Subject: Compulsory Mathematics

Solid objects, as shown below are the cones.

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.

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.

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.

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

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.

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

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The base area of a cone is 125 so. cm and its height are 9 cm. Find its volume.

Base are of the cone = \(\pi\)r^{2} = 125 cm^{2}

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*}

Find the volume of the given solid cone.

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*}

The slant height of a cone is 50 cm and its height 48 cm. Find the volume of the cone.

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*}

If the volume of the given cone is 1848 cu. cm. and its radius is 14 cm, what is its height?

Here,

r = 14 cm,

V = 1848 cm^{3}

h = ?

By formula,

V = \(\frac 13\)\(\pi\)r^{2}h

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}

Find the curved surface area of given solid cone.

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*}

A cone have slant height 5 cm, and height 4 cm, find the curved surface area of that cone.

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*}

The circumference of base of a cone is 8 cm and slant height 30 cm. Find the curved surface area of the cone.

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*}

The circumference of base of a cone is 44 cm, and the sum of its radius and slant height is 32 cm, find the total surface area of the cone.

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*}

Find the total surface area of the given cone where the radius is 9 cm and the slant height 15 cm.

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*}

The curved surface area of the given cone is 8800 sq. cm. If the slant height is 100 cm, determine the height PQ of the cone.

Here,

l = 100 cm

Curved Surface Area = \(\pi\)rl

or, 8800 cm^{2} = \(\frac {22}7\)× 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*}

The total surface area of a cone is 4928 square cm. If the sum of the radius of the base and the slant height of the cone is 32 cm, find the radius of its base.

Here,

r + l = 32 cm

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

By Question,

Total Surface Area = 4928 cm^{2}

or, \(\pi\)r (r + l) = 4928 cm^{2}

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

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

∴ r = 49 cm_{Ans}

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

The radius and height of a right circular cone are in the ratio 7: 12. If its volume is 616 cm^{3} find the area of its curved surface.

Suppose,

r = 7x

h = 12x

Volume (V) = 616 cm^{3}

By formula,

V = 616

or, \(\frac 13\) \(\pi\)r^{2}h = 616

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

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

or, 616x^{3} = 616

or, x^{3} = 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*}

The total surface area and the curved surface area of cone are 704 sq. cm and 550 sq. cm respectively. Find the radius of the base of the cone.

Here,

TSA = 704 cm^{2}

CA = 550 cm^{2}

By formula,

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

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

or, 704 = \(\pi\)r^{2} + \(\pi\)rl...........................(i)

Again,

CA = \(\pi\)rl

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

From (i) and (ii)

704 = \(\pi\)r^{2} + 550

or, 154 = \(\pi\)r^{2}

or, r^{2} = \(\frac {154}{\pi}\)

or, r^{2} = \(\frac {154 × 7}{22}\)

or, r^{2} = 49

∴ r = 7 cm_{Ans}

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