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
.
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*}
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 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}
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 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*}
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: Here, r + l = 32< 49 = r, which is impossible. So, the question is wrong.
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*}
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}
A cone has its base radius 7 cm and height 21 cm, Find its volume.
1050 cm^{3}
1078 cm^{3}
970 cm^{3}
1025 cm^{3}
Determine the volume of a cone having height 48 cm and the slant height 50 cm.
9950 cm^{3}
9856 cm^{3}
8856 cm^{3}
9956 cm^{3}
A cone has the diameter of its base 14 cm and slant height 25 cm.Fina the volume.
1240 cm^{3}
1232 cm^{3}
1200 cm^{3}
1230 cm^{3}
A cone whose radius of base is 7 cm has a volume of 1232 cm^{3}. Find the height of the cone.
20 cm
24 cm
21cm
30 cm
A cone has a volume of 1078 cm^{3}, If the height of the cone is 21 cm, calculate the radius of its base.
7 cm
4 cm
9 cm
12 cm
The radius of the base of a cone is 7 cm and height 25 cm.calculate the curved surface area of the cone.
45.14 cm^{2}
47.47 cm^{2}
47.14 cm^{2}
31.13 cm^{2}
The curved surface area of a cone is 550 cm^{2} and its slant height is 25 cm.Find the height of the cone.
88 cm
86 cm
99 cm
96 cm
The radius of the base of a cone is 6 cm and slant height is 8 cm.Find its total surface area.
264 cm^{2}
284 cm^{2}
274 cm^{2}
260 cm^{2}
The radius of the base of a cone is 5 cm and slant height is 9 cm. Find the total surface area.
225 cm^{2}
235 cm^{2}
230 cm^{2}
220 cm^{2}
If the total surface area of the cone is 704 cm^{2} and radius of its base 7 cm, find the slant height of the cone.
22 cm
30 cm
25 cm
28 cm
The total surface area and the diameter of the base of a cone are.594 sq.cm and 18 cm respectively. find the slant height of the cone.
13 cm
12 cm
15 cm
14 cm
The total surface area and the diameter of a base of the cone are 300π sq. cm and 24 cm respectively.Find the slant height of the cone.
13 cm
16 cm
15 cm
14 cm
The total surface area of a cone is 814 square cm. If the sum of its slant height and the radius of its base is 37 cm, find the slant height and the radius of its base is 37 cm, find the slant height of the cone.
48 cm
35 cm
20 cm
30 cm
The total surface area of a cone is 704 square cm.If the sum of its slant height and the radius of its base is32 cm, find the slant height of the cone.
33 cm
25 cm
30 cm
44 cm
If the circumference of base of a cone is 33 cm.and the slant height is 10 cm.Find the curved surface area.
165 cm^{2}
135 cm^{2}
125 cm^{2}
145 cm^{2}
A cylinder of height 40 cm and diameter 14 cm is melted to form a right circular cone of height 30 cm.what is the radius of the cone.
19 cm
14 cm
17 cm
15 cm
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