CBSE Class 10 Mathematics Surface Area and Volume Assignment Set 15

NCERT Exercise

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

Volume of each cube \( a^3 = 64 \Rightarrow a = 4 \) cm. Joining two cubes end to end gives a cuboid of length \( 2 \times 4 = 8 \) cm, breadth 4 cm and height 4 cm.
Surface area \( = 2(lb+bh+hl) = 2(8 \times 4 + 4 \times 4 + 4 \times 8) = 2 \times 80 = 160 \) cm².
Hence, the surface area of the resulting cuboid is 160 cm².
In simple words: The joined faces disappear, so use the cuboid formula directly.

Teacher's Note:
1. Find the edge of the cube from its volume first.
2. The two cubes make an 8 cm by 4 cm by 4 cm cuboid.

Question 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.
Answer:

Radius of the hemisphere and cylinder \( r = 7 \) cm. Height of the cylindrical part \( h = 13 - 7 = 6 \) cm.
Inner surface area \( = \) C.S.A. of cylinder \( + \) C.S.A. of hemisphere \( = 2\pi rh + 2\pi r^2 = 2\pi r(h+r) = 2 \times \frac{22}{7} \times 7 \times 13 = 572 \) cm².
Hence, the inner surface area of the vessel is 572 cm².
In simple words: Add the curved surfaces of the cylinder and the hemisphere.

Teacher's Note:
1. The height of the hemispherical part equals its radius.
2. The cylinder part is hollow and open at the top.

Question 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.
Answer:

Radius \( r = 3.5 \) cm. Height of the cone \( h = 15.5 - 3.5 = 12 \) cm. Slant height \( l = \sqrt{12^2 + 3.5^2} = \sqrt{156.25} = 12.5 \) cm.
C.S.A. of the cone \( = \pi rl = \frac{22}{7} \times 3.5 \times 12.5 = 137.5 \) cm². C.S.A. of the hemisphere \( = 2\pi r^2 = 2 \times \frac{22}{7} \times 3.5^2 = 77 \) cm².
Total surface area \( = 137.5 + 77 = 214.5 \) cm².
Hence, the total surface area of the toy is 214.5 cm².
In simple words: Only the curved surfaces of the cone and hemisphere are outside the toy.

Teacher's Note:
1. The base of the cone is joined to the hemisphere, so it is not counted.
2. Use l squared = h squared + r squared.

Question 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.
Answer:

The hemisphere must rest on the top face, so its greatest diameter equals the edge of the cube, 7 cm. Then \( r = 3.5 \) cm.
Total surface area \( = \) surface area of the cube \( - \) area of the base of the hemisphere \( + \) C.S.A. of the hemisphere \( = 6l^2 - \pi r^2 + 2\pi r^2 = 6l^2 + \pi r^2 \).
\( = 6 \times 49 + \frac{22}{7} \times 3.5 \times 3.5 = 294 + 38.5 = 332.5 \) cm².
Hence, the greatest diameter is 7 cm and the surface area is 332.5 cm².
In simple words: The hemisphere covers a circle on the top face but adds its own curved surface.

Teacher's Note:
1. The hemisphere cannot overhang the cube's edge.
2. Subtract the covered circle once, add the curved surface once.

Question 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.
Answer:
Edge of the cube \( = l \), so the radius of the hemisphere \( r = \frac{l}{2} \).
Surface area of the remaining solid \( = \) surface area of the cube \( - \) area of the circular opening \( + \) C.S.A. of the hemisphere \( = 6l^2 - \pi r^2 + 2\pi r^2 = 6l^2 + \pi r^2 \).
\( = 6l^2 + \pi\frac{l^2}{4} = \frac{l^2}{4}(24 + \pi) \) square units.
Hence, the surface area is \( \frac{l^2}{4}(24+\pi) \) square units.
In simple words: The depression removes a circle of the face but adds a hemispherical surface.

Teacher's Note:
1. Write the radius in terms of l first.
2. The inside of the hollow counts as surface.

Question 6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Answer:

Radius \( r = \frac{5}{2} \) mm. Length of the cylindrical part \( h = 14 - 2 \times \frac{5}{2} = 9 \) mm.
Surface area \( = 2 \times \) C.S.A. of a hemisphere \( + \) C.S.A. of the cylinder \( = 2 \times 2\pi r^2 + 2\pi rh = 4\pi\left(\frac{5}{2}\right)^2 + 2\pi\left(\frac{5}{2}\right)(9) = 25\pi + 45\pi = 70\pi \) mm².
\( = 70 \times \frac{22}{7} = 220 \) mm².
Hence, the surface area of the capsule is 220 mm².
In simple words: The two hemispheres together form a sphere's surface.

Teacher's Note:
1. Subtract two radii from the length to get the cylinder's length.
2. Keep all lengths in mm.

Question 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.)
Answer:

Radius \( r = 2 \) m, height of the cylinder \( h = 2.1 \) m, slant height of the cone \( l = 2.8 \) m.
Area of canvas \( = \) C.S.A. of the cone \( + \) C.S.A. of the cylinder \( = \pi rl + 2\pi rh = \pi r(l + 2h) = \frac{22}{7} \times 2 \times (2.8 + 4.2) = 44 \) m².
Cost \( = 44 \times 500 = \) Rs. 22000.
Hence, the canvas needed is 44 m² and its cost is Rs. 22000.
In simple words: Only curved surfaces are covered with canvas.

Teacher's Note:
1. The base is not covered, so do not add pi r squared.
2. Cost = area times rate.

Question 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.
Answer:

Radius \( r = 0.7 \) cm, height \( h = 2.4 \) cm. Slant height of the cone \( l = \sqrt{0.7^2 + 2.4^2} = \sqrt{6.25} = 2.5 \) cm.
Total surface area of the remaining solid \( = \) C.S.A. of cylinder \( + \) C.S.A. of cone \( + \) area of the lower base \( = 2\pi rh + \pi rl + \pi r^2 \).
\( = \frac{22}{7} \times 0.7 \times (4.8 + 2.5 + 0.7) = 2.2 \times 8 = 17.6 \) cm².
Hence, the total surface area of the remaining solid is 17.6 cm² \( \approx \) 18 cm².
In simple words: The cone's base takes away the whole upper face, leaving the cone's curved surface inside.

Teacher's Note:
1. The upper circular face is completely removed by the cavity.
2. Round to the nearest cm² at the end.

Question 9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. 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.
Answer:

Radius \( r = 3.5 \) cm, height \( h = 10 \) cm. The two flat ends are completely removed and replaced by hemispherical hollows.
Total surface area \( = \) C.S.A. of cylinder \( + 2 \times \) C.S.A. of hemisphere \( = 2\pi rh + 2 \times 2\pi r^2 = 2\pi r(h + 2r) \).
\( = 2 \times \frac{22}{7} \times 3.5 \times (10 + 7) = 374 \) cm².
Hence, the total surface area of the article is 374 cm².
In simple words: Each hollow hemisphere adds its curved surface while the flat ends vanish.

Teacher's Note:
1. The scooped hemispheres have the same radius as the cylinder.
2. Do not count the flat circular ends.