Question 1

Question 1Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Accepted Answer

Given,Diameter of cone (d) = 10.5 cmSlant height of cone (l) = 10 cmBy formula,Radius of cone (r) = Diameter2=10.52\dfrac{\text{Diameter}}{2} = \dfrac{10.5}{2}2Diameter​=210.5​ = 5.25 cmBy formula,Curved surface area of cone (C.S.A.) = πrlSubstituting values we get :C.S.A.=227×5.25×10=11557=165 cm2\text{C.S.A.} = \dfrac{22}{7} × 5.25 × 10 \[1em] = \dfrac{1155}{7} \[1em] = 165 \text{ cm}^2C.S.A.=722​×5.25×10=71155​=165 cm2Hence, curved surface area of the cone = 165 cm2.

Question 2

Question 2Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Accepted Answer

Given,Diameter of cone (d) = 24 mRadius of cone (r) = Diameter2=242\dfrac{\text{Diameter}}{2} = \dfrac{24}{2}2Diameter​=224​ = 12 mSlant height of the cone (l) = 21 mTotal surface area of the cone = Curved surface area of cone + area of the baseBy formula,Total surface area of the cone (T.S.A.) = πrl + πr2 = πr(l + r)Susbtituting values we get :=227×12×(12+21)=227×12×33=87127=1244.57 m2= \dfrac{22}{7} \times 12 \times (12 + 21) \[1em] = \dfrac{22}{7} \times 12 \times 33 \[1em] = \dfrac{8712}{7} \[1em] = 1244.57 \text{ m}^2=722​×12×(12+21)=722​×12×33=78712​=1244.57 m2Hence, total surface area of the cone = 1244.57 m2.

Question 3

Question 3Curved 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.

Accepted Answer

(i) Let the radius be r cm.Given,Curved surface area of cone = 308 cm2⇒πrl=308⇒r=308πl⇒r=308227×14⇒r=30822×2⇒r=30844⇒r=7 cm\Rightarrow πrl = 308 \[1em] \Rightarrow r = \dfrac{308}{πl} \[1em] \Rightarrow r = \dfrac{308}{\dfrac{22}{7} \times 14} \[1em] \Rightarrow r = \dfrac{308}{22 \times 2} \[1em] \Rightarrow r = \dfrac{308}{44} \[1em] \Rightarrow r = 7 \text{ cm}⇒πrl=308⇒r=πl308​⇒r=722​×14308​⇒r=22×2308​⇒r=44308​⇒r=7 cmHence, radius of the base of cone = 7 cm.(ii) By formula,Total surface area of cone (T.S.A.) = πr(l + r)Substituting values we get :T.S.A. = 227\dfrac{22}{7}722​ × 7 × (7 + 14)= 227\dfrac{22}{7}722​ × 7 x 21= 22 × 21= 462 cm2.Hence, the total surface area of the cone is 462 cm2.

Question 4

Question 4A 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 ₹ 70.

Accepted Answer

(i) Given,Radius of conical tent (r) = 24 mHeight of conical tent (h) = 10 mBy formula,Slant height of conical tent (l) = r2+h2\sqrt{r^2 + h^2}r2+h2​Substituting values we get :l=(24)2+(10)2=576+100=676=26 ml = \sqrt{(24)^2 + (10)^2} \[1em] = \sqrt{576 + 100} \[1em] = \sqrt{676} = 26 \text{ m}l=(24)2+(10)2​=576+100​=676​=26 mHence, slant height of the conical tent is 26 m.(ii) Given,Curved surface area of the cone = πrl= 227\dfrac{22}{7}722​ × 24 × 26= 137287\dfrac{13728}{7}713728​ m2The cost of the canvas required to make the tent, at ₹ 70 per m2 = Curved surface area of the cone x ₹ 70= 137287\dfrac{13728}{7}713728​ × 70= ₹ 137280.Hence, the cost of the canvas is ₹ 137280.

Question 5

Question 5What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? 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).

Accepted Answer

Given,Height of tent (h) = 8 mRadius of tent (r) = 6 mBy formula,Slant height of tent (l) = r2+h2\sqrt{r^2 + h^2}r2+h2​Substituting values we get :l=(6)2+(8)2=36+64=100=10m.l = \sqrt{(6)^2 + (8)^2} \[1em] = \sqrt{36 + 64} \[1em] = \sqrt{100} = 10 m.l=(6)2+(8)2​=36+64​=100​=10m.By formula,Curved surface area of cone = πrl= 3.14 × 6 m × 10 m= 188.4 m2.Area of tarpaulin used to make tent will be equal to curved surface area of tent.So,Area of the tarpaulin = Width of the tarpaulin × Length of the tarpaulin⇒ 188.4 = 3 × length of the tarpaulin⇒ Length of the tarpaulin = 188.43\dfrac{188.4}{3}3188.4​ = 62.8 mExtra length of the material = 20 cm = 20100\dfrac{20}{100}10020​ m = 0.2 mTotal length required = 62.8 m + 0.2 m = 63 m.Hence, the required length of the tarpaulin is 63 m.

Question 6

Question 6The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per 100 m2.

Accepted Answer

Given,Diameter of conical tomb (d) = 14 mRadius of conical tomb (r) = Diameter2=142\dfrac{	ext{Diameter}}{2} = \dfrac{14}{2}2Diameter​=214​ m = 7 m.Slant height of conical tomb (l) = 25 mBy formula,Curved surface area of conical tomb = πrl= 227\dfrac{22}{7}722​ × 7 × 25= 550 m2.Given,Cost of white-washing 100 m2 area = ₹ 210Cost of white-washing 550 m2 area = ₹ 210×550100210 	imes \dfrac{550}{100}210×100550​ = ₹ 1155.Hence, the cost of white-washing the conical tomb is ₹ 1155.

Question 7

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

Accepted Answer

Given,Radius of conical cap (r) = 7 cmHeight of conical cap (h) = 24 cmBy formula,Slant height of conical cap (l) = r2+h2\sqrt{r^2 + h^2}r2+h2​Substituting values we get :l=(7)2+(24)2=49+576=625=25 cm.l = \sqrt{(7)^2 + (24)^2} \[1em] = \sqrt{49 + 576} \[1em] = \sqrt{625} = 25 \text{ cm}.l=(7)2+(24)2​=49+576​=625​=25 cm.By formula,Curved surface area of one conical cap = πrl= 227\dfrac{22}{7}722​ × 7 × 25= 38507\dfrac{3850}{7}73850​= 550 cm2Curved surface area of 10 such conical caps = 10 × 550 cm2 = 5500 cm2.Hence, area of sheet required for making 10 caps = 5500 cm2.

Question 8

Question 8A 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 ₹ 12 per m2, what will be the cost of painting all these cones? (Use π = 3.14 and take 1.04\sqrt{1.04}1.04​ = 1.02)

Accepted Answer

Given,Diameter of cone (d) = 40 cm = 40100\dfrac{40}{100}10040​ m = 0.4 mRadius of cone (r) = Diameter2=0.42\dfrac{\text{Diameter}}{2} = \dfrac{0.4}{2}2Diameter​=20.4​ = 0.2 mHeight of cone (h) = 1 mSlant height of cone (l) = r2+h2\sqrt{r^2 + h^2}r2+h2​Substituting values we get :l=(0.2)2+(1)2=0.04+1=1.04=1.02 m.l = \sqrt{(0.2)^2 + (1)^2} \[1em] = \sqrt{0.04 + 1} \[1em] = \sqrt{1.04} = 1.02 \text{ m}.l=(0.2)2+(1)2​=0.04+1​=1.04​=1.02 m.By formula,Curved surface area of each cone = πrl= 3.14 × 0.2 m × 1.02 m= 0.64056 m2Curved surface area of 50 cones = 50 × 0.64056 m2 = 32.028 m2.Given,Cost of painting = ₹ 12 per m2.Cost of painting of 32.028 m2 area = ₹ (32.028 × 12)= ₹ 384.34 (approx.)Hence, the cost of painting 50 such hollow cones = ₹ 384.34 (approx.)

Question 9

Question 1(i)Find the surface area of a sphere of radius 10.5 cm.

Accepted Answer

Given,Radius of sphere (r) = 10.5 cmBy formula,Surface area of the sphere (A) = 4πr2Substituting values we get :A=4×227×(10.5)2=4×227×10.5×10.5=4×22×1.5×10.5=1386 cm2.A = 4 \times \dfrac{22}{7} \times (10.5)^2 \[1em] = 4 \times \dfrac{22}{7} \times 10.5 \times 10.5 \[1em] = 4 \times 22 \times 1.5 \times 10.5 \[1em] = 1386 \text{ cm}^2.A=4×722​×(10.5)2=4×722​×10.5×10.5=4×22×1.5×10.5=1386 cm2.Hence, surface area of the sphere = 1386 cm2.

Question 10

Question 1(ii)Find the surface area of a sphere of radius 5.6 cm.

Accepted Answer

Given,Radius of sphere (r) = 5.6 cmBy formula,Surface area of the sphere (A) = 4πr2Substituting values we get :A=4×227×(5.6)2=4×227×5.6×5.6=4×227×31.36=4×22×4.48=394.24 cm2A = 4 \times \dfrac{22}{7} \times (5.6)^2 \[1em] = 4 \times \dfrac{22}{7} \times 5.6 \times 5.6 \[1em] = 4 \times \dfrac{22}{7} \times 31.36 \[1em] = 4 \times 22 \times 4.48 \[1em] = 394.24 \text{ cm}^2A=4×722​×(5.6)2=4×722​×5.6×5.6=4×722​×31.36=4×22×4.48=394.24 cm2Hence, surface area of sphere = 394.24 cm2.

Question 11

Question 1(iii)Find the surface area of a sphere of radius 14 cm.

Accepted Answer

Given,Radius of sphere (r) = 14 cmBy formula,Surface area of the sphere (A) = 4πr2Substituting values we get :A=4×227×(14)2=4×227×14×14=4×227×196=4×22×28=2464 cm2.A = 4 \times \dfrac{22}{7} \times (14)^2 \[1em] = 4 \times \dfrac{22}{7} \times 14 \times 14 \[1em] = 4 \times \dfrac{22}{7} \times 196 \[1em] = 4 \times 22 \times 28 \[1em] = 2464 \text{ cm}^2.A=4×722​×(14)2=4×722​×14×14=4×722​×196=4×22×28=2464 cm2.Hence, surface area of sphere = 2464 cm2.

Question 12

Question 2(i)Find the surface area of a sphere of diameter 14 cm.

Accepted Answer

Given,Diameter of sphere (d) = 14 cmRadius of sphere (r) = Diameter2=142\dfrac{\text{Diameter}}{2} = \dfrac{14}{2}2Diameter​=214​ = 7 cm.By formula,Surface area of a sphere (A) = 4πr2Substituting values we get :A=4×227×(7)2=4×227×7×7=4×22×7=616 cm2.A = 4 \times \dfrac{22}{7} \times (7)^2 \[1em] = 4 \times \dfrac{22}{7} \times 7 \times 7 \[1em] = 4 \times 22 \times 7 \[1em] = 616 \text{ cm}^2.A=4×722​×(7)2=4×722​×7×7=4×22×7=616 cm2.Hence, the surface area of a sphere = 616 cm2.

Question 13

Question 2(ii)Find the surface area of a sphere of diameter 21 cm.

Accepted Answer

Given,Diameter of sphere (d) = 21 cmRadius of sphere (r) = Diameter2=212\dfrac{\text{Diameter}}{2} = \dfrac{21}{2}2Diameter​=221​ = 10.5 cmBy formula,Surface area of a sphere (A) = 4πr2A=4×227×(10.5)2=4×227×10.5×10.5=4×22×1.5×10.5=1386 cm2.A = 4 \times \dfrac{22}{7} \times (10.5)^2 \[1em] = 4 \times \dfrac{22}{7} \times 10.5 \times 10.5 \[1em] = 4 \times 22 \times 1.5 \times 10.5 \[1em] = 1386 \text{ cm}^2.A=4×722​×(10.5)2=4×722​×10.5×10.5=4×22×1.5×10.5=1386 cm2.Hence, the surface area of a sphere = 1386 cm2.

Question 14

Question 2(iii)Find the surface area of a sphere of diameter 3.5 m.

Accepted Answer

Given,Diameter of sphere (d) = 3.5 mRadius of sphere (r) = Diameter2=3.52\dfrac{\text{Diameter}}{2} = \dfrac{3.5}{2}2Diameter​=23.5​ = 1.75 mBy formula,Surface area of a sphere (A) = 4πr2Substituting values we get :A=4×227×(1.75)2=4×227×1.75×1.75=4×22×0.25×1.75=38.5 m2.A = 4 \times \dfrac{22}{7} \times (1.75)^2 \[1em] = 4 \times \dfrac{22}{7} \times 1.75 \times 1.75 \[1em] = 4 \times 22 \times 0.25 \times 1.75 \[1em] = 38.5 \text{ m}^2.A=4×722​×(1.75)2=4×722​×1.75×1.75=4×22×0.25×1.75=38.5 m2.Hence, the surface area of a sphere = 38.5 m2.

Question 15

Question 3Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Accepted Answer

Given,Radius of the hemisphere (r) = 10 cmBy formula,Total Surface area of hemisphere (A) = 3πr2Substituting values we get :A=3×3.14×(10)2=3×3.14×100=942 cm2.A = 3 \times 3.14 \times (10)^2 \[1em] = 3 \times 3.14 \times 100 \[1em] = 942 \text{ cm}^2.A=3×3.14×(10)2=3×3.14×100=942 cm2.Hence, the total surface area of the hemisphere is 942 cm2.

Question 16

Question 4The 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.

Accepted Answer

Given,Radius of the balloon before pumping air (r1) = 7 cmRadius of the balloon after pumping air (r2) = 14 cmInitial surface area = 4πr12Surface area after pumping air into ballon = 4πr22Ratio = Initial surface areaSurface area after pumping air into ballon\dfrac{\text{Initial surface area}}{\text{Surface area after pumping air into ballon}}Surface area after pumping air into ballonInitial surface area​=4πr124πr22=r12r22=(r1r2)2=(714)2=(12)2=14=1:4.= \dfrac{4πr_1^2}{4πr_2^2} \[1em] = \dfrac{r_1^2}{r_2^2} \[1em] = \Big(\dfrac{r_1}{r_2}\Big)^2 \[1em] = \Big(\dfrac{7}{14}\Big)^2 \[1em] = \Big(\dfrac{1}{2}\Big)^2 \[1em] = \dfrac{1}{4} \[1em] = 1 : 4.=4πr22​4πr12​​=r22​r12​​=(r2​r1​​)2=(147​)2=(21​)2=41​=1:4.Hence, the ratio of the surface area of the balloons = 1 : 4.

Question 17

Question 5A 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 ₹ 16 per 100 cm2.

Accepted Answer

Given,Inner diameter of hemispherical bowl (d) = 10.5 cmInner radius of hemispherical bowl (r) = Diameter2=10.52\dfrac{\text{Diameter}}{2} = \dfrac{10.5}{2}2Diameter​=210.5​ = 5.25 cmCurved surface area of hemispherical bowl = 2πr2=2×227×(5.25)2=2×227×27.5625=2×22×3.9375=173.25 cm2.= 2 \times \dfrac{22}{7} \times (5.25)^2 \[1em] = 2 \times \dfrac{22}{7} \times 27.5625 \[1em] = 2 \times 22 \times 3.9375 \[1em] = 173.25 \text{ cm}^2.=2×722​×(5.25)2=2×722​×27.5625=2×22×3.9375=173.25 cm2.Given,The cost of tin-plating 100 cm2 of the bowl = ₹ 16∴ The cost of tin-plating 1 cm2 of the bowl = ₹ 16100\dfrac{16}{100}10016​∴ The cost of tin-plating 173.25 cm2 area of the bowl = ₹ 16100×173.25\dfrac{16}{100} \times 173.2510016​×173.25 = ₹ 27.72Hence, the cost of tin-plating the hemispherical bowl is ₹ 27.72.

Question 18

Question 6Find the radius of a sphere whose surface area is 154 cm2.

Accepted Answer

Given,The surface area of the sphere = 154 cm2∴4πr2=154⇒r2=1544π⇒r2=1544×227⇒r2=154×74×22⇒r2=494⇒r=494⇒r=72=3.5 cm\therefore 4πr^2 = 154 \[1em] \Rightarrow r^2 = \dfrac{154}{4π} \[1em] \Rightarrow r^2 = \dfrac{154}{4 \times \dfrac{22}{7}} \[1em] \Rightarrow r^2 = \dfrac{154 \times 7}{4 \times 22} \[1em] \Rightarrow r^2 = \dfrac{49}{4} \[1em] \Rightarrow r = \sqrt{\dfrac{49}{4}} \[1em] \Rightarrow r = \dfrac{7}{2} = 3.5 \text{ cm}∴4πr2=154⇒r2=4π154​⇒r2=4×722​154​⇒r2=4×22154×7​⇒r2=449​⇒r=449​​⇒r=27​=3.5 cmHence, the radius of the sphere = 3.5 cm.

Question 19

Question 7The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Accepted Answer

Given,⇒ Diameter of the moon (d) = 14\dfrac{1}{4}41​ × diameter of the earth (D)⇒ d2=14×D\dfrac{d}{2} = \dfrac{1}{4} \times D2d​=41​×D⇒ Radius of the moon (r) = 14\dfrac{1}{4}41​ × radius of the earth (R)⇒ r = 14\dfrac{1}{4}41​ × R⇒ rR=14\dfrac{r}{R} = \dfrac{1}{4}Rr​=41​ .......(1)Now,Surface area of earth = 4πR2Surface area of moon = 4πr2The ratio of their surface areas = Surface area of moonSurface area of earth\dfrac{\text{Surface area of moon}}{\text{Surface area of earth}}Surface area of earthSurface area of moon​=4πr24πR2=r2R2=(rR)2=(14)2=116.= \dfrac{4πr^2}{4πR^2} \[1em] = \dfrac{r^2}{R^2} \[1em] = \Big(\dfrac{r}{R}\Big)^2 \[1em] = \Big(\dfrac{1}{4}\Big)^2 \[1em] = \dfrac{1}{16}.=4πR24πr2​=R2r2​=(Rr​)2=(41​)2=161​.Hence, the ratio of their surface area = 1 : 16.

Question 20

Question 8A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Accepted Answer

Given,The inner radius of the bowl (r) = 5 cmThickness of steel = 0.25 cmOuter radius of the bowl (R) = inner radius of the bowl + thickness of steel = 5 cm + 0.25 cm = 5.25 cmOuter curved surface area of the hemisphere = 2πR2= 2 × 227\dfrac{22}{7}722​ × (5.25)2= 447×27.5625\dfrac{44}{7} 	imes 27.5625744​×27.5625= 44 × 3.9375= 173.25 cm2.Hence, the outer curved surface area of the bowl is 173.25 cm2.

Surface Areas and Volumes

Exercise 111

Exercise 112

Exercise 113

Exercise 114