KSEEB SSLC Solutions for Class 10 Maths – Mensuration

KSEEB SSLC Solutions for Class 10 Maths – Mensuration (English Medium)

Exercise 16.1:

Question 1:

The height of a right circular cylinder is 14 cm and the radius of its base is 2 cm. Find its (i) CSA (ii) TSA

Solution :

Given:
Height of the cylinder = 14 cm
Radius of the cylinder = 2 cm

Question 2:

The CSA of a cylindrical pipe is 550 sq.cm. If the height of the pipe is 25 cm find the diameter of the base.

Solution :

Given:
CSA of cylindrical pipe = 550 cm²
Height of the pipe = 25 cm
To find the diameter of the pipe, find the radius using the formula.

Question 3:

An iron pipe 20 cm long has external radius equal to 12.5 cm and internal radius equals to 11.5 cm. Find the TSA of the pipe.

Solution :

Given:
h = 20 cm, internal radius (r)= 11.5 cm, External radius (R)= 12.5 cm

Question 4:

The radii of two right circular cylinders are in the ratio 2:3 and the ratio of their curved surface areas is 5:6. Find the ratio of their heights.

Solution :

Let the radius of cylinder A = r₁ = 2x
and radius of cylinder B = r₂ = 3x
Let the heights of the cylinders be h₁ and h₂ respectively.
Given:
r₁: r₂ = 2 : 3
CSA₁ : CSA₂ = 5 : 6

Question 5:

Find the ratio between TSA of a cylinder to its CSA given its height and radius are 7 cm and 3.5 cm respectively.

Solution :

Given:
r = 3.5 cm, h = 7cm, TSA:CSA = ?

Question 6:

The inner diameter of a circular well is 2.8 m. It is 10 m deep. Find its inner curved surface area. Also find the cost of plastering this curved surface at the rate of
Rs. 42 per m²?

Solution :

Given:
d = 2.8 m, r = 1.4 m, h =10m
Cost of plastering per metre = Rs. 42 per m²

Question 7:

Craft teacher of a school taught the students to prepare cylindrical pen holders out of card board. In a class of strength 42, if each child prepared a pen holder of radius 5 cm and height 14 cm, how much cardboard was consumed?

Solution :

Given:
Radius of each pen holder = 5 cm
Height of pen holder = 14 cm

Question 8:

A solid cylinder has total surface area of 462 cm². If its curved surface area is one third of its total surface area, find the radius and height of the cylinder?

Solution :

Question 9:

A cylindrical vessel without lid has to be tin coated on its outside. If the radius of the base is 70 cm and its height is 1.4 m, Calculate the cost of tin coating at the rate of Rs. 3.50 per 1000 cm²

Solution :

Given:
r = 70 cm, h = 1.4m = 140 cm

Question 10:

The diameter of a garden roller is 1.4 m and is 2 m long. How much area will it cover in 5 revolutions?

Solution :

Question 11:

Find the volume of a right circular cylinder whose radius is 10.5 cm and height is 16 cm

Solution :

Question 12:

The inner diameter of cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The pipe is 35 cm long. Find the mass of the pipe if 1 cm³ of wood has a mass of 0.06 gm

Solution :

Given:
Inner diameter = 24 cm, ∴ r₁ = 12 cm
Outer diameter= 28 cm, ∴ r₂ = 14 cm

Question 13:

The lateral surface area of cylinder of height 5 cm is 94.2 cm². Find (i) radius of its base (ii) Volume of the cylinder.

Solution :

For the cylinder, the lateral surface area means area of curved surface not including top or bottom.
Hence, for the cylinder LSA = CSA
Given:
LSA= 94.2 cm²
h = 5 cm, r =? v =?

Question 14:

Two circular cylinders of equal volumes have their heights in the ratio 1:2. Find the ratio of their radii

Solution :

Question 15:

A rectangular sheet of paper, 44 cm x 20 cm is rolled along its length to form a cylinder. Find the volume of the cylinder so formed.

Solution :

Given:
Length of the rectangle = 44 cm
Breadth of the rectangle = 20 cm
∴ Circumference of the base of the cylinder = 44 cm
Height of cylinder = 20 cm

Question 16:

The trunk of a tree is cylindrical. Its circumferences is 176 cm. If the length of the trunk is 3 m, find the volume of the timber that can obtained from the trunk.

Solution :

Given:
C = 176 cm, h = 3, V = ?,

Question 17:

A well of diameter 14 m is dug 8m deep. The earth taken out of it has been evenly spread all around in to width of 21 m to form an embankment. Find the height of the embankment

Solution :

Question 18:

In a village fair, a stall keeper has kept a large cylindrical vessel of base radius 15 cm filled upto a height of 32 cm with orange juice. He sells them at Rs. 3 per glass in a cylindrical glass of radius 3 cm and height 8 cm. How much money does the shop keeper earn by selling the juice completely?

Solution :

Radius of large cylindrical vessel = 15 cm
Height of the juice in the cylindrical vessel = 32 cm
Radius of small cylindrical glass = 3 cm
Height of the juice in the cylindrical glass = 8 cm

Let r₁ and r₂ be the radii of the cylindrical vessel and cylindrical glass respectively.

Let h₁ and h₂ be the heights of the cylindrical vessel and cylindrical glass respectively

Money earned by stall keeper by selling 100 glasses of juices at Rs. 3 per glass = 100 × 3 = Rs. 300

Exercise 16.2:

Question 1:

Find the curved surface area of cone, if its slant is 60 cm and the radius of its base is 21 cm

Solution :

Question 2:

The radius of the cone is 7 cm and area of curved surface is 176 cm². Find its slant height.

Solution :

Question 3:

The area of the curved surface of a cone is 60π cm². If the slant height of the cone is 8 cm, find the radius of the base.

Solution :

Question 4:

Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.

Solution :

Question 5:

A clown’s cap is in the form of right circular cone of base radius 7cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Solution :

Given:
Radius of the base, r = 7 cm, height = 24 cm

550 cm² sheet is required for making 1 cap
∴ Area of the sheet required for making 10 such caps
= 10 × 550
= 5500 cm²

Question 6:

Find the ratio of the surface areas of two cones if their diameters of the bases are equal and slant heights are in the ratio 4:3

Solution :

Given:
Let the slant heights of two cones be l₁ and l₂
Given:
l₁: l₂ = 4:3

∴let slant height of cone 1 be 4x and cone 2 be 3x.
∵ Diameters of the cones are equal
∴The radii of both the cones also equal.

Question 7:

A cylinder and a cone have equal of their bases and equal heights. If their curved surface areas are in the ratio 8:5, show that the ratio of the radius of each to the height of each is 3:4

Solution :

Given:
Height of cylinder = Height of cone (h)
Base of cylinder and base of cone are equal.
∴ radius of cylinder = radius of cone (r)
CSA of a cylinder = 2πrh
CSA of cone = πrl
CSA of a cylinder: CSA of cone = 8:5
As per the relation between height, slant height and radius of the cone,
l² = h² + r²

Question 8:

Find the volume of a right circular cone with (i) radius 5 cm, height 7 cm (ii) radius 10.5 cm, height 20 cm (iii) height 21 cm, slant height 28 cm.

Solution :

Question 9:

Two cones have their heights in the ratio 1:3 and the radii of their bases in the ratio 3:1. Find the ratio of their volumes.

Solution :

Given:
Let height of the two cones be h₁ and h₂
As per given information,
h₁: h₂ = 3:1
∴ The heights one cone be x and the other cone be 3x
Let radii of the bases of two cones be r₁ and r₂
As per given information,
r₁: r₂ = 3:1
∴ The radius one cone be 3y and the other cone be y.

Question 10:

A right angled triangle, of which the sides containing the right angle are 6.3 cm and 10 cm in length, is made to turn around on the longer side. Find the volume of the solid thus generated.

Solution :

Given:
R = 6.3 cm, h = 10 cm, V = ?

Question 11:

A right circular cone is of height 81 cm and radius of its base is 16 cm is melted and recast into a right circular cylinder of height 48 cm. Find the radius of the base of the cylinder

Solution :

Given:
For cone : h₁ = 81 cm, r₁ = 16 cm
For cylinder : h₂ =48 cm, r₂ = ?
Volume of the cylinder =
Volume of the cylinder formed = volume of the cone

Question 12:

A right circular cone is of height 3.6 cm and radius of its base is 1.6 cm. It is melted and recast into a right circular cone with radius of its base 1.2 cm. Find the height of the cone so formed.

Solution :

Cone – A (original)
h₁ = 3.6 cm, r₁ = 1.6 cm,

Cone – B (Formed)
h₂ = ?, r₂ = 1.2 cm

∴ Volume of cone A = Volume of cone B

Question 13:

Solution :

Question 14:

A tent is of the shape of a right circular cylinder up to a height of 3 m and then becomes a right circular cone with a maximum height of 13.5m above the ground. Calculate the cost of paining the inner side of the tent at the rate of Rs.2 per sq m, if the radius of the base is 14 m

Solution :

Given:
Height  of the tent that is cylinder be = 3m
∴ height of the cone = 13.5 – 3 = 10.5 m
Radius of both = 14 m
∴ Area of the inner side of the tent
= CSA of the cylinder + CSA of the cone

Exercise 16.3:

Question 1:

A flower vase is in the form of a frustum of a cone. The perimeters of the ends are 44 cm and 8.4πcm. If the depth is 14 cm, find how much water it can hold?

Solution :

Question 2:

A bucket is in the shape of a frustum with the top and bottom circles of radii 15 cm and 10 cm. Its depth is 12 cm. Find its curved surface area and total surface area. (Express the answer in terms of π)

Solution :

Question 3:

From the top of a cone of base radius 24 cm and height 45 cm, a cone of slant height 17cm is cut off. What is the volume of the remaining frustum of the cone?

Solution :

Question 4:

Solution :

Question 5:

A container, opened from the top and made up of metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill in the container, at the rate of Rs. 40 per litre. Also find the cost of metal sheet used to make the container. It costs Rs. 8 per 100 sq cm (take π = 3.14)

Solution :

Exercise 16.4:

Question 1:

Find the surface area of a sphere of radius

(i) 14 cm
(ii) 2.8 m
(iii) 6.3 cm

Solution :

Question 2:

Find the TSA of a hemisphere of radius 5 cm.

Solution :

Question 3:

A hemispherical bowl made of wood has inner diameter of 10.5 cm. Find the cost of painting it on the inside at the rate of Rs.12 per 100 sq. cm.

Solution :

Question 4:

Calculate the surface area of the largest sphere that can be cut out of a cube of side 15 cm.

Solution :

Question 5:

The surface area of a solid hemisphere is 432π cm². Calculate its radius.

Solution :

Question 6:

A hemispherical bowl made of steel is 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Solution :

Given:
Inner radius = 5cm, thickness = 0.25 cm
∴ Outer radius = 5 + 0.25 = 5.25 cm

Question 7:

Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 10.5 cm

Solution :

Question 8:

The diameter of a metal ball is 3.5 cm. What is the mass of the ball, if the density of the metal is 8.9 g/cm³ [Hint : Mass = Volume x density]

Solution :

Question 9:

Find the volume of sphere whose surface area is 154 cm²

Solution :

Question 10:

The outer and inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume.

Solution :

Given:
r₁ (external radius) = 12 cm
r₂ (Internal radius) = 10 cm
Volume of hollow sphere = Volume of whole sphere – Volume with internal radius.

Question 11:

The volume of a solid hemisphere is 1152 π cm³. Find its curved surface area.

Solution :

Question 12:

A capsule of medicine is in the shape if a sphere of diameter 3.5 mm. how much medicine is needed to fill this capsule?

Solution :

Question 13:

A solid hemisphere of wax of radius 12 cm is melted and made into a cone of base radius 6 cm. calculate the height of the cone.

Solution :

Given:
Radius the hemisphere, r₂= 12 cm
Radius of the base of the cone, r₁ = 6 cm
Volume of the wax cone = Volume of the wax hemisphere

Question 14:

A right circular metallic cone of height 20 cm and base radius 5 cm is melted and recast into a sphere. Find the radius of the sphere.

Solution :

Question 15:

The diameter of a metallic sphere is 18 cm. It is melted and drawn into a wire having diameter of cross section 0.4 cm. Find the length of the wire.

Solution :

Question 16:

Solution :

Question 17:

Solution :

Given:
Diameter of the sphere = 28 cm
∴ Its radius = 14 cm = r₁
Height of the cone = 3 cm

Exercise 16.5:

Question 1:

A petrol tank is in the shape of a cylinder with hemispheres of same radius attached to both ends. If the total length of the tank is 6m and the radius is 1m, what is the capacity of the tank in litres.

Solution :

Question 2:

A rocket is in the shape of a cylinder with a cone attached to one end and a hemisphere attached to the other. All of them are of the same radius of 1.5m. The total length of the rocket is 7m and height of the cone is 2m. Calculate the volume of the rocket.

Solution :

Question 3:

A cup is in the form of a hemisphere surmounted by a cylinder. The height of the cylindrical portion is 8cm and the total height of the cup is 11.5 cm. Find the TSA of the cup.

Solution :

Given:
Total height of the cup = 11.5 cm and
Height of the cylinder = 8 cm
∴ radius of the hemisphere = 11.5 – 8 = 3.5 cm
∴ TSA of the cup
= CSA of the cylinder + CSA of the hemisphere
= 2πrh + 2πr²
= 2πrh + 2πr²
= 2πr (h + r)

Question 4:

A storage tank consists of a circular cylinder with a hemisphere adjoined on either ends. The external diameter of the cylinder is 1.4m and length is 8m, find the cost of painting it on the outside at the rate of Rs. 10 per m²

Solution :

Question 5:

A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16 cm. and its height is 15 cm. Find the cost of painting the toy at Rs. 7 per 100 cm².

Solution :

Given:
Diameter of the hemisphere = 16 cm
∴Radius = 8 cm
∴QR = 8 cm
Height of the cone = 15 cm
∴ PQ = 15 cm
Let l be the Slant height of the cone:
Here PR = l
In ∆PQR,
PR² = PQ² + QR²
∴ l² = (15)² + (8)²
∴l² = 225 + 64
∴ l² = 289

Question 6:

A circus tent is cylindrical up to a height of 3 m and conical above it. If the diameter of the base is 105 m and the slant height of the conical part is 53 m, find the total cost of canvas used to make the tent of the cost of the canvas per sq. m is Rs. 10

Solution :

Exercise 16.6:

Question 1:

Draw a plan and calculate the area of a level ground using the information given below.

	Metre to C
To D 120 To E 180	220 210 R 120 Q 80 P	200 to B
	From A

Solution :

Steps of construction:

1. Draw the vertical base line AC = 11 cm
2. Mark points P, Q, R, such that AP = 4 cm, AQ = 6 cm, AR = 10.5 cm.
3. Draw perpendiculars from P, Q and R with the given measurements.
   EP = 9 cm, DR = 6 cm, BQ = 10 cm.
4. Join the points to get the figure ABCDE.
5. Record the measurements.
6. Calculate the area of ABCDE as follows:
   Area of ABCDE = area of ∆AQB + area of ∆BQC + area of ∆CRD + area of □PRDE+ area of ∆APE

Question 2:

Plan out and find the area of the field from the data given from the Surveyor’s field book.

	Metre to E
To D 100 To C 75 To B 50	350 300 S 250 R 150 Q 50 P	150 to F 100 to G
	From A

Solution :

Steps of construction:

1. Draw the vertical base line AE = 7 cm
2. Mark points P, Q, R, S such that AP = 1cm, AQ = 3 cm, AR = 5 cm, AS = 6 cm
3. Draw perpendiculars from P, Q, R and S with the given measurements.
   PB = 1 cm, QG = 2 cm, RC = 1.5 cm, SD = 2 cm, SF = 3 cm.
4. Join the points to get the figure ABCDEFG.
5. Record the measurements.
6. Calculate the area of ABCDEFG as follows:
   Area of ABCDEFG = area of ∆AQG + area of □QSFG + area of ∆ESF + area of ∆ESD + area of □DSRC+ area of □CRPB + area of ∆APB

Question 3:

Sketch a rough plan and calculate the area of the field ABCDEFG from the following data.

	Metre to D
To E 90 To F 60 To G 15	225 175 T 125 S 100 R 80 Q 60 P	20 to C 70 to B
	From A

Solution :

Steps of construction:

1. Draw the vertical base line AD = 9 cm
2. Mark points P, Q, R, S, T such that AT = 7 cm, AS = 5 cm, AR = 4 cm, AQ = 3.2 cm, AP = 2.4 cm.
3. Draw perpendiculars from P, Q, R, S and T with the given measurements.
   PB = 2.8 cm, QG = 0.6 cm, RF = 2.4 cm, SC = 0.8 cm, TE = 3.6 cm.
4. Join the points to get the figure ABCDEFG.
5. Record the measurements.
6. Calculate the area of ABCDEFG as follows:
   Area of ABCDEFG
   = Area of ∆APB + area of □BPSC + area of ∆DSC + area of ∆DTE + area of □ETRF+ area of □FRQG + area of ∆AQG

Question 4:

Solution :