Question 13 Marks
The diameters of two cones are equal. If their slant height are in the ratio $5 : 4$, find the ratio of their curved surfaces.
AnswerIt is given that: Diameters of two cones are equal Therefore their radius are also equal
i.e $r_1= r_2$ Let the ratio of slant height be $x$ Therefore$ l_1 = 5x l_2 = 4x$
Therefore Ratio of curved surface area
$=\frac{\text{c}_1}{\text{c}_2}$$\Rightarrow\frac{\text{c}_1}{\text{c}_2}=\frac{\pi\text{r}_1\text{l}_1}{\pi\text{r}_2\text{l}_2}$
$=\frac{\pi\text{r}_1\times\text{5x}}{\pi\text{r}_2\times\text{4x}}=\frac{5}{4}$
$\therefore$ Ratio of curved surface area is $5 : 4$.
View full question & answer→Question 23 Marks
There are two cones. The curved surface area of one is twice that of the other. The slant height of the later is twice that of the former. Find the ratio of their radii.
AnswerLet the curved surface area of $1^{st} cone = 2x$
$C.S.A$ of $2^{nd} cone = x$ Slant height of $1^{st} cone = h$
Slant height of $2^{nd} cone = 2h$
Therefore $\frac{\text{C.S.A }\text{of } 1^\text{st}\text{cone}}{\text{C.S.A }\text{of } 2^\text{nd}\text{cone}}$
$=\frac{\text{2x}}{\text{x}}$
$\Rightarrow\frac{\pi\text{r}_1\text{l}_1}{\pi\text{r}_2\text{l}_2}=\frac{2}{1}$
$\Rightarrow\text{r}_1\times\frac{\text{h}}{\text{r}_2}\times\text{h}=\frac{2}{1}$
$\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{4}{1}$ Therefore ratio of $r_1$_ and $r_2$_ is $4 : 1$
View full question & answer→Question 33 Marks
Find the total surface area of a cone, if its slant height is $21\ m$ and diameter of its base is $24\ m.$
AnswerIt is given that: Diameter of the cone $(d)$ = 24m Radius of the cone $(r)$
$=\frac{24}{2}=12\text{m}$ Slant height of the cone $(l) = 21m$
$T.S.A$ = Curved Surface Area of Cone + Area of Circular Base
$=\pi\times\text{r}\times\text{l}+\pi\times\text{r}^2$
$=\Big(\frac{22}{7}\times12\times21\Big)+\Big(\frac{22}{7}\times12\times12\Big)$
$=\frac{22}{7}\times12(12+21)=1244.57\text{m}^2.$
Therefore the total surface area of the cone is $1244.57m^2$.
View full question & answer→Question 43 Marks
A conical pit of top diameter $3.5m$ is $12m$ deep. What is its capacity in kilolitres?
AnswerIt is given that: Diameter of the conical pit $(d) = 3.5m$
Height of the conical pit $(h) = 12m$
Radius of the conical pit $(r) = ?$
Volume of the conical pit $(v) = ?$
Radius of the conical pit (r) $=\frac{\text{d}}{2}=\frac{3.5}{2}=1.75\text{m}$
Volume of the cone (v) $=\frac{1}{3}\pi\text{r}^2\text{h}$$=\frac{1}{3}\times3.14\times1.75^2\times12=38.5\text{m}^3$
Capacity of the pit $= (38.5 \times 1)$ kilolitres $= 38.5$ kilolitres
View full question & answer→Question 53 Marks
The radius of a cone is $7\ cm$ and area of curved surface is $176cm^2.$ Find the slant height.
AnswerIt is given that: Radius of cone $(r) = 7\ cm$
Curved Surface Area $(C.S.A)$ $= 176cm^2$
Slant Height of tent $(l) = ?$
Now we know, $C.S.A$ $=\pi\text{rl}$
$\Rightarrow\pi\text{rl}=176$
$\Rightarrow\frac{22}{7}\times7\times\text{l}=176$
$\Rightarrow\text{l}=\frac{176\times7}{22\times7}=8\text{cm}$
Therefore the slant height of the cone is $8\ cm$.
View full question & answer→Question 63 Marks
If the height and slant height of a cone are $21\ cm$ and $28\ cm$ respectively. Find its volume.
AnswerThe formula of the volume of a cone with base radius $‘r’$ and vertical height $‘h’$ is given as:
Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
The vertical height is given as $‘h’ = 21\ cm$, and the slant height is given as $‘l’ = 28\ cm$.
To find the base radius $‘r’$ we use the relation between $r, l$ and $h$.
We know that in a cone $\text{l}^2=\text{r}^2+\text{h}^2$
$\text{r}^2=\text{l}^2-\text{h}^2$
$\text{r}=\sqrt{\text{l}^2-\text{h}^2}$
$\text{r}=\sqrt{\text{28}^2-\text{21}^2}$
$\text{r}=\sqrt{784-441}$
$\text{r}=\sqrt{343}$
Therefore the base radius is, $\text{r}=\sqrt{343}\text{cm}.$
Substituting the values of $\text{r}=\sqrt{343}\text{cm}$ and $h = 21\ cm$ in the formula for volume of a cone.
Volume $=\frac{\pi\text{r}^2\text{h}}{3}=\frac{\pi\big(\sqrt{343}\big)^2\times21}{3}$
$=2401\pi$
Hence the volume of the given cone with the specified dimensions is $2401\pi\text{cm}^3.$
View full question & answer→Question 73 Marks
A circus tent is cylindrical to a height of $3m$ and conical above it. If its diameter is $105m$ and the slant height of the conical portion is $53m$, calculate the length of the canvas $5m$ wide to make the required tent.
AnswerGiven diameter $= 105\ m$,
Radius $=\frac{105}{2}\text{m}=52.5\text{m}$
Therefore curved surface area of circus tent $=\pi\text{rl}+2\pi\text{rh}$
$=\Big(\frac{22}{7}\times52.5\times53\Big)+\Big(2\times\frac{22}{7}\times52.5\times3\Big)$
$=8745+990=9735\text{m}^2$
Therefore length of the canvas required for tent$=\frac{\text{Area of canvas}}{\text{Width of canvas}}$
$=\frac{9735}{5}\text{m}=1947\text{m}$
View full question & answer→Question 83 Marks
The slant height and base diameter of a conical tomb are $25\ m$ and $14\ m$ respectively.
Find the cost of white washing its curved surface area at the rate of $Rs. 210$ per $100 \mathrm{~m}^2$.
AnswerIt is given that
Slant height of conical tomb $(I)=25 \mathrm{~m}$
Base radius ( $r$ ) of tomb $=\frac{14}{2}=7 \mathrm{~m}$
Curved surface area of conical length tomb $=\pi \mathrm{rl}$
$=\frac{22}{7} \times 7 \times 25$
$=550 \mathrm{~m}^2$
Cost of white washing $100 \mathrm{~m}^2$ area $=$ $Rs. 210$
Cost of white washing $550 \mathrm{~m}^2$ area $=$ $RS.$ $\left(\frac{210 \times 550}{100}\right)=$ $Rs. 1155$
Therefore the cost of white washing the whole tomb is $Rs. 1155$
View full question & answer→Question 93 Marks
The height of a cone $21\ cm$. Find the area of the base if the slant height is $28\ cm.$
AnswerIt is given that:
Height of the traffic cone $(h) = 21\ cm$
Slant height of the traffic cone $(l) = 28\ cm$
Now we know that, $l^2 = r^2 + h^2 28^2 = r^2 + 21^2 r^2 = 28^2 − 21^2$
$\text{r}=\sqrt{77}\text{cm}$
Area of the circular base $=\pi\text{r}^2$
$=\frac{22}{7}\times\big(7\sqrt{7}\big)^2=1078\text{cm}^2$
Therefore the area of the base is $1078\ cm^2.$
View full question & answer→Question 103 Marks
Find the total surface area of a right circular cone with radius $6\ cm$ and height $8\ cm$.
AnswerIt is given that:
Radius of the cone $(r)=6 \mathrm{~cm}$
Height of the cone $(h)=8 \mathrm{~cm}$
Total Surface Area of the Cone $(T . S . A)=$ ?
$l^2=r^2+h^2=6^2+8^2=36+64=100 \mathrm{I}=10 \mathrm{~cm}$
$\text { T.S.A = Curved Surface Area of Cone }+ \text { Area of Circular Base }$
$=\pi \times r \times 1+\pi \times r^2$
$=\left(\frac{22}{7} \times 6 \times 10\right)+\left(\frac{22}{7} \times 6 \times 6\right)$
$=\frac{1320+792}{7}$
$=301.171 \mathrm{~cm}^2$
Therefore the area of the base is $301.71 \mathrm{~cm}^2$.
View full question & answer→Question 113 Marks
A conical tent is $10\ m$ high and the radius of it base is $24\ m$. Find the slant height of the tent. If the cost of $1m^2$ canvas is $Rs. 70$, find the cost of canvas required for the tent.
Answer$i.$It is given that:
Height of the conical tent $(h) = 10\ m$
Radius of conical tent $(r) = 24\ m$
Let slant height of conical tent be $l$
$l^2 = h^2 + r^2$
$= 10^2 + 24^2 = 100 + 576$
$= 676\ m^2$
$\Rightarrow l = 26\ m$
Thus, the slant height of the conical tent is $26\ m$.
$ii.$ It is given that:
Radius $(r) = 24\ m$
Slant height $(l) = 26\ m$
$\text{C.S.A}$ of tent $=\pi\text{rl}$
$\frac{22}{7}\times24\times26$
$=\frac{22}{7}\times24\times26$
$=\frac{1378}{7}\text{m}^2$
Cost of $1m^2$ canvas $= Rs. 70$
Cost of $\frac{1378}{7}\text{m}^2$ canvas $=\text{Rs.}\ \frac{1378}{7}\times70$
$= Rs. 1,37,280$
Thus the cost of canvas required to make the tent is $= Rs. 1,37,280$
View full question & answer→Question 123 Marks
Find the ratio of the curved surface area of two cones if their diameter of the bases are equal and slant heights are in the ratio $4 : 3.$
AnswerGiven that, Diameter of two coins are equal. Therefore their radius are equal.
Let $r_1 = r_2 = r$
Let ratio be $x$ Therefore slant height $l_1$_ of $1^{st}$
cone =$4x$ Similarly slant height $l_2 $ of $2^{nd}$ cone $= 3x$
Therefore $\frac{\text{C.S.A}_1}{\text{C.S.A}_2}$
$=\frac{\pi\times\text{r}_1\times\text{l}_2}{\pi\times\text{r}_2\times\text{l}_2}$
$=\frac{\pi\times\text{r}\times\text{4x}}{\pi\times\text{r}\times\text{3x}}$
$=\frac{4}{3}$
Hence ratio is $4 : 3$.
View full question & answer→Question 133 Marks
The height of a cone is $15\ cm$. If its volume is $500\pi\text{cm}^3,$ then find the radius of its base.
AnswerThe formula of the volume of a cone with base radius $‘r’$ and vertical height $‘h’$ is given as:
Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
It is given that the height of the cone is $‘h’ = 15cm$ and that the volume of the cone is $500\pi\text{cm}^3$
We can now find the radius of base $‘r’$ by using the formula for the volume of a cone.
$\text{r}^2=\frac{3(\text{Volume of the cone})}{\pi\text{h}}$
$=\frac{3(500\pi)}{\pi(15)}$
$\text{r}^2=100$
$\text{r}=10$
Hence the radius of the base of the given cone is $10\ cm$.
View full question & answer→Question 143 Marks
If the volume of a right circular cone of height $9\ cm$ is $48\pi\text{cm}^3,$ find the diameter of its base.
AnswerThe formula of the volume of a cone with base radius $‘r’$ and vertical height $‘h’$ is given as:
Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$ It is given that the height of the cone is $‘h’ = 9\ cm$ and that the volume of the cone is $48\pi\text{cm}^3$ We can now find the radius of base $‘r’$ by using the formula for the volume of a cone.
$\text{r}^2=\frac{3(\text{Volume of the cone})}{\pi\text{h}}$
$=\frac{3(48\pi)}{\pi(9)}$
$\text{r}^2=16$
$\text{r}=4$
Hence the radius of the base of the cone with given dimensions is $‘r’ = 4\ cm$.
The diameter of base is twice the radius of the base.
Hence the diameter of the base of the cone is $8\ cm$.
View full question & answer→Question 153 Marks
A cylinder and a cone have equal radii of their bases and equal heights. Show that their volumes are in the ratio $3 : 1.$
AnswerIt's given that: A cylinder and a cone are having equal radii of their bases and heights
Let the radius of the cone = radius of the cylinder $= r$
Height of the cone = height of the cylinder $= h$
Let the volume of cone $= v_x$_ Volume of cylinder $= v_y$
$\Rightarrow\frac{\text{v}_\text{x}}{\text{v}_\text{y}}=\frac{\frac{1}{3}\pi\text{r}^2\text{h}}{\pi\text{r}^2\text{h}}=\frac{1}{3}$
$\Rightarrow\frac{\text{v}_\text{y}}{\text{v}_\text{x}}=\frac{3}{1}$
Therefore the ratio of their volumes is $3 : 1.$
View full question & answer→Question 163 Marks
The radius and slant height of a cone are in the ratio $4 : 7$. If its curved surface area is $792 \mathrm{~cm}^2$, find its radius.
AnswerIt is given that: Curved surface area $=\pi\text{r}\text{l}=792$
Let the radius $(r) = 4x$ Height $(h) = 7x$
Now, $C.S.A$ $= 792$ $\frac{22}{7}\times\text{4x}\times\text{7x}=792$
$\Rightarrow\text{88x}^2=792$
$\Rightarrow\text{x}^2=\frac{792}{88}=9$
$\Rightarrow\text{x}=3$
Therefore Radius $=\text{4x}=4\times3=12\text{cm}$
View full question & answer→Question 173 Marks
The curved surface area of a cone is $4070cm^2$ and diameter is $70\ cm$ .What is its slant height?
AnswerDiameter of the cone $(d) = 70\ cm$
Radius of the cone $(r) = d^2 = 35cm$
Slant height of the cone $(l) = ?$
Now, Curved Surface Area $= 4070cm^2$
$\Rightarrow\pi\text{r}\text{l}=4070$
Where, $r$ = Radius of the cone $l$ = Slant height of the cone
Therefore $\pi\text{r}\text{l}=4070$
$\Rightarrow\frac{22}{7}\times35\times\text{l}=4070$
$\Rightarrow\text{l}=\frac{4070\times7}{22\times35}=37\text{cm}$
Therefore slant height of the cone is $37\ cm$.
View full question & answer→Question 183 Marks
If the radius of the base of a cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone?
AnswerLet the radius of the cone be $r_x $ and height be $h_x$
Then the volume of cone$\text{v}_\text{x}=\frac{1}{3}\pi\text{r}_\text{x}^2\text{h}_\text{x}$
Now, Radius of the reduced cone $=\frac{\text{r}_\text{x}}{2}$
Therefore volume of reduced cone vy$=\frac{1}{3}\pi\Big(\frac{\text{r}}{2}\Big)^2\times\text{h} _\text{x}$
$=\frac{1}{12}\times\pi\times\text{r}_\text{x}\text{x}^2\times\text{h}_\text{x}$
$\therefore\frac{\text{v}_\text{y}}{\text{v}_\text{x}}=\frac{\frac{1}{2}\pi\text{r}^2\text{h}}{\frac{1}{3}\pi\text{r}^2\text{h}}$
$=\frac{3}{12}=\frac{1}{4}$
Therefore the ratio between the volumes of the reduced and the original cone is $1 : 4.$
View full question & answer→Question 193 Marks
The radius of a cone is $5\ cm$ and vertical height is $12\ cm$. Find the area of the curved surface.
AnswerIt is given that: Radius of cone $(r) = 5\ cm$
Height of the tent $(h) = 12\ cm$
Slant Height of tent $(l) = ?$
Curved Surface Area$ (C.S.A) = ?$
Now we know we that, $l = r^2 + h^2 l^2= r^2 + h^2 l^2$
$= 5^2 + 12^2 l^2= 25 + 144$
$ \Rightarrow l = 13cm$
Now, C.S.A $=\pi\text{rl}$
$= 3.14 \times 5 \times 12 = 204.28cm^2$
Therefore the curved surface area of the cone is $204.28cm^2$
View full question & answer→Question 203 Marks
Find the volume of the largest right circular cone that can be fitted in a cube whose edge is $14\ cm$.
AnswerRadius of the base of the largest cone $=\frac{1}{2}\times$ edge of the cube $=\frac{1}{2}\times14=7\text{cm}$
Height of the cone = Edge of the cube $= 14\ cm$
Therefore, Volume of cone $(v)$ $=\frac{1}{3}\pi\text{r}^2\text{h}$ $=\frac{1}{3}\times3.14\times7^2\times14=718.66\text{cm}^3$
View full question & answer→Question 213 Marks
Curved surface area of a cone is $308\ cm^2$ and its slant height is $14\ cm$. Find the radius of the base and total surface area of the cone.
Answer$i.$ It is given that
Slant height of cone $= 14\ cm$
Let radius of circular end of cone $= r$
Curved surface area of cone $=\pi\text{rl}$
$\Rightarrow308\text{cm}^2=\frac{22}{7}\times\text{r}\times14$
$\Rightarrow\text{r}=\frac{308}{44}=7\text{cm}$
Thus radius of circular end of cone $= 7\ cm$.
$ii.$ It is given that $\text{C.S.A}$
$= 308\ cm^2$
We know that total surface area of a cone
$=$ Curved surface area of a cone $+$ Area of base
$=\pi\text{rl}+\pi\text{r}^2$
$=\bigg[308+\Big(\frac{22}{7}\times7^2\Big)\bigg]$
$=308+154$
$=462\text{ cm}^2$
Thus total surface area of the cone is $462\ cm^2$
View full question & answer→Question 223 Marks
A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is $24\ m$. The height of the cylindrical portion is $11\ m$ while the vertex of the cone is $16\ m$ above the ground. Find the area of the canvas required for the tent.
AnswerIt is given that Diameter of cylinder $= 24\ m$
Therefore $\text{Radius}=\frac{\text{Diameter}}{2}$
$=\frac{24}{2}=12\text{cm}$
Also radius of cone $= 12m$
Height of cylinder $= 11m$
Height of cone $= 16 - 11 = 5m$
Slant height of cone$=\sqrt{5^2+12^2}$
$=13\text{m}$
Therefore area of canvas required for the tent
$=\pi\text{rl}+2\pi\text{rh}$
$=\frac{22}{7}\big[(12+13)+(2\times12\times11)\big]$
$=490.286+829.741$
$=1320\text{m}^3$
View full question & answer→Question 233 Marks
The area of the curved surface of a cone is $60\pi\text{cm}^2.$ If the slant height of the cone be $8\ cm$, find the radius of the base.
AnswerIt is given that: Curved surface area $(C.S.A)$ $=60\pi\text{cm}^2$
Slant height of the cone $(l) = 8cm$
Radius of the cone $(r) = ?$
Now we know, Curved Surface Area $(C.S.A)$ $=\pi\text{rl}$
$\Rightarrow\pi\text{rl}=60\pi\text{cm}^2$
$\Rightarrow\text{r}\times8=60$
$\Rightarrow\text{r}=7.5\text{cm}$
Therefore the radius of the base of the cone is $7.5\ cm$.
View full question & answer→Question 243 Marks
A joker's cap is in the form of a right circular cone of base radius $7\ cm$ and height $24\ cm$. Find the area of the sheet required to make $10 $such caps.
AnswerGiven that, Radius of conical cap $(r)=7 \mathrm{~cm}$
Height of the conical cap $(h)=24 \mathrm{~cm}$
Slant height $(l)$ of conical cap
$=\sqrt{\mathrm{r}^2+\mathrm{h}^2}$
$=\sqrt{7^2+24^2}$
$=25 \mathrm{~cm}$
$C.S.A$ of $1$ conical cap $=\pi \mathrm{rl}=\frac{22}{7} \times 7 \times 25$
$=550 \mathrm{~cm}^2$
Curved surface area of $0$ such conical caps $=5500 \mathrm{~cm}^2$
Thus, $5500 \mathrm{~cm}^2$ sheet will be required for making $10$ caps.
View full question & answer→Question 253 Marks
Find the weight of a solid cone whose base is of diameter $14\ cm$ and vertical height $51\ cm$, supposing the material of which it is made weighs $10$ grams per cubic $cm$.
AnswerIt is given that:
Diameter $(d) = 14\ cm$
Height of the cone $(h) = 51\ cm$
Radius of the cone$(r)$
$=\frac{\text{d}}2{}$
$=\frac{14}{2}=7\text{cm}$
Therefore, Volume of cone $(v)$ $=\frac{1}{3}\pi\text{r} ^2\text{h}$
$=\frac{1}{3}\times3.14\times7\times5\times51=2618\text{cm}^3$
Now it is given that $1\ cm^3$ material wheighs $10\ gm$.
Therefore, $2618 cm^3$ weighs $=2618 \times 10=26180$ grams or $26.180\ kg$ .
View full question & answer→Question 263 Marks
A heap of wheat is in the form of a cone of diameter $9m$ and height $3.5\ m$. Find its volume. How much is canvas cloth required to just cover the heap? $($Use $\pi=3.14).$
AnswerIt is given that Diameter of heap $(d) = 9m$
Therefore, Radius of the heap $(r)$$=\frac{\text{d}}2{}$
$=\frac{9}{2}=4.5\text{m}$
Height of the heap $(h) = 3.5m$ Therefore,
Volume of the heap $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\frac{1}{3}\times3.14\times4.5^2\times3.5$
$=74.18\text{m}^3$
Now,$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$\text{l}=\sqrt{\text{4.5}^2+\text{3.5}^2}=5.70\text{m}$
Area to be covered by the cloth = Curved surface area of the heap
$=\pi\text{rl}=3.14\times4.5\times5.70=80.54\text{m}^3$
View full question & answer→