
01. Real Numbers
9
Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9


02. Polynomials
11
Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10

Lecture2.11


03. Linear Equation
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Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7

Lecture3.8

Lecture3.9


04. Quadratic Equation
8
Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6

Lecture4.7

Lecture4.8


05. Arithmetic Progressions
11
Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7

Lecture5.8

Lecture5.9

Lecture5.10

Lecture5.11


06. Some Applications of Trigonometry
7
Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6

Lecture6.7


07. Coordinate Geometry
17
Lecture7.1

Lecture7.2

Lecture7.3

Lecture7.4

Lecture7.5

Lecture7.6

Lecture7.7

Lecture7.8

Lecture7.9

Lecture7.10

Lecture7.11

Lecture7.12

Lecture7.13

Lecture7.14

Lecture7.15

Lecture7.16

Lecture7.17


08. Triangles
15
Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

Lecture8.5

Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9

Lecture8.10

Lecture8.11

Lecture8.12

Lecture8.13

Lecture8.14

Lecture8.15


09. Circles
8
Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8


10. Areas Related to Circles
10
Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7

Lecture10.8

Lecture10.9

Lecture10.10


11. Introduction to Trigonometry
7
Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6

Lecture11.7


12. Surface Areas and Volumes
9
Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5

Lecture12.6

Lecture12.7

Lecture12.8

Lecture12.9


13. Statistics
12
Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5

Lecture13.6

Lecture13.7

Lecture13.8

Lecture13.9

Lecture13.10

Lecture13.11

Lecture13.12


14. Probability
9
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9


15. Construction
7
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

NCERT Solutions – Surface Areas and Volumes
Q.1 2 cubes each of volume 64cm3 are joined end to end. Find the surface area of the resulting cuboid.
Sol.
Let the length of each edge of the cube be a cm.
Then, Volume =64cm3
⇒ a3=64
⇒ a = 4
When two cubes of equal volumes (i.e., equal edges are joined end to end, we get a cuboid such that its.
l = Length = 4cm + 4cm = 8 cm
b = Breadth = 4 cm
and h = Height = 4 cm
Therefore Surface area of the cuboid
= 2(lb + bh + hl)
= 2(8 × 4 + 4 × 4 + 4 × 8) cm2
= 2(32 + 16 + 32) cm2
= (2 × 80) cm2 = 160 cm2
Q.2 A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Sol.
Here r, the radius of hemisphere = 7 cm ,
h , the height of cylinder = (13 – 7) cm = 6 cm
Clearly, radius of the base of cylindrical part is also r cm .
Surface area of the vessel = Curved surface area of the cylindrical part + Curved surface area of hemispherical part
=(2πrh+2πr2)cm2=2πr(h+r)cm2
=2×227×7×13cm2=572cm2
Q.3 A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Sol.
We have VO = 15.5 cm,
OA = OO’ = 3.5 cm
Let r be the radius of the base of cone and h be the height of conical part of the by.
Then r = OA = 3.5 cm
h = VO = VO’ – OO = (15.5 – 3.5) cm = 12 cm
Also radius of the hemisphere = OA = r = 3.5 cm
Total surface area of the toy
= Curved surface area of cone + Curved surface area of hemisphere
=πrℓ+2πr2
where l = Slant height =OA2+OV2−−−−−−−−−−√=(3.5)2+122−−−−−−−−−−√
=12.25+144−−−−−−−−−√=156.25−−−−−√=12.5cm
=πr(ℓ+2r)
=227×3.5×(12.5+2×3.5)cm2
=227×3.5×19.5cm=214.5cm2
Q.4 A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter of the hemisphere can have? Find the surface area of the solid?
Sol.
The greatest diameter that a hemisphere can have = 7 cm .
Surface area of the solid after surmounted hemisphere
=6ℓ2−πR2+2πR2
=6ℓ2+πR2
=6(7)2+227×(72)2
=6×49+11×72
= 294 + 38.5 = 332.5 cm2
Q.5 A hemispherical depression is cut of one face of a cubical wooden block such the diameter l of the hemisphere is equal to the edge of the cube. Determine th surface area of the remaining solid.
Sol.
Edge of the cube = ℓ
Diameter of the hemisphere = ℓ
Therefore Radius of the hemisphere = ℓ2
Therefore Area of the remaining solid after cutting out the hemispherical depression.
=6ℓ2−π(ℓ2)2+2π(ℓ�2)2
=6ℓ2+π(ℓ2)2
=6ℓ2+π×ℓ24=ℓ24(24+π)
Q.6 A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Sol.
Let r cm be the radius and h cm be the height of the cylinder. Then.
r=52mm=2.5mm
and h=(14−2×52)mm
= (14 – 5)mm = 9mm
Also the radius of hemisphere = 52cm=rcm
Now, surface area of the capsule = Curved surface of cylinder + Surface area of two hemispheres
=(2πrh+2×2πr2)mm2
=2πr(h+2r)mm2
=2×227×52×(9+2×52)mm2
=227×5×14mm2=220mm2
Q.7 A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not covered with canvas).
Sol.
We have, Total canvas used = Curved surface area of cylinder + Curved surface area of cone
=(2πrh+πrℓ)m2
=πr(2h+ℓ)m2
=227×2×(2×2.1+2.8)m2
=227×2×7m2
=44m2
Now, cost of 1m2 the canvas for the tent = Rs 500
So, cost of 44m2 the canvas for the tent = Rs 44 × 500 = Rs 22000
Q.8 From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Sol.
Radius of the cylinder =1.42cm=7cm
Height of the cylinder = 2.4 cm
Radius of the cone = .7 cm
Height of the cone = 2.4 cm
Slant height of the cone
=(.7)2+(2.4)2−−−−−−−−−−−√cm
=0.49+5.76−−−−−−−−−√cm
=6.25−−−−√cm=2.5cm
Surface area of the remaining solid = Curved surface area of cylinder + Curved surface of the cone + Area of upper circular base of cylinder
=2πrh+πrℓ+πr2=πr(2h+ℓ+r)
=227×.7×(2×2.4+2.5+.7)cm2
=2×.1×(4.8+2.5+.7)cm2
2.2×8.0cm2=17.6cm2=18cm2
Q.9 A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm,and its base is of radius 3.5 m, find the total surface area of the article.
Sol.
Surface area of the article when it is ready = Curved surface area of cylinder + 2 × Curved surface area of hemisphere.
=2πrh+2×2πr2
=2πr(h+2r)
where r = 3.5 m and h = 10 cm
=2×227×3.5×(10+2×3.5)cm2
=2×227×3.5×17cm2=374cm2
Q.1 A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
Sol.
Volume of the solid = Volume of the cone + Volume of the hemisphere
=13πr2h+23πR3
=13πr2×r+23πr3 [Since h = r and R = r]
=π3r3(1+2)=πz3r3×3
=πr3=π(1)3=π×1=πcm3 [Since r = 1 cm]
Q.2 Rachel an engineering student was asked to make a model shaped like a cylinder with two cones attached at its two ends by using thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm,find the volume of air contained in the model the Rachel made. (Assume the outer and inner dimensions of the model be nearly the same).
Sol.
Volume of the air contained in the model = Volume of the cylindrical portion of the model + Volume of its two conical ends.
=πr2h1+2×13πr2h2
=πr2(h1+23h2)
where r=32cm
h1=8cmandh2=2cm
=227×(32)2×(8+23×2)cm3
=227×94×24+43cm3
=(227×94×283)cm3
=66cm3(approx)
Q.3 A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemsipherical ends with length 5 cm and diameter is 2.8 cm (see figure).
Sol.
Volume of the gulab jamun = Volume of the cylindrical portion + Volume of the hemispherical ends
=πr2h+2(23πr3)
=πr2(h+43r), where r = 1.4 cm, h = 2.2 cm
=227×(1.4)2×(2.2+43×1.4)cm3
=227×1.96×(6.6+5.63)cm3
=227×1.96×12.23cm3
Volume of 45 gulab jamuns
=45×227×1.96×12.23cm3
Quantity of syrup in gulab jamuns = 30% of their volume
=30100×45×227×1.96×12.23cm3
9×11×1.96×12.27cm3=338.184cm3
=338cm3(approx)
Q.4 A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 m and the depth is 1.4 cm. Find the volume of wood in the entire stand (see figure).
Sol.
Volume of wood in the entire stand = Volume of the cuboid – 4 × Volume of a depression (i.e., cone)
=15×10×3.5cm3−4×13×227×0.5×0.5×1.4cm3
=(525−1.47)cm3=523.53cm3
Q.5 A vessel 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 brim. 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.
Sol.
Height of the conical vessel, h = 8 cm.
Its radius r = 5 cm
Volume of cone = Volume of water in cone
=13πr2h
=13×227×5×5×8cm3
=440021cm3
Volume of water flows out = Volume of lead shots
=14 of the volume of water in the cone
=14×440021cm3=110021cm3
Radius of the lead shot = 0.5 cm
Volume of one spherical lead shot =43πr3
=43×227×510×510×510cm3=1121cm3
Therefore Number of lead shots dropped into the vessel
=VolumeofwaterflowsoutVolumeofoneleadshot
=110021÷1121=110021×2111=100
Q.6 A solid iron pole consists of a cylindrical height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use π=3.14)
Sol.
Volume of the solid iron pole = Volume of the cylindrical portion + Volume of the other cylindrical portion
=πr12h1+πr22h2
=(3.14×(12)2×220+3.14×(8)2×60)cm3
=(3.14×144×220+3.14×64×60)cm3
=(99475.2+12057.6)cm3
=111532.8cm3
Hence, the mass of the pole = (111532.8 × 8) grams
=(111532.8×81000)kg=892.26kg
Q.7 A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water circular such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Sol.
Volume of the cylinder =πr2h
=227×(60)2×180cm3
=22×3600×1807cm3
=142560007cm3
Volume of the solid = Volume of cone + Volume of hemisphere
=(13×227×602×120+23×227×603)cm3
=(31680007+31680007)cm3
=63360007cm3
Volume of water left in the cylinder = Volume of the cylinder – Volume of the solid
=(142560007−63360007)cm3
=(79200007)cm3
=1131428.57142cm3
=1.131m3(approx)
Q.1 A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Sol.
Volume of the sphere
=43πr3=43×π×(4.2)3cm3
If h is the height of a cylinder of radius 6 cm. Then its volume,
=π(6)2hcm3=36πhcm3
Since, the volume of metal in the form of sphere and cylinder remains the same , we have
36πh=43×π×4.2×4.2×4.2
⇒ h=136×43×4.2×4.2×4.2
⇒ h = 2.744
Q.2 Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Sol.
Sum of the volumes of 3 gives spheres.
=43π(r13+r23+r33)
=43π(63+83+103)cm3
=43π(216+512+1000)cm3
=43π(1728)cm3
Let R be the radius of the new spheres whose volume is the sum of the volumes of 3 given spheres.
Therefore =43πR3=43π(1728)
⇒ R3=1728
⇒ R3=(12)3
⇒ R = 12
Hence, the radius of the resulting sphere is 12 cm.
Therefore 1728 = 2^{3} × 6^{3} = (2 × 6)^{3} = 12)^{3}
Q.3 A 20 cm deep well with diameter 7 cm is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Sol.
Let h m be the required height of the platform.
The shape of the platform will be like the shape of a cuboid 22 m × 14 m × h with a hole in the shape of cylinder of radius 3.5 m and depth h m.
The volume of the platform will be equal to the volume of the earth dug out from the well.
Now, the volume of the earth = Volume of the cylindrical well
=πr2h
=227×12.25×20m3
=770m3
Also, the volume of the platform = 22 × 14 × h m3
But volume of the platform = Volume of the well
i.e., 22 × 14 × h = 770
h=77022×14=2.5
Therefore Height of the platform = 2.5 m
Q.4 A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Sol.
Let h be the required height of the embankment.
The shape of the embankment will be like the shape of a cylinder of internal radius 1.5 m and external radius (4 + 1.5)m = 5.5 m (see figure).
The volume of the embankment will be equal to the volume of earth dug out from the well.
Now, the volume of the earth = Volume of the cylindrical well
=π×(1.5)2×14m3=31.5πm3
Also the volume of the embankment
=π(5.52−1.52)hm3
=π(5.5+1.5)(5.5−1.5)hm3
=π×7×4hm3=28πhm3
Hence, we have
28πh=31.5π
⇒ h=31.528=1.125
Hence, the required height of the embankment = 1.125 m
Q.5 A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height12 cm and diameter 6 cm, having hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Sol.
Volume of the cylinder
=πr2h
=π(122)2×15
=π×62×15
Volume of a cone having hemispherical shape on the top
=13πr2h+23πr3=13πr2(h+2r)
=13π(62)2(12+2×62)
=13π×32×18
Let the number of cone that can be filled with ice cream be h.
Then 13π×32×18×n=π×62×15
n=π×6×6×15π×3×3×18×3=10
Q.6 How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Sol. The shape of the coin will be like the shape of a cylinder of radius 1.752cm
= 0.875 cm and of height 2mm =210cm=.2cm
Its volume
=πr2h=227×0.875×0.875×.2cm3
=0.48125cm3
Volume of the cuboid =5.5×10×3.5cm3=192.5cm3
Number of coins required to form the cuboid
=VolumeofthecuboidVolumeofthecoin=192.50.48125=400
Hence, 400 coins must be melted to form a cuboid.
Q.1 A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Sol.
Capacity of the glass,
V=π×h3×(R2+r2+Rr)
Here, R = 2 cm , r = 1 cm , h = 14 cm
Therefore V=2273×(22+12+2×1)
=443×(4+1+2)
=443×7=308310223cm3
Q.2 The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Sol.
Slant height ,l = 4 cm (Given)
2πr1=6 ⇒ πr1=3
and 2πr2=18 ⇒ πr2=9
Curved surface of the frustum
=(πr1+πr2)ℓ
=(3+9)×4cm2
=12×4cm2
=48cm2
Q.3 A fez, the (cap) used by the Turks, is shaped like the frustum of a cone (see figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cmand its slant height is 15 cm, find the area of material used for making it.
Sol.
Here R = 10 cm, r = 4 cm and l = 15 cm
Area of the material used for making the feza = Surface area of frustum
+ The surface of top circular section
=π(R+r)ℓ+πr2
=227(10+4)15+227×4×4
=(227×14×15+3527)cm2
=4620+3527cm2=49727cm2
=71027cm2
Q.4 A container opened from the top and made up of a 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 milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2. (Take π=3.14).
Sol.
Here R = 20 cm, r = 8 cm and h = 16 cm Capacity of the container
= Volume of the frustum
=13πh(R2+r2+Rr)
=13×3.14×16(202+82+20×8)cm3
=50.243×(400+64+160)cm3
=50.243×624cm3=50.24×208cm3
=10449.921000litres
Cost of milk @ Rs 20 per litre
=Rs(20×10449.921000)
=Rs208.99≈Rs209
To find the slant height
ℓ=162+122−−−−−−−−√
=256+144−−−−−−−−√
400−−−√=20cm
=π(R+r)ℓ
=227×(20+8)×20cm2
=227×28×20cm2=1758.4cm2
and area of the bottom
=πr2=227×(8)2=227×64=200.96cm2
Therefore Total area of metal required
= 1758.4 cm2+200.96cm2=1959.36cm2
Cost of metal sheet used to manufacture the container @ Rs 8 per 100cm2
=Rs(8100×1959.36)=Rs156.75
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