Exercise 11.1
Q1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area (Assume π = 22 / 7)
Ans:
Diameter of the base of the cone is 10.5 cm. To find the radius, we need to divide the diameter by 2.
Radius of the base of the cone, r = diameter / 2 = 10.5 / 2 = 5.25 cm
The slant height of the cone is given as 10 cm. Let's denote it as l.
Slant height of the cone, l = 10 cm
Now, we can find the curved surface area (CSA) of the cone using the formula:
CSA = π r l
where π (pi) is approximately equal to 22 / 7.
CSA = (22 / 7) × 5.25 × 10
CSA = 165 cm²
Hence, the curved surface area of the cone is 165 cm².
Q2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. (Assume π = 22 / 7)
Solution:
The radius of the cone, r = 24 / 2 m = 12m
Slant height, l = 21 m
To find the total surface area of a cone, we use the formula: Total Surface area of the cone = πr(l + r)
Total Surface area of the cone = (22 / 7) × 12 × (21 + 12) m2
= (22 / 7) × 12 × 33 m2
= 1244.57 m2
Thus, the total surface area of the cone is 1244.57 m2.
Q3. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find
(i) radius of the base and
(ii) total surface area of the cone. (Assume π = 22 / 7)
Ans:Slant height of cone, l = 14 cm
Let the radius of the cone be r.
(i)We know, CSA of cone = πrl
Given: Curved surface area of a cone is 308 cm2
(308) = (22 / 7) × r × 14
308 = 44r
r = 308 / 44 = 7cm
The radius of the cone base is 7 cm.
(ii) To find the total surface area of the cone, we need to add the curved surface area (CSA) and the area of the base (πr2).
Total surface area of cone = CSA of cone + Area of base (πr2)
Total surface area of cone = 308 + (22 / 7) × 72 = 308 + 154
Therefore, the total surface area of the cone is 462 cm2.
Q4. A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70. (Assume π = 22 / 7)
Ans: Let ABC be a conical tent with vertex A, base center O, and base circumference point B.
Height of conical tent (AO), h = 10 m
Radius of conical tent (BO), r = 24m
Let the slant height of the tent (AB) be l.
(i)In right triangle ABO, we have
AB2 = AO2 + BO2 (using Pythagoras theorem)
l2 = h2 + r2
= (10)2 + (24)2
= 100 + 576
= 676
l = √676 = 26m
Therefore, the slant height of the tent is 26 m.
(ii) The curved surface area (CSA) of the conical tent can be calculated using the formula:
CSA = πrl
= (22/7) × 24 × 26 m2
=13728/7 m2
Now, let's calculate the cost of the canvas required to make the tent.
Cost of 1 m2 canvas = Rs 70
Cost of (13728 / 7) m2 canvas is equal to Rs (13728 / 7) × 70 = Rs 137280
Therefore, the cost of the canvas required to make such a tent is Rs 137280.
Q5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. [Use π = 3.14]
Ans:Height of conical tent, h = 8m
Radius of base of tent, r = 6m
Slant height of tent, l2 = (r2 + h2)
l2 = (62 + 82) = (36 + 64) = (100)
or l = 10
Again, CSA of conical tent = πrl
= (3.14 × 6 × 10) m2
= 188.4m2
Let the length of tarpaulin sheet required be L
As 20 cm will be wasted, therefore,
Effective length will be (L - 0.2m).
Breadth of tarpaulin = 3m (given)
Area of sheet = CSA of tent
[(L – 0.2) × 3] = 188.4
L - 0.2 = 62.8
L = 63
Therefore, the length of the required tarpaulin sheet will be 63 m.
Q6. The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2. (Assume π = 22 / 7).
Ans: Slant height of conical tomb, l = 25m
Base radius, r = diameter / 2 = 14 / 2 m = 7m
CSA of conical tomb = πrl= (22 / 7) × 7 × 25 = 550
CSA of conical tomb = 550m2
Cost of white-washing 550 m2 area, which is Rs (210 × 550) / 100
= Rs.1155
Therefore, cost will be Rs. 1155 while white-washing tomb.
Q.7. A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. (Assume π = 22 / 7)
Ans: Radius of conical cap, r = 7 cm
Height of conical cap, h = 24cm
Slant height, l2 = (r2 + h2)
= (72 + 242)
= (49 + 576)
= (625)
Or l = 25 cm
CSA of 1 conical cap = πrl
= (22 / 7) × 7 × 25
= 550
CSA of 10 caps = (10 × 550) cm2 = 5500 cm2
Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.
Q8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m2, what will be the cost of painting all these cones? (Use π = 3.14 and take √(1.04) = 1.02).
Ans:Given:
Radius of cone, r = diameter / 2 = 40 / 2 cm = 20cm = 0.2 m
Height of cone, h = 1m
Slant height of cone is l, and l2 = (r2 + h2)
Using given values, l2 = (0.22 + 12)
= (1.04)
Or l = 1.02
Slant height of the cone is 1.02 m
Now,
CSA of each cone = πrl
= (3.14 × 0.2 × 1.02)
= 0.64056m2
CSA of 50 such cones = (50 × 0.64056)
CSA of 50 such cones = 32.028 m2
Again,
Cost of painting 1 m2 area = Rs 12 (given)
Cost of painting 32.028 m2 area
= Rs (32.028 × 12)
= Rs.384.336
= Rs.384.34 (approximately)
Therefore, the cost of painting all these cones is Rs. 384.34.
Exercise 11.2
Q1. Find the surface area of a sphere of radius:
(i) 10.5cm
(ii) 5.6cm
(iii) 14cm
(Assume π = 22 / 7)
Ans: Formula:Surface area of sphere (SA) = 4πr2
(i) Radius of sphere, r = 10.5 cm
SA = 4 × (22 / 7) × 10.52 = 1386
Surface area of sphere is 1386 cm2
(ii) Radius of sphere, r = 5.6cm
Using formula, SA = 4 × (22 / 7) × 5.62 = 394.24
Surface area of sphere is 394.24 cm2
(iii)Radius of sphere, r = 14cm
SA = 4πr2
= 4 × (22 / 7) × (14)2
= 2464
Surface area of sphere is 2464 cm2.
Q2. Find the surface area of a sphere of diameter:
(i) 14cm
(ii) 21cm
(iii) 3.5cm
(Assume π = 22 / 7)
Ans:(i)Radius of sphere, r = diameter / 2 = 14 / 2 cm = 7 cm
Formula for Surface area of sphere = 4πr2
= 4 × (22 / 7) × 72= 616
Surface area of a sphere is 616 cm2
(ii) Radius (r) of sphere = 21/2 = 10.5 cm
Surface area of sphere = 4πr2
= 4 × (22 / 7) × 10.52 = 1386
Surface area of a sphere is 1386 cm2
Therefore, the surface area of a sphere having diameter 21cm is 1386 cm2
(iii) Radius(r) of sphere = 3.5 / 2 = 1.75 cm Surface area of sphere = 4πr2
= 4 × (22 / 7) × 1.752 = 38.5
Surface area of a sphere is 38.5 cm2.
Q3. Find the total surface area of a hemisphere of radius 10 cm. [Use π = 3.14]
Ans: Radius of hemisphere, r = 10cm
Formula: Total surface area of hemisphere = 3πr2
= 3 × 3.14 × 102 = 942
The total surface area of given hemisphere is 942 cm2.
Q4. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Ans: Let r1 and r2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So
r1 = 7cm
r2 = 14 cm
Now, Required ratio = (initial surface area)/(Surface area after pumping air into balloon)
= 4r12 / 4r22
= (r1 / r2)2
= (7 / 14)2 = (1 / 2)2 = 1 / 4
Therefore, the ratio between the surface areas is 1 : 4.
Q5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. (Assume π = 22 / 7)
Ans: Inner radius of hemispherical bowl, say r = diameter / 2 = (10.5) / 2 cm = 5.25 cm
Formula for Surface area of hemispherical bowl = 2πr2
= 2 × (22 / 7) × (5.25)2 = 173.25
Surface area of hemispherical bowl is 173.25 cm2
Cost of tin-plating 100 cm2 area = Rs 16
Cost of tin-plating 1 cm2 area = Rs 16 /100
Cost of tin-plating 173.25 cm2 area = Rs. (16 × 173.25) / 100 = Rs 27.72
Therefore, the cost of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm2 is Rs 27.72.
Q6. Find the radius of a sphere whose surface area is 154 cm2. (Assume π = 22/7)
Ans: Let the radius of the sphere be r.
Surface area of sphere = 154 (given)
Now,
4πr2 = 154
r2= (154 × 7) / (4 × 22) = (49 / 4)
r = (7 / 2) = 3.5
The radius of the sphere is 3.5 cm.
Q7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Ans: If diameter of earth is said d, then the diameter of moon will be d / 4 (as per given statement)
Radius of earth = d / 2
Radius of moon = 1 / 2 × d / 4 = d / 8
Surface area of moon = 4π(d / 8)2
Surface area of earth = 4π(d / 2)2
Ratio of their Surace areas = 4π(d / 8)2 / 4π (d / 2)2= 4 / 64 = 1 / 16
The ratio between their surface areas is 1 : 16.
Q8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. (Assume π = 22 / 7)
Ans:Given:
Inner radius of hemispherical bowl = 5cm
Thickness of the bowl = 0.25 cm
Outer radius of hemispherical bowl = (5 + 0.25) cm = 5.25 cm
Formula for outer CSA of hemispherical bowl = 2πr2, where r is radius of hemisphere
= 2 × (22 / 7) × (5.25)2 = 173.25
Therefore, the outer curved surface area of the bowl is 173.25 cm2.
Q9. A right circular cylinder just encloses a sphere of radius r (see Fig). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in(i) and (ii).
Ans:(i) Surface area of sphere = 4πr2, where r is the radius of sphere
(ii) Height of cylinder, h = r + r = 2r
Radius of cylinder = r
CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h)
= 4πr2
(iii) Ratio between areas = (Surface area of sphere) / CSA of Cylinder)
= 4r2 / 4r2
= 1 / 1
Ratio of the areas obtained in (i) and (ii) is 1 : 1.
Exercise 11.3
Q1. Find the volume of the right circular cone with
(i) radius 6cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm (Assume π = 22 / 7)
Ans: Volume of cone = (1 / 3) πr2h cube units
Where r be radius and h be the height of the cone
(i) Radius of cone, r = 6 cm
Height of cone, h = 7cm
Say, V be the volume of the cone, we have
V = (1 / 3) × (22 / 7) × 36 × 7
= (12 × 22)
= 264
The volume of the cone is 264 cm3.
(ii) Radius of cone, r = 3.5cm
Height of cone, h = 12cm
Volume of cone = (1 / 3) × (22 / 7) × 3.52 × 7 = 154
Hence, The volume of the cone is 154 cm3.
Q2. Find the capacity in litres of a conical vessel with
(i) radius 7cm, slant height 25 cm
(ii) height 12 cm, slant height 12 cm (Assume π = 22 / 7)
Ans: (i) Radius of cone, r = 7 cm
Slant height of cone, l = 25 cm
or h = 24
Height of the cone is 24 cm
Now,
Volume of cone, V = (1/3) πr2h (formula)
V = (1 / 3) × (22 / 7) × 72 × 24
= (154 × 8)
= 1232
So, the volume of the vessel is 1232 cm3
Therefore, capacity of the conical vessel = (1232/1000) liters (because 1L = 1000 cm3)
= 1.232 Liters.
(ii)Height of cone, h = 12 cm
Slant height of cone, l = 13 cm
r = 5
Hence, the radius of cone is 5 cm.
Now, Volume of cone,
V = (1 / 3)πr2h V = (1 / 3) × (22 / 7) × 52 × 12 cm3
= 2200 / 7
Volume of cone is 2200 / 7 cm3
Now, Capacity of the conical vessel= 2200 / 7000 litres (1L = 1000 cm3)
= 11 / 35 litres.
Q3. The height of a cone is 15cm. If its volume is 1570cm3, find the diameter of its base. (Use π = 3.14)
Ans: Height of the cone, h = 15 cm
Volume of cone = 1570 cm3
Let r be the radius of the cone
As we know: Volume of cone, V = (1 / 3) πr2h
So, (1 / 3) πr2h = 1570
(1 / 3) × 3.14 × r2 × 15 = 1570
r2 = 100
r = 10
Radius of the base of cone 10 cm.
Q4. If the volume of a right circular cone of height 9cm is 48πcm3, find the diameter of its base.
Ans: Height of cone, h = 9cm
Volume of cone = 48π cm3
Let r be the radius of the cone.
As we know: Volume of cone, V = (1 / 3) πr2h
So, 1/3 π r2(9) = 48 π
r2= 16
r = 4
Radius of cone is 4 cm.
So, diameter = 2 × Radius = 8
Thus, diameter of base is 8cm.
Q5. A conical pit of top diameter 3.5m is 12m deep. What is its capacity in kiloliters? (Assume π = 22 / 7)
Ans: Diameter of conical pit = 3.5 m
Radius of conical pit, r = diameter / 2 = (3.5 / 2)m = 1.75m
Height of pit, h = Depth of pit = 12m
Volume of cone, V = (1 / 3) πr2h
V = (1 / 3) × (22 / 7) × (1.75)2 × 12 = 38.5
Volume of cone is 38.5 m3
Hence, capacity of the pit = (38.5 × 1) kiloliters = 38.5 kiloliters.
Q6. The volume of a right circular cone is 9856cm3. If the diameter of the base is 28cm, find
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone (Assume π = 22 / 7)
Ans: Volume of a right circular cone = 9856 cm3
Diameter of the base = 28 cm
(i) Radius of cone, r = (28 /2) cm = 14 cm
Let the height of the cone be h
Volume of cone, V = (1 / 3) πr2h
(1 / 3) πr2h = 9856
(1 / 3) × (22 / 7) × 14 × 14 × h = 9856
h = 48
The height of the cone is 48 cm.
Slant height of the cone is 50 cm.
(iii)curved surface area of cone = πrl
= (22 / 7) × 14 × 50 = 2200
curved surface area of the cone is 2200 cm2.
Q7. A right triangle ABC with sides 5cm, 12cm and 13cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Ans:Height (h)= 12 cm
Radius (r) = 5 cm, and
Slant height (l) = 13 cm
Volume of cone, V = (1 / 3) πr2h
V = (1 / 3) × π × 52 × 12 = 100π
Volume of the cone so formed is 100π cm3.
Q8. If the triangle ABC in the Question 7 is revolved about the side 5cm, then find the volume of the solids so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Ans:
A right-angled ΔABC is revolved about its side 5cm, a cone will be formed of radius as 12 cm, height as 5 cm, and slant height as 13 cm.
Volume of cone = (1/3) πr2h; where r is the radius and h be the height of cone
= (1/3) × π × 12 × 12 × 5
= 240 π
The volume of the cones of formed is 240π cm3.
So, required ratio = (result of question 7) / (result of question 8) = (100π) / (240π) = 5 / 12 = 5 : 12.
Q9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas. (Assume π = 22 / 7)
Ans: Radius (r) of heap = (10.5 / 2) m = 5.25
Height (h) of heap = 3m
Volume of heap = (1 / 3)πr2h
= (1 / 3) × (22 / 7) × 5.25 × 5.25 × 3
= 86.625
The volume of the heap of wheat is 86.625 m3. Again,
Area of canvas required = CSA of cone = πrl, where l = 
After substituting the values, we have
= (22 / 7) × 5.25 × 6.05
= 99.825
Therefore, the area of the canvas is 99.825 m2.
Exercise 11.4
Q1. Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m (Assume π = 22 / 7)
Ans: (i) Radius of sphere, r = 7 cm
Using, Volume of sphere = (4 / 3) πr3
= (4 / 3) × (22 / 7) × 73 = 4312 / 3
Hence, volume of the sphere is 4312 / 3 cm3
(ii)Radius of sphere, r = 0.63 m
Using, volume of sphere = (4 / 3) πr3
= (4 / 3) × (22 / 7) × 0.633 = 1.0478
Hence, volume of the sphere is 1.05 m3 (approx).
Q2. Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm
(ii) 0.21 m (Assume π = 22 / 7)
Ans: (i)Diameter = 28 cm
Radius, r = 28 / 2 cm = 14cm
Volume of the solid spherical ball = (4 / 3) πr3
Volume of the ball = (4 / 3) × (22 / 7) × 143 = 34496 / 3
Hence, volume of the ball is 34496 / 3 cm3
(ii) Diameter = 0.21 m
Radius of the ball = 0.21 / 2 m = 0.105 m
Volume of the ball = (4 / 3)πr3
Volume of the ball = (4 / 3) × (22 / 7) × 0.1053
Hence, volume of the ball = 0.004851 m3
Q3. The diameter of a metallic ball is 4.2cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3? (Assume π = 22 / 7)
Ans: Given,
Diameter of a metallic ball = 4.2 cm
Radius(r) of the metallic ball, r = 4.2 / 2 cm = 2.1 cm
Volume formula = 4 / 3 πr3
Volume of the metallic ball = (4 / 3) × (22 / 7) × 2.13
Volume of the metallic ball = 38.808 cm3
Now, using relationship between, density, mass and volume,
Density = Mass / Volume
Mass = Density × volume
= (8.9 × 38.808) g
= 345.3912 g
Mass of the ball is 345.39 g (approx).
Q4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Ans: Let the diameter of earth be “d”. Therefore, the radius of earth will be will be d/2
Diameter of moon will be d / 4 and the radius of moon will be d/8
Find the volume of the moon:
Volume of the moon = (4 / 3) πr3 = (4 / 3) π (d / 8)3 = 4 / 3π(d3 / 512)
Find the volume of the earth:
Volume of the earth = (4 / 3) πr3 = (4 / 3) π (d / 2)3 = 4 / 3π(d3 / 8)
Fraction of the volume of the earth is the volume of the moon
Volume of moon is of the 1 / 64 volume of earth.
Q5. How many litres of milk can a hemispherical bowl of diameter 10.5cm hold? (Assume π = 22 / 7)
Ans: Diameter of hemispherical bowl = 10.5 cm
Radius of hemispherical bowl, r = 10.5 / 2 cm = 5.25 cm
Formula for volume of the hemispherical bowl = (2 / 3) πr3
Volume of the hemispherical bowl = (2 / 3) × (22 / 7) × 5.253 = 303.1875
Volume of the hemispherical bowl is 303.1875 cm3
Capacity of the bowl = (303.1875) / 1000 L = 0.303 litres(approx.)
Therefore, hemispherical bowl can hold 0.303 litres of milk.
Q6. A hemi spherical tank is made up of an iron sheet 1cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. (Assume π = 22/7)
Ans: Inner Radius of the tank, (r) = 1m
Outer Radius (R) = 1.01m
Volume of the iron used in the tank = (2 / 3) π (R3 – r3)
Put values,
Volume of the iron used in the hemispherical tank = (2 / 3) × (22 / 7) × (1.013 – 13) = 0.06348
So, volume of the iron used in the hemispherical tank is 0.06348 m3.
Q7. Find the volume of a sphere whose surface area is 154 cm2. (Assume π = 22 / 7)
Ans: Let r be the radius of a sphere.
Surface area of sphere = 4πr2
4πr2 = 154 cm2 (given)
r2 = (154 × 7) / (4 × 22) r = 7 / 2
Radius is 7 / 2 cm
Now, Volume of the sphere = (4 / 3) πr3
Volume of the sphere = (4 / 3) x (22 / 7) x (7 / 2)3 = 179 x 2 / 3
Volume of the sphere is 179 x 2 / 3 cm3
Q8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs. 4989.60. If the cost of white-washing isRs20 per square meter, find the
(i) Inside surface area of the dome
(ii) volume of the air inside the dome (Assume π = 22 / 7)
Ans: (i) Cost of white-washing the dome from inside = Rs 4989.60
Cost of white-washing 1m2area = Rs 20
CSA of the inner side of dome = 498.96 / 2 m2 = 249.48 m2
(ii) Let the inner radius of the hemispherical dome be r.
CSA of inner side of dome = 249.48 m2 (from (i))
Formula to find CSA of a hemi sphere = 2πr2
2πr2 = 249.48
2 × (22 / 7) × r2 = 249.48
r2 = (249.48 × 7) / (2 × 22)
r2 = 39.69
r = 6.3
So, radius is 6.3 m
Volume of air inside the dome = Volume of hemispherical dome
Using formula, volume of the hemisphere
= 2 / 3 πr3 = (2 / 3) × (22 / 7) × 6.3 × 6.3 × 6.3
= 523.908
= 523.9(approx.)
Volume of air inside the dome is 523.9 m3.
Q9. Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’. Find the
(i) radius r’ of the new sphere,
(ii) ratio of Sand S’.
Ans: Volume of the solid sphere = (4 / 3)πr3
Volume of twenty seven solid sphere = 27 × (4 / 3)πr3 = 36 πr3
(i)New solid iron sphere radius = r’
Volume of this new sphere = (4/3)π(r’)3
(4 / 3)π(r’)3 = 36 πr3
(r’)3 = 27r3
r’= 3r
Radius of new sphere will be 3r (thrice the radius of original sphere)
(ii) Surface area of iron sphere of radius r, S = 4πr2
Surface area of iron sphere of radius r’= 4π (r’)2
Now
S / S’ = (4πr2) / ( 4π (r’)2)
S / S’ = r2 / (3r’)2 = 1 / 9
The ratio of S and S’ is 1: 9.
Q10. A capsule of medicine is in the shape of a sphere of diameter 3.5mm. How much medicine (in mm3) is needed to fill this capsule? (Assume π = 22 / 7)
Ans: Diameter of capsule = 3.5 mm
Radius of capsule, say r = diameter / 2 = (3.5 / 2) mm = 1.75mm
Volume of spherical capsule = 4 / 3 πr3
Volume of spherical capsule = (4 / 3) × (22 / 7) × (1.75)3 = 22.458
The volume of the spherical capsule is 22.46 mm3.
