11 MENSURATION - Testlabz · CBSEPracticalSkills.com ©Edulabz International CBSEPracticalSkills.com ©Edulabz International 1 11 MENSURATION Exercise 11.1 Q.1. A square and a rectangular

CBSEPracticalSkills.com ©Edulabz International


1

11 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)

⇒ 2402

= 80 + b

⇒ 120 = 80 + b ⇒ 120 – 80 = b ⇒ 40 = b or, b = 40 m Area of rectangle = l × b = 80 m × 40 m = 3200 m2 ---------------- (ii)



2

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].



3

Ans. 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 = 12

Diameter

∴ Radius of semicircular ends = 3.5 m Hence, perimeter of semicircular ends = 2 × πr

= 2 × 227

× 3.5

= 22 m ------------ (i) 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 × 12

πr2

= 227

× 3.5 × 3.5

= 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



4

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 or, 1 m2 = 100 × 100 × cm2 Area of the floor = 1080 m2

= 1080 × 100 × 100 × cm2

= 10800000 cm2

Number of tiles required = Area of floorArea of flooring tiles

= 10800000240

= 45000.

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



5

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.

Ans. Area of the trapezium = 12

h(a + b) [W

= 12

(1.2 + 1) × 0.8

= 12

× 2.2 × 0.8 = 0.88 m2

here h is height andand are parallel sides]a b



6

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.

Ans. Area of trapezium = 12

h(a + b)

⇒ 34 = 12

× 4 × (10 + b)

⇒ 34 = 2 × (10 + b) ⇒ 34 = 20 + 2b ⇒ 2b = 34 – 20 ⇒ 2b = 14

⇒ b = 142

or, b = 7 cm 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 = 12

(a + b) × h



7

= 12

× (48 + 40) × 15

= 12

× 88 × 15

= 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

= 12

× AC × MB + 12

AC × ND

= 12

AC (MB + ND)

= 12

× 24(13 + 8)

= 12(21) = 12 × 21 = 252 m2

Q.5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Ans. Area of rhombus = 12

× d1 × d2

Length of one diagonal (d1) = 7.5 cm Length of other diagonal (d2) = 12 cm



8

∴ Area of rhombus = 12

× 7.5 × 12

= 7.5 × 6 = 45.0 = 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 = 12

× d1 × d2

Also, area of the given rhombus = 6 × 4 = 24 cm2

So, 12

d1 × d2 = 6 × 4

⇒ 12

× 8 × d2 = 6 × 4

⇒ 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



9

= 12

d1 × d2

= 12

× 45 × 30

= 45 × 15 = 675 cm2

∴ Area of 1 tile = 675 cm2

Area of 3000 tiles = 3000 × 675 cm2

= 22025000

100 × 100cm

= 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

Area of trapezium = 12

(a + b) × h

= 12

(x + 2x) × 100



10

= 12

× 3x × 100

∴ 10500 m2 = 150x

⇒ 10500150

= x

⇒ 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

= 12

(a + b) × h

= 12

(11 + 5) × 4

= 12

× 16 × 4 = 32 m2

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



11

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 = 12

(a + b) × h

= 12

(15 + 30) × 7.5

= 12

× 45 × 7.5 = 168.75 m2

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 = 12

× base × height

Area of square = (side)2



12

∴ Area of the given figure = 12

× 15 × 15 + (15)2

= 225 225 225 + 450 + = 2 1 2

= 6752

= 337.5 m2

Hence, area of figure using Kavita’s way = 337.5 m2

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

Area of section AEFB = 12

h (a + b)

= 12

× 4 (24 + 16)

= 12

× 4 × 40

= 80 cm Similarly,

Area of section DCGH = 12

× 4(24 + 16)



13

= 12

× 4 × 40

= 80 cm2

Area of section DHEA = Area of trapezium = 12

h(a + b)

= 12

× 4(28 + 20) = 96 cm2

Similarly,

Area of section BFGC = 12

h(a + b)

= 12

× 4(28 + 20)

= 12

× 4 × 48

= 96 cm2

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



14

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 suitcaseWidth of tarpaulin

= 1382496

= 144 cm

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



15

⇒ 6006

= l2

⇒ 100 = l2

⇒ 100 = l (Taking square root on both side) ⇒ 10 = l 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



16

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 = 500100

= 5

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

= 2 × 22 7 7 2

× × 7

where d = 7 cm7 = 2

= 7

r cm

h cm

⎡ ⎤⎢ ⎥⎢ ⎥∴⎢ ⎥⎢ ⎥⎣ ⎦

= 154 cm2

Lateral surface area of cube = 4 × (side2)

= 4 × (7)2 [∴ side = 7 cm]



17

= 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)

= 2 × 227

× 7 (7 + 3)

= 2 × 22 × 10 = 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.



18

Ans. Diameter of road roller = 84 cm

∴ Radius of road roller (r) = 842

= 42 cm

Here, l = h = 1 m = 100 cm Surface area of road roller = 2πrh

= 2 × 227

× 42 × 100 = 26400 cm2

So, area of the road = 26400 × 750 = 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)

∴ r = 2d = 14

2 = 7 cm

The label is placed 2 cm from top and bottom, then required height = 20 – 4 = 16 cm

⇒ Lateral surface area of the label

= 2πrh = 2 × 227

× 7 × 16

= 2 × 22 × 16 = 704 cm2 Hence, area of the label = 704 cm2



19

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

∴ Radius = r = 7= 2 2d cm

Also, height = h = 14 cm Volume of cylinder = πr2h

= 222 7 ×

7 2⎛ ⎞⎜ ⎟⎝ ⎠

× 14



20

= 22 7 7× × 7 2 2

× 14

= 539 cm3. -------------- (i) (B) Diameter of cylinder B = 14 cm

∴ Radius = r = 142

= 7 cm

Also, height = h = 7 cm Volume of cylinder = πr2h

= 227

× (7)2 × 7

= 227

× 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

= 2 × 22 7× 7 2

× 14

= 308 cm2

Curved surface area of cylinder B = 2 × 227

× 7 × 7

= 308 cm2

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



21

Base area = l × b = 180 cm2

∴ Volume of cuboid = 180 × h ⇒ 180 × h = 900 cm3

⇒ h = 900180

= 5 cm

Hence, Height of the cuboid = 5 cm

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

Number of small cubes = 97200216

= 450 cubes

Hence, 450 cubes of side 6 cm can be placed in the given cuboid.

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 = 1402

= 70 cm = 70100

m

Volume of cylinder = πr2h

⇒ 1.54 m3 = 227

× 70100

× 70100

× h



22

⇒ 1.54 = 22 × 10100

× 70100

× h

⇒ 1.54 × 100 × 100 15400

= 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

= 227

× 1.5 × 1.5 × 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



23

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

Required time = 10800060

= 1800 minutes

= 180060

= 30 hours (∵ 1 hr = 60 minutes)

Documents

11 MENSURATION - Testlabz · CBSEPracticalSkills.com ©Edulabz International CBSEPracticalSkills.com ©Edulabz International 1 11 MENSURATION Exercise 11.1 Q.1. A square and a rectangular