In this chapter, we provide KSEEB SSLC Class 10 Maths Solutions Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 10 Maths Solutions Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 pdf, free KSEEB SSLC Class 10 Maths Solutions Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 pdf download. Now you will get step by step solution to each question.

**Karnataka State Syllabus Class 10 Maths Solutions** **Chapter 15 Surface Areas and Volumes Ex 15.4**

(Use π = 227 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 = ?

Volume of glass = Radius with 2 cm + Volume of frustum with radius 1 cm.

Question 2.

The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its ciruclar ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Solution:

Slant height of a i frustum of a cone, l = 4 cm.

Circumference of its circular ends = 18 cm. and 6 cm

2πr =18 cm.

∴ R = 9π cm.

2πr = 6

∴ r = 3π cm.

∴ Lateral surface area of a frustum of a Cone

Question 3.

A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the figure).

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

R = 10 cm, r = 4 cm, l = 15 cm

∴ Curved surface = π × l × (R + r)

= π × 15 × (10 + 4)

= 227 × 15 × 14

= 660 cm^{2}

∴ Area of the cap at the end

∴ Area of material used for making it.

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 Rs. 20 per litre. Also find the cost of the metal sheet used to make the container, if it costs Rs. 8 per 100 cm2. (Take π = 3.14).

Solution:

R = 20 cm

r = 8 cm

h = 20 cm

∴ Slant height,

∴ Volume of metallic sheet,

Quantity of milk in the container

= 10449.821000

Cost of 1 litre of milk is Rs. 20,

Cost of 10.45 litres of milk ……. ??

∴ 20 × 10.45 = Rs. 209.

Cost of metal sheet = π(R + r) + πr^{2}

= π {20 × (20 + 8) + (8)^{2}}

= 3.14 × 624

= 1959.36 cm^{2}.

∴ Cost of preparing metallic container: For 100 cm^{2} Rs. 8

∴ For 1959.36 cm^{2} ……?

= 8 × 1159.36100

= Rs. 156.75.

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 116 cm., find the length of the wire.

Solution:

Height of a cone, h = 20 cm.

Vertical angle = 60°

Diameter of frustum wire = 116 cm.

i) In ⊥∆AQC, ∠Q = 90°

ii) In ⊥∆PRA, ∠R = 90°

Height of frustum, h = 10 cm.

∴ Volume of metallic frustum, ABCD,

Let the length of wire which has diameter 116 cm be ‘x’ cm

∴ Length of wire = 7964.4 m

**All Chapter KSEEB Solutions For Class 10 Maths**

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