Exercise E13.7 Surface Areas and Volumes NCERT Solutions Class 9

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Chapter 13 Ex.13.7 Question 1

Assume \(\begin{align}\pi = \frac{{22}}{7} \end{align}\) unless stated otherwise

Find the volume of the right circular cone with

(i) radius \(6\rm\, cm,\) height \(7\rm\, cm\)

(ii) radius \(3.5\rm\, cm,\) height \(12\rm\, cm\)

Solution

Video Solution

Reasoning:

Volume of the cone \(\begin{align} = \frac{1}{3}\pi {r^2}h \end{align}\)

Where \(r\) is base radius and \(h\) is height.

What is known?

Radius and height of two cones.

What is unknown?

Volume of the cone.

Steps:

Radius \((r) = 6\rm\, cm\)

Height \((h) = 7\rm\, cm\)

Volume of cone
\[\begin{align}  & = \frac{1}{3}\pi {r^2}h \\ &= \frac{1}{3} \times \frac{{22}}{7} \times 6 \times 6 \times 7\\ &= \,\,264\,\,\rm c{m^3} \end{align}\]

Radius \((r) = 3.5\rm\, cm\)

Height \((h) = 12\rm\, cm\)

Volume of the cone 
\[\begin{align}&= \frac{1}{3}\pi {r^2}h \\ &= \frac{1}{3} \times \frac{{22}}{7} \times 3.5 \times 3.5 \times 12\\ &= 154\,\,\rm c{m^3} \end{align}\]

Answer:

Volume of the cone \(\begin{align} = 264\,\,\rm c{m^3} \end{align}\)

Volume of the cone \(\begin{align} = 154\,\,\rm c{m^3} \end{align}\)

Chapter 13 Ex.13.7 Question 2

Find the capacity in litres of a conical vessel with

(i) radius \(7\rm\, cm,\) slant height \(25\rm\, cm\)

(ii) height \(12\rm\, cm,\) slant height \(13\rm\, cm\)

Solution

Video Solution

Reasoning:

Capacity of a conical vessel is nothing but the Volume of the cone \(\begin{align} = \frac{1}{3}\pi {r^2}h \end{align}\)

What is  known?

Radius \((r) = 7\rm\, cm\)

Slant height \((1) = 25\rm\, cm\)

What is  unknown?

Capacity of the vessel in litres.

Steps:

Capacity of the conical vessel \(=\) volume of cone \(\begin{align} = \frac{1}{3}\pi {r^2}h \end{align}\)

Where \(\begin{align}h = \sqrt {{l^2} - {r^2}} \end{align}\)

Radius \(\begin{align}(r) = 7\rm \,\rm\,cm \end{align}\)

Slant height \(\begin{align}(l) = 25\rm\,\rm \,cm \end{align}\)

\[\begin{align}\therefore h &= \sqrt {{{(25)}^2} - {7^2}} \\ &= \sqrt {(25 + 7)(25 - 7)}\\ {\rm{ [ (}}{{\rm{a}}^2} - {b^2}) &= (a + b)(a - b)]\\ &= \sqrt {32 \times 18} \\ &=\! \sqrt {2 \!\times\! 2\! \times\! 2\! \times \!2\! \times \!2 \!\times\! 3 \!\times \!3\! \times\! 2} \\ &= \sqrt {576} \\ &= 24\,\,\rm\,cm \end{align}\]

Volume
\[\begin{align}& = \frac{1}{3} \times \frac{{22}}{7} \times 7 \times 7 \times 24\\ &= 1232\,\,\rm\,c{m^3}\,\,or\,\,1.232\,\text{l} \end{align}\]

\[\begin{align} [\therefore 1000\,\,c{m^3} = 1\,l] \end{align}\]

What is  known?

Radius \((r) = 12 \rm\,cm\)

Slant height \(\begin{align}(l) = 13\,\,\rm\,cm \end{align}\)

What is  unknown?

Capacity of the vessel in litres.

Steps:

Capacity of the conical vessel \(=\) volume of cone \(\begin{align} = \frac{1}{3}\pi {r^2}h \end{align}\)

Height \((h) = 12\rm\, cm\)

Slant height \(\begin{align}(l) = 13\,\,\rm\,cm \end{align}\)

\[\begin{align}&{r^2} + {h^2} = {l^2} \\ &{r^2} = {l^2} - {h^2} \\ &r=\sqrt{l^{2}-h^{2}} \\ &= \sqrt{(13)^{2}-(12)^{2}} \\ &=\sqrt{(13+12)(13-12)} \\&= \sqrt {25} \\ &= 5\,\,\rm\,cm \end{align}\]

Capacity

\[\begin{align}&= \frac{1}{3} \times \frac{{22}}{7} \times 5 \times 5 \times 12\\&= \frac{{2200}}{7}\\& = \frac{2200}{7000}\\& = \frac{11}{35}l \end{align}\]

Answer:

Capacity \(=1.232\,\, l\)

Capacity \(\begin{align}=\frac{{11}}{{35}} \,l \end{align}\)

Chapter 13 Ex.13.7 Question 3

The height of a cone is \(15\rm\, cm.\) If its volume is \(\begin{align}1570\,\,c{m^3} \end{align}\), find the radius of the base.(Use \(p = 3.14\)).

Solution

Video Solution

Reasoning:

Volume of the cone is \(\begin{align} \frac{1}{3} \end{align}\) times of the volume of a cylinder having radius r and height h i.e. \(\begin{align} = \frac{1}{3}\pi {r^2}h \end{align}\)

What is  known?

Volume of the cone and the height of the cone.

What is  unknown?

Radius of the base.

Steps:

Height \((h) = 15\,\rm cm\)

Radius \((r) = ?\)

Volume of cone

\[\begin{align}&= \frac{1}{3}\pi {r^2}h \\1570&= \frac{1}{3}\pi {r^2}h \\ 1570 &= \frac{1}{3} \times 3.14 \times {r^2} \times 15 \\{r^2} &= \frac{{1570 \times 3}}{{15 \times 3.14}} = \frac{{4710}}{{47.1}} \\{r^2} &= 100 \\ r &= \sqrt {100} = 10\,\,\rm\,cm \end{align}\]

Answer:

Radius of the base \(= 10 \rm\,cm\)

Chapter 13 Ex.13.7 Question 4

If the volume of a right circular cone of height \(9\rm\, cm\) is \(\begin{align}48\pi \,\,\rm\,c{m^3} \end{align}\), find the diameter of its base.

Solution

Video Solution

Reasoning:

Volume of the cone is \(\begin{align}\frac{1}{3}\end{align}\) time of the volume of a cylinder having radius \(r\) and height \(h\) \(\begin{align}\frac{1}{3}\pi {r^2}h \end{align}\)

What is known?

Volume and height of the cone.

What is unknown?

Base diameter.

Steps:

Height \((h) = 9 \rm\,cm\)

\[\begin{align} \text{Volume of cone} &= \frac{1}{3}\pi {r^2}h \\48\pi &= \frac{1}{3}\pi {r^2}h \\48\pi &= \frac{1}{3}\pi \times {r^2} \times 9 \\{r^2} &= \frac{{48 \times 3}}{{3 \times 3}} \\&= 16\\r &= \sqrt {16} \\&= 4\,\,\rm\,cm \end{align}\]

Base diameter \(\begin{align} = 2\,r = 2 \times 4 = 8\,\,\rm\,cm \end{align}\)

Answer:

The diameter of the box of the night circular cone is \(8\rm\, cm.\)

Chapter 13 Ex.13.7 Question 5

A conical pit of top diameter \(3.5\rm\, m\) is \(12\rm\, m\) deep. What is its capacity in kiloliters?

Solution

Video Solution

Reasoning:

Volume of the cone is \(\begin{align}\frac{1}{3} \end{align}\)times of the volume of a cylinder having radius r and height h \(\begin{align} = \frac{1}{3}\pi {r^2}h \end{align}\).

What is  known?

On a meter and the depth of the cone.

What is  unknown?

Volume in kilometer.

Steps:

Volume of cone \(\begin{align} = \frac{1}{3}\pi {r^2}h \end{align}\)

Height of the cone \((h) =\) depth of the pit.

Height \((h) = 12\rm\, cm\)

Diameter \(= 2r = 3.5\rm\, m\)

\(\begin{align}\therefore r = \frac{{3.5}}{2} \end{align}\)

Volume

\(\begin{align} = \frac{1}{3} \times \frac{{22}}{7} \times {(1.75)^2} \times 12 = 38.5\,\,\rm\,c{m^3} \end{align}\)

\[\begin{align}1\,\,\,\rm\,c{m^3} &= 1000\,\,l = 1\,\,k\,\,l \\\therefore\,\, 38.5\,\,\rm\,c{m^3} &= 38500\,\,l = 38.5\,\,kl \end{align}\]

Answer:

Capacity in kilolitre \(\begin{align} = 38.5\,\,kl \end{align}\).

Chapter 13 Ex.13.7 Question 6

The volume of a right circular cone is \(9856\)\(\begin{align}c{m^3} \end{align}\). If the diameter of the base is \(28\rm\, cm,\) find

(i) height of the cone

(ii) slant height of the cone

(iii) curved surface area of the cone

Solution

Video Solution

Reasoning:

Volume of the cone is \(\begin{align}\frac{1}{3} \end{align}\)times of the volume of a cylinder having radius \(r\) and height \(h\)\(\begin{align} = \frac{1}{3}\pi {r^2}h \end{align}\).Curved surface area of the cone is \(\begin{align}\pi rl \end{align}\)

What is  known?

Volume of the right circular cone and diameter of the base.

i.

What is unknown?

Height of the cone.

Steps:

Radius \((r) = 28/2 \rm\,cm\)

Volume of cone 

\[\begin{align} &= \frac{1}{3}\pi {r^2}h \\9856 &= \frac{1}{3} \times \frac{{22}}{7} \times \frac{{28}}{2} \times \frac{{28}}{2} \times h \\h &= \frac{{9856 \times 3 \times 7 \times 2 \times 2}}{{22 \times 28 \times 28}} \\&= 48\,\,\rm\,cm \end{align}\]

ii.

What is  unknown?

Slant height of the cone.

Steps:

Radius \((r) = 14 \rm\,cm\)

Height \((h) = 48 \rm\,cm\)

\[\begin{align}l &= \sqrt {{r^2} + {h^2}} \\ &= \sqrt {{{14}^2} + {{48}^2}} \\ &= \sqrt {196 + 2304} \\ &= \sqrt {2500} \\ &= 50\,\,\,\rm\,cm \end{align}\]

iii.

What is unknown?

Curved surface area of the cone.

Steps:

Radius \((r) = 14 \rm\,cm\)

Slant height \((l) = 50\rm\, cm\)

Curved surface area \(\begin{align}\, = \pi rl \end{align}\)

\[\begin{align} &= \frac{{22}}{7} \times 14 \times 50 \\ &= 2200\,\,\rm\,c{m^2} \end{align}\]

Answer:

Height of the cone \(= 48\rm\,cm\)

Slant height of the cone \(= 50\rm\,cm\)

Surface area\(\begin{align}\, = 2200\,\,\rm\,c{m^2} \end{align}\)

Chapter 13 Ex.13.7 Question 7

A right triangle \(ABC\) with sides \(5\rm\, cm,\) \(12 \rm\,cm\) and \(13\rm\, cm\) is revolved about the side \(12 \rm\,cm.\) Find the volume of the solid so obtained.

Solution

Video Solution

Reasoning:

Volume of the cone is \(\begin{align}\frac{1}{3} \end{align}\) times of the volume of a cylinder \(\begin{align} = \frac{1}{3}\pi {r^2}h \end{align}\).

What is known?

Sides of the right triangle.

What is unknown?

Volume of the solid obtained.

Steps:

Radius \((r) = 5\rm\, cm\)

height \((h) = 12 \rm\,cm\)

Volume of cone 

\[\begin{align}&= \frac{1}{3}\pi {r^2}h \\ &= \frac{1}{3} \times \frac{{22}}{7} \times 5 \times 5 \times 12\\ &= \frac{{2200}}{7}\,\,or\,\,100\pi \end{align}\]

Answer:

Volume of the cone \(\begin{align}100\pi \,\,\rm\,c{m^3} \end{align}\).

Chapter 13 Ex.13.7 Question 8

If the triangle \(ABC\) in the Question 7 above is revolved about the side \(5 \rm\,rm \) 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

Video Solution

Reasoning:

Volume of the right circular cone is \(\begin{align}\frac{1}{3} \end{align}\)time of the volume of a cylinder\(\begin{align} = \frac{1}{3}\pi {r^2}h \end{align}\).

What is  known?

Sides of the right triangle.

What is  unknown?

Volume of the cone and ratio between the volumes.

Solution:

Radius of the cone \(= 12\rm\, cm\)

Height of the cone \(= 5 \rm\,cm\)

Volume of cone 

\[\begin{align}&= \frac{1}{3}\pi {r^2}h \\ &= \frac{1}{3} \times \pi \times {(12)^2} \times 5 \\ & = 240\pi \,\,\,\rm\,c{m^3} \end{align}\]

Volume in question \(7\)\(\begin{align} = 100\pi \,\,\rm\,c{m^3} \end{align}\)

Ratio \(\begin{align}100:240 = 5:12 \end{align}\)

Answer:

Volume of the cone is \(\begin{align} = 240\pi \,\,\rm\,c{m^3} \end{align}\)

Ratio between the volume \(\begin{align}=5:12 \end{align}\)

Chapter 13 Ex.13.7 Question 9

A heap of wheat is in the form of a cone whose diameter is \(10.5\rm\,m\) and height is \(3\rm\, m.\) Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Solution

Video Solution

Reasoning:

Volume of the right circular cone is \(\begin{align}\frac{1}{3} \end{align}\)times of the volume of a cylinder. Curved surface area of the cone =\(\begin{align}\pi rl \end{align}\)

What is  known?

Diameter and height of the cone.

What is  unknown?

Volume of the cone and area of the canvas to cover the heap.

Steps:

Diameter \(= 2r = 10.5 \rm\,m\)

\(\begin{align}r = \frac{{10.5}}{2} = 5.25m \end{align}\)

Height \((h) = 3 \rm\,m\)

Volume of cone 

\[\begin{align}&= \frac{1}{3}\pi {r^2}h \\ &= \frac{1}{3} \times \frac{{22}}{7} \times {(\frac{{10.5}}{2})^2} \times 3 \\ &= 86.625\,\,\,\rm\,{m^3} \end{align}\]

Slant height 

\[\begin{align} &= \sqrt {{r^2} + {h^2}} \\&= \sqrt {{{(5.25)}^2} + {{(3)}^2}} \\ &= \sqrt {27.5625 + 9} \\ &= \sqrt {36.5625}\\ &= 6.05\,\,\rm\,m \end{align}\]

The area of the canvas to cover the heap of wheat \(=\) surface area of the cone \(=\) \(\begin{align}\pi rl \end{align}\)

\[\begin{align}& = \frac{{22}}{7} \times \frac{{10.5}}{2} \times 6.05 \\ &= 99.825\,\,\rm\,{m^2} \end{align}\]

Answer:

Volume of the cone \(\begin{align} = 86.625\,\,\rm\,{m^3} \end{align}\)

Area of the canvas required \(\begin{align} = 99.825\,\,\rm\,{m^2} \end{align}\)

  
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