10th Maths 13.1

Exercise 13.1

Q1.

$2$ cubes each of volume

$text{64 cm}^3$ are joined end to end. Find the surface area of the resulting cuboid.

$text{Sol. :}$

$text{Given :}$

$text{Volume of cube 64 cm}^3$

$text{To find :}$

$text{The Surface area of the resulting cuboid.}$

$text{Solve :}$

$text{Volume of cube = (side)}^3$

$64 = text{(side)}^3$

$(4)^3 = text{(side)}^3$

$text{side = 4 cm}$

After join

$2$ cubes the resulting cuboid

$text{Height = 4cm, Length = 8 cm, Breadth = 4 cm}$

$text{Surface Area of Cuboid = 2(LB + BH + HL)}$

$text{S A of Cuboid = 2}(8 times 4 + 4 times 4 + 4 times 8)$

$text{S A of Cuboid = 2}(32 + 16 + 32)$

$text{S A of Cuboid = 2}(80)$

$text{S A of Cuboid = 160 cm}^2$

$text{Answer:}$

$text{The Surface Area of resulting Cuboid is 160 cm}^2.$

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Q2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is

$text{14 cm}$ and the total height of the vessel is

$text{13 cm}$ . Find the inner surface area of the vessel.

$text{Sol. :}$

$text{Given :}$

$text{The radius of the vessel is }14/2 = 7 text{cm}$

$text{The height of the Cylinder is 6 cm}$

$text{To find :}$

$text{inner S A of the vessel}$

$text{Solve :}$

$text{inner S A of the vessel}$

$text{= SA of hemisphere + SA of Cylinder}$

$= 2pir^2 + 2pirh$

$= 2pir(r + h)$

$= 2 times 22/7 times 7(7 + 6)$

$= 44(13)$

$= 572 text{cm}^3$

$text{Answer :}$

$text{inner S A of the vessel 572 cm}^3$ .

Q3. A toy is in the form of a cone of radius

$text{3.5 cm}$ mounted on a hemisphere of same radius. The total height of the toy is

$text{15.5 cm}$ . Find the total surface area of the toy.

$text{Sol. :}$

$text{Given :}$

$text{Radius of toy is 3.5 cm}$

$text{height of toy is 15.5 cm}$

$text{To find :}$

$text{The total Surface Area of the toy}$

$text{Solve :}$

$text{Slant height of cone }l = sqrt((h)^2+(r)^2)$

$=sqrt((12)^2+(3.5)^2)$

$=sqrt((144 + 12.25)$

$=sqrt(156.25)$

$text{= 12.5 cm}$

$text{Total SA of the toy}$

$text{= SA of Cone + SA of hemisphere}$

$= pirl + 2pir^2$

$= pir(l + 2r)$

$= 22/7 times 3.5(12.5 + 2times3.5)$

$= 11(12.5 + 7)$

$= 11 times 19.5$

$= 214.5 cm^2$

$text{Answer :}$

$text{Total SA of the toy is 214.5 cm}^2$ .

Q4. A cubical block of side

$text{7 cm}$ is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

$text{Sol. :}$

$text{Given :}$

$text{Side of Cubical block = 7 cm}$

$text{To find :}$

$text{The Surface Area of the solid}$

$text{Solve :}$

$text{Diameter of sphere d = 7cm}$

$text{Radius of sphere r = }7/2 text{ cm}$

$text{Total S A of solid}$

$text{= T S A of the Cube + C S A of the hemisphere}$

$text{ - base area of hemisphere}$

$= 6a^2 + 2pir^2 - pir^2$

$= 6a^2 + pir^2$

$= 6 times (7)^2 + 22/7 times(7/2)^2$

$= 6 times 49 + 22/7 times 7/2 times 7/2$

$= 294 + 11 times 7/2$

$= 294 + 38.5$

$= 332.5 text{cm}^2$

$text{Answer :}$

$text{ T S A of solid is 332.5 cm}^2.$

Q5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter $text{l}$ of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

$text{Sol. :}$

$text{Given :}$

$text{Let the Side of the Cube } a = l$

$text{Let the diameter of the hemispherical bore } d = l$

$text{Radius of the hemisphere bore } r = l/2$

$text{To find :}$

$text{Surface Area of the remaining solid}$

$text{Solve :}$

$text{Surface Area of the remaining solid}$

$text{= Total S A of cube + C S A of the hemisphere}$

$text{ - Area of the circular base}$

$= 6l^2 + 2pir^2 - pir^2$

$= 6l^2 + pir^2$

$= 6l^2 + pi(l/2)^2$

$= (24l^2 + pil^2)/4$

$= l^2/4(24 + pi)$

$text{Answer}$

$text{Surface Area of remaining solid }l^2/4(24 + pi).$

Q6. 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

$text{14 mm}$ and the diameter of the capsule is

$text{5 mm}$ . Find its surface area.

$text{Sol. :}$

$text{Given :}$

$text{Length of the entire capsule is 14 mm}$

$text{Diamter of hemisphere & cylinder }d = 5 text{mm}$

$text{Radius of hemisphere & cylinder }r = 5/2 text{mm}$

$text{To find :}$

$text{Surface Area of the capsule}$ .

$text{Solve :}$

$text{Length of the capsule = length of the cylinder }(l)$

$text{ + 2}times text{radius of the hemisphere}$

$14 = l + 5$

$l = 14 - 5$

$l = 9$

$text{S A of capsule = C S A Cylinder }$

$text{ + 2 (C S A of hemisphere)}$

$text{S A of capsule} = 2pirl + 2 times 2pir^2$

$text{S A of capsule} = 2pir(l + 2r)$

$text{S A of capsule} = 2times22/7times5/2(9 + 2times5/2)$

$text{S A of capsule} = 110/7(14)$

$text{S A of capsule} = 220 text{ mm}^2$

$text{Answer :}$

$text{S A of capsule is 220 mm}^2$ .

Q7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are

$text{2.1 m}$ and

$text{4 m}$ respectively, and the slant height of the top is

$text{2.8 m,}$ 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

$text{500 per m}^2$ . ( Note that the base of the tent will not be covered with canvas.)

$text{Sol. :}$

$text{Given :}$

$text{Height of Cylinder part of tent }h = 2.1text{ m}$

$text{Diameter of Cylinder & Cone part }d = 4text{ m}$

$text{Radius of Cylinder & Cone part }r = 2text{ m}$

$text{Slant height of Cone part }l = 2.8text{ m}$

$text{To find :}$

$text{Find the area of the canvas used for making a tent.}$

$text{Find the cost of the canvas of the tent}$

$text{Solve :}$

$text{Area of canvas = C S A of Cylinder}$

$text{ + C S A of Cone}$

$text{Area of canvas} = 2pirh + pirl$

$text{Area of canvas} = pir(2h + l)$

$text{Area of canvas} = 22/7times2(2times2.1 + 2.8)$

$text{Area of canvas} = 44/7times(4.2 + 2.8)$

$text{Area of canvas} = 44/7times 7.0$

$text{Area of canvas} = 44.0 text{m}^2$

$text{cost of the canvas = Area}times text{Rate}$

$text{ = 44}times text{₹ 500}$

$text{ = ₹ 22000}$

$text{Answer :}$

$text{Area of canvas 44 m}^2, text{and cost ₹ 22000}$

Q8. From a solid cylinder whose height is

$text{2.4 cm}$ and diameter

$text{1.4 cm,}$ 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

$text{cm}^2$ .

$text{Sol. :}$

$text{Given:}$

$text{Height of Cylinder }h = 2.4 text{ cm}$

$text{Diameter }d = 1.4 text{ cm}$

$text{Radius } r = 1.4/2 = 0.7 text{cm}$

$text{To find :}$

$text{The total Surface Area of the remaining solid}$

$text{Solve :}$

$text{slant height } l = sqrt((h)^2 + (r)^2)$

$l = sqrt((2.4)^2 + (0.7)^2)$

$l = sqrt(5.76 + 0.49)$

$l = sqrt(6.25)$

$l = 2.5 text{cm}$

$text{T S A of remaining solid = S A of Cylinder}$

$text{ + S A of cone + area of base}$

$text{ = } 2pirh + pirl + pir^2$

$text{ = } pir(2h + l + r)$

$text{ = } 22/7times0.7(2times2.4 + 2.5 + 0.7)$

$text{ = } 2.2(4.8 + 3.2)$

$text{ = } 2.2(8)$

$text{ = } 17.6 text{ cm}^2$

$text{ = } 18 text{ cm}^2$

$text{Answer :}$

$text{T S A of remaining solid 18 cm}^2$ .

Q9. 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

$text{10 cm}$ , and its base is of radius

$text{3.5 cm,}$ find the total surface area of the article.

$text{Sol. :}$

$text{Given :}$

$text{Height of the Cylinder } h = 10 text{ cm}$

$text{Radius of the base} r = 3.5 text{ cm}$

$text{To find :}$

$text{The total surface area of the artical}$

$text{Solve :}$

$text{T S A of the article = C S A of Cylinder}$

$text{ + 2(C S A of hemisphere)}$

$text{ = }2pirh + 2times2pir^2$

$text{ = }2pir(h + 2r)$

$text{ = }2times22/7times3.5(10 + 2times3.5)$

$text{ = }22 times 17$

$text{ = }374 text{cm}^2$

$text{Answer :}$

$text{T S A of the article is 374 cm}^2$ .

Comments

Anonymous said…

Very good

19 December 2021 at 22:22

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