HBSE 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1

Haryana State Board HBSE 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1

Unless stated otherwise, take π = \(\frac{22}{7}\).

Question 1.
2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Let a be length of an edge of the cube,
Volume of cube = 64 cm³ (given)
⇒ a³ = 64
⇒ a = \(\sqrt[3]{64}\)
⇒ a = 4 cm

Now, dimensions of the resulting cuboid made by joining two cubes are 8 cm × 4 cm × 4 cm.
∴ Length of cuboid = 8 cm
Breadth of cuboid = 4 cm
Height of cuboid = 4 cm
Surface area of the cuboid= 2 (lb + bh + hl)
= 2(8 × 4 + 4 × 4 + 4 × 8)
= 2 (32 + 16 + 32)
= 2 × 80 = 160 cm²
Hence, surface area of the cuboid = 160 cm².

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 total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:
We have,
Diameter of the hemispherical portion = 14 cm
Radius of the hemispherical portion = \(\frac{14}{2}\) = 7 cm

Total height of the vessel = 13 cm
Height of the cylindrical portion = 13 – 7 = 6 cm
Total inner surface area of the vessel
= C.S.A. of hemisphrical portion + C.S.A. of cylindrical portion
= 2πr² + 2πrh
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 7 (7 + 6)
= 44 × 13 = 572 cm²
Hence, total inner surface area of the vessel = 572 cm²

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.
Solution:
We have,
Radius (r) = 3.5 cm
Total height of the toy = 15.5 cm
height of the conical part (h) = 15.5 – 3.5 = 12 cm

Slant height of the conical part (l) = \(\sqrt{h^2+r^2}\)
= \(\sqrt{12^2+(3 \cdot 5)^2}\)
= \(\sqrt{(12)^2+(3 \cdot 5)^2}\)
= \(\sqrt{144+12 \cdot 25}\)
= \(\sqrt { 1 5 6 \cdot 2 5 }\) = 12.5 cm
Total surface area of the toy = 2πr² + πrl = πr (2r + l)
= \(\frac{22}{7}\) × 3.5 (2 × 3.5 + 12.5)
= 11 × 19.5 = 214.5 cm².
Hence, total surface area of the toy = 214.5 cm².

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.
Solution:
We have,
Edge of cubical block = 7 cm
Greatest diameter of the hemisphere = edge of cube = 7 cm

Surface area of the solid = The surface area of the cube + C.S.A. of the hemisphere – base area of the hemisphere
= 6 × (edge)² + 2πr² – πr²
= 6 × (7)² + πr²
= 6 × 49 + \(\frac{22}{7}\) × (3.5)²
= 294 + 38.5 = 332.5 cm²
Hence, surface area of the solid = 332.5 cm².

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.
Solution:
We have,
Diameter of the hemisphere = l
Edge of the cube = diameter of the hemsphere = l
radius of the hemisphere = \(\frac{l}{2}\)

Surface Area of the remaining solid = surface area of the cube + Inner C.S.A. of the hemisphere – Area of the top of the hemisphere
= 6l² + 2πr x (\(\frac{l}{2}\))² – π (\(\frac{l}{2}\))²
= 6l² + π × (\(\frac{l}{2}\))²
= 6l² + \(\frac{\pi l^2}{4}\)
= \(\frac{1}{4}\) (24l² + πl²)
= \( \frac{l^2}{4}\) (24 + π)
Hence, surface area of the remaining solid = \(\frac{1}{4}\) l² (24 + π).

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

Solution:
We have,
Radius of the cylindrical part and hemispherical part (r) = \(\frac{5}{2}\) = 2.5 mm
Length of the cylindrical part (h) = 14 – (2.5 + 2.5)
= 14 – 5 = 9 mm
Surface area of the capsule = C.S.A. of the cylindrical part + C.S.A. of two hemispherical parts
= 2πrh + 2 × 2πr²
= 2πr (h + 2r)
= 2 × \(\frac{22}{7}\) × 2.5 (9 + 2 × 2.5)
= \(\frac{22}{7}\) × 5 × 14
= 220 mm².
Hence, surface area of the capsule = 220 mm².

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 ₹ 500 per m². (Note that the base of the tent will not be covered with canvas.)
Solution.
We have,
Radius of conical portion radius (r) = \(\frac{4}{2}\) = 2 m
Slant height (l) = 2.8 m
For cylindrical portion Radius (r) = 2 m
Height (h) = 2.1 m
Area of canvas used for making the tent = Surface area of tent = C.S.A. of conical part + C.S.A. of cylindrical part
= πrl + 2πrh
= πr(l + 2h)

= \(\frac{22}{7}\) × 2(2.8 + 2 × 2.1)
= \(\frac{44}{7}\) × 7 = 44 m².
Cost of canvas at the rate of ₹ 500/m² = 44 × 500 = ₹ 22000
Hence, area of the canvas = 44 m²
and cost of canvas = Rs 22000.

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.
Solution.
We have,
Height of the cylinder = 2.4 cm
Diameter of the cylinder = 1.4 cm
∴ Radius of the cylinder = \(\frac{1.4}{2}\) = 0.7 cm

Slant height of the cone (l) = \(\sqrt{h^2+r^2}\)
= \(\sqrt{(2 \cdot 4)^2+(0 \cdot 7)^2}\) = 2.5 cm
Total surface area of the remaining solid = C.S.A. of the cylinder + Area of base of the cylinder + C.S.A. of the cone
= 2πrh + πr² + πrl
= πr (2h + r + l)
= \(\frac{22}{7}\) × 0.7 (2 × 2.4 + 0.7 + 2.5)
= 2.2 (4.8 + 3.2)
= 2.2 × 8 = 17.6 cm²
= 18 cm², (apporx)
Hence, the total surface area of the remaining solid to the nearest cm² = 18 cm².

Question 9.
A wooden article was made by scooping out 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.
We have,
Height of the cylinder (A) = 10 cm
Radius of cylinder and hemisphere (r) = 3.5 cm

Total surface area of the article = C.S.A of the cylinder + 2 (surface area of a hemisphere)
= 2πrh + 2 × 2πr²
= 2πr (h + 2r)
= 2 × \(\frac{22}{7}\) × 3.5 (10 + 2 × 3.5)
= 22 × 17
= 374 cm²
Hence, total surface area of the article = 374 cm².