πR2-πr2=2πrh+πR2
Hence, the correct answer is option (c).
 

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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

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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% .

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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 × aa × 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 d2d1×d22

Here, the diagonals of a rhombus get doubled.
New area=2d1×2d22=4×d1×d22=4×Original area

Hence, if the diagonals of a rhombus get doubled, then the area of the rhombus becomes four times its original area.

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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 = πr2h

Now, if its radius becomes half of the original radius, then
 Volume of cylinder=π12r2h=14πr2h=14×Original Volume

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 = 2πrh

If the height is half of the original height,
New curved surface area=2π12rh=πrh

Now,
Percentage reduction in the CSA of cylinder=Final CSA-Initial CSAInitial CSA×100=πrh-2πrh2πrh×100=-πrh2πrh×100=-50
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 a2.

Now,
Volume of a cylinder with radius r and height hπr2h

Here,
Volume of the cylinder=πa22a=πa34
Hence, the volume of a cylinder which exactly fits in a cube of side a is πa34.

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 b2.

Now,
TSA of a cylinder with radius r and height h2πrh+r

Here,
TSA of the cylinder=2πb2b+b2=πb3b2=3πb22

Hence, the surface area of a cylinder which exactly fits in a cube of side b is 3πb22.

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' = h12 and h22 respectively.

Therefore,
Area of the quadrilateral=12×2d×h12+12×2d×h22          Area of triangle=12×b×h=2d2h1+h22=d2h1+h2

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 d2h1+h2.

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 2πrh.

Here, 
Curved surface area of a cylinder of radius h and height r is 2πhr.

Hence, the curved surface area of a cylinder of radius h and height r is 2πhr.

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 = 2πrh+r 

Here,
Total surface area of a cylinder of radius h and height r = 2πhr+h

Hence, the total surface area of a cylinder of radius h and height r is 2πhr+h.

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 = πr2h

Here,
Volume of a cylinder with radius h and height r = πh2r

Hence, the volume of a cylinder with radius h and height r is πh2r.



Page No 351:

Question 46:

Fill in the blanks to make the statements true.

Area of a rhombus = 12 product of _________.

Answer:

Let d1 and d2 be two diagonals of a rhombus.
Area of a rhombus = 12×d1×d2

Hence, area of a rhombus = 12 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 hπr2h

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.
2πr1=20 and h1 = 10
r1=10π

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.
2πr2=10 and h2 = 20
r2=5π

Now,
Ratio of their volumes=Volume of Cylinder AVolume of Cylinder B=πr12h1πr22h2=r12h1r22h2=10π×10π×105π×5π×20=42=2
Hence,  volume of A is twice of volume of B.
 

Page No 351:

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 = 2πrh

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.
2πr1=20 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.
2πr2=10 and h2 = 20

Now,
Ratio of their CSAs=CSA of Cylinder ACSA of Cylinder B=2πr1h12πr2h2=πr1h1πr2h2=10×105×20=1
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 hπr2h

Let r1 and r2 be the radii of the two cylinders.

Ratio of the volume of two cylinders=Volume of first cylinderVolume of second cylinder1=πr12hπr22h9=r1r2232=r1r22r1r2=31

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 = πr2h

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.
πr12h1=πr22h2r1r22=h2h1h2h1=x6x2h2h1=136h1h2=361

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.                (∵ Volume of cuboid = lbh)

Surface area of the first cuboid = 2(5 × 4 + 4 × 6 + 6 × 5)                          [∵ 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)                          [∵ 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 = 12×sum of parallel sides×h

Here, the height gets doubled.
New area of a trapezium=12×sum of parallel sides×2h=2×Original area of the trapezium

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 = 271 = 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 = πr2h

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 = π22×9=π32×4 = 36π cm3

The volume of both the cylinders is equal.

Now,
Surface area of first cylinder=2πrh=2π×2×9=36π cm2
and
Surface area of second cylinder=2πrh=2π×3×4=24π cm2

Now, 36π24π. 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.
s=r

Now,
Area of circleArea of square=πr2s×s=πr2r×r=π11π
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,
AC2=AB2+BC2AC2=62+82AC2=36+64=100AC=10 m        length of a side is always positive
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.
3x+5=4x3x+15=4x4x-3x=15x=15

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 = 12×70=35 m


Now, the four horses will graze areas in shape of quadrants of a circle of radius 35 m.
Area grazed by four horses=4×14π×352                Area of a quadrant of a circle=14πr2=1225π=1225×227=3850 m2                     .....1

Area of the entire field=70×70                   Area of square=s×s=4900 m2               .....2

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.
Total area to be plastered=2bh+hl+lb=2×3×3.5+3.5×4.5+4.5×3=2×26.25+13.5=52.5+13.5=66 m2

Now, the cost of plastering at the rate of ₹8 per m2.
Therefore,
Total cost of plastering = 66 × 8
                                      = ₹528

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)
                              =12×22+20×16.8+12×20×22.3+12×22×22.3=352.8+223+245.3=821.1 m2

(ii) Area of the sails = ar(ABC) + ar(CDE)
                                =12×23.9×8.6+12×19.5×10.9=102.77+106.275=209.045 m2209 m2

(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)
                                  =12×9.6×16.8+12×25×12.4=80.64+155=235.64 m2235.6 m2 

 

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 = 12×Sum of parallel sides×height
168=12×Sum of parallel sides×height=12×36+20×h=28×h
h=6 m

Applying Pythagoras theorem in AED,
AD2=AE2+DE2AD2=82+62AD2=64+36=100AD=10 m          Length is always positive

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
Distance covered by him in 1 round=2×227×14                   Circumference of circle=2πr=88 m
Distance covered in 20 rounds=88×20=1760 m

Since, he covers 4000 m in 60 mins.
Therefore,
Time taken to cover 1760 m=604000×1760=26.4 mins



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.
Area of C1Area of C2=4964πr12πr22=4964                 Area of circle=πr2r12r22=4964r1r22=782r1r2=78

Now, ratio of their circumference is given as:
Circumference of C1Circumference of C2=2πr12πr2               Circumference of circle=2πrCircumference of C1Circumference of C2=r1r2=78

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.
264=2πrr=13222×7=42 m

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,
Circumference of semi circular ends=2×πr                       Circumference of semi circle=πr=2×227×63=396 m

Now, the perimeter of the track is 1000 m.
Length of the straight paths=1000-396=604 m

Since, the two straight paths are identical.
Therefore,
Length of each straight path = 6042 = 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 × 2πr4                 Circumference of a circle=2πr
                                     =25.2+227×6.3=25.2+19.8=45 cm

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, 1000500=2 m are covered.

Now,
Circumference of a circle=2πr2=2×227×r2=2r×227d=2×722                2r=dd=0.63 m=63 cm

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,
Circumference of a circle=2πr=2r×227=60×227=60100×1000×227 km 

In 140 revolutions made by the wheels in an hour,
Distance covered=140×60100×1000×227 km

Speed of the bicycle=Distance covered by the bicycleTime taken=140×60100×1000×227 km1 hr=0.264 km/hr


 

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,
AF2=AC2+CF2
AF2=122+422+32=144+162+9=1602+9=160+9=169
⇒ AF = 13 m            (∵ 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 = 12×b×h

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)
                                     =12×40×80+100×160+12×40×120+60×120+12×60×60+12×100×160+40×160+12×160×120=1600+16000+2400+7200+1800+8000+6400+9600=53000 m2

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 = 12×b×h
Area of trapezium = 12×sum of parallel sides×height

Area of the figure = ar(FCD) + ar(FCBH) + ar(BHA) + ar(AGE) + ar(EGD)
                              =12×100×100+12×100+50×110+12×50×50+12×80×120+12×120×180=5000+8250+1250+4800+10800=30100 m2

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,
Area of triangle=12×b×h=12×36×24=432 m2

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 = 12×b×h
and
Area of rectangle = l × b

Therefore,
Area of the shaded region=12×40×16-10×8=320-80=240 m2

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 = 12×b×h

Therefore,
Area of the shaded region=40×30-12×40×30=12×40×30=600 cm2

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 = 12×sum of parallel sides×height
Area of triangle = 12×b×h
Area of circle = πr2

Therefore,
Area of the shaded figure = Area of the trapezium ABCD − Area of rectangle − Area of circle
                                         =12×120+160×100-40×20-227×7×7=14000-800-154=13046 cm2



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 = 12×sum of parallel sides×height

Therefore,
Area of the shaded portion=12×6+9×4+12×8+12×3+4×3=12×15×4+12×20×3+12=30+30+12=72 cm2

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 = πr2
Area of a square = s × s
Area of triangle = 12×b×h

Thus,
Area of the shaded region = Area of the circle − Area of four triangles − Area of a square
                                          =227×212×212-4×12×7×7-7×7=346.5-98-49=199.5 cm2
 

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 = 12×πr2
Area of a square = s × s
Area of a quadrant = 14×πr2

Area of the given figure=12×12+12×227×122×122+14×227×6×6=144+56.57+28.28=228.85 cm2

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:
52=32+h2h2=25-9=16h=4 cm            length of side cannot be negative

Now,
Area of triangle = 12×b×h
Area of a square = s × s
Area of semi circle = 12×πr2

Therefore,
Area of the shaded region=2×12×6×4+6×6+2×12×227×3×3=24+36+28.28=88.28 cm2 

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
                                       =2×12×x×x2×2x=x3 c.u.

(b) Volume of the cuboid = base area × height
                                         =y×2y×3y=6y3 c.u.

(c) Volume of the figure = base area × height
                                       =2p×2p×2p=8p3 c.u.

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

Number of smaller cubes that can be obtained=Volume of larger cubeVolume of a smaller cube=1251=125

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

Ratio of their SA=Surface area of the original cubeSurface area of 125 smaller cubes=150750=15
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,
s=2πr                   Circumference=2πrr=s2π

Required ratio=Base radius of the cylinderSide of the square=s2πs=12π

Hence, the ratio of the base radius to the side of the square is 12π.

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 = πr2h

Here,
r = 1.4 m
h = 7 m
Volume of the well=πr2h=227×1.4×1.4×7=43.12 m3

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 = πr2h
19404=227×3x×3x×2xx3=19404×722×3×3×2=343x=7 cm

Therefore,
Radius = 3= 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 = πhR2-r2

So,
Volume of metal=227×70×142-122=22×10×196-144=220×52=11440 cm3

Now, the metal weighs 8 g per cm3.
Total weight of the metal=11440×8=91520 g=91.520 kg

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 = πr2h

(i) Let the new height be 2h.
New volume of the cylinder=πr22h=2πr2h=2×original volume
Hence, the volume of the cylinder becomes double the initial volume.

(ii) Let the new height be 2h and the new radius be r2.
New volume of the cylinder=πr222h=πr2h2=12×original volume
Hence, the volume of the cylinder becomes half the initial volume.

(iii) Let the new radius be r2.
New volume of the cylinder=πr22h=πr2h4=14×original volume
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 = x3

Now, if the length of each edge of a cube is tripled, then the new length becomes 3x.
So,
New volume of the cube=3x×3x×3x=27x3=27×Original volume of the cube

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 cmand base has an area of 670 cm2.

Thus,
lb = 670

Now, volume of a cuboid = lbh

So,
13400=lbh13400=670hh=20 cm

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.

Area of metal sheet required for 10 boxes=10×2bh+hl+lb=10×216×14+14×20+20×16=10×2224+280+320=10×1008+320=13280 cm2

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
                                        =4.2×3×1.8=22.68 m3=22.68×1000              1 m3=1000 L=22680 L

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

Number of cuber required=Volume of the larger cubeVolume of the smaller cubes=80.125=64

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,
Volume of the wood=Volume of the outer box-Volume of the inner box=40×34×30-38×32×28=40800-34048=6752 cm3

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.
Volume of water flowing in one minute=lbh=50×2×45=4500 m3



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
                              =2lb+bh+hl-2×πr2=215×12+12×20+20×15-2×227×2.82=2180+240+300-49.28=1440-49.28=1390.72 cm2

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.

7.8=lbh7.8=l×5×2l=0.78 m=78 cm

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,
Volume of iron used=πR2-r2h=227×212-192×152=227×441-361×152=227×80×152=38217.142 cm3
 

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,
Volume of water flowing through pipes in figure A=2×πr2h=2×227×8×8×h=28167h
and
Volume of water flowing through pipe in figure B=πR2h=227×15×15×h=49507h

Now,
49507h>28167h
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,

Volume of the cylindrical tank=πr2h=227×5×5×7=550 m3=550000 L
                                                  

Therefore,
Number of days water will last=Volume of tankDaily requirement of water=550000550000=1
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 = πr2h
                                          =3.14×0.75×0.75×0.2

Thus,
Number of discs melted=Total volume of metal obtainedNumber of discs melted=508.683.14×0.75×0.75×0.2=508.680.35325=1440

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,
Volume of cylinder=πr2h 
12936=227×2x×2x×3xx3=12936×722×2×2×3=343x=7

Therefore,
TSA of the cylinder=2πrr+h=2×227×2727+37=88×35=3080 cm2

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 ₹5 per dm2 is ₹11,750.
Total area painted=Total cost of painting the boxTotal area of the box=117505=2350 dm2                  .....2

Equating (1) and (2),
94x2=2350x2=25x=5             length cannot be negative

Therefore,
length of the box be 5x = 5 × 5 = 25 dm,
the breadth be 4x = 4 × 5 = 20 dm and
the height be 3= 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 = 11000×15.4=0.0154 m3

So,
0.0154=πr2h                                   Volume of cylinder=πr2h 0.0154=227×r2×1r2=0.0154×722=0.049r=0.07 m

Now, a metal sheet would be needed to make this vessel.
TSA of metal sheet required=2πrh+r=2×227×0.071+0.07=0.4708 m2

 

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 a4.
Therefore,
Volume of the cube=a43=a364
Hence, its volume becomes 164 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.

2πr=252×227×r=25r=25×744
So,
Volume of the cylinder=πr2h=227×25×744×25×744×7=348.01 cm3

Also,
SA of the cylinder=2πrh=2×227×25×744×7=175 cm2

 

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 = πr2h

In one second, 7 m of water flows through the pipe.
So, in one hour, 7×60×60 m of water flows through the pipe. Thus, this is the height of the water flowing through the pipe in one hour.

Volume of water flowing in one hour=πr2h=227×0.0075×0.0075×7×60×60=4.455 m3=4.455×1000 L                                1 m3=1000 L=4455 L

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,
4×2πrh=6×2πr24h=6r6r=4×12=48            h=12 cmr=8 cm

Now,
CSA of the cylinder=2πrh=2×227×8×12=603.42 cm2

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).

Increase in volume of water=πr2h=227×1.54×1.54×1.5=11.18 m3=11180 L            1 m3=1000 L=11.18 kL

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            (∵ 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 110thpart 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 110th part of the wall.

Now,
Volume of the wall=1100×350×40
Volume of the wall with bricks=910×1100×350×40

Also,
Volume of a brick = 22 × 10 × 7

Therefore,
Number of bricks used for the wall=Volume of the bricks in the wallVolume of a brick=910×1100×350×4022 × 10 × 7=9000

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              (∵ Area of the floor = lb)
b = 10 m

Also,
2000 = lbh
⇒ 2000 = 25 × 10 × h
h=200025×10=8 m 

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, CSA of the cylinderTSA of the cylinder=12.

Now,
CSA of the cylinderTSA of the cylinder=2πrh2πrh+r12=2πrh2πrh+r2h=h+rh=r

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            (∵ Area of rectangle = lb)
                                          = 288 cm2

Area of the back of the solid = 18 × (3 + 2 + 18)            (∵ 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

CSA of the cylinder=2πrh=2×227×42×20=251.2 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
252000=60×60×hh=70 cm

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.
33=2πr1                              Circumferene of circle=2πrr1=33×72×22=214 cm

So,
Volume of cylinder=πr12h1=227×214×214×44=3811.5 cm3

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.
44=2πr2                              Circumferene of circle=2πrr2=44×72×22=7 cm

So,
Volume of cylinder=πr22h2=227×7×7×33=5082 cm3

Hence, volume of cylinder is 5082 cm3.
 



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