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MENSURATION
Exercise 11.1
Q.1. A square and a rectangular field with measurements as
given in the figure have the same perimeter. Which field
has a larger area?
Ans. (a)
Side = 60 m
(Given)
Perimeter of square = 4 × side = 4 × 60 = 240 m
Area of square = (Side)2 = (60)2 = 3600 m2 ----- (i)
(b) Here, length (l) = 80 m (Given)
breadth (b) = ?
Perimeter of rectangle = 2(l + b)
⇒
240 m = 2(80 + b)
⇒
240
= 80 + b
2
⇒
120 = 80 + b
⇒
120 – 80 = b
⇒
40 = b
or,
b = 40 m
Area of rectangle = l × b = 80 m × 40 m
= 3200 m2 ---------------- (ii)
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From eqn. (i) and (ii), we have
Area of square > Area of rectangle
Hence, field (a) has larger area.
Q.2. Mrs. Kaushik has a square plot with the measurement
as shown in the figure. She wants to construct a house in
the middle of the plot. A garden is developed around the
house. Find the total cost of developing a garden around
the house at the rate of Rs 55 per m2.
Ans. We know that area of a square = (Side)2
= (25)2 = 625 m2 ---------- (i)
Area of rectangular plot = l × b
= 20 × 15 = 300 m2 -------- (ii)
Subtracting eqn. (ii) from (i), we get
Area of garden = Area of square plot – Area of
middle plot (Rectangular plot)
= (625 – 300) m2 = 325 m2
Hence, the total cost of developing the garden
= 325 m2 × Rs 55 = Rs 17,875
Q.3. The shape of a garden is rectangular in the middle and
semicircular at the ends as shown in the diagram. Find
the area and the perimeter of this garden [Length of
rectangle is 20 – (3.5 + 3.5) metres].
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Ans.
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Length of rectangle = 20 – (3.5 + 3.5) m
= (20 – 7) m = 13 m
Breadth of rectangle = 7 m
Also, breadth of rectangle = Diameter of semi-circle = 7 m
As radius =
1
Diameter
2
∴ Radius of semicircular ends = 3.5 m
Hence, perimeter of semicircular ends = 2 × πr
22
× 3.5
7
= 22 m ------------ (i)
=2×
Perimeter of the rectangular portion = 2(l + b)
= 2 × (13 + 7)
= 2 × 20 = 40 m ------------ (ii)
Adding eqn. (i) and (ii), we get
Perimeter of the garden = (22 m + 40 m) = 62 m
Area of the rectangular portion = l × b = 13 × 7 = 91 m2
Area of semicircular ends = 2 ×
1 2
πr
2
22
× 3.5 × 3.5
7
= 38.5 m2
=
Area of the garden = 38.5 cm2 + 91 cm2
= 129.5 m2
Hence, perimeter of the garden = 62 m
Area of the garden = 129.5 m2
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Q.4. A flooring tile has the shape of a parallelogram whose
base is 24 cm and the corresponding height is 10 cm.
How many such tiles are required to cover a floor of
area 1080 m2? (If required you can split the tiles in
whatever way you want to fill up the corners).
Ans.
Area of a tile = base × height
= b×h
= 24 × 10 = 240 cm2
We know, 1 m = 100 cm
1 m2 = 100 × 100 × cm2
or,
Area of the floor = 1080 m2
= 1080 × 100 × 100 × cm2
= 10800000 cm2
Number of tiles required =
Area of floor
Area of flooring tiles
10800000
= 45000.
240
Hence, 45000 tiles are required.
=
Q.5. An ant is moving around a few food pieces of different
shapes scattered on the floor. For which
food-piece would the ant have to take a
longer round? Remember, circumference
of a circle can be obtained by using the
expression c = 2πr, where r is the radius of the circle.
(a)
(b)
(c)
Ans. (a) Given diameter of semi-circle = 2.8 cm
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So, radius of semi-circle = 1.4 cm
∴ Circumference of semi-circle = 4.4 cm
(b) Perimeter of the given shape = Circumference of
semicircle + Perimeter of square
= 4.4 cm + 6.0 cm
= 10.4 cm
(c) Perimeter of the given shape = Circumference of
semicircle + 2 cm + 2 cm.
= 4.4 cm + 2 cm + 2 cm
= 8.4 cm
Hence, the ant have to take a longer round for the (b)
food-piece.
Exercise 11.2
Q.1. The shape of the top surface of a table is a trapezium.
Find its area if its parallel sides are 1 m and 1.2 m and
perpendicular distance between them is 0.8 m.
1
h(a + b) [Where h is height and
2
a and b are parallel sides]
1
(1.2 + 1) × 0.8
=
2
1
=
× 2.2 × 0.8 = 0.88 m2
2
Ans. Area of the trapezium =
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Q.2. The area of a trapezium is 34 cm2 and the length of one
of the parallel sides is 10 cm and its height is 4 cm. Find
the length of the other parallel side.
1
h(a + b)
2
1
× 4 × (10 + b)
34 =
2
Ans. Area of trapezium =
⇒
⇒
34 = 2 × (10 + b)
⇒
34 = 20 + 2b
⇒
2b = 34 – 20
⇒
2b = 14
⇒
or,
14
2
b = 7 cm
b =
Hence, another parallel side is 7 cm
Q.3. Length of the fence of a trapezium shaped filed ABCD is
120 m. If BC = 48 m, CD = 17 m and
AD = 40 m, find the area of this field.
Side AB is perpendicular to the parallel
sides AD and BC.
Ans. Perimeter of the trapezium ABCD = AB + BC + CD + AD
⇒
120 = AB + 48 + 17 + 40
= AB + 105
⇒
AB = 120 – 105
∴
AB = 15 m
Area of trapezium ABCD =
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1
(a + b) × h
2
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1
× (48 + 40) × 15
2
1
× 88 × 15
=
2
= 660 m2
=
Q.4. The diagonal of a quadrilateral shaped field is 24 m and
the perpendiculars dropped on it from the
remaining opposite vertices are 8 m and
13 m. Find the area of the field.
Ans. Area of quadrilateral ABCD
= Area of ΔABC + Area of ΔADC
1
1
× AC × MB + AC × ND
2
2
1
AC (MB + ND)
=
2
1
=
× 24(13 + 8)
2
= 12(21)
=
= 12 × 21
= 252 m2
Q.5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find
its area.
Ans.
1
× d1 × d2
2
Length of one diagonal (d1) = 7.5 cm
Area of rhombus =
Length of other diagonal (d2) = 12 cm
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∴
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1
× 7.5 × 12
2
= 7.5 × 6 = 45.0
Area of rhombus =
= 45 cm2
Q.6. Find the area of a rhombus whose side is 6 cm and
whose altitude is 4 cm. If one of its diagonals is 8 cm
long, find the length of the other diagonal.
Ans.
Area of rhombus =
1
× d1 × d2
2
Also, area of the given rhombus = 6 × 4 = 24 cm2
So,
⇒
1
d1 × d2 = 6 × 4
2
1
× 8 × d2 = 6 × 4
2
⇒ 4 × d2 = 6 × 4
⇒ d2 = 6 cm
Q.7. The floor of a building consists of 3000 tiles which are
rhombus shaped and each of its diagonals are 45 cm and
30 cm in length. Find the total cost of polishing the floor,
if the cost per m2 is Rs 4.
Ans.
Area of each tiles = Area of rhombus
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1
d1 × d2
2
1
× 45 × 30
=
2
= 45 × 15
=
= 675 cm2
∴
Area of 1 tile = 675 cm2
Area of 3000 tiles = 3000 × 675 cm2
2025000 cm 2
=
100 × 100
= 202.5 m2
∴ Total cost of polishing the floor = 202.5 × 4
= Rs 810.0
= Rs 810
Hence, the cost of polishing the floor = Rs 810
Q.8. Mohan wants to buy a trapezium shaped field. Its side
along the river is parallel to and twice the side along
the road. If the area of this field is
10500 m2 and the perpendicular
distance between the two parallel
sides is 100 m, find the length of the
side along the river.
Ans. Let the length of one of the parallel sides (a) = xm
Length of the other parallel side (b) = 2xm
1
(a + b) × h
2
1
(x + 2x) × 100
=
2
Area of trapezium =
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=
1
× 3x × 100
2
∴
10500 m2 = 150x
⇒
10500
= x
150
⇒
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x = 70 m
Hence,
a = 70 m
b = 2 × 70 = 140 m
Q.9. Top surface of a raised platform is in the
shape of a regular octagon as shown in the
figure. Find the area of the octagonal surface.
Ans. Area of trapeziumABCH = Area of trapezium GDEF
1
(a + b) × h
2
1
=
(11 + 5) × 4
2
1
× 16 × 4 = 32 m2
=
2
Area of rectangle HCDG = length × breadth
=
= 11 × 5 = 55 m2
Hence, area of the octagonal surface
= Area of trapezium ABCH
+ Area of rectangle HCDG
+ Area of trapezium GDEF.
= 32 m2 + 55 m2 + 32 m2
= 119 m2
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Q.10. There is a pentagonal shaped park as shown in the
figure. For finding its area Jyoti and Kavita divided it in
two different ways.
Find the area of this park using both ways. Can you
suggest some other way of finding its area?
Ans. Jyoti’s way :
We know area of trapezium =
1
(a + b) × h
2
1
(15 + 30) × 7.5
2
1
× 45 × 7.5 = 168.75 m2
=
2
=
Hence, area of figure using Jyoti’s way = 2 × 168.75
= 337.5 m2
Kavita’s way :
Area of the given figure
= Area of triangle + Area of square
We know area of triangle =
1
× base × height
2
Area of square = (side)2
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1
× 15 × 15 + (15)2
2
225 225 225 + 450
=
+
=
2
1
2
675
=
= 337.5 m2
2
Hence, area of figure using Kavita’s way = 337.5 m2
∴ Area of the given figure =
Q.11. Diagram of the adjacent picture frame
has outer dimensions = 24 cm × 28 cm
and inner dimensions 16 cm × 20 cm.
Find the area of each section of the frame,
if the width of each section is same.
Ans.
Given KE = FL = HI = GJ = 4 cm
1
h (a + b)
2
1
× 4 (24 + 16)
=
2
1
=
× 4 × 40
2
= 80 cm
Area of section AEFB =
Similarly,
Area of section DCGH =
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1
× 4(24 + 16)
2
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1
× 4 × 40
2
= 80 cm2
=
Area of section DHEA = Area of trapezium =
=
1
h(a + b)
2
1
× 4(28 + 20) = 96 cm2
2
Similarly,
1
h(a + b)
2
1
× 4(28 + 20)
=
2
1
× 4 × 48
=
2
= 96 cm2
Area of section BFGC =
Exercise 11.3
Q.1. There are two cuboidal boxes as shown in the adjoining
figure. Which box requires the lesser amount of
material to make?
Ans. (a)
l = Length = 60 cm
b = breadth = 40 cm
h = height = 50 cm
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Total surface area of cuboidal box = 2(lb + bh + hl)
= 2 × (60 × 40 + 40 × 50 + 50 × 60)
= 2 × (2400 + 2000 + 3000)
= 2 × (7400)
= 14800 cm2
(b)
Here, l = b = h = 50 cm
Total surface area of cube = 6 × (side2)
= 6 × (50)2
= 6 × 2500
= 15000 cm2
Hence, box (a) requires the lesser amount of material to
make.
Q.2. A suitcase with measures 80 cm × 48 cm × 24 cm is to be
covered with a tarpaulin cloth. How many metres of
tarpaulin of width 96 cm is required to cover 100 such
suitcases?
Ans. Total surface area of suitcase = 2(lb + bh + hl)
= 2(80 × 48 + 48 × 24 + 24 × 80)
= 2 × (3840 + 1152 + 1920)
= 2 × 6912 = 13824 cm2
Length of required tarpaulin =
=
Surface area of suitcase
Width of tarpaulin
13824
= 144 cm
96
Q.3. Find the side of a cube whose surface area is 600 cm2.
Ans.
Surface area of cube = 6 × side2 (l = side)
⇒
600 cm2 = 6l2
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⇒
600
= l2
6
⇒
100 = l2
⇒
100 = l
⇒
10 = l
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(Taking square root on both side)
Hence, the required side of the cube = 10 cm.
Q.4. Rukhsar painted the outside of the cabinet of measure
1 m × 2 m × 1.5 m. How much surface area did she cover
if she painted all except the bottom of the cabinet.
Ans.
Here,
l = 2 m,
b = 1 m and h = 1.5 m
Lateral surface area of the cabinet = 2h(l + b)
= 2 × 1.5 (2 + 1)
= 2 × 3 × 1.5 = 9 m2
Area of the top of the cabinet = 1 × 2 m2 = 2 m2
Hence, required covered surface area = 9 m2 + 2 m2 = 11 m2
Q.5. Daniel is painting the walls and ceiling of a cuboidal hall
with length, breadth and height of 15 m, 10 m and 7 m
respectively. From each can of paint 100 m2 of area is
painted. How many cans of paint will she need to paint
the room?
Ans.
Here, Length = (l) = 15 m
Breadth = (b) = 10 m
Height = (h) = 7 m
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Lateral surface area of the hall
= 2h(l + b)
= 2 × 7(15 + 10)
= 2 × 7 × 25 = 350 m2
Area of ceiling = l × b = 15 × 10 = 150 m2
Hence, the total area to be painted
= 350 m2 + 150 m2 = 500 m2
Since, each can of paint covers 100 m2 of area
So, number of cans that will be required to paint =
500
=5
100
Hence, 5 cans will be needed to paint the room.
Q.6. Describe how the two figures at the right are alike and
how they are different. Which box has larger lateral
surface area?
One of the given figure is cylinder while other is cube
edge having an edge of 7 cm. The height of both figures
are same.
Ans. Lateral surface area of cylinder
= 2πrh
22
7
×7
×
7
2
= 154 cm2
= 2×
Lateral surface area of cube
= 4 × (side2)
= 4 × (7)2
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⎡ where d = 7 cm ⎤
⎢
7 ⎥
⎢∴
r = cm ⎥
2 ⎥
⎢
⎢
h = 7 cm ⎥⎦
⎣
[∴ side = 7 cm]
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= 4 × 49 = 196 cm2
Hence, the cube has larger lateral surface area.
Q.7. A closed cylindrical tank of radius 7 m and height 3 m is
made from a sheet of metal. How much sheet of metal is
required?
Ans.
Total surface area of cylindrical tank
= 2πr (r + h)
22
× 7 (7 + 3)
7
= 2 × 22 × 10
= 2×
= 440 m2
Hence, the sheet of metal required = 440 m2.
Q.8. The lateral surface area of a hollow cylinder is 4224 cm2.
It is cut along its height and formed a rectangular sheet
of width 33 cm. Find the perimeter of rectangular sheet?
Ans.
Surface area of cylinder = 4224 cm2
⇒
2πrh = 4224 cm2
⇒
33 × h = 4224 cm2
⇒
(∵ 2πr = 33 cm)
h = 128 cm
Perimeter of rectangle = 2(l + b)
= 2(128 + 33)
= 2 × 161
= 322 cm
Q.9. A road roller takes 750 complete revolutions
to move once over to level a road. Find the
area of the road if the diameter of a road
roller is 84 cm and length is 1 m.
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Ans. Diameter of road roller = 84 cm
84
= 42 cm
2
l = h = 1 m = 100 cm
∴ Radius of road roller (r) =
Here,
Surface area of road roller = 2πrh
22
× 42 × 100 = 26400 cm2
7
So, area of the road = 26400 × 750
= 2×
= 19800000 cm2
= 1980 m2 (∵ 1 m2 = 10000 cm2)
Hence, area of the road = 1980 cm2
Q.10. A company packages its milk powder in cylindrical
container whose base has a diameter of 14 cm
and height 20 cm. Company places a label
around the surface of the container (as shown
in the figure). If the label is placed 2 cm from
top and bottom, what is the area of the label.
Ans.
Let r cm be the radius of the label.
We know diameter (d) = 2 × radius (r)
d 14
=
= 7 cm
2
2
The label is placed 2 cm from top and bottom, then required
height = 20 – 4 = 16 cm
∴
r =
⇒ Lateral surface area of the label
22
× 7 × 16
7
= 2 × 22 × 16 = 704 cm2
= 2πrh = 2 ×
Hence, area of the label = 704 cm2
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Exercise 11.4
Q.1. Given a cylindrical tank, in which situation
will you find surface area and in which
situation volume.
(a) To find how much it can hold.
(b) Number of cement bags required to
plaster it.
(c) To find the number of smaller tanks that can be
filled with water from it.
Ans. (a) Volume
(b) Surface area
(c) Volume
Q.2. Diameter of cylinder A is 7 cm, and the
height is 14 cm. Diameter of cylinder
B is 14 cm and height is 7 cm. Without
doing any calculations can you suggest
whose volume is greater? Verify it by
finding the volume of both the cylinders. Check whether
the cylinder with greater volume also has greater
surface area?
Ans. (A)Diameter of cylinder A = 7 cm
∴
Also,
d
7
= cm
2
2
height = h = 14 cm
Radius = r =
Volume of cylinder = πr2h
2
22 ⎛ 7 ⎞
× ⎜ ⎟ × 14
=
7
⎝2⎠
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22 7 7
× × × 14
7
2 2
= 539 cm3. -------------- (i)
=
(B) Diameter of cylinder B = 14 cm
∴
Also,
14
= 7 cm
2
height = h = 7 cm
Radius = r =
Volume of cylinder = πr2h
22
× (7)2 × 7
7
22
=
×7×7×7
7
= 1078 cm3 ------------ (ii)
=
Hence, the volume of cylinder (B) is greater than the
volume of cylinder (A)
Curved surface area of cylinder A = 2πrh
22 7
× × 14
7
2
= 308 cm2
= 2×
22
×7×7
7
= 308 cm2
Curved surface area of cylinder B = 2 ×
Hence, surface area of cylinder A and cylinder B are equal.
Q.3. Find the height of a cuboid whose base area is 180 cm2
and volume is 900 cm3?
Ans.
Let h be the height of cuboid.
Volume of cuboid = l × b × h
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Base area = l × b = 180 cm2
∴
Volume of cuboid = 180 × h
⇒
180 × h = 900 cm3
900
= 5 cm
180
Hence, Height of the cuboid = 5 cm
⇒
h =
Q.4. A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How
many small cubes with side 6 cm can be placed in the
given cuboid?
Ans.
Volume of cuboid = 60 × 54 × 30
= 97200 cm3
Volume of cube = (6 cm)3 = 216 cm3
97200
= 450 cubes
216
Hence, 450 cubes of side 6 cm can be placed in the given
cuboid.
Number of small cubes =
Q.5. Find the height of the cylinder whose volume is 1.54 m3
and diameter of the base is 140 cm?
Ans.
Let h be the height of the cylinder
Volume of cylinder = 1.54 m3
Diameter of cylinder = 140 cm
∴
Radius = r =
140
70
= 70 cm =
m
2
100
Volume of cylinder = πr2h
⇒
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1.54 m3 =
21
22
70
70
×
×
×h
7
100 100
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⇒
⇒
1.54 = 22 ×
1.54 × 100 × 100
15400
⇒
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10
70
×
×h
100 100
= h
1 = h
Hence, height of cylinder = 1 m
Q.6. A milk tank is in the form of cylinder
whose radius is 1.5 m and length is
7 m. Find the quantity of milk in
litres that can be stored in the tank?
Ans.
Radius of milk tank = 1.5 m
Height of milk tank = 7m
Volume of milk tank = πr2h = Volume of cylinder
22
× 1.5 × 1.5 × 7
7
= 49.5 cm3
=
Since, 1 m3 = 1000 litres
∴
49.5 m3 = 1000 × 49.5 m3
= 49500 litres
Hence, the quantity of milk in tank = 49500 litres.
Q.7. If each edge of a cube is doubled,
(a) how many times will its surface area increase?
(b) how many times will its volume increase?
Ans.
(a) Let the edge of a cube be x units
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If each edge of a cube is doubled then the new edge will
becomes 2x units
Surface area of cube = 6 × (side)2 = 6x2 sq. units
If edge is doubled, then surface area of cube
= 6 × (2x)2
= 6 × 4x2 = 24x2 sq. units
Hence, its surface area will becomes four times.
(b) Volume of cube = (l)3 = x3 cube units
If edge is doubled, then volume of cube = (2x)3
= 8x3 cubic units
= x3 × 8 = 8x3 cubic units
Hence, its volume will become eight times.
Q.8. Water is pouring into a cuboidal reservoir
at the rate of 60 litres per minute. If the
volume of reservoir is 108 m3, find the
number of hours it will take to fill the reservoir.
Ans.
Volume of cuboidal reservoir = 108 m3
Volume of cuboidal reservoir in litres
= 108 × 1000 litres (∵ 1 m3 = 1000l)
= 108000 litres
108000
60
= 1800 minutes
Required time =
=
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1800
= 30 hours (∵ 1 hr = 60 minutes)
60
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