GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

Gujarat Board GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

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:
Here
E:\New folder\gujrat bord\10th maths\img\13.1\13.2\GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5 - 12.png
d1 = 4 cm
r1 = \(\frac {4}{2}\) = 2 cm
and d2 = 2 cm
r2 = 1 cm
Height h = 14
Hence, capacity of glass
= \(\frac {1}{3}\) π (r21 + r21 + r1r2) x h
= \(\frac {1}{3}\) x \(\frac {22}{7}\) [22 + 12 + 2 x 1] x 14
= \(\frac {1}{3}\) x \(\frac {22}{7}\) x [4 + 1 + 2] x 14
= \(\frac {22}{3 x 7}\) x 7 x 14 = \(\frac {308}{3}\) cm3 = 102 \(\frac {2}{3}\) cm2

GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

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:
Here,
Slant height l = 4 cm
and let r1 r2 be the radii of frustum and r1 > r2
E:\New folder\gujrat bord\10th maths\img\13.1\13.2\GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5 - 12.png
Perimeter of circular ends = 2Ï€r1
2Ï€r1 = 18 cm
⇒  Ï€r1 = 9 cm ……(1)
and 2Ï€r2 = 6 cm
Ï€r2 = 3 cm …….(2)
Curved surface area of frustum
= Ï€(r1 + r2) l = (Ï€r1 + Ï€r2)l ……(3)
Putting value of eqn. (1) and (2) in eqn. (3),
Curved surface area of frustum = (9 + 3) x 4 = 12 x 4 = 48 cm2

Question 3.
A fez, the cap used by the Turks, is shaped like the frustum of a cone. 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.
E:\New folder\gujrat bord\10th maths\img\13.1\13.2\GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5 - 12.png
Solution:
Here
r1 = 10 cm (Radius on the open side)
and r1 = 4 cm (Radius of the upper base)
Surface area of cap = CSA of frustum + Area of upper circular end
= π(r1 + r2) l + πr22
E:\New folder\gujrat bord\10th maths\img\13.1\13.2\GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5 - 12.png

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:
Here
h = 16 cm
r1 = 20 cm and r2 = 8 cm
GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4
Volume of frustum shaped container
= \(\frac {1}{3}\)Ï€h (r21 + r22 + r12)
= \(\frac {1}{3}\) x 3.14 x 16(202 + 82 + 20 x 8)nh
= \(\frac {1}{3}\) x 3.14 x 16 [400 + 64 + 160]
= \(\frac {1}{3}\) x 3.14 x 16 x 624
= 3.14 x 16 x 208
= 10449.92 cm3 (1000 cm3 = 1 litre)
= \(\frac{10449.92}{1000}\) = 10.44992 litre
Cost of milk = 10.44992 x 20
= ₹ 208.9984 = ₹ 209
Surface area of frustum
= Ï€ (r1 + r2)l + Ï€r22 ……….(1)
l = \(\frac{10449.92}{1000}\)
= \(\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}\)
= \(\sqrt{(20-8)^{2}+16^{2}}\)
= \(\sqrt{12^{2}+16^{2}}\)
= \(\sqrt{144+256}\)
= \(\sqrt{400}\)
l = 20 cm
Area of metal sheet = Surface area of frustum
= π(r1 + r2)l + πr22 [From (1)]
= 3.14 (20 + 8) x 20 + 3.14 x 82
= 3.14 x 28 x 20 + 3.14 x 64
= 1758.4 + 200.96
= 1959.36 cm2
Cost of metal sheet
= 1959.36 x \(\frac{8}{100}\) = ? 156.75

GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

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:
Here height h of cone = 20 cm
Cone is cut into the middle then height of frustum
= \(\frac{20}{2}\) = 20 cm
Hence AC = 10 cm
GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4
In rt. ΔACE
tan 300 = \(\frac{CE}{AC}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{r_{2}}{10}\)
⇒ r2 = \(\frac{10}{\sqrt{3}}\) cm
Now rt. ΔAOB
tan 300 = \(\frac{OC}{OA}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{r_{1}}{20}\)
⇒ r1 = \(\frac{20}{\sqrt{3}}\) cm
Volume of frustum
GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4
Diameter of wire which is drawn = \(\frac{1}{16}\)cm
Radius of wire r = \(\frac{1}{2}\) x \(\frac{1}{16}\) = \(\frac{1}{32}\)cm
Volume of wire drawn = Volume of frustum
Ï€r2h = \(\frac{22000}{9}\)
GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4
= 7964.4 m
Hence, the length of wire is 7964.4 in.

GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

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