GSEB Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 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.
Answer: Here, the height \( h = 14 \) cm.
The first diameter \( d_1 = 4 \) cm, which means its radius \( r_1 = \frac{4}{2} = 2 \) cm.
The second diameter \( d_2 = 2 \) cm, so its radius \( r_2 = \frac{2}{2} = 1 \) cm.
The capacity (volume) of the glass, shaped like a frustum, is calculated using the formula:
\( = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2) \)
\( = \frac{1}{3} \times \frac{22}{7} \times 14 (2^2 + 1^2 + 2 \times 1) \)
\( = \frac{1}{3} \times \frac{22}{7} \times 14 (4 + 1 + 2) \)
\( = \frac{1}{3} \times \frac{22}{7} \times 14 \times 7 \)
\( = \frac{22 \times 14}{3} \)
\( = \frac{308}{3} \) cm³
\( = 102 \frac{2}{3} \) cm³
In simple words: We need to determine how much a frustum-shaped glass can hold. We use its given height and the radii of its top and bottom circles in a special formula to calculate its volume.

Exam Tip: Always remember the specific formula for the volume of a frustum and ensure you correctly convert diameters to radii before beginning your calculations.

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.
Answer: Here, the slant height \( l = 4 \) cm.
The perimeter of one circular end \( = 2 \pi r_1 = 18 \) cm.
This means \( \pi r_1 = 9 \) cm ......(1)
The perimeter of the other circular end \( = 2 \pi r_2 = 6 \) cm.
This means \( \pi r_2 = 3 \) cm ......(2)
The curved surface area of the frustum is given by the formula:
\( = \pi (r_1 + r_2) l \)
\( = (\pi r_1 + \pi r_2) l \)
By putting the values from equations (1) and (2) into the formula:
\( = (9 + 3) \times 4 \)
\( = 12 \times 4 \)
\( = 48 \) cm²
In simple words: To find the curved surface area of a frustum, we add the half-perimeters (which are \( \pi r \)) of its two circular ends and then multiply by the slant height.

Exam Tip: Remember that the curved surface area formula for a frustum can be simplified to \( (\pi r_1 + \pi r_2) l \) when the circumferences are provided, making calculations easier.

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.
Answer: Here,
The radius on the open side (bottom) \( r_1 = 10 \) cm.
The radius at the upper base (top) \( r_2 = 4 \) cm.
The slant height \( l = 15 \) cm.
The surface area of the cap (material used) is the sum of the curved surface area of the frustum and the area of the upper circular end (since the bottom is open).
Area of material \( = \pi (r_1 + r_2) l + \pi r_2^2 \)
\( = \frac{22}{7} (10 + 4) \times 15 + \frac{22}{7} \times 4^2 \)
\( = \frac{22}{7} \times 14 \times 15 + \frac{22}{7} \times 16 \)
\( = 22 \times 2 \times 15 + \frac{352}{7} \)
\( = 660 + \frac{352}{7} \)
\( = \frac{4620 + 352}{7} \)
\( = \frac{4972}{7} \) cm²
\( = 710 \frac{2}{7} \) cm²
In simple words: To determine the material needed for the cap, we add the area of its curved side to the area of its top circular part, because the bottom part remains open.

Exam Tip: For objects like a fez cap, which are open at one end, always remember to include only the area of the closed circular base (the upper one in this case) and the curved surface area in your total calculation.

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 metal sheet used to make the container, if it costs Rs 8 per 100 cm². (Take \( \pi = 3.14 \))
Answer: Here,
Height \( h = 16 \) cm
Lower radius \( r_L = 8 \) cm
Upper radius \( r_U = 20 \) cm
Volume of frustum shaped container:
\( = \frac{1}{3} \pi h (r_U^2 + r_L^2 + r_U r_L) \)
\( = \frac{1}{3} \times 3.14 \times 16 (20^2 + 8^2 + 20 \times 8) \)
\( = \frac{1}{3} \times 3.14 \times 16 (400 + 64 + 160) \)
\( = \frac{1}{3} \times 3.14 \times 16 \times 624 \)
\( = 3.14 \times 16 \times 208 \)
\( = 10449.92 \) cm³
Since \( 1000 \) cm³ \( = 1 \) litre,
Volume in litres \( = \frac{10449.92}{1000} = 10.44992 \) litres
Cost of milk \( = 10.44992 \times 20 \)
\( = 208.9984 \approx 209 \) Rs
Now, for the surface area of the frustum (metal sheet):
First, find the slant height \( l \):
\( l = \sqrt{(r_U - r_L)^2 + h^2} \)
\( = \sqrt{(20 - 8)^2 + 16^2} \)
\( = \sqrt{12^2 + 16^2} \)
\( = \sqrt{144 + 256} \)
\( = \sqrt{400} \)
\( l = 20 \) cm
The area of the metal sheet used = Curved surface area of frustum + Area of the lower circular end (since it's open from the top).
Area \( = \pi (r_U + r_L) l + \pi r_L^2 \)
\( = 3.14 (20 + 8) \times 20 + 3.14 \times 8^2 \)
\( = 3.14 \times 28 \times 20 + 3.14 \times 64 \)
\( = 1758.4 + 200.96 \)
\( = 1959.36 \) cm²
Cost of metal sheet \( = 1959.36 \times \frac{8}{100} \)
\( = 156.7488 \approx 156.75 \) Rs
In simple words: First, calculate the container's volume to find the milk cost. Next, calculate the metal surface area (curved part plus the bottom base, as the top is open) to find the metal cost.

Exam Tip: When dealing with open containers, always be careful to include only the necessary base areas in your total surface area calculations. Also, remember to convert volume to litres correctly for liquid cost problems.

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.
Answer: Here, the total height of the cone \( H = 20 \) cm.
The vertical angle is \( 60^\circ \), so the half-vertical angle is \( 30^\circ \).
The cone is cut exactly in the middle of its height, which means the height of the frustum \( h_f = \frac{20}{2} = 10 \) cm.
Let \( r_1 \) be the radius of the original cone's base (larger radius of frustum) and \( r_2 \) be the radius of the cut section (smaller radius of frustum).
For the original cone, using trigonometry (with height \( H = 20 \) cm):
\( \tan 30^\circ = \frac{r_1}{H} \)
\( \frac{1}{\sqrt{3}} = \frac{r_1}{20} \)

\( \implies r_1 = \frac{20}{\sqrt{3}} \) cm
For the smaller cone (top part), using trigonometry (with height \( h_s = 10 \) cm):
\( \tan 30^\circ = \frac{r_2}{h_s} \)
\( \frac{1}{\sqrt{3}} = \frac{r_2}{10} \)

\( \implies r_2 = \frac{10}{\sqrt{3}} \) cm
Now, calculate the volume of the frustum:
Volume \( = \frac{1}{3} \pi h_f (r_1^2 + r_2^2 + r_1 r_2) \)
\( = \frac{1}{3} \times \frac{22}{7} \times 10 \left( \left(\frac{20}{\sqrt{3}}\right)^2 + \left(\frac{10}{\sqrt{3}}\right)^2 + \frac{20}{\sqrt{3}} \times \frac{10}{\sqrt{3}} \right) \)
\( = \frac{220}{21} \left( \frac{400}{3} + \frac{100}{3} + \frac{200}{3} \right) \)
\( = \frac{220}{21} \left( \frac{400 + 100 + 200}{3} \right) \)
\( = \frac{220}{21} \times \frac{700}{3} \)
\( = \frac{22 \times 10 \times 700}{21 \times 3} \)
\( = \frac{22 \times 10 \times 100}{3 \times 3} \)
\( = \frac{22000}{9} \) cm³
This frustum is transformed into a wire. The volume of the frustum will be equal to the volume of the wire.
Diameter of the wire \( = \frac{1}{16} \) cm
Radius of the wire \( r_{wire} = \frac{1}{2} \times \frac{1}{16} = \frac{1}{32} \) cm
The volume of the wire (cylinder) \( = \pi r_{wire}^2 h_{wire} \)
Set the volumes equal:
\( \frac{22}{7} \times \left(\frac{1}{32}\right)^2 \times h_{wire} = \frac{22000}{9} \)

\( \implies h_{wire} = \frac{22000}{9} \times \frac{7}{22} \times 32^2 \)

\( \implies h_{wire} = \frac{1000}{9} \times 7 \times 1024 \)

\( \implies h_{wire} = \frac{7168000}{9} \) cm
\( h_{wire} \approx 796444.44 \) cm
To convert to meters: \( \frac{796444.44}{100} = 7964.4444 \) m
Hence, the length of the wire is approximately \( 7964.4 \) m.
In simple words: We first find the volume of the frustum. Then, we equate this volume to the volume of the wire (a cylinder) into which it is drawn. Knowing the wire's diameter, we can then calculate its length.

Exam Tip: When converting one shape into another, remember that the volume of the material remains constant. Also, ensure you use accurate trigonometric ratios to determine dimensions.