# KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.1.

## Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.1

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

Question 1.
2 cubes each of volume 64cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution: Let the length of each edge of the cube of volume 64 cm3 be ‘a’ cm
Volume = 64 cm3
a3 = 64
a = $$\sqrt{64}$$
a = 4 cm
The dimensions of cuboid so formed are l = length = (4 + 4) cm = 8 cm
b = Breadth = 4cm and h = Height = 4 cm
∴ Surface area of the cuboid
= 2(lb + bh + lh)
= 2(8 × 4 + 4 × 4 + 8 × 4)
= 2(32 + 16 + 32)
= 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 the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:
For hemispherical part,
radius (r)= $$\frac{14}{2}$$ = 7cm
∴ Curved surface area = 2πr2
= 2 × $$\frac{22}{7}$$ × 7 × 7cm2
= 308cm2 7
Total height of vessel = 13 cm ∴ Height of cylinder = (13 – 7)cm = 6 cm and radius(r) = 7 cm
∴ Curved surface area of cylinder = 2πrh
= 2 × $$\frac{22}{7}$$ × 7 × 6cm2 = 264cm2 7
∴ Inner surface area of vessel = (308 + 264)cm2 = 572 cm2 Question 3.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
(i) Radius of the base of the circular toy,
r = 3.5 cm.
∴ Surface Area of the hemisphere, = 2πr2  = 77 sq.cm
(ii) Let slant height of cone is l cm. then ∴ Curved surface area of a cone = πrl  = 137.5 sq.cm.
∴ Total surface area of a toy = Surface area of hemisphere + Curved surface area of a cone
= 77.0 + 137.5
= 214.5 sq.cm.

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:
Let side of the block, l = 7 cm ∴ The greatest diameter of the hemisphere = 7 cm
Surface area of the solid = [Total surface area of the cubical block] + [C.S.A. of the hemisphere] – [Base area of the hemisphere]
= 6 × l2 + 2πr2 – πr2
[where l = 7 cm and r = $$\frac{7}{2}$$ cm]  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 cube. Determine the surface area of the remaining solid.
Solution:
Let, Length of each edge of cubical block = Diameter of hemisphere = l
∴ Radius of hemisphere, $$r=\frac{l}{2} unit$$. The total surface area of newly formed cube = Curved surface area of the cube – Upper part of Cube + Area of the hemisphere which is depressed Question 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: Curved surface area of one hemispherical part = 2πr2
∴ Surface area of both hemispherical parts
= 2(2πr2) = 4πr2 = [4 × $$\frac{22}{7} \times\left(\frac{25}{10}\right)^{2}$$] mm2
= $$\left(4 \times \frac{22}{7} \times \frac{25}{10} \times \frac{25}{10}\right)$$ mm2
Entire length of capsule = 14 mm
∴ Length of cylindrical part = [Length of capsule – Radius of two hemispherical part]
= (14 – 2 × 2.5)mm = 9mm Area of cylindrical part = 2πrh
= (2 × $$\frac{22}{7}$$ × 2.5 × 9 ]mm2 = (2 × $$\frac{22}{7} \times \frac{25}{10}$$ × 9) mm2
Total surface area
= [Surface area of cylindrical part + Surface area of both hemispherical parts]  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 2.1m and 4m 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:
(i) Radius of base of cylinder, r= 2 mm.
Height of base of cylinder, h= 2.1 m.
∴ Curved Surface area of cylinder = 2πrh
= 2π × 2 × (2.1)
$$=4 \times \frac{22}{7} \times 2.1$$
= 26.4 m2. (ii) radius = 2m and slant height = l = 2.8 m
Total area of the canvas = CSA of cylinder + CSA of cone
= 2πrh + πrl
= πr(2h + l)
= $$\frac{22}{7}$$ × 2 (2 × 2.1 + 2.8)
= $$\frac{44}{7}$$ × 2 (4.2 + 2.8)
= $$\frac{44}{7}$$ × 7.0 = 44m2
Total cost of canvas used
= ₹ 500 × 44
= ₹ 22,000

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:
Height of the cylinder, h = 2.4 m.
Diameter of the cylinder, d = 1.4 m.
∴ Radius, r = 0.7 m.
(i) Outer Curved surface area of cylinder = 2πrh $$=2 \times \frac{22}{7} \times(0.7) \times 2.4$$
= 44. × 0.24
= 10.56 cm2.
(ii) Area of base of = πr2 = 1.54 cm2.
(iii) Inner surface area of cylinder, = πrl = 5.5 cm2
∴ Total surface area of newly formed cube = 10.56 + 1.54 + 5.5 = 17.6 cm2.
∴ Nearest value is 18 sq.cm. Question 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:
Radius of the cylinder (r) = 3.5 cm
Height of the cylinder (h) = 10 cm
∴ Curved surface area = 2πrh
= 2 × $$\frac{22}{7} \times \frac{35}{10}$$ × 10cm2 = 220cm2
Curved surface area of a hemisphere = 2πr2
∴ Curved surface area of both hemispheres
= 2 × 2πr2 = 4πr2 = 4 × $$\frac{22}{7} \times \frac{35}{10} \times \frac{35}{10}$$ cm2
= 154 cm2
Total surface area of the remaining solid = (220 + 154) cm2 = 374 cm2.

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