Exercise-11.3
Maths - Exercise
Q1. Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm
Let the formula for the volume of a right circular cone be used:
\[ V = \frac{1}{3} \pi r^2 h \] where \( r \) is the radius of the base, and \( h \) is the height of the cone.
For part (i), the radius \( r = 6 \) cm and the height \( h = 7 \) cm.
Substituting these values:
\[ \begin{aligned} V &= \frac{1}{3} \times \pi \times (6)^2 \times 7 \\ &= \frac{1}{3} \times \frac{22}{7} \times 36 \times 7 \\ &= \frac{1}{3} \times \frac{22}{7} \times 252 \\ &= \frac{1}{3} \times 792 \\ &= 264~\text{cm}^3 \end{aligned} \]
Thus, the volume of the cone with radius 6 cm and height 7 cm is \( 264~\text{cm}^3 \).
For part (ii), the radius \( r = 3.5 \) cm and the height \( h = 12 \) cm.
Substituting these values:
\[ \begin{aligned} V &= \frac{1}{3} \times \pi \times (3.5)^2 \times 12 \\ &= \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 12 \\ &= \frac{1}{3} \times \frac{22}{7} \times 147 \\ &= \frac{1}{3} \times 462 \\ &= 154~\text{cm}^3 \end{aligned} \]
Therefore, the volume of the cone with radius 3.5 cm and height 12 cm is \( 154~\text{cm}^3 \).
Q2. 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
Solution:
To find the capacity of a conical vessel, we first need to determine the height if it is not given, using the Pythagorean theorem relating the slant height \( l \), radius \( r \), and height \( h \) of the cone:
\[ l^{2} = r^{2} + h^{2} \]
For part (i), the radius \( r = 7 \) cm and the slant height \( l = 25 \) cm. Using the relation, we calculate the height \( h \):
\[ \begin{aligned} h^2 &= l^2 - r^2 \\ &= 25^2 - 7^2 \\ &= 625 - 49 \\ &= 576 \\ h &= \sqrt{576} \\&= 24~\text{cm} \end{aligned} \]
Next, we calculate the volume of the conical vessel using the formula:
\[ \begin{aligned} V &= \frac{1}{3} \pi r^{2} h \\\\&= \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24 \\&= 1232~\text{cm}^3 \end{aligned} \]
Since 1 litre equals 1000 cubic centimeters, the capacity in litres is:
\[ \begin{aligned} 1232~\text{cm}^3 &= \frac{1232}{1000} \\\\&= 1.232~\text{litres} \end{aligned} \]
For part (ii), the height \( h = 12 \) cm and slant height \( l = 13 \) cm. Let the radius be \( r \). Using the Pythagorean theorem:
\[ \begin{aligned} r^2 &= l^2 - h^2 \\ &= 13^2 - 12^2 \\ &= (13 + 12)(13 - 12) \\ &= 25 \times 1 \\&= 25 \\ r &= \sqrt{25} \\&= 5~\text{cm} \end{aligned} \]
Now, the volume of the vessel:
\[ \begin{aligned} V &= \frac{1}{3} \pi r^{2} h \\\\&= \frac{1}{3} \times \frac{22}{7} \times 5^2 \times 12 \\\\&= \frac{22}{7} \times 100 \\\\&= \frac{2200}{7}~\text{cm}^3 \end{aligned} \]
Converting to litres:
\[ \begin{aligned} V &= \frac{2200}{7} \times \frac{1}{1000} \\\\&= \frac{22}{70} \\\\&= \frac{11}{35} \\\\&\approx 0.314~\text{litres} \end{aligned} \]
Hence, the capacity of the conical vessel is approximately 1.232 litres for part (i) and 0.314 litres for part (ii).
Q3. The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base. (Use π = 3.14)
Solution:
Height of the cone \((h)\) = 15 5 cm
volume of the cone \((V)\) = 1570 cm³
Let Radius of the cone = \(r\)
Q4. If the volume of a right circular cone of height 9 cm is 48 π cm3, find the diameter of its base.
Solution:
Given the volume \( V \) of the right circular cone as \( 48\pi \) cubic centimeters and the height \( h = 9 \) cm, the formula for the volume of a cone is applied:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting the known values, we get:
\[ 48 \pi = \frac{1}{3} \pi r^2 \times 9 \]
Dividing both sides by \( \pi \) and simplifying:
\[ \begin{aligned} 48 &= \frac{1}{3} \times 9 \times r^2 \\&= 3 r^2 \end{aligned} \]
From this, the radius squared is:
\[ \begin{aligned} r^2 &= \frac{48}{3} \\&= 16 \end{aligned} \]
Taking the positive square root (since radius is positive):
\[ \begin{aligned} r &= \sqrt{16} \\&= 4~\text{cm} \end{aligned} \]
The diameter \( D \) of the base of the cone is twice the radius:
\[ \begin{aligned} D &= 2r \\&= 2 \times 4 \\&= 8~\text{cm} \end{aligned}\]
Therefore, the diameter of the base of the cone is 8 cm.
Q5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Solution:
The diameter of the conical pit is given as 3.5 meters, so the radius \( r \) is half of that:
\[ r = \frac{3.5}{2} = 1.75~\text{m} \]
The height \( h \) of the pit is 12 meters. Using the formula for the volume of a cone:
\[ V = \frac{1}{3} \pi r^{2} h \]
Substituting the values into the formula:
\[\small V = \frac{1}{3} \times \frac{22}{7} \times (1.75)^2 \times 12 \]
Calculating step by step:
\[\small \begin{aligned} V &= \frac{1}{3} \times \frac{22}{7} \times 3.0625 \times 12 \\&= 38.5~\text{m}^3 \end{aligned} \]
Since 1 cubic meter is equal to 1 kilolitre, the capacity of the conical pit is:
\[ 38.5~\text{kilolitres} \]
Q6. 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
Solution:
The volume of the cone is given as \( 9856 \, \text{cm}^3 \) and the diameter of its base is 28 cm, so the radius \( r \) is half of that:
\[ r = \frac{28}{2} = 14 \, \text{cm} \]
Using the formula for the volume of a cone:
\[ V = \frac{1}{3} \pi r^{2} h \]
Substituting the known values:
\[ 9856 = \frac{1}{3} \times \frac{22}{7} \times 14^{2} \times h \]
To find the height \( h \), rearrange the formula:
\[\begin{aligned} h &= \frac{9856 \times 3 \times 7}{22 \times 14^{2}} \\\\&= \frac{9856 \times 21}{22 \times 196} \end{aligned}\]
Calculating the value:
\[ h = 48 \, \text{cm} \]
For the slant height \( l \), use the Pythagorean theorem:
\[ \begin{aligned} l^{2} &= r^{2} + h^{2} \\&= 14^{2} + 48^{2} \\&= 196 + 2304 \\&= 2500 \end{aligned}\]
Taking the square root:
\[ \begin{aligned} l &= \sqrt{2500} \\&= 50 \, \text{cm} \end{aligned}\]
The curved surface area (CSA) is given by:
\[ \begin{aligned} \text{CSA} &= \pi r l \\\\&= \frac{22}{7} \times 14 \times 50 \\\\&= 2200 \, \text{cm}^2 \end{aligned}\]
Thus, the height of the cone is 48 cm, the slant height is 50 cm, and the curved surface area is 2200 cm².
Q7. 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.
Solution:
The given triangle ABC is a right triangle with sides 5 cm, 12 cm, and 13 cm. Since the triangle is revolved about the side measuring 12 cm, this side acts as the height \( h \) of the cone formed.
The other side perpendicular to the axis of revolution, which is 5 cm, becomes the radius \( r \) of the base of the cone.
Using the formula for the volume of a cone:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting the values:
\[ \begin{aligned} V &= \frac{1}{3} \times \pi \times 5^2 \times 12 \\\\&= \frac{1}{3} \times \pi \times 25 \times 12 \\\\&= 100 \pi~\text{cm}^3 \end{aligned}\]
Thus, the volume of the solid obtained by revolving the triangle about the side of length 12 cm is \( 100 \pi \, \text{cm}^3 \), which is approximately \( 314.16 \, \text{cm}^3 \).
Q8. If the triangle ABC in the 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.
Solution:
If the right triangle ABC with sides 5 cm, 12 cm, and 13 cm is revolved about the side measuring 5 cm, this side becomes the height \( h \) of the cone formed.
The other perpendicular side, 12 cm, serves as the radius \( r \) of the base of the cone.
Using the volume formula for a cone:
\[ V = \frac{1}{3} \pi r^{2} h \]
Substituting the known values:
\[ \begin{aligned} V &= \frac{1}{3} \times \pi \times 12^{2} \times 5 \\\\&= \frac{1}{3} \times \pi \times 144 \times 5 \\\\&= 240 \pi~\text{cm}^3 \end{aligned} \]
The volume of the solid formed by revolving the triangle about the 5 cm side is \( 240\pi \, \text{cm}^3 \).
From Question 7, the volume was \( 100 \pi \, \text{cm}^3 \).
The ratio of the volumes of the two solids from Questions 7 and 8 is therefore:
\[ 100 \pi : 240 \pi = 5 : 12 \]
This means the second solid's volume is \(\frac{12}{5}\) times the first solid's volume.
Q9. 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.
Solution:
The diameter of the base of the wheat heap is 10.5 meters, so the radius \( r \) is half of that:
\[ r = \frac{10.5}{2} = 5.25 \, \text{m} \]
The height \( h \) of the heap is 3 meters. Using the formula for the volume of a cone:
\[ V = \frac{1}{3} \pi r^{2} h \]
Substituting the given values, the volume is calculated as:
\[ \begin{aligned} V &= \frac{1}{3} \times \frac{22}{7} \times (5.25)^2 \times 3 \\\\&= \frac{1}{3} \times \frac{22}{7} \times 27.5625 \times 3 \\\\&= 86.625 \, \text{m}^3 \end{aligned}\]
For the canvas area required, which is the curved surface area of the cone, first find the slant height \( l \) using the Pythagorean theorem:
\[ \begin{aligned} l &= \sqrt{r^2 + h^2} \\&= \sqrt{(5.25)^2 + 3^2} \\&= \sqrt{27.5625 + 9} \\&= \sqrt{36.5625} \\&\approx 6.047 \, \text{m} \end{aligned}\]
The curved surface area (CSA) is given by:
\[ \begin{aligned} CSA &= \pi r l \\\\&= \frac{22}{7} \times 5.25 \times 6.047 \\\\&\approx 99.78 \, \text{m}^2 \end{aligned} \]
Therefore, the volume of the wheat heap is approximately 86.625 cubic meters and the area of canvas required to cover the heap is approximately 99.78 square meters.
