GSEB Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.7

Question 1. Find the volume of the right circular cone with (i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm
Answer:
(i) Given: radius \( r = 6 \) cm, height \( h = 7 \) cm
Volume of the right circular cone \( = \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \frac{22}{7} \times (6)^2 \times 7 \)
\( = \frac{1}{3} \times \frac{22}{7} \times 36 \times 7 \)
\( = 22 \times 12 \)
\( = 264 \) cm\(^3 \)
(ii) Given: radius \( r = 3.5 \) cm, height \( h = 12 \) cm
Volume of the right circular cone \( = \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 12 \)
\( = \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 12 \)
\( = 22 \times 12.25 \times \frac{4}{7} \)
\( = 22 \times 1.75 \times 4 \)
\( = 22 \times 7 \)
\( = 154 \) cm\(^3 \)
In simple words: To find the volume of a cone, we use the formula \( \frac{1}{3} \pi r^2 h \). We just need to put in the given radius and height values and then do the calculations for each part.

Exam Tip: Remember to use the correct units (cm\(^3\)) for volume and double-check your calculations, especially with decimal numbers.

Question 2. Find the capacity in litres of a conical vessel with (i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm
Answer:
(i) Given: radius \( r = 7 \) cm, slant height \( l = 25 \) cm
We know that \( r^2 + h^2 = l^2 \)
\( \implies (7)^2 + h^2 = (25)^2 \)
\( \implies 49 + h^2 = 625 \)
\( \implies h^2 = 625 - 49 \)
\( \implies h^2 = 576 \)
\( \implies h = \sqrt{576} \)
\( \implies h = 24 \) cm
Capacity (Volume) \( = \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \frac{22}{7} \times (7)^2 \times 24 \)
\( = \frac{1}{3} \times \frac{22}{7} \times 49 \times 24 \)
\( = 22 \times 7 \times 8 \)
\( = 1232 \) cm\(^3 \)
To convert to litres, we know that 1000 cm\(^3 = 1\) L.
So, \( 1232 \) cm\(^3 = \frac{1232}{1000} \) L \( = 1.232 \) L

(ii) Given: height \( h = 12 \) cm, slant height \( l = 13 \) cm
We know that \( r^2 + h^2 = l^2 \)
\( \implies r^2 + (12)^2 = (13)^2 \)
\( \implies r^2 + 144 = 169 \)
\( \implies r^2 = 169 - 144 \)
\( \implies r^2 = 25 \)
\( \implies r = \sqrt{25} \)
\( \implies r = 5 \) cm
Capacity (Volume) \( = \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \frac{22}{7} \times (5)^2 \times 12 \)
\( = \frac{1}{3} \times \frac{22}{7} \times 25 \times 12 \)
\( = \frac{22 \times 25 \times 4}{7} \)
\( = \frac{2200}{7} \) cm\(^3 \)
To convert to litres:
\( = \frac{2200}{7 \times 1000} \) L \( = \frac{22}{70} \) L \( = \frac{11}{35} \) L
In simple words: First, we use the slant height and the given radius or height to find the missing dimension (height or radius) using Pythagoras' theorem. Then, we use the cone's volume formula to get the volume in cubic centimeters. Finally, we change the volume from cubic centimeters into litres by dividing by 1000.

Exam Tip: Remember the relationship \( r^2 + h^2 = l^2 \) for a right cone. Pay attention to unit conversions, especially from cm\(^3\) to litres (1 L = 1000 cm\(^3\)).

Question 3. The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use \( \pi = 3.14 \))
Answer:
Given: height \( h = 15 \) cm, Volume \( V = 1570 \) cm\(^3 \)
We know the formula for the volume of a cone is \( V = \frac{1}{3} \pi r^2 h \)
Substitute the given values:
\( 1570 = \frac{1}{3} \times 3.14 \times r^2 \times 15 \)
\( 1570 = 3.14 \times r^2 \times 5 \)
\( 1570 = 15.70 \times r^2 \)
\( r^2 = \frac{1570}{15.70} \)
\( r^2 = 100 \)
\( r = \sqrt{100} \)
\( r = 10 \) cm
Hence, the radius of the base of the cone is 10 cm.
In simple words: We know the cone's height and total volume. By using the volume formula \( V = \frac{1}{3} \pi r^2 h \) and plugging in the known numbers, we can solve for \( r^2 \). Taking the square root of \( r^2 \) gives us the radius of the base.

Exam Tip: When \( \pi \) is given as 3.14, use that value instead of \( \frac{22}{7} \). Be careful with decimal calculations during division.

Question 4. If the volume of a right circular cone of height 9 cm is 48\( \pi \) cm³, find the diameter of its base.
Answer:
Given: height \( h = 9 \) cm, Volume \( V = 48 \pi \) cm\(^3 \)
We know the formula for the volume of a cone is \( V = \frac{1}{3} \pi r^2 h \)
Substitute the given values:
\( 48 \pi = \frac{1}{3} \pi r^2 \times 9 \)
Divide both sides by \( \pi \):
\( 48 = \frac{1}{3} r^2 \times 9 \)
\( 48 = 3 r^2 \)
\( r^2 = \frac{48}{3} \)
\( r^2 = 16 \)
\( r = \sqrt{16} \)
\( r = 4 \) cm
The diameter of the base is \( 2r \).
Diameter \( = 2 \times 4 = 8 \) cm
Hence, the diameter of the base of the right circular cone is 8 cm.
In simple words: We are given the height and the volume (in terms of \( \pi \)) of a cone. We use the volume formula \( V = \frac{1}{3} \pi r^2 h \) to first find the radius \( r \). Since the volume already contains \( \pi \), we can easily cancel it out on both sides. Once we have the radius, we simply multiply it by two to get the diameter.

Exam Tip: If \( \pi \) is present on both sides of the equation, you can cancel it out, simplifying calculations. Remember that diameter is twice the radius.

Question 5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Answer:
Given: Diameter of the conical pit \( = 3.5 \) m
Radius \( r = \frac{\text{Diameter}}{2} = \frac{3.5}{2} = 1.75 \) m
Depth (height) \( h = 12 \) m
Capacity (Volume) of the conical pit \( = \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \frac{22}{7} \times (1.75)^2 \times 12 \)
\( = \frac{1}{3} \times \frac{22}{7} \times 3.0625 \times 12 \)
\( = 22 \times 3.0625 \times \frac{4}{7} \)
\( = 22 \times 0.4375 \times 4 \)
\( = 22 \times 1.75 \)
\( = 38.5 \) m\(^3 \)
To convert cubic meters to kilolitres, we use the conversion: \( 1 \) m\(^3 = 1 \) KL.
So, capacity \( = 38.5 \) KL
In simple words: First, we find the pit's radius from its given diameter. Then, using the cone's volume formula \( V = \frac{1}{3} \pi r^2 h \) with the radius and depth (height), we calculate the total capacity in cubic meters. Finally, we change cubic meters to kilolitres, remembering that one cubic meter is equal to one kilolitre.

Exam Tip: Be mindful of units: diameter is given in meters, and capacity needs to be in kilolitres. Remember the conversion 1 m\(^3\) = 1 KL.

Question 6. The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find (i) height of the cone, (ii) slant height of the cone, (iii) curved surface area of the cone.
Answer:
Given: Volume \( V = 9856 \) cm\(^3 \)
Diameter of the base \( = 28 \) cm
Radius \( r = \frac{\text{Diameter}}{2} = \frac{28}{2} = 14 \) cm

(i) To find the height of the cone:
We use the formula for the volume of a cone: \( V = \frac{1}{3} \pi r^2 h \)
Substitute the known values:
\( 9856 = \frac{1}{3} \times \frac{22}{7} \times (14)^2 \times h \)
\( 9856 = \frac{1}{3} \times \frac{22}{7} \times 196 \times h \)
\( 9856 = \frac{22 \times 28}{3} \times h \)
\( 9856 = \frac{616}{3} \times h \)
\( h = \frac{9856 \times 3}{616} \)
\( h = 16 \times 3 \)
\( h = 48 \) cm
Hence, the height of the cone is 48 cm.

(ii) To find the slant height of the cone:
We use the relationship \( l^2 = r^2 + h^2 \)
\( l^2 = (14)^2 + (48)^2 \)
\( l^2 = 196 + 2304 \)
\( l^2 = 2500 \)
\( l = \sqrt{2500} \)
\( l = 50 \) cm
Hence, the slant height of the cone is 50 cm.

(iii) To find the curved surface area of the cone:
We use the formula: Curved surface area \( = \pi r l \)
\( = \frac{22}{7} \times 14 \times 50 \)
\( = 22 \times 2 \times 50 \)
\( = 22 \times 100 \)
\( = 2200 \) cm\(^2 \)
Hence, the curved surface area of the cone is 2200 cm\(^2 \).
In simple words: First, we use the given volume and radius to figure out the cone's height. Next, using that height and the radius, we calculate the slant height. Finally, we use the radius and slant height to determine the curved surface area of the cone.

Exam Tip: It is often necessary to find the height or slant height first before calculating other properties. Ensure you remember all three key formulas: volume, slant height, and curved surface area.

Question 7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Answer:
When a right triangle ABC with sides 5 cm, 12 cm, and 13 cm is revolved about the side 12 cm, the solid formed is a right circular cone. In this case, the side that is revolved becomes the height (h), and the other leg becomes the radius (r). The hypotenuse becomes the slant height (l).
So, for the cone obtained:
Height \( h = 12 \) cm
Radius \( r = 5 \) cm
Slant height \( l = 13 \) cm
Volume of the cone \( = \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \pi \times (5)^2 \times 12 \)
\( = \frac{1}{3} \times \pi \times 25 \times 12 \)
\( = \pi \times 25 \times 4 \)
\( = 100 \pi \) cm\(^3 \)
Hence, the volume of the solid obtained is \( 100 \pi \) cm\(^3 \).
In simple words: When a right triangle spins around one of its shorter sides, it forms a cone. The side it spins around becomes the cone's height, and the other shorter side becomes the cone's radius. We then use the cone volume formula \( V = \frac{1}{3} \pi r^2 h \) with these measurements.

Exam Tip: Clearly identify which side of the right triangle becomes the height and which becomes the radius when it's revolved to form a cone. The side of revolution is the height.

Question 8. If the triangle ABC in question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Answer:
When the right triangle ABC is revolved about the side 5 cm, the solid formed is another right circular cone.
For this new cone:
Height \( h = 5 \) cm
Radius \( r = 12 \) cm
Slant height \( l = 13 \) cm
Volume of this cone \( V_2 = \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \pi \times (12)^2 \times 5 \)
\( = \frac{1}{3} \times \pi \times 144 \times 5 \)
\( = \pi \times 48 \times 5 \)
\( = 240 \pi \) cm\(^3 \)

From Question 7, the volume of the first solid was \( V_1 = 100 \pi \) cm\(^3 \).
Now, we find the ratio of the volumes of the two solids \( V_1 : V_2 \):
\( V_1 : V_2 = 100 \pi : 240 \pi \)
\( = 100 : 240 \)
Divide by 20:
\( = 5 : 12 \)
The ratio of the volumes of the two solids is 5:12.
In simple words: When the same triangle spins around its other shorter side (5 cm), it makes a different cone with that side as its height and the 12 cm side as its radius. We calculate this new cone's volume. Then, we compare this volume to the first cone's volume from Question 7 to find their simple ratio.

Exam Tip: Be careful to switch the roles of height and radius correctly when the axis of revolution changes. Simplifying ratios involves finding the greatest common divisor of the two numbers.

Question 9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Answer:
Given: Diameter of the conical heap \( = 10.5 \) m
Radius \( r = \frac{\text{Diameter}}{2} = \frac{10.5}{2} = 5.25 \) m
Height \( h = 3 \) m

(i) To find the volume of the heap:
Volume of the cone \( = \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \frac{22}{7} \times (5.25)^2 \times 3 \)
\( = \frac{22}{7} \times 27.5625 \)
\( = 22 \times 3.9375 \)
\( = 86.625 \) m\(^3 \)
The volume of the heap is 86.625 m\(^3 \).

(ii) To find the area of the canvas required (curved surface area):
First, we need to find the slant height \( l \).
\( l = \sqrt{r^2 + h^2} \)
\( l = \sqrt{(5.25)^2 + (3)^2} \)
\( l = \sqrt{27.5625 + 9} \)
\( l = \sqrt{36.5625} \)
\( l \approx 6.046 \) m \( \approx 6.05 \) m (approximately)
Area of canvas required \( = \) Curved surface area \( = \pi r l \)
\( = \frac{22}{7} \times 5.25 \times 6.05 \)
\( = 22 \times 0.75 \times 6.05 \)
\( = 16.5 \times 6.05 \)
\( = 99.825 \) m\(^2 \)
Hence, the area of the canvas required is 99.825 m\(^2 \).
In simple words: First, we calculate the volume of the wheat heap using the cone's volume formula with its given diameter and height. Then, to find how much canvas is needed to cover it, we calculate the cone's slant height using its radius and height. After that, we use the slant height and radius to find the curved surface area, which tells us the canvas area needed.

Exam Tip: Always remember that the canvas required to cover a conical heap corresponds to its curved surface area, not its total surface area (as the base is on the ground). Pay close attention to calculations involving square roots and decimals.