KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 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 15.4.

Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.4

(Use π = \(\frac{22}{7}\) unless stated otherwise.)

Question 1.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:
Height of drinking glass, h = 14 cm.
Diameter of circular ends
D = 4 cm ∴ r = 2 cm.
d = 2 cm ∴ r = 1 cm.
Capacity of glass = ?
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 1
Volume of glass = Radius with 2 cm + Volume of frustum with radius 1 cm.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 1.1
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 1.2

Question 2.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:
We have,
Slant height (l) = 4 cm
Circumference of one end = 2πr1 = 18 cm
and Circumference of other end = 2πr2 = 6 cm
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q2
⇒ πr1 = \(\frac{18}{2}\) = 9 cm
and πr2 = \(\frac{6}{2}\) = 3 cm
∴ Curved surface area of the frustum of the cone
= π(r1 + r2) l = (πr1 + πr2) l = (9 + 3 ) × 4 cm2
= 12 × 4 cm2 = 48 cm2.

KSEEB Solutions

Question 3.
A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the figure).
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 3
If its radius on the open side is 10 cm, radius at the upper base is 4 cm, and its slant height is 15 cm, find the area of material used for making it.
Solution:
Radius on the open side of the cap, R = 10 cm.
Radius of upper end, r = 4 cm.
Slant height l = 15 cm
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 3.1
R = 10 cm, r = 4 cm, l = 15 cm
∴ Curved surface = π × l × (R + r)
= π × 15 × (10 + 4)
= \(\frac{22}{7}\) × 15 × 14
= 660 cm2
∴ Area of the cap at the end
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 3.2
∴ Area of material used for making it.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 3.3

KSEEB Solutions

Question 4.
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ₹ 20 per litre. Also find the cost of metal sheet used to make the container, if it costs ₹ 8 per 100 cm2. (Take π = 3.14)
Solution:
We have, r1 = 20 cm, r2 = 8 cm and h = 16 cm
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q4
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 7
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q4 1
Area of the bottom = πr22
= (\(\frac{314}{100}\) × 8 × 8) cm2 = 200.96 cm2
∴ Total area of metal required
= 1758.4 cm2 + 200.96 cm2 = 1959.36 cm2
Cost of metal required for 100 cm2 = ₹ 8
∴ Cost of metal required for 1959.36 cm2
= ₹ \(\frac{8}{100}\) × 1959.36 = ₹ 156.75

KSEEB Solutions

Question 5.
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \(\frac{1}{16}\) cm., find the length of the wire.
Solution:
Height of a cone, h = 20 cm.
Vertical angle = 60°
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 5
Diameter of frustum wire = \(\frac{1}{16}\) cm.
i) In ⊥∆AQC, ∠Q = 90°
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 5.1
ii) In ⊥∆PRA, ∠R = 90°
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 5.2
Height of frustum, h = 10 cm.
∴ Volume of metallic frustum, ABCD,
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 5.3
Let the length of wire which has diameter \(\frac{1}{16}\) cm be ‘x’ cm
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 5.4
∴ Length of wire = 7964.4 m

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