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Note on Cone

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Solid objects, as shown below are the cones.

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As we see above, a cone is a solid object whose base is a circle and another part is a smooth curved surface that symmetrically ends at a point in space. The point is called the vertex of the cone. The line segments that joins the centre of the base and the vertex is called the height of the cone. Line segments that join the vertex to the point of the circumference of the base circle are called the generator. The length of the generator is called the slant height of the cone. If the axis is perpendicular to the base of the circle is called the generator. The length of the generator is called the slant height of the cone. If the axis is perpendicular to the base circle the cone is called right circular cone. The surface beside the base circle is called the curved surface of the cone. A right circular cone can be formed by rotating a right angled triangle along with its vertical height.

Surface area of a cone

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The cone can be formed from the sector of a circle. In the figure given below, a sector of the circle joining along its cut edges.

In this case, the curved surface area of the cone is equal to the area of the sector of the circle.

Here, \begin{align*} \text {radius of circle} &= l \: units \\ \text {Central angle} &= \theta \\ \text {Arc length} &= 2 \pi r\\ \text {Where r} &= \text {radius of base circle} \end{align*}

[ Circumference of base circle of a cone is equal to the length of the sector. ]
We know,

\begin{align*} \theta &= \frac {2 \pi r} {l} \left[ \theta = \frac {l} {r} , l = 2 \pi r, r = l \right] \\ Also, \\Area \: of \: sector &= \frac {\theta \pi l^2}{2 \pi}\\ &= \frac {\theta \pi l^2} {2 \pi} [ \because 360º = 2 \pi ^c , \theta ^c = central \: angle ] \\ &= \frac {\theta l^2} {2} \\ &= \frac {2 \pi r} {l} . \frac {l^2} {2} [\because \theta = \frac {2 \pi r} {l} ] \\&= \pi r l \end{align*}

Therefore, curved surface area of the cone (CSA) =πrl square units, where r is the radius of the base and l is the slant height of cone.

Alternatively,

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We can demonstrate the following materials in the classroom to show the curved surface area of the cone. Take a cone and color the curved surface area of the cone by any indices having equal parts. Take the half circle whose radius is equal to the slant height of the cone. The radius of the base circle is r. Roll the cone above the half circle such that vertex of the cone is fixed at the center of a half-circle. One complete roll of cone exactly fits in half circle. In this case, the area of half circle is equal to the curved surface area of the cone. Since the radius of the half circle is same as the slant height of the cone, so that area of half circle is \( \frac {\pi l^2} {2} \) square units.

From above experiment was see that circumference of the base circle with radius r units is equal to the circumference of half circle with radius l units.
i.e. \begin{align*} 2 \pi r &= \pi l \\ or, l &= \pi r \\ or, r &= \frac {l} {2} \end{align*}

\begin{align*} \therefore Area \: of \: half \: circle &= \frac {\pi l^2} {2} sq. \: units \\ &= \pi \frac {l} {2} . l \: sq. \: units \\ &= \pi r l \: sq. \: units \end{align*}

Hence, Curved surface area of cone ( CSA ) =πrl square units.
Total surface Area of the cone (TSA) = CSA + area of circular base.
i.e. \begin{align*} TSA &= CSA + A_b \\ &= \pi r l + \pi r^2 \\ \pi r \: (l + r) and \:l^2 &= h^2 + r^2 \\ &=\pi rl\end{align*}

There 'h' is the height of the cone, l is the slant height of the cone and 'r' is the radius of base circle.

Curved surface area of cone can also be illustrated by the following activities

Let's take a hollow cone made up of paper. Cut the cone along its slant height. We get a sector of a circle whose radius is l units and arc length 2πr units as shown above in the last figure. The last figure is now approximate rectangle whose length is half of 2πr and breadth is l units.

i.e. Length =πr
Breadth = l
\begin{align*} Area \: of \: rectangle \: (A) = l \times b \\ &= \pi r \times l \\ &= \pi r l \: square \: units \end{align*}

\( \therefore \text {Curved surface area of cone} (A) = \pi r l \)
where l = slant height
r = radius of the base circle

Volume of a cone

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A cone is a circular pyramid. As we know the volume of a pyramid is one-third of the base area times height. So, this fact can be generalized in case of cone also. Therefore, the volume of a cone is one-thirdof the base area times height.

i.e. \begin{align*} \text {Volume of cone} (v) &= \frac {1} {3} \times base\: area \times height \\ V &= \frac {1} {3} \pi r^2 h \\ \end{align*}

Where r is the radius of the base and h is the height of the cone.

Alternatively,

Take a conical pot whose radius is 'r' units and height are 'h' units. Take a measuring cylinder having the radius of base r units and height h units. Fill the cylinder with water from full of a cone. The water which filled the circular cylinder of radius 'r' and height 'h' could also fill exactly three conical pots of the radius 'r' and height 'h'.

From the above experiment, we conclude that volume of a right circular cone of radius 'r' and height 'h' is one-third of the volume of a right circular cylinder of the same radius and height.

\begin{align*} \text{Volume of a cone} &= \frac {1} {3} \times base \: area \times height \\ &= \frac {1} {3} \pi r^2 \times h \\ &= \frac {1} {3} \pi r^2 h \: cubic \: units \end{align*}

where r = radius of base circle, h = height of cone .

In our context, the cone is right circular cone only.

Total surface area of the cone = \(\pi\)r(i + r) square on

Curbed surface area (CSA) = \(\pi\)rl

.

Very Short Questions

Base are of the cone = \(\pi\)r2 = 125 cm2

Height of the cone (h) = 9 cm

\begin{align*} Volume (V) &= \frac 13 {\pi}r^2h\\ &= \frac 13 × 125 × 9 cm^3\\ &= 125 × 3 cm^3\\ &= 375 cm^3_{Ans}\\ \end{align*}

Here,

\begin{align*} {\text{Radius of cone (r)}} &=\sqrt {l^2 - h^2}\\ &= \sqrt {(25 cm)^2 - (24 cm)^2}\\ &= \sqrt {625 cm^2 - 576 cm^2}\\ &= \sqrt {49 cm^2}\\ &= 7 cm\\ \end{align*}

Now,

\begin{align*} {\text{Volume of the cone (V)}} &=\frac 13 {\pi}r^2h\\ &= \frac 13 × \frac {22}7 × (7 cm)^2 × 24 cm\\ &= 22 × 7 × 8 cm^3\\ &= 1232 cm^3_{Ans}\\ \end{align*}

Here,

l = 50 cm

h = 48 cm

Then,

\begin{align*} r &= \sqrt {l^2 - h^2}\\ &= \sqrt {50^2 - 48^2} cm\\ &= \sqrt {196} cm\\ &= 14 cm\\ \end{align*}

\begin{align*} \therefore {\text{Volume of the cone}} &=\frac 13 {\pi}r^2h\\ &= \frac 13 × \frac {22}7 × 14^2 × 48 cm^3\\ &=\ \frac {22 × 196 × 48}{21} cm^3\\ &= 9856 cm^3_{Ans} \end{align*}

Here,

r = 14 cm,

V = 1848 cm3

h = ?

By formula,

V = \(\frac 13\)\(\pi\)r2h

or, 1848 = \(\frac 13\)× \(\frac {22}7\)× (14)2× h

or, 1848× 21 = 22× 196 h

or, h = \(\frac {1848 × 21}{22 × 196}\) cm

∴ h = 9 cm

Hence, the height of cone is 9 cm.Ans

Here,

l = 25 cm

h = 24 cm

If r be the radius of base, then:

\begin{align*} r &= \sqrt {l^2 - h^2}\\ &= \sqrt {(25)^2 - (24)^2}\\ &= \sqrt {49}\\ &= 7 cm\\ \end{align*}

Now,

\begin{align*} {\text{Curved Surface Area}} &= {\pi}rl\\ &= \frac {22}7 × 7 × 25 cm^2\\ &= 550 cm^2_{Ans}\\ \end{align*}

Here,

r = 5 cm

h = 4 cm

Curved Surface Area of cone (S) = ?

Here,

\begin{align*} r &= \sqrt {l^2 - h^2}\\ &= \sqrt {5^2 - 4^2}\\ &= \sqrt {25 - 16}\\ &= \sqrt 9\\ &= 3 cm\\ \end{align*}

\begin{align*} S &= {\pi}rl\\ &= \frac {22}7× 3× 5 cm^2 \\ &= 47.14 cm^2_{Ans}\\ \end{align*}

Here,

2\(\pi\)r = 88 cm

or, \(\pi\) = \(\frac {88 × 7}{2 × 22} cm\)

or, r = 2× 7 cm

∴ r = 14 cm

l = 30 cm

Now,

\begin{align*} {\text{Curved Surface Area of the cone}} &= {\pi}rl\\ &= \frac {22}7× 14× 30 cm^2\\ &= 44× 30 cm^2\\ &= 1320 cm^2_{Ans}\\ \end{align*}

Here,

(l + r) = 32 cm

We have,

Circumference = 2\(\pi\)r

or, 44 cm = 2\(\pi\)r

or, \(\pi\)r = \(\frac {44}2\)

∴ \(\pi\)r = 22 cm

Now,

\begin{align*} {\text{Total Surface Area of Cone}} &= {\pi}r (l + r)\\ &= 22 cm × 32 cm\\ &= 704 cm^2_{Ans}\\ \end{align*}

Here,

r = 9 cm

l = 15 cm

Now,

\begin{align*} {\text{Total Surface Area}} & = {\pi}r^2 + {\pi}rl\\ &= {\pi}r (r + l)\\ &= \frac {22}7 × 9 (9 + 15) cm^2\\ &= \frac {22}7 × 9 × 24 cm^2\\ &= 678.857 cm^2_{Ans}\\ \end{align*}

Here,

l = 100 cm

Curved Surface Area = \(\pi\)rl

or, 8800 cm2 = \(\frac {22}7\)× r× 100 cm

or, r = \(\frac {8800 cm^2 × 7}{22 × 100 cm}\)

∴ r = 28 cm

Radius (OR) = 28 cm

In right angled \(\triangle\)POR,

\begin{align*} PO &= \sqrt {PR^2 - OR^2}\\ &= \sqrt {100^2 - 28^2}cm\\ &= 96 cm_{Ans}\\ \end{align*}

Here,

r + l = 32 cm

Toral surface area of the cone = \(\pi\)r (r + l)

By Question,

Total Surface Area = 4928 cm2

or, \(\pi\)r (r + l) = 4928 cm2

or, \(\frac {22r}7\) (32 cm) = 4928 cm2

or, r = 4928× \(\frac {7 cm^2}{22 × 32 cm}\)

∴ r = 49 cmAns

Note: Here, r + l = 32< 49 = r, which is impossible. So, the question is wrong.

Suppose,

r = 7x

h = 12x

Volume (V) = 616 cm3

By formula,

V = 616

or, \(\frac 13\) \(\pi\)r2h = 616

or, \(\frac 13\)× \(\frac {22}7\)× (7x)2 . 12x = 616

or, \(\frac {22 × 49x^2 × 12x}{21}\) = 616

or, 616x3 = 616

or, x3 = 1

∴ x = 1

∴ r = 7x = 7 cm

∴ h = 12x = 12 cm

Now,

\begin{align*} l &=\sqrt {r^2 + h^2}\\ &= \sqrt {7^2 + 12^2}\\ &= \sqrt {193}\\ \end{align*}

\begin{align*} {\text{Curved Surface Area (CA)}} &= {\pi}rl\\ &= \frac {22}7 × 7 × \sqrt {193}\\ &= 305.63 cm^2_{Ans}\\ \end{align*}

Here,

TSA = 704 cm2

CA = 550 cm2

By formula,

TSA = \(\pi\)r (r + l)

or, 704 = \(\pi\)r (r + l)

or, 704 = \(\pi\)r2 + \(\pi\)rl...........................(i)

Again,

CA = \(\pi\)rl

or, 550 = \(\pi\)rl...........................(ii)

From (i) and (ii)

704 = \(\pi\)r2 + 550

or, 154 = \(\pi\)r2

or, r2 = \(\frac {154}{\pi}\)

or, r2 = \(\frac {154 × 7}{22}\)

or, r2 = 49

∴ r = 7 cmAns

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