## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere (Surface Area and Volume) Ex 20E

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21E

Other Exercises

Question 1.
A cone of height 15 cm and diameter 7 cm is mounted on a hemisphere of same diameter. Determine the volume of the solid thus formed.
Solution:
Height of cone = 15 cm
and radius of base = $$\frac { 7 }{ 2 }$$cm.

Question 2.
A buoy is made in the form of hemisphere surmounted by a right, cone whose circular base coincides with the plane surface of hemisphere. The radius of the base of the cone is 3.5 metres and its volume is two-thirds of the hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two places of decimal.
Solution:

Question 3.
From a rectangular solid of metal 42 cm by 30 cm by 20 cm, a conical cavity of diam­eter 14 cm and depth 24 cm is drilled out. Find:
(i) the surface area of remaining solid,
(ii) the volume of remaining solid,
(iii) the weight of the material drilled out if it weighs 7 gm per cm3.
Solution:
Length of rectangular solid (l) = 30 cm
and height (h) = 42 cm
Diameter of the cone = 14 cm

(i)  Surface area of remaining solid Surface area of rectangular solid + Surface area of curved surface of cone – Surface area of the base of the cone
= 2 (lb + bh + hl) + πrl – πr2

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere of the largest size. Find the surface area of the resulting solid.
Solution:
Side of a cubical block = 7 cm
Radius of the hemisphere = $$\frac { 7 }{ 2 }$$cm
Now total surface area of the block

Question 5.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the rim. When lead shots each of which is a sphere of radius 0.5 cm are dropped into the vessel, one- fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
Radius of conical vessel (R) = 5 cm
and height (h) = 8 cm

Question 6.
A hemi-spherical bowl has neligible thickness and the length of its circumference is 198 cm. Find the capacity of the bowl.
Solution:
Upper circumference of the hemi-spherical bowl = 198 cm

Question 7.
Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r cm.
Solution:
Radius of solid hemisphere = r
Radius of the cone carved out of the hemisphere = r
and height (h) = r

Question 8.
The radii of the bases of two solid right circular cones of same height are r1 and r2 The cones are melted and recast into a solid sphere of radius R. Find the height of each cone in terms of r1, r2 • and R.
Solution:

Question 9.
A solid metallic hemisphere of diameter 28 cm is melted and recast into a number of identical solid cones, each of diameter  14 cm and height 8 cm. Find the number * of cones so formed.
Solution:
Diameter of solid hemisphere = 28 cm

Question 10.
A cone and a hemisphere have the same base and the same height. Find the ratio between their volumes.
Solution:
Let radius of the base of cone = r
and height = h
Then radius of hemisphere = r

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20E are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere (Surface Area and Volume) Ex 20D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21D

Other Exercises

Question 1.
A solid sphere of radius 15 cm is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate the number of cones recast.
Solution:

Question 2.
A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Find the height of the cone. [2002]
Solution:
External diameter = 8cm
∴ Radius (R) = $$\frac { 8 }{ 2 }$$ = 4 cm
Internal diameter = 4 cm
∴ Radius (r) = $$\frac { 4 }{ 2 }$$ = 2cm.
Volume of metal used in hollow sphere

Question 3.
The radii of the internal and external surfaces of a metallic spherical shell are 3 cm and 5 cm respectively. It is melted and recast into a solid right circular cone of height 32 cm. Find the diameter of the base of the cone.
Solution:
Inner radius of spherical shell (r) = 3 cm
and outer radius (R) = 5 cm

Question 4.
Total volume of three identical cones is the same as that of a bigger cone whose height is 9 cm and diameter 40 cm. Find the radius of the base of each smaller cone, if height of each is 108 cm.
Solution:
Height of bigger cone (H) = 9 cm
Diameter 40 cm

Question 5.
A solid rectangular block of metal 49 cm by 44 cm by 18 cm is melted and formed into a solid sphere. Calculate the radius of the sphere.
Solution:
Dimensions of rectangular block of metal = 49cm x 44 cm x 18 cm.
∴ Volume = 49 x 44 x 18 cm3 = 38808 cm3
Let radius of solid sphere = r

Question 6.
A hemisphere bowl of internal radius 9 cm is full of liquid. This liquid is to be filled into conical shaped small containers each of diameter 3 cm and height 4 cm. How many containers are necessary to empty the bowl?
Solution:
Internal radius of hemispherical bowl (r) = 9 cm

Question 7.
A hemispherical bowl of diameter 7.2 cm is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius 4.8 cm. Find the height of the cone if it is completely filled. [2010]
Solution:
Diameter of hemispherical bowl = 7.2 cm

Question 8.
A solid cone of radius 5 cm and height 8 cm is melted and made into small spheres of radius 0.5 cm. Find the number of spheres formed. [2011]
Solution:
Radius of solid cone = (r) = 5 cm
and height (h) = 8 cm

Question 9.
The total area of a solid metallic sphere is 1256 cm2. It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate:,
(i)  the radius of the solid sphere,
(ii) the number of cones recast.   Take π = 3.14              [2000]
Solution:
Total area of solid sphere = 125

Question 10.
A solid metallic cone, with radius 6 cm and height 10 cm, is made of some heavy metal A. In order to reduce its weight, a conical hole is made in the cone as shown and it is completely filled with a lighter metal B. The conical hole has a diameter of 6 cm and depth 4 cm. Calculate the ratio of the volume of metal A to the volume of the metal B in the solid.

Solution:
Radius of solid metallic cone A(R) = 6 cm
and height (H) = 10 cm

Question 11.
A hollow sphere of internal and external radii 6 cm and 8 cm respectively is melted and recast into small cones of base radius 2 cm and height 8 cm. Find the number of cones.           [2012]
Solution:
Inner radius of a hollow sphere (r) = 6 cm
and outer radius (R) = 8 cm

Question 12.
The surface area of a solid metallic sphere is 2464 cm2. It is melted and recast into solid right circular cones of radius 3.5 cm and height 7 cm. Calculate:
(i) the radius of the sphere.
(ii) the number of cones recast. (Take π = 22/7)
Solution:
(i) Surface area=4πr2=2464 cm2 (given)

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20D are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere (Surface Area and Volume) Ex 20C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C

Other Exercises

Question 1.
The surface area of a sphere is 2464 cm2, find its volume.
Solution:
Surface area of sphere = 2464 cm2

Question 2.
The volume of a sphere is 38808 cm3; find its diameter and the surface area.
Solution:
Volume of sphere = 38808 cm3
Let radius of shpere = r

Question 3.
A spherical ball of lead has been melted and made into identical smaller balls with radius equal to half the radius of the original one. How many such balls can be made ?
Solution:
Let the radius of spherical ball = r

Question 4.
How many balls each of radius 1 cm can be made by melting a bigger ball whose diameter is 8 cm.
Solution:

Question 5.
Eight Metallic sphere; each of radius 2 mm, are melted and cast into a single sphere. Calculate the radius of the new sphere.
Solution:
Radius of metallic sphere = 2mm = $$\frac { 1 }{ 5 }$$ cm

Question 6.
The volume of one sphere is 27 times that of another sphere. Calculate the ratio of their:
(ii) surface areas.
Solution:
Volume of first sphere = 27 x volume of second sphere.
Let radius of first sphere = r1
and radius of second sphere = r2

Question 7.
If the number of square centimetres on the surface of a sphere is equal to the number of cubic centimetres in its volume, what is the diameter of the sphere ?
Solution:

Question 8.
A solid metal sphere is cut through its centre into 2 equal parts. If the diameter of the sphere is 3$$\frac { 1 }{ 2 }$$ cm, find the total surface area of each part correct to two decimal places.
Solution:
A solid sphere is cut into two equal hemispheres.

Question 9.
The internal and external diameters of a hol­low spectively. Find:
(i) internal curved suface area,
(ii) external curved surface area,
(iii) total surface area,
(iv) volume of material of the vessel.
Solution:
Internal diameter of hollow hemispher = 21cm
and external diameter = 28 cm

Question 10.
A solid sphere and a solid hemi-sphere have the same total surface area. Find the ratio between their volumes.
Solution:
Let radius of a sphere = R
∴  Surface area = 4πR2
and radius of hemi-sphere = r
∴ Surface area = 3πr2
∵ Their surface area are equal
4πR2 = 3πr2 ⇒ 4R2 = 3r2

Question 11.
Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted and recasted into a single solid sphere. Taking π = 3.1, find the surface area of solid sphere formed.
Solution:
Radius of first sphere (r1) = 6 cm
Radius of second sphere (r2) = 8 cm
Radius of third sphere (r3) = 10 cm

Question 12.
The surface area of a solid sphere is increased by 21% without changing its shape. Find the percentage increase in its:

(ii) volume
Solution:
(i)  Let r be the radius of the solid sphere then surface area = 4πr2
Increase in area = 21%

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20C are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere (Surface Area and Volume) Ex 20B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B.

Other Exercises

Question 1.
Find the volume of a cone whose slant height is 17 cm and radius of base is 8 cm.
Solution:
Slant height (L) = 17 cm
But l2= r2 + h2
⇒  h2 = l2-r2 = 172 – 82
⇒ h2 = 289 – 64 = 225 = (15)2
∴   h=15 cm.

Question 2.
The curved suface area of a cone is 12320 cm2. If the radius of its base is 56 cm, find its height.
Solution:
Curved surface area = 12320 cm2
Radius of base (r) = 56 cm.
Let slant height = l.

Question 3.
The circumference of the base of a 12 m high conical tent is 66m. Find the volume of the air contained in it.
Solution:
Circumference of conical tent = 66 m
and height (h) = 12 m.

Question 4.
The radius and the height of a right circular cone are in the ratio 5 :12 and its volume is 2512 cubic Cm. Find the radius and slant height of the cone. (Take π = 3.14)
Solution:
The ratio between radius and height = 5 : 12
Volume =2512 cm3
Let radius (r) = 5x and
height (h) = 12x
and slant height = l

Question 5.
Two right circular cones x and y are made, x having three times the radius of y and y having half the volume of x. Calculate the ratio between the heights of x and y.
Solution:
Let radius of cone y = r
∴ radius of cone x = 3r
Let volume of cone y = V
Then volume of x = 2V
Let h1 be the height of x and h2 be the height of y.

Question 6.
The diameters of two cones are equal, if their slant heights are in the ratio of 5:4, find the ratio of their curved surface area.
Solution:
Let radius of each one = r
and ratio between their slant heights =5:4
Let slant height of the first cone = 5x
and slant height of second = 4 x
∴  Curved surface area of the first cone
= πr = πr x 5x = 5πrx.
and curved suface area of second cone
= πr x 4x = 4πrx
∴ Ratio between them = 5 πrx : 4 πrx
= 5:4

Question 7.
There are two cones. The curved surface area of one is twice that of the other. The slant height of the latter is twice that of the former. Find the ratio of their radii.
Solution:
Let the slant height of first cone = l
then slant height of the second cone = 2l
and let r1  be the radius of the first cone and r2 be the radius of the second cone.
Then curved surface area of the first cone = πr1l
and that of second cone = πr22l= 2πr2l.
According to the condition,

Question 8.
A heap of wheat is in the form of a cone of diameter 16.8 m and height 3.5m. Find its volume. How much cloth is required to just cover the heap?
Solution:

Question 9.
Find what length of canvas, 1.5m in width, is required to make a conical tent 48 m in diameter and 7m in height Given that 10% of the canvas is used in folds and stritchings. Also, find the cost of the canvas at the rate of ₹24 per metre.
Solution:

Question 10.
A solid cone of height 8 cm and base radius 6 cm is melted and recast into identical cones, each of height 2 cm and diameter 1 cm. Find the number of cones formed.
Solution:
Height of solid cone (h) = 8 cm.

Question 11.
The total surface area of a right circular cone of slant height 13 cm is 90π cm2. Calculate:
(ii) its volume in cm3. [Take π = 3.14]
Solution:
Total surface area of cone = 90π cm2
slaint height (l) = 13 cm
Let r be its radius, then
Total surface area = πrl + πr2 = πr (l + r)

Question 12.
The area of the base of a conical solid is 38.5 cm2 and its volume is 154 cm3. Find curved surface area of the solid.
Solution:
Area of base of a solid cone = 38.5 cm2
and volume  = 154 cm3

Question 13.
A vessel, in the form of an invested cone, is filled with water to the brim. Its height is 32 cm and diameter of the base is 25.2 cm. Six equal solid cones are dropped in it, so that they are fully submerged. As a result, one-fourth of water in the original cone overflows. What is the volume of each of the solid cones submerged ?
Solution:
Diameter of the base of cone = 25.2 cm

Question 14.
The volume of a conical tent is 1232 m3 and the area of the base floor is 154 m2. Calculate the:
(ii) height of the tent.
(iii) length of the canvas required to cover this conical tent if its width is 2 m. (2008)
Solution:
Volume of conical tent = 1232 m3
Area of base floor = 154 m2
(i)  Let r be the radius of the floor

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere (Surface Area and Volume) Ex 20A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A

Other Exercises

Question 1.

The height of a circular cylinder is 20 cm and the radius of its base is 7 cm. Find :
(i) the volume
(ii) the total surface area.
Solution:
Height of cylinder (h) =  20cm
and radius of its base (r) = 7 cm
(î) Volurne=πr²h

Question 2.
The inner radius of a pipe is 2.1 cm. How much water can 12 m of this pipe hold ?
Solution:
Inner radius of a pipe (r) = 2.1 cm
and length of pipe (h) = 12 m = 1200 cm
∴ Volume of water in it = πr2h

Question 3.
A cylinder of circumference 8 cm and length 21 cm roils without sliding for 4 $$\frac { 1 }{ 2 }$$ seconds at the rate of 9 complete rounds per second. Find:
(i) the distance travelled by the cylinder in 4 $$\frac { 1 }{ 2 }$$ seconds, and
(ii) the area covered by the cylinder in 4 $$\frac { 1 }{ 2 }$$ seconds.
Solution:
Circumference of a cylinder = 8 cm

Length of cylinder (h) = 21 cm
It takes 9 complete rounds per second
∴ Curved surface area = 2πrh

Question 4.
How many cubic metres of earth must be dug out to make a well 28 m deep and 2.8 m in diameter ? Also, find the cost of plastering its inner surface at ₹4.50 per sq. metre.
Solution:

Question 5.
What length of solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of external diameter 20 cm, 0.25 cm thick and 15 cm long ?
Solution:
Diameter of solid cylinder = 2 cm
∴ Radius (r) = $$\frac { 2 }{ 2 }$$ = 1 cm
Let length (h) = x cm
∴ Volume = πr²h = π x 1 x 1 x x
= πx cm3       …(i)
External diameter of hollow cylinder = 20cm
∴ External radius =  $$\frac { 20 }{ 2 }$$ = 10 cm
Thickness of cylinder = 0.25 cm
∴ Innerradius= 10-0.25 = 9.75 cm
Length = 15 cm
∴ Volume = π(R2 – r2) x h
= π(R + r)(R-r) x h
= π(10 + 9.75)(10-9.75) x 15 cm3
= πx 19.75 x 0.25 x 15 cm3    ………(ii)
Comparing (i) and (ii), we get
∴ π x 19.75 x 0.25 x 15 = π x
x = 19.75 x 0.25 x 15 cm
= 74.0625 = 74.06 cm

Question 6.
A cylinder has a diameter of 20 cm. The area of the curved surface is 100 cm2 (sq. cm). Find:
(i) the height of the cylinder correct to one decimal place.
(ii) the volume of the cylinder correct to one decimal place.
Solution:

Question 7.
A metal pipe has a bore (inner diameter) of 5 cm. The pipe is 5 mm thick all round. Find the weight, in kilogram, of 2 metres of the pipe if 1 cm3 of the metal weighs 7.7 g.
Solution:

Question 8.
A cylindrical container with diameter of base 42 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 22 cm x 14 cm x 10.5 cm. Find the rise in the. level of the water when the solid is submerged.
Solution:
Diameter of base of a cylindrical container =42 cm
∴ Radius = $$\frac { 42 }{ 2 }$$ = 21 cm
Size of rectangular solid = 22cmx 14cmx 10.5 cm
∴ Volume of solid = 3234 cm3
∴ Height of water level raise in the container

Question 9.
A cylindrical container with infernal radius of its base 10 cm’, contains water up”to a height of 7 cm. Find the area of the wet surface of the cylinder.
Solution:
Internal radius of cylindrical container (r) = 10cm
Water upto height (h) = 7 cm
∴ Area of wet surface by the water of the container = 2πrh
= π x 19.75 x 0.25 x 15 cm3                       …(ii)
Comparing (i) and (ii), we get
∴ π x 19.75 x 0.25 x 15 = πx
x = 19.75×0.25x 15 cm
= 74.0625
= 74.06 cm

Question 10.
Find the total surface area of an open pipe of length 50 cm, external diameter 20 cm and internal diameter 6 cm.
Solution:
Length of open pipe (h) = 50 cm
External diameter=20 cm
and internal diameter = 6 cm
∴ External radius (R) = $$\frac { 20 }{ 2 }$$ = 10 cm
and internal radius (r) = $$\frac { 6 }{ 2 }$$ = 3 cm
∴ Total surface area of the open pipe

Question 11.
The height and the radius of the base of a cylinder are in the ratio 3:1. If its volume is 1029 πcm3; find its total surface area.
Solution:
Ratio in height and radius of cylinder = 3:1
Let height = 3x
∴ Volume = πr2h = π x 3x x x2 = 3πx3
∴ 3πt3= 1029π

∴ x = 7
∴ Height = 7 x 3=21 cm
Now total surface area = 2πr2 + 2 πrh
= 2 πr (r + h)
= 2 x $$\frac { 22 }{ 7 }$$ x 7(7 + 21)
=44 x 28 = 1232 cm2

Question 12.
The radius of a solid right circular cylinder increases by 20% and its height decreases by 20%. Find the percentage change in its volume.
Solution:
Let radius of the cylinder (r) = 100 cm
and height (h) = 100 cm
∴ Volume = πr2h =π( 100)2 x 100 cm3 = 1000000π cm3
Now radius (R) = 100 + 20 = 120 cm
and new height (h) = 100 – 20 = 80 cm
∴ Volume = π(120)2 x 80
= π x 14400 x 80cm3= 152000π cm3
Percentage change (increase) in the volume

Question 13.
The radius of a solid right circular cylinder decreases by 20% and its height increases by 10%. Find the percentage change in its:
(i) volume
(ii) curved surface area
Solution:
Let radius of the cylinder = 100 cm
and height = 100 cm
∴ Volume = πr2h
= πx 100 x 100 x 100 cm3
= 1000000πcm3
and Curved surface area = 2πrh
= 2 x π x 100 x 100 cm2
=20000π cm2
Decreased radius = 100 – 20 = 80 cm
and increased height = 100 + 10 = 110 cm
∴ Increased volume = π x 80 x 80 x 110 cm3 = 704000π cm3
and increased curved surface =2 x π x 80 x 110 cm2 = 17600πcm2
⇒  Decrease in volume = 1000000π – 704000π = 296000π cm3

Question 14.
Find the minimum length in cm and correct to nearest whole number of the thin metal sheet required to make a hollow and closed cylindrical box of diameter 20 cm and height 35 cm. Given that the width of the metal sheet is 1 m. Also, find the cost of the sheet at the rate of ₹56 per m.
Find the area of metal sheet required, if 10% of it is wasted in cutting, overlapping, etc.
Solution:
Diameter of the hollow closed cylinder =20 cm
∴ Radius (r) = $$\frac { 20 }{ 2 }$$
and height (h) = 35 cm

Question 15.
3080 cm3 of water is required to fill a cylindrical vessel completely and 2310 cm3 of water is required to fill it upto 5 cm below the top. Find:
(ii) height of the vessel.
(iii) wetted surface area of the vessel when it is half-filled with water.
Solution:
Volume of water to fill a cylindrical vessel = 3080 cm3
Volume of water to fill it upto 5 cm below = 2310 cm2
Volume of water for 5 cm height =3080- 2310 = 770cm3

Question 16.
Find the volume of the largest cylinder formed when a rectangular piece of paper 44 cm by 33 cm is rolled along it:
(i) shorter side.
(ii) longer side.
Solution:
Length of rectangular sheet = 44 cm

(i) Folding along shorter side i.e., 33cm
∴ Circumference of cylinder = 33cm

Question 17.
A metal cube of side 11 cm is completely submerged in water contained in a cylindrical vessel with diameter 28 cm. Find the rise in the level of water.
Solution:
Side of a cube = 11 cm
∴ Volume = (Side)3 = 11 x 11 x 11 cm3 = 1331 cm3
Diameter of cylinder = 28 cm

Question 18.
A circular tank of diameter 2 m is dug and the earth removed is spread uniformly all around the tank to form an embankment 2 m in width and 1.6 m in height. Find the depth of the circular tank.
Solution:
Diameter of circular tank = 2m
Width of embankment = 2 m
Height = 1.6 m

Question 19.
The sum of the inner and the outer curved surfaces of a hollow metallic cylinder is 1056 cm2 and the volume of material in it is 1056 cin3. Find its internal and external radii. Given that the height of the cylinder is 21 cm.
Solution:
Sum of outer and inner surface area of a hollow cylinder = 1056 cm2
Volume of metal used =1056 cm3
Height of cylinder (h) = 21 cm
Height = 21 cm
We are given
Outer surface + Inner surface = 1056 cm2
Volume = πR2h – πr2h =1056 cm3
Now, 2πRh + 2πrh =1056

Question 20.
The difference between the outer curved surface area and the inner curved surface area of a hollow cylinder is 352 cm2. If its height is 28 cm and the volume of material in it is 704 cm3; find its external curved surface area.
Solution:
Difference in outer and inner curved surface of a hollow cylinder = 352 cm2
Height (h) = 28 cm
Volume of material used = 704 cm3
∴ 2πRh-2πrh = 352
2πh(R-r) = 352

Question 21.
The sum of the height and the radius of a solid cylinder is 35 cm and its total surface area is 3080 cm2; find the volume of the cylinder.
Solution:

Question 22.
The total surface area of a solid cylinder is 616 cm2. If the ratio between its curved surface area and total surface area is 1:2; find the volume of the cylinder.
Solution:
Total surface area of a cylinder = 616 cm2

Question 23.
A cylindrical vessel of height 24 cm and diameter 40 cm is full of water. Find the exact number of small cylindrical bottles, each of height 10 cm and diameter 8 cm, which can be filled with this water.
Solution:
Height of cylinder (6) = 24 cm
Radius (r)= $$\frac { 40 }{ 2 }$$ = 20 cm
∴ Volume of water filled in it = πr2h
= π x 20 x 20 x 24 cm3
= 9600π cm3
Radius of small cylindrical bottle = $$\frac { 8 }{ 2 }$$ = 4 cm
and height (6) = 10 cm
∴  Volume of one small bottle = πr2h
π x 4 x 4 x 10 cm3 = 160π cm3
∴ Number of small bottles

Question 24.
Two solid cylinders, one with diameter 60 cm and height 30 cm and the other with radius 30 cm and height 60 cm, are melted and recasted into a third solid cylinder of height 10 cm. Find the diameter of the cylinder formed.
Solution:
Diameter of first cylinder = 60 cm
∴ Radius (R)= $$\frac { 60 }{ 2 }$$ =30 cm
and height (h) = 30 cm
Radius of second cylinder = 30 cm
and height = 60 cm
Volume of first cylinder = πR2h
= π30 x 30 x 30 cm3 = 27000π cm3
and volume of second cylinder = π x 30 x 30 x 60 cm3 = 54000πcm3
Total volume of both cylinders
= 27000π+ 54000π cm3
= 81000π cm3
Volume of third cylinder = 81000π cm3
Height of third cylinder = 10 cm

Question 25.
The total surface area of a hollow cylinder, which is open from both the sides, is 3575 cm2; area of its base ring is 357.5 cm2 and its height is 14 cm. Find the thickness of the cylinder.
Solution:
Total surface area of an opened hollow cylinder = 3575 cm2
Area of ring of its base = 357.5 cm2
Height = 14 cm
Let R and r be the outer and inner radius of the ring.
∴ π(R2 – r2) = 357.5

Question 26.
The given figure shows a solid formed of a solid cube of side 40 cm and a solid cylinder of radius 20 cm and height 50 cm attached to the cube as shown.
Find the volume and the total surface area of the whole solid [Take π=3.14]

Solution:
Side of the cube = 40 cm
Radius of cylinder = 20 cm
Height (h) = 50 cm
Volume of cube = (40)3 = 64000 cm3
and volume of cylinder = πr2h
= 3.14 x 20 x 20 x 50 cm3
=314×200 = 62800 cm3
∴ Total volume = 64000 + 62800 =126800cm3
Total surface area = (6a2 + 2πrh)
= 6 x 40 x 40 + 2 x 3.14 x 20 x 50 =9600+6280
= 15880 cm2

Question 27.
Two right circular solid cylinders have radii in the ratio 3: 5 and heights in the ratio 2:3. Find the ratio between their:
(i) curved surface areas.
(ii) volumes.
Solution:
The ratio is the radii of two right circular solid cylinder = 3:5
and ratio in their heights = 2:3
(i) Let radius of first cylinder = 3x
and height = 2y
∴ Curved surface area = 2πrh
= 2π(3x) (2y) = 12 πxy
and radius of second cylinder = 5x
and height = 3y
∴  Curved surface = 2πrh
= 2π x 5x 3y
= 30πxy
∴ Ratio in their curved surface
= 12πxy : 30πxy
= 2 :5
(ii) Volume of first cylinder = πr2h
= π(3x)2 x 2y = 18πx2y
and volume of second cylinder = π(5x)2 x 3y
= 75πx2y
∴ Ratio = 18πx2y : 75πx2y
= 6:25

Question 28.
A closed cylindrical tank, made of thin iron sheet, has diameter = 8.4 m and height 5.4 m. How much metal sheet, to the nearest m2, is used in making this tank, if $$\frac { 1 }{ 15 }$$ of the sheet actually used was wasted in making the tank ?
Solution:
Diameter of a closed cylindrical tank=8.4 cm
∴ Radius (r) = $$\frac { 8.4 }{ 2 }$$ = 4.2 m
and height (h) = 5.4 m
∴ Total surface area = 2πr(r + h)
= 2 x $$\frac { 22 }{ 7 }$$ x 4.2(4.2+ 5.4) m2
=26.4 x 9.6 m2=253.44 m2
Area of sheet used in wastage
= $$\frac { 1 }{ 15 }$$ of 253.44= 16.896 m2
Total sheet required = 16.896 +253.44 m2
= 270.336 m2
=270.34 m2 (approx)

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20A are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity (With Applications to Maps and Models) Ex 15E

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E.

Other Exercises

Question 1.
In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm.

Solution:
In the given figure,

Question 2.
In the following figure, ABCD to a trapezium with AB // DC. If AB = 9 cm, DC = 18 cm, CF= 13.5 cm, AP = 6 cm and BE = 15 cm. Calculate:
(i) EC
(ii) AF
(iii) PE

Solution:
In the figure,
ABCD is a trapezium
AB || DC
AB = 9 cm, DC = 18 cm, CF = 13.5 cm AP = 6 cm and BE = 15 cm

Question 3.
In the following figure, AB, CD and EF are perpendicular to the straight line BDF.

Solution:
In the given figure,
AB, CD and EF are perpendicular to the line BDF
AB = x, CD = z, EF = y

Question 4.
Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that:
$$\frac { AB }{ PQ }$$ = $$\frac { AD }{ PM }$$
Solution:
∆ABC ~ ∆PQR
AD and PM are the medians of ∆ABC and ∆PQR respectively.

Question 5.
Triangle ABC is similar to triangle PQR. If AD and PM are altitudes of the two triangles, prove that: $$\frac { AB }{ PQ }$$ = $$\frac { AD }{ PM }$$.
Solution:
Given, ∆ABC ~ ∆PQR
AD and PM are altitude of these two triangles

Question 6.
Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that: $$\frac { AB }{ PQ }$$ = $$\frac { AD }{ PM }$$
Solution:
Given, ∆ABC ~ ∆PQR
AD and PM are the angle bisectors of ∠A and ∠P respectively.

Question 7.

Solution:

But ∠AXY = ∠AYX is given
∠B = ∠C
AC = AB (Side opposite to equal angles)
∆ABC is an isosceles triangle.

Question 8.
In the following diagram, lines l, m and n are parallel to each other. Two transversals p and q intersect the parallel lines at points A, B, C and P, Q, R as shown.

Solution:
In the given figure,
l || m || n
Transversal p and q intersects them at A, B, C and P, Q, R respectively as shown in the given figure.

Question 9.

Solution:

Question 10.
In the figure given below, AB // EF // CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm.
Calculate:
(i) EF
(ii) AC

Solution:
In the given figure,
AB || EF || CD
AB = 22.5 cm, EP = 7.5 cm
PC = 15 cm and DC = 27 cm

Question 11.
In quadrilateral ABCD, its diagonals AC and BD intersect at point O such that
$$\frac { OC }{ OA }$$ = $$\frac { OD }{ OB }$$ = $$\frac { 1 }{ 3 }$$
Prove that:
(i) ∆OAB ~ ∆OCD
(ii) ABCD is a trapezium.
Further if CD = 4.5 cm; find the length of AB.
Solution:
In quadrilateral ABCD, diagonals AC and BD intersect each other at O and

Question 12.
In triangle ABC, angle A is obtuse and AB = AC. P is any point in side BC. PM ⊥ AB and PN x AC.
Prove that: PM x PC = PN x PB
Solution:
Given, AB = AC
Since equal sides has equal angle opposite to it
∠B = ∠C …(i)
In ∆PMB and ∆PNC, we have
∠B = ∠C [using (i)]
∠PMB = ∠PNC (each 90°)

Question 13.
In triangle ABC, AB = AC = 8 cm, BC = 4 cm and P is a point in side AC such that AP = 6 cm. Prove that ∆BPC is similar to ∆ABC. Also, find the length of BP.
Solution:
In ∆ABC,
AB = AC = 8 cm
BC = 4cm
P is a point on AC such that AP = 6 cm
PB and PC are joined
To prove: ∆BPC ~ ∆ABC
and find length of BP
Proof: AC = 8 cm and AP = 6 cm
PC = AC – AP = 8 – 6 = 2 cm

Question 14.
In ∆ABC, ∠ABC = ∠DAC. AB = 8 cm, AC = 4 cm, AD = 5 cm.
(i) Prove that ∆ACD is similar to ∆BCA.
(ii) Find BC and CD.
(iii) Find area of ∆ACD : area of ∆ABC. (2014)
Solution:
In ∆ACD and ∆BCA
∠C = ∠C (common)
∆ACD = ∆BCA (by AA axiom)

Question 15.
In the given triangle P, Q and R are the mid-points of sides AB, BC and AC respectively. Prove that triangle PQR is similar to triangle ABC.

Solution:
Given : P and R the mid points of AB and AC respectively.
PR || BC and PR = $$\frac { 1 }{ 2 }$$ BC = BQ.
PRQB is a || gm.
∠B = ∠PRQ ….(i)
Similarly, Q and R are the mid points of sides. BC and AC respectively
RQ || AB and QR = $$\frac { 1 }{ 2 }$$ AB = AP
APQR is a ||gm.
∠A = ∠PQR ….(ii)
Similarly, we can prove that ∠C = ∠RPQ.
Now in ∆PQR and ∆ABC,
∠PQR = ∠A , ∠PRQ = ∠B and ∠RPQ = ∠C
(i) In ∆BCE
D is mid-point of BC and DF || CE
∆PQR ~ ∆ABC (AAA criterion of similarity)

Question 16.
In the following figure, AD and CE are medians of A ABC. DF is drawn parallel to CE. Prove that:
(i) EF = FB
(ii) AG : GD = 2 : 1

Solution:
Proof:
(i) In ∆BCE
D is the mid point of BC and DF || CE
E is mid-point of BE and EF = FB
(ii) AE = EB (E is mid point of AB)
and EF = FB (Proved)

Question 17.
In the given figure, triangle ABC is similar to triangle PQR. AM and PN are altitudes whereas AX and PY are medians.

Solution:

Question 18.
The two similar triangles are equal in area. Prove that the triangles are congruent.
Solution:
Given : ∆ABC ~ ∆PQR and are equal in area

Question 19.
The ratio between the altitudes of two similar triangles is 3 : 5; write the ratio between their:
(i) medians
(ii) perimeters
(iii) areas
Solution:
∆ABC ~ ∆PQR
AL ⊥ BC and PM ⊥ QR

Question 20.
The ratio between the areas of two similar triangles is 16 : 25. Find the ratio between their:
(i) perimeters
(ii) altitudes
(iii) medians.
Solution:
∆ABC ~ ∆DEF,
AL ⊥ BC and DM ⊥ EF
and AP and DQ are the medians and also
area ∆ABC : area ∆DEF = 16 : 25

Question 21.
The following figure shows a triangle PQR in which XY is parallel to QR. If PX : XQ = 1 : 3 and QR = 9 cm, find the length of XY. Further, if the area of ∆PXY = x cm²; find in terms of x, the area of :
(i) triangle PQR.
(ii) trapezium XQRY.

Solution:
In ∆PQR, XY || QR and PX : XQ = 1 : 3, QR = 9 cm.

Question 22.
On a map, drawn to a scale of 1 : 20000, a rectangular plot of land ABCD has AB = 24 cm, and BC = 32 cm. Calculate :
(i) The diagonal distance of the plot in kilometre
(ii) The area of the plot in sq. km.
Solution:

Question 23.
The dimensions of the model of a multistoreyed building are 1 m by 60 cm by 1.20 m. If the scale factor is 1 : 50,. find the actual dimensions of the building. Also, find :
(i) the floor area of a room of the building, if the floor area of the corresponding room in the model is 50 sq cm.
(ii) the space (volume) inside a room of the model, if the space inside the corresponding room of the building is 90 m3.
Solution:
The scale factor is 1 : 50 or k = $$\frac { 1 }{ 50 }$$
Dimension of the building = 100 cm x 60 cm x 120 cm.
k x actual dimensions of the building = Dimension of the model.

Question 24.
In a triangle PQR, L and M are two points on the base QR, such that ∠LPQ = ∠QRP and ∠RPM = ∠RQP. Prove that:
(i) ∆PQL ~ ∆RPM
(ii) QL x RM = PL x PM
(iii) PQ² = QR x QL [2003]

Solution:
(i) In ∆PQL and ∆RPM
∠PQL = ∠RPM (Given)
∠LPQ = ∠MRP (Given)
∆PQL ~ ∆RPM (AA criterion of similarity)
(ii) ∆PQL ~ ∆RPM (Proved)

Question 25.

Solution:

Question 26.
A triangle ABC with AB = 3 cm, BC = 6 cm and AC = 4 cm is enlarged to ∆DEF such that the longest side of ∆DEF = 9 cm. Find the scale factor and hence, the lengths of the other sides of ∆DEF.
Solution:
In ∆ABC.
AB = 3 cm. BC = 6 cm and AC = 4 cm
In ∆DEF,
Longest side EF = 9 cm
and longest side in ∆ABC is BC = 6 cm

Question 27.
Two isosceles triangles have equal vertical angles. Show that the triangles are similar. If the ratio between the areas of these two triangles is 16 : 25, find the ratio between their corresponding altitudes.
Solution:
Let in two ∆ABC and ∆DEF
The vertical angles of two isosceles triangles are equal i.e. ∠A = ∠D
But AB = DE and AC = DF (isosceles ∆s)
Then base angles are also equal (Angles opposite to equal sides)
The two triangles are similar.
∆ABC ~ ∆DEF
Let AL ⊥ BC and DM ⊥ EF

Question 28.
In ∆ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA.

Find: (i) area ∆APO : area ∆ABC.
(ii) area ∆APO : area ∆CQO.
Solution:
In ∆ABC,
AP : PB = 2 : 3
PQ || BC and CQ || BA

Question 29.
The following figure shows a triangle ABC in which AD and BE are perpendiculars to BC and AC respectively.

Show that:
(ii) CA x CE = CB x CD
(iii) ∆ABC ~ ∆DEC
(iv) CD x AB = CA x DE
Solution:
In ∆ABC, AD ⊥ BC and BE ⊥ AC, DE is joined
To prove:
(ii) CA x CE = CB x CD
(iii) ∆ABC ~ ∆DEC
(iv) CD x AB = CA x DE
Proof:
∠C = ∠C (common)
∠ABE = ∠BEC (each 90°)

Question 30.
In the given figure, ABC is a triangle-with ∠EDB = ∠ACB. Prove that ∆ABC ~ ∆EBD.
If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of ∆BED = 9 cm². Calculate the
(i) length of AB
(ii) area of ∆ABC

Solution:
In ∆ABC and ∆EBD
∠1 = ∠2 (given)
∠B = ∠B (common)

Question 31.
In the given figure, ABC is a right-angled triangle with ∠BAC = 90°.
(ii) If BD = 18 cm, CD = 8 cm, find AD.
(iii) Find the ratio of the area of ∆ADB is to area of ∆CDA.

Solution:
(i) In ∆ADB and ∆CDA :
∆ADB ~ ∆CDA [by AA similarity axiom]

Question 32.
In the given figure, AB and DE are perpendicular to BC.
(i) Prove that ∆ABC ~ ∆DEC
(ii) If AB = 6 cm, DE = 4 cm and AC = 15 cm. Calculate CD.
(iii) Find the ratio of the area of ∆ABC : area of ∆DEC.

Solution:
(i) To prove : ∆ABC ~ ∆DEC
In ∆ABC and ∆DEC
∠ABC = ∠DEC = 90°
∠C = ∠C (common)
∆ABC ~ ∆DEC (by AA axiom)

Question 33.
ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that:

(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.
Solution:
In the given figure,
∆ABC is right angled triangle right angle at B.
D is any point on AB and DE ⊥ AC
To prove:
(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm.Find DE and AD.
Proof:
∠A = ∠A (common)
∠E = ∠B (each = 90°)
(ii) AC = 13 cm, BC = 5 cm, AE = 4 cm

Question 34.

Solution:
In the given figure, AB || DE, BC || EF

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15E are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity (With Applications to Maps and Models) Ex 15D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15D.

Other Exercises

Question 1.
A triangle ABC has been enlarged by scale factor m = 2.5 to the triangle A’ B’ C’. Calculate:
(i) the length of AB, if A’ B’ = 6 cm.
(ii) the length of C’ A’ if CA = 4 cm.
Solution:
Scale factor (k) = 2.5
∆ABC is enlarged to ∆A’B’C’
(i) A’B’ = 6 cm

Question 2.
A triangle LMN has been reduced by scale factor 0.8 to the triangle L’ M’ N’. Calculate:
(i) the length of M’ N’, if MN = 8 cm.
(ii) the length of LM, if L’ M’ = 5.4 cm.
Solution:
∆LMN has been reduced by the scale factor

Question 3.
A triangle ABC is enlarged, about the point O as centre of enlargement, and the scale factor is 3. Find:
(i) A’ B’, if AB = 4 cm.
(ii) BC, if B’ C’ = 15 cm.
(iii) OA, if OA’= 6 cm.
(iv) OC’, if OC = 21 cm.
Also, state the value of:
(a) $$\frac { OB’ }{ OB }$$
(b) $$\frac { C’A’ }{ CA }$$
Solution:
∆ABC is enlarged to ∆A’B’C’ about the point O as its centre of enlargement.
Scale factor = 3 = $$\frac { 3 }{ 1 }$$

Question 4.
A model of an aeroplane is made to a scale of 1 : 400. Calculate:
(i) the length, in cm, of the model; if the length of the aeroplane is 40 m.
(ii) the length, in m, of the aeroplane, if length of its model is 16 cm.
Solution:
Model of an aeroplane to the actual = 1 : 400

Question 5.
The dimensions of the model of a multistorey building are 1.2 m x 75 cm x 2 m. If the scale factor is 1 : 30; find the actual dimensions of the building.
Solution:
Dimensions of a model of multistorey building = 1.2 m x 75 cm x 2 m

Question 6.
On a map drawn to a scale of 1 : 2,50,000; a triangular plot of land has the following measurements : AB = 3 cm, BC = 4 cm and angle ABC = 90°.
Calculate:
(i) the actual lengths of AB and BC in km.
(ii) the area of the plot in sq. km.
Solution:
Scale of map drawn of a triangular plot = 1 : 2,50,000
Measurement of plot AB = 3 cm, BC = 4 cm
and ∠ABC = 90°

Question 7.
A model of a ship of made to a scale 1 : 300
(i) The length of the model of ship is 2 m. Calculate the lengths of the ship.
(ii) The area of the deck ship is 180,000 m². Calculate the area of the deck of the model.
(iii) The volume of the model in 6.5 m3. Calculate the volume of the ship. (2016)
Solution:

Question 8.
An aeroplane is 30 in long and its model is 15 cm long. If the total outer surface area of the model is 150 cm², find the cost of painting the outer surface of the aeroplane at the rate of ₹ 120 per sq.m. Given that 50 sq. m of the surface of the aeroplane is left for windows.
Solution:
Length of aeroplane = 30 m = 3000 cm
and length of its model = 15 cm
Surface area of model = 150 cm²
Scale factor (k) = $$\frac { 3000 }{ 15 }$$ = $$\frac { 200 }{ 1}$$
Area of plane = k² x area of model = (200)² x 150 cm² = 40000 x 150 cm²
$$\frac { 40000 x 150 }{ 10000 }$$ = 600 m² (1 m² = 10000 cm²)
Shape left for windows = 50 sq. m
Balance area = 600 – 50 = 550 sq. m
Race of painting the outer surface = ₹ 120 per sq.m
Total cost = ₹ 550 x 120 = ₹ 66000

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15D are helpful to complete your math homework.

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## Chapter 21 Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Ex 21E

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21E.

Other Exercises

Question 1.
Prove the following identities :

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Question 2.
If sin A + cos A = p and sec A + cosec A = q then prove that: q(p² – 1) 2p
Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.
If tan A=n tan B and sin A=m sin B, prove that:

Solution:

Question 6.
(i) If 2 sin A-1 = 0, show that:
sin 3 A = 3 sin A – 4 sin3 A.             [2001]
(ii) If 4cos2 A-3 = 0, show that:
cos 3A = 4 cos3 A – 3 cos A
Solution:

Question 7.
Evaluate:

Solution:

Question 8.
Prove that:

Solution:

Question 9.
If A and B are complementary angles, prove that:
(i) cot B + cos B sec A cos B (1 + sin B)
(ii) cot A cot B – sin A cos B – cos sin B = 0
(iii) cosec2 A + cosec2 B = cosec2 A cosec2 B

Solution:

Question 10.
Prove that:

Solution:

Question 11.
If 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that :
(i) sin 3A = 3 sinA – 4 sin3A
(ii) cos 3A = 4 cos3 A – 3 cos A
Solution:

Question 12.
Find A, if 0° ≤ A ≤ 90° and :
(i) 2 cos2 A – 1 = 0
(ii) sin 3A – 1 = 0
(iii) 4 sin2 A – 3 = 0
(iv) cos2 A – cos A = 0
(v) 2cos2 A + cos A – 1 = 0
Solution:

Question 13.
If 0° < A < 90° ; find A, if :

Solution:

Question 14.
Prove that : (cosec A – sin A) (sec A – cos A) sec2 A = tan A. (2011)
Solution:

Question 15.
Prove the identity : (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ. (2014)
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21E are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity (With Applications to Maps and Models) Ex 15C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C.

Other Exercises

Question 1.
(i) The ratio between the corresponding sides of two similar triangles is 2 is to 5. Find the ratio between the areas of these triangles.
(ii) Areas of two similar triangles are 98 sq.cm and 128 sq.cm. Find the ratio between the lengths of their corresponding sides.
Solution:

Question 2.
A line PQ is drawn parallel to the base BC, of ΔABC which meets sides AB and AC at points P and Q respectively. If AP = $$\frac { 1 }{ 3 }$$ PB; find the value of:

Solution:

Question 3.
The perimeters of two similar triangles are 30 cm and 24 cm. If one side of first triangle is 12 cm, determine the corresponding side of the second triangle.
Solution:
Let we are given ΔABC and ΔPQR are similar.
Perimeter of ΔABC = 30 cm.
and perimeter of ΔPQR = 24 cm.
and side BC = 12 cm.
Now we have to find the length of QR, the corresponding side of ΔPQR
ΔABC ~ ΔPQR

Question 4.
In the given figure AX : XB = 3 : 5

Find :
(i) the length of BC, if length of XY is 18 cm.
(ii) ratio between the areas of trapezium XBCY and triangle ABC.
Solution:
We are given in the ΔABC, AX : XB = 3 : 5
XY || BC.
Let AX = 3x and XB = 5x
AB = 3x + 5x = 8x.
Now in ΔAXY and ΔABC,
∠AXY = ∠ABC (corresponding angles)
∠A = ∠A (common)
ΔAXY ~ ΔABC (AA Postulate)

Question 5.
ABC is a triangle. PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides triangle ABC into two parts equal in area. Find the value of ratio BP : AB.
Given- In ΔABC, PQ || BC in such away that area APQ = area PQCB
To Find- The ratio of BP : AB

Solution:
In ΔABC, PQ || BC.
Z APQ = Z ABC (corresponding angles)
Now in ΔAPQ and ΔABC,
∠APQ = ∠ABC (proved)
∠A = ∠A (common)
ΔAPQ ~ ΔABC (AA postulate)

Question 6.
In the given triangle PQR, LM is parallel to QR and PM : MR = 3 : 4

Solution:
In ΔPQR, LM || QR in such away that PM : MR = 3 : 4
(i) In ΔPQR, LM || QR

Question 7.
The given diagram shows two isosceles triangles which are similar also. In the given diagram, PQ and BC are not parallel: PC = 4, AQ = 3, QB = 12, BC = 15 and AP = PQ.

Calculate-
(i) the length of AP
(ii) the ratio of the areas of triangle APQ and triangle ABC.
Solution:

Question 8.
In the figure, given below, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2. DP produced meets AB produced at Q. Given the area of triangle CPQ = 20 cm². Calculate
(i) area of triangle CDP
(ii) area of parallelogram ABCD [1996]

Solution:
(i) Join QC.
In ΔBPQ and ΔCPD,
∠DPC = ∠BPQ (vertically opposite angles.)
∠PDC = ∠BQP (Alternate angles.)
ΔBPQ ~ ΔCDP (AA postulate)

⇒ area ΔCDP = 4 (area ΔBPQ)
⇒ 2 (2 area ΔBPQ) = 2 x 20 = 40 cm² (2 area A BPQ = 20 cm²)
(ii) Area || gm ABCD = area ΔCPD + area ΔADQ – area ΔBPQ
= 40 + 9 (area BPQ) – area BPQ [(AD = CB = 3 BP)]
= 40 + 8 (area ΔBPQ)
= 40 + 8 (10) cm²
= 40 + 80
= 120 cm²

Question 9.
In the given figure. BC is parallel to DE. Area of triangle ABC = 25 cm². Area of trapezium BCED = 24 cm² and DE = 14 cm. Calculate the length of BC. Also. Find the area of triangle BCD.

Solution:
Area of ΔABC = 25 cm²
and area of trapezium BCED = 24 cm²
Area of ΔADE = 25 + 24 = 49 cm²
DE = 14 cm,
Let BC = x cm.
∠A = ∠A (common)

Question 10.
The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersect at point P. If AP : CP = 3 : 5.

Find-
(i) ΔAPB : ΔCPB
(ii) ΔDPC : ΔAPB
Solution:
AP : CP = 3 : 5 ⇒ $$\frac { AP }{ CP }$$ = $$\frac { 3 }{ 5 }$$
(i) Now in ΔAPB and ΔCPB,
These triangles have same vertex and their bases are in the same straight line
area ΔAPB : area ΔCPB = AP : PC = 3 : 5 or ΔAPB : ΔCPB = 3 : 5
(ii) In ΔAPB and ΔDPC,
∠APB = ∠DPC (vertically opposite angles)
∠PAB = ∠PCD (alternate angles)
ΔAPB ~ ΔDPC (AA postulate)

⇒ area ΔDPC : area ΔAPB = 25 : 9 or ΔDPC : ΔAPB = 25 : 9
There have the same vertex and their bases arc in the same straight line.
area ΔADP : area ΔAPB = DP : PB
But PC : AP = 5 : 3
ΔADP : ΔAPB = 5 : 3
(iv) Similarly area ΔAPB : area ΔADB = PB : DB = 3 : (3 + 5) = 3 : 8

Question 11.
In the given figure, ARC is a triangle. DE is parallel to BC and $$\frac { AD }{ DB }$$ = $$\frac { 3 }{ 2 }$$.
(i) Determine the ratios $$\frac { AD }{ AB }$$ , $$\frac { DE }{ BC }$$
(ii) Prove that ΔDEF is similar to ΔCBF.
Hence, find $$\frac { EF }{ FB }$$
(iii) What is the ratio of the areas of ΔDEF and ΔBFC?

Solution:

Question 12.
In the given figure, ∠B = ∠E, ∠ACD = ∠BCE, AB = 10.4 cm and DE = 7.8 cm. Find the ratio between areas of the ΔABC and ΔDEC.

Solution:
In the figure DE = 7.8 cm, AB = 10.4 cm
∠ACD = ∠BCE (given)
∠ACD + ∠DCB = ∠DCB + ∠BCE
∠ACB = ∠DCE
Now in ΔABC and ΔDCE
∠B = ∠E (given)
∠ACB = ∠DCE (proved)
ΔABC ~ ΔDCE (AA axiom)

Question 13.
Triangle ABC is an isosceles triangle in which AB = AC = 13 cm and BC = 10 cm. AD is perpendicular to BC. If CE = 8 cm and EF ⊥ AB, find:

Solution:
In the figure, ΔABC is an isosceles triangle in which
AB = AC = 13 cm, BC = 10 cm,
AD ⊥ BC, CE = 8 cm and EF ⊥ AB
(i) Now in ΔADC and ΔFEB
∠C = ∠B (AB = AC)
∠ADC = ∠EFB (each = 90°)

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C are helpful to complete your math homework.

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## Chapter 21 Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Ex 21D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21D.

Other Exercises

Question 1.
Use tables to find sine of:
(i) 21°
(ii) 34°42′
(iii) 47° 32′
(iv) 62°57′
(v) 10°20′ + 20° 45′
Solution:
From tables of sine of angles, we find that:
(i) sin 21°= 0.3584,
(ii) sin 34°42’= .5693
(iii) sin 47° 32′ = 0.7377
(iv) sin 62° 57′ = 0.8906
(v) sin 10° 20′ + 20°45′ = sin 31°5′
= 0.5162

Question 2.
Use tables to find cosine of:
(i) 2°4′
(ii) 8°12′
(iii) 26°32’
(iv) 65°41′
(v) 9°23′ +15°54′
Solution:
From tables of cosine of angle, we find that:
(i) cos 2°4′ = 0.9993
(ii) cos 8° 12’ = 0.9898
(iii) cos 26°32′ = 0.8946
(iv) cos 65°41′ = 0.4118
(v) cos 9°23′ + 15°54′ = cos 25° 17′
= 0.9042

Question 3.
Use trigonometrical tables to find tangent of:
(i) 37°
(ii) 42°18′
(iii) 17°27′
Solution:
From the tables of tangents, we find that
(i) tan 35° = 0.7536
(ii) tan 42°18’= 0.9099
(iii) tan 17°27’= 0.3144

Question 4.
Use tables to find the acute angle θ, if the value of sin θ
(i) 4848
(ii) 0.3827
(iii) 0.6525
Solution:
From the tables of series, we find that of :
(i) sinθ = 0.4848, then θ = 29°
(ii) sinθ = 0.3827, then θ = 20° 30′
(iii) sin θ = 0.6525, then θ = 40° 42’ + 2′ = 40°44′

Question 5.
Use tables to find the acute angle θ, if the value of cos θ is :
(i) 0.9848
(ii) 0.9574
(iii) 0.6885
Solution:
From the tables of cosines, we find that if :
(i) cos θ = 0.9848, then θ = 10°
(ii) cos θ = 0.9574, then θ = 16°48′- 1’=16°47’
(iii) cos θ = 0.6885, then θ = 46° 30′ or 46°30′
= 46° 29’

Question 6.
Use tables to find the acute angle θ, if the value of tan θ is :
(i) 2419
(ii) 0.4741
(iii) 0.7391
(iv) 1.06
Solution:
From the table of tangents, we find that if:
(i) tan θ = 0.2419, then θ=13° 36’
(ii) tan θ = 0.4741, then θ = 25° 18’ + 4’ = 25°22′
(iii) tan θ = 0.7391, then θ= 36°24’+ 4′ = 36°28′
(iv) tan θ = 1.06, then θ = 46°36′ + 4′ = 46°40′

Question 7.
If sin θ=0.857; find:
(i) θ
(ii) tan θ
Solution:
From the tables of T. Ratio’s we find this :
(i) If sin θ = 0.857, then θ = 58°54′ + 4.5′ = 58° 58′ or 58°59’
(ii) tan 58°58’= 1.6577 +43 = 1.662 or tan 58° 59′ = 1.6577 + 53 = 1.663

Question 8.
If θ is the acute angle and cos θ = 0.7258; find:
(i) θ
(ii) 2 tan θ – sin θ
Solution:
From the tables of T-ratio’s, we find that:
(i) If cos θ = 0.7258, then θ= 43° 30′ -2′ = 43°28’
(ii) Now 2 tan θ – sin θ= 2 tan 43°28′ – sin 43°28′
2 tan 43°28’ = 2 x (0.9457 + 0.0022)
= 0:9479 x 2 = 1.8958
and sin 43°28′ = 0.6871 + 0.0008 = 0.6879
∴ 2 tan 43°28′ – sin 43° 28′ = 1.8958 – 0.6879 = 1.2079

Question 9.
Let θ be an acute angle and tan θ = 0.9490 find:
(i) θ
(ii) cos θ
(iii) sin θ – cos θ

Solution:
From the tables of T-raios, we find that:
(i) if tan θ = 0.9490 , then θ = 43°30′
(ii) cos θ = cos 43°30′ = 0.7254
(iii) sin θ = sin 43°50′ = 0.6884
∴ sin θ – cos θ = 0.6884 – 0.7254 = -0.0370 = -0.037

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21D are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity (With Applications to Maps and Models) Ex 15B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B.

Other Exercises

Question 1.
In the following figure, point D divides AB in the ratio 3 : 5. Find:

Solution:

Question 2.
In the given figure, PQ // AB;
CQ = 4.8 cm, QB = 3.6 cm and AB = 6.3 cm. Find:
(i) $$\frac { CP }{ PA }$$
(ii) PQ
(iii) If AP = x, then the value of AC in terms of x.

Solution:
In the given figure,
PQ || AB
CQ = 4.8 cm, QB = 3.6 cm, AB = 6.3 cm

Question 3.
A line PQ is drawn parallel tp the side BC of ΔABC which cuts side AB at P and side AC at Q. If AB = 9.0 cm, CA = 6.0 cm and AQ = 4.2 cm, find the length of AP.
Solution:
In ΔABC, PQ || BC
AB = 9.0 cm, CA = 6 cm, AQ = 4.2 cm

Question 4.
In ΔABC, D and E are the points on sides AB and AC respectively.
Find whether DE // BC, if:
(i) AB = 9 cm, AD = 4 cm, AE = 6 cm and EC = 7.5 cm
(ii) AB = 63 cm, EC = 11.0 cm, AD = 0.8 cm and AE = 1.6 cm.
Solution:
In ΔABC, D and E are the points on sides AB and AC respectively.

Question 5.
In the given figure, ΔABC ~ ΔADE. If AE : EC = 4 : 7 and DE = 6.6 cm, find BC. If ‘x’ be the length of the perpendicular from A to DE, find the length of perpendicular from A to BC in terms of ‘x’.

Solution:
In the given figure,

AE : EC = 4 : 7, DE = 6.6 cm, BC = ?
Draw AL ⊥ DE and AM ⊥ BC and AL = x cm
Find AM in terms of x

Question 6.
A line segment DE is drawn parallel to base BC of ΔABC which cuts AB at point D and AC at point E. If AB = 5 BD and EC = 3.2 cm, find the length of AE.
Solution:
In ΔABC DE || BC
AB = 5 BD, EC = 3.2 cm

Question 7.
In the figure, given below, AB, CD and EF are parallel lines. Given AB = 7.5 cm, DC = y cm, EF = 4.5 cm, BC = x cm and CE = 3 cm, calculate the values of x and y.

Solution:
(i) In ΔACB and ΔFCE, we have
∠ACB = ∠FCE (vertically opposite angles)
∠CBA = ∠CEF (alternate angles)
ΔACB ~ ΔFCE (AA Axiom of similarity)
Thus their corresponding sides are proportional.

Question 8.
In the figure, given below, PQR is a right angle triangle right angled at Q. XY is parallel to QR, PQ = 6 cm, PY = 4 cm and PX : XQ = 1 : 2. Calculate the lengths of PR and QR.

Solution:
Given, PQ = 6 cm; PY = 4 cm;
PX : XQ = 1 : 2
Since a line drawn || to one side of triangle divide the other two sides proportionally.

Question 9.
In the following figure, M is mid-point of BC of a parallelogram ABCD. DM intersects the diagonal AC at P and AB produced at E. Prove that: PE = 2PD.

Solution:
In the given figure, ABCD is a ||gm
AB || CD, AD || BC
M is mid point of BC
DM intersect AB produced at E and AC at P
To prove: PE = 2PD
Proof: In ΔDEA,
AD || BC (Opposite sides of || gm)
M is mid-point of CB B is mid-point of AE
AB = BE ⇒ AE = 2AB or 2CD
In ΔPAE and ΔPCD
∠APE = ∠CPD (Vertically opposite angles)
∠PAE = ∠PCD (Alternate angles)
ΔPAE ~ ΔPCD

Question 10.
The given figure shows a parallelogram ABCD. E is a point in AD and CE produced meets BA produced at point F. If AE = 4 cm, AF = 8 cm and AB = 12 cm, find the perimeter of the parallelogram ABCD.

Solution:
In the given figure, ABCD is a ||gm E is a point on AD
CE is produced to meet BA produced at point F
AE = 4 cm, AF = 8 cm, AB = 12 cm
To find the perimeter of ||gm ABCD
In ΔFBC,
AD or AE || BC (Opposite sides of ||gm)

Perimeter of ||gm ABCD = 2 (AB + BC) = 2 (12 + 10) cm = 2 x 22 = 44 cm

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.