NCERT Solutions Maths Class 10 Exercise 13.1

NCERT Solutions Maths Class 10 Exercise 13.1

 

Maths Class 10 Exercise 13.1

 

1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

 

Solution: According to question, volume of the cube = 64 cm3

Volume of a cube = a3

64 = a3

(a)3 = 43

 Side = 4 cm

For the resulting cuboid, length (l) = 4 + 4 = 8 cm, breadth (b) = 4 cm and height (h) = 4 cm.

Surface area of the resulting cuboid = 2(lb + bh + hl)

                                                                 = 2(8 × 4 + 4 × 4 + 4 × 8)

                                                                 = 2(32 + 16 + 32)

                                                                 = 2 × 80

                                                                 = 160 cm2

 

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.

 

Solution: Given, the diameter of the hollow hemisphere = 14 cm

Therefore, the radius of the hollow hemisphere = 14/2 = 7 cm

Total height of the vessel = 13 cm

Therefore, the height of the hollow cylinder (h) = 13 – 7 = 6 cm

So, the inner surface area of the vessel = Inner surface area of the hollow hemisphere + Inner curved surface area of the hollow cylinder

= 2π(7)2 + 2π(7)(6)

= 98π + 84π

= 182π

= 182 × 22/7  

= 26 × 22

= 572 cm2

 

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.

 

Solution: Here, the radius of the cone = 3.5 cm

Therefore, the radius of the hemisphere = 3.5 cm

Total height of the toy = 15.5 cm

So, the height of the cone = 15.5 – 3.5 = 12 cm

Slant height of the cone (l) = 

=

=   

= 12.5 cm

Now, the total surface area of the toy = Curved surface area of the hemisphere + Curved surface area of the cone

= 2πr2rl

= 2π(3.5)2 + π(3.5)(12.5)

= 24.5π + 43.75π  

= 68.25π

= 68.25 × 22/7 

= 214.5 cm2

 

4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Solution: The greatest diameter of the hemisphere = Edge of the cubical block = 7 cm

Now, the total surface area of the solid = External surface area of the cubical block + Curved surface area of the hemisphere

= [6(7)2 – π(7/2)2] + 2π(7/2)2 

= 294 – 49π/4 + 49π/2 

= 294 + 49π/4  

= 294 + 49/4 × 22/7 

= 294 + 77/2  

= 294 + 38.5

= 332.5 cm2

 

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

SolutionGiven, the diameter of the hemisphere = l

Therefore, the radius of the hemisphere = l/2 

Also, the length of the edge of the cube = l

Now, the total surface area of the remaining solid = External surface area of the cubical wooden block + Inner curved surface area of the hemisphere

= 6l2 – π(l/2)2 + 2π(l/2)2   

= π(l/2)2 + 6l2  

= πl2/4 + 6l2

= ¼ l2(π + 24)  

 

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.

Solution: Given, the diameter of the capsule = 5 mm

Then, radius of the hemisphere and cylinder (r) = 2.5 mm

Cylindrical height (h) = Total height – 2 × Radius of hemisphere = 14 – 2 × 2.5 = 9 mm

Surface area of the capsule = Curved surface area of the cylinder + 2 × Surface area of the hemisphere

= 2πrh + 2(2πr2

= 2π(2.5)(9) + 2[2π(2.5)2

= 45π + 25π 

= 70π

= 70 × 22/7  

220 mm2

 

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 be covered with canvas.)

 

Solution: Given, the diameter of the cylindrical part = 4 m

Therefore, the radius of the cylindrical part (r) = 2 m

Height of the cylindrical part (h) = 2.1 m

Slant height of the conical part (l) = 2.8 m

Total surface area of the tent = Curved surface area of the cylindrical part + Curved surface area of the conical part

= 2πrh + πrl

= 2π(2)(2.1) + π(2)(2.8)

= 8.4π + 5.6 π

= 14π

= 14 × 22/7

= 44 m2

Therefore, the cost of the canvas of the tent at the rate of Rs. 500 per m2

= 44 × 500 

= Rs. 22000

 

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.

 

Solution: Given, the diameter of the solid cylinder = 1.4 cm

Therefore, the radius of the solid cylinder (r) = 0.7 cm

Now, the radius of the base of the conical cavity = 0.7 cm

Height of the solid cylinder (h) = 2.4 cm

Therefore, the height of the conical cavity = 2.4 cm

Therefore, the slant height of the conical cavity = 

= 

= √6.25 
= 2.5 cm

Then, the total surface area of the remaining solid = 2π(0.7)(2.4) + π(0.7)2 + π(0.7)(2.5)

= 3.36π + 0.49π + 1.75π

= 5.6π

= 5.6 × 22/7  

= 17.6 cm2

18 cm2     (to the nearest cm2)

 

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 cm, find the total surface area of the article.

Solution: Given, the height of the solid cylinder (h) = 10 cm and the base radius of the cylinder (r) = 3.5 cm.

Therefore, the radius of the hemisphere = 3.5 cm

Now, the total surface area of the article = 2πrh + 2(2πr2)

= 2π(3.5)(10) + 2[2π(3.5)2] 

= 70π + 49π 

119π

= 119 × 22/7 

= 374 cm2

 

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