## The diameter of a sphere is decreased by 25%.

[question] Question. The diameter of a sphere is decreased by $25 \%$. By what per cent does its curved surface area decrease? [/question] [solution] Solution: Let the diameter of the sphere be d. Radius $\left(r_{1}\right)$ of sphere $=\frac{d}{2}$ New radius $\left(\mathrm{r}_{2}\right)$ of sphere $=\frac{d}{2}\left(1-\frac{25}{100}\right)=\frac{3}{8} d$ $\operatorname{CSA}\left(S_{1}\right)$ of sphere $=4 \pi r_{1}^{2}$ $=4 \pi\left(\frac{d}{2}\right)^{2}=\pi d^{2}$ CSA $\left(S_{2}\right)$ o...

## A capsule of medicine is in the shape of a sphere of diameter 3.5 mm.

[question] Question. A capsule of medicine is in the shape of a sphere of diameter $3.5 \mathrm{~mm}$. How much medicine (in $\mathrm{mm}^{3}$ ) is needed to fill this capsule? $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Radius $(r)$ of capsule $=\left(\frac{3.5}{2}\right) \mathrm{mm}=1.75 \mathrm{~mm}$ Volume of spherical capsule $=\frac{4}{3} \pi r^{3}$ $=\left[\frac{4}{3} \times \frac{22}{7} \times(1.75)^{3}\right] \mathrm{mm}^{3}$ $=22.458 \mathrm{...

## Twenty seven solid iron spheres,

[question] Question. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the (i) radius r' of the new sphere, (ii) ratio of S and S'. [/question] [solution] Solution: (i)Radius of 1 solid iron sphere $=r$ Volume of 1 solid iron sphere $=\frac{4}{3} \pi r^{3}$ Volume of 27 solid iron spheres $=27 \times \frac{4}{3} \pi r^{3}$ 27 solid iron spheres are melted to form 1 iron sphere. Therefore, the volume of this iron sphere wil...

## A dome of a building is in the form of a hemisphere. From inside,

[question] Question. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs $498.96$. If the cost of white-washing is Rs $2.00$ per square meter, find the (i) inside surface area of the dome, (ii) volume of the air inside the dome. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: (i) Cost of white-washing the dome from inside $=$ Rs $498.96$ Cost of white-washing $1 \mathrm{~m}^{2}$ area $=$ Rs 2 Therefore, C...

## Find the volume of a sphere whose surface area is 154 cm2.

[question] Question. Find the volume of a sphere whose surface area is $154 \mathrm{~cm}^{2} .\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Let radius of sphere be r. Surface area of sphere $=154 \mathrm{~cm}^{2}$ $\Rightarrow 4 \pi r^{2}=154 \mathrm{~cm}^{2}$ $\Rightarrow r^{2}=\left(\frac{154 \times 7}{4 \times 22}\right) \mathrm{cm}^{2}$ $\Rightarrow r=\left(\frac{7}{2}\right) \mathrm{cm}=3.5 \mathrm{~cm}$ Volume of sphere $=\frac{4}{3} \pi r^{3}$ $=\l...

## A hemispherical tank is made up of an iron sheet 1 cm thick.

[question] Question. A hemispherical tank is made up of an iron sheet $1 \mathrm{~cm}$ thick. If the inner radius is $1 \mathrm{~m}$, then find the volume of the iron used to make the tank. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Inner radius $\left(r_{1}\right)$ of hemispherical tank $=1 \mathrm{~m}$ Thickness of hemispherical tank $=1 \mathrm{~cm}=0.01 \mathrm{~m}$ Outer radius $\left(r_{2}\right)$ of hemispherical tank $=(1+0.01) \mathrm{m}=1.01...

## How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

[question] Question. How many litres of milk can a hemispherical bowl of diameter $10.5 \mathrm{~cm}$ hold? $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Radius $(r)$ of hemispherical bowl $=\left(\frac{10.5}{2}\right) \mathrm{cm}=5.25 \mathrm{~cm}$ Volume of hemispherical bowl $=\frac{2}{3} \pi r^{3}$ $=\left[\frac{2}{3} \times \frac{22}{7} \times(5.25)^{3}\right] \mathrm{cm}^{3}$ $=303.1875 \mathrm{~cm}^{3}$ Capacity of the bowl $=\left(\frac{303.1875}...

## The diameter of the moon is approximately one-fourth of the diameter of the earth.

[question] Question. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon? [/question] [solution] Solution: Let the diameter of earth be $d$. Therefore, the radius of earth will be $\frac{d}{2}$. Diameter of moon will be $\frac{d}{4}$ and the radius of moon will be $\frac{d}{8}$. Volume of moon $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi\left(\frac{d}{8}\right)^{3}=\frac{1}{512} \times \frac{4}{3} \pi d^...

## The diameter of a metallic ball is 4.2 cm.

[question] Question. The diameter of a metallic ball is $4.2 \mathrm{~cm}$. What is the mass of the ball, if the density of the metal is $8.9 \mathrm{~g}$ per $\mathrm{cm}^{3} ?\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Radius $(r)$ of metallic ball $=\left(\frac{4.2}{2}\right) \mathrm{cm}=2.1 \mathrm{~cm}$ Volume of metallic ball $=\frac{4}{3} \pi r^{3}$ $=\left[\frac{4}{3} \times \frac{22}{7} \times(2.1)^{3}\right] \mathrm{cm}^{3}$ $=38.808 \mathrm{~...

## Find the volume of a sphere whose radius is

[question] Question. Find the volume of a sphere whose radius is (i) $7 \mathrm{~cm}$ (ii) $0.63 \mathrm{~m}$ $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: (i) Radius of sphere $=7 \mathrm{~cm}$ Volume of sphere $=\frac{4}{3} \pi r^{3}$ $=\left[\frac{4}{3} \times \frac{22}{7} \times(7)^{3}\right] \mathrm{cm}^{3}$ $=\left(\frac{4312}{3}\right) \mathrm{cm}^{3}$ $=1437 \frac{1}{3} \mathrm{~cm}^{3}$ Therefore, the volume of the sphere is $1437 \frac{1}{3} \m...

## If the triangle ABC in the Question 7 above is revolved about the side 5 cm,

[question] Question. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, 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. [/question] [solution] Solution: When right-angled $\triangle A B C$ is revolved about its side $5 \mathrm{~cm}$, a cone will be formed having radius $(r)$ as $12 \mathrm{~cm}$, height $(h)$ as $5 \mathrm{~cm}$, and slant height ( $l$ ) as $13 \mathrm{~cm}$. Volume of co...

## A right triangle $A B C$ with sides $5 \mathrm{~cm}$,

[question] Question. A right triangle $A B C$ with sides $5 \mathrm{~cm}, 12 \mathrm{~cm}$ and $13 \mathrm{~cm}$ is revolved about the side $12 \mathrm{~cm}$. Find the volume of the solid so obtained. [/question] [solution] Solution: When right-angled $\triangle \mathrm{ABC}$ is revolved about its side $12 \mathrm{~cm}$, a cone with height $(h)$ as $12 \mathrm{~cm}$, radius $(r)$ as $5 \mathrm{~cm}$, and slant height $(l) 13 \mathrm{~cm}$ will be formed. Volume of cone $=\frac{1}{3} \pi r^{2} h$...

## The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres.

[question] Question. The capacity of a closed cylindrical vessel of height $1 \mathrm{~m}$ is $15.4$ litres. How many square metres of metal sheet would be needed to make it? $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Let the radius of the circular end be r. Height (h) of cylindrical vessel = 1 m Volume of cylindrical vessel $=15.4$ litres $=0.0154 \mathrm{~m}^{3}$ $\pi r^{2} h=0.0154 \mathrm{~m}^{3}$ $\left(\frac{22}{7} \times r^{2} \times 1\right) \...

## It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep.

[question] Question. It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel $10 \mathrm{~m}$ deep. If the cost of painting is at the rate of Rs 20 per $\mathrm{m}^{2}$, find (i) Inner curved surface area of the vessel (ii) Radius of the base (iii) Capacity of the vessel Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: (i) Rs 20 is the cost of painting $1 \mathrm{~m}^{2}$ area. Rs 2200 is the cost of painting $=\left(\frac{1}{20} \times 2200\right) \math...

## If the lateral surface of a cylinder is $94.2 \mathrm{~cm}^{2}$

[question] Question. If the lateral surface of a cylinder is $94.2 \mathrm{~cm}^{2}$ and its height is $5 \mathrm{~cm}$, then find (i) radius of its base (ii) its volume. [Use $\pi=3.14]$ [/question] [solution] Solution: (i) Height $(h)$ of cylinder $=5 \mathrm{~cm}$ Let radius of cylinder be r. CSA of cylinder $=94.2 \mathrm{~cm}^{2}$ $2 \pi r h=94.2 \mathrm{~cm}^{2}$ $(2 \times 3.14 \times r \times 5) \mathrm{cm}=94.2 \mathrm{~cm}^{2}$ $r=3 \mathrm{~cm}$ (ii) Volume of cylinder $=\pi r^{2} h$ ...

## The circumference of the base of cylindrical vessel is 132 cm and its height is 25 cm.

[question] Question. The circumference of the base of cylindrical vessel is $132 \mathrm{~cm}$ and its height is $25 \mathrm{~cm}$. How many litres of water can it hold? (1000 cm $^{3}=1 /$ ) $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Let the radius of the cylindrical vessel be r. Height (h) of vessel = 25 cm Circumference of vessel $=132 \mathrm{~cm}$ $2 \pi r=132 \mathrm{~cm}$ $r=\left(\frac{132 \times 7}{2 \times 22}\right) \mathrm{cm}=21 \mathrm{~...

## A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour.

[question] Question. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute? [/question] [solution] Solution: Rate of water flow = 2 km per hour $=\left(\frac{2000}{60}\right) \mathrm{m} / \mathrm{min}$ $=\left(\frac{100}{3}\right) \mathrm{m} / \mathrm{min}$ Depth (h) of river = 3 m Width (b) of river = 40 m Volume of water flowed in $1 \mathrm{~min}=\left(\frac{100}{3} \times 40 \times 3\right) \mathrm{m}^{3}=4000 \mathrm{~m}^{3...

## A solid cube of side 12 cm is cut into eight cubes of equal volume.

[question] Question. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas. [/question] [solution] Solution: Side (a) of cube = 12 cm Volume of cube $=(a)^{3}=(12 \mathrm{~cm})^{3}=1728 \mathrm{~cm}^{3}$ Let the side of the smaller cube be $a_{1}$. Volume of 1 smaller cube $=\left(\frac{1728}{8}\right) \mathrm{cm}^{3}=216 \mathrm{~cm}^{3}$ $\left(a_{1}\right)^{3}=216 \mathrm{~cm}^{3}$ $\Rightarr...

## A godown measures 40 m × 25 m × 15 m.

[question] Question. A godown measures $40 \mathrm{~m} \times 25 \mathrm{~m} \times 15 \mathrm{~m}$. Find the maximum number of wooden crates each measuring $1.5 \mathrm{~m} \times 1.25 \mathrm{~m} \times 0.5 \mathrm{~m}$ that can be stored in the godown. [/question] [solution] Solution: The godown has its length $\left(l_{1}\right)$ as $40 \mathrm{~m}$, breadth $\left(b_{1}\right)$ as $25 \mathrm{~m}$, height $\left(h_{1}\right)$ as $15 \mathrm{~m}$, while the wooden crate has its length $\left...

## A village, having a population of 4000,

[question] Question. A village, having a population of 4000 , requires 150 litres of water per head per day. It has a tank measuring $20 \mathrm{~m} \times 15 \mathrm{~m} \times 6 \mathrm{~m}$. For how many days will the water of this tank last? [/question] [solution] Solution: The given tank is cuboidal in shape having its length (l) as 20 m, breadth (b) as 15 m, and height (h) as 6 m. Capacity of tank $=1 \times b \times h$ $=(20 \times 15 \times 6) \mathrm{m}^{3}=1800 \mathrm{~m}^{3}=1800000$...

## The capacity of a cuboidal tank is 50000 litres of water.

[question] Question. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m. [/question] [solution] Solution: Let the breadth of the tank be b m. Length (l) and depth (h) of tank is 2.5 m and 10 m respectively. Volume of $\operatorname{tank}=1 \times b \times h$ $=(2.5 \times b \times 10) \mathrm{m}^{3}$ $=25 b \mathrm{~m}^{3}$ Capacity of tank $=25 b \mathrm{~m}^{3}=25000 \mathrm{~b}$ litres $\therefore 250...

## Find the cost of digging a cuboidal pit 8 m long,

[question] Question. Find the cost of digging a cuboidal pit $8 \mathrm{~m}$ long, $6 \mathrm{~m}$ broad and $3 \mathrm{~m}$ deep at the rate of Rs 30 per $\mathrm{m}^{3}$. [/question] [solution] Solution: The given cuboidal pit has its length $(l)$ as $8 \mathrm{~m}$, width $(b)$ as $6 \mathrm{~m}$, and depth $(h)$ as $3 \mathrm{~m}$. Volume of pit $=1 \times b \times h$ $=(8 \times 6 \times 3) \mathrm{m}^{3}=144 \mathrm{~m}^{3}$ Cost of digging per $\mathrm{m}^{3}$ volume $=$ Rs 30 Cost of dig...

## A cuboidal vessel is 10 m long and 8 m wide.

[question] Question. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid? [/question] [solution] Solution: Let the height of the cuboidal vessel be h. Length (l) of vessel = 10 m Width (b) of vessel = 8 m Volume of vessel $=380 \mathrm{~m}^{3}$ $\therefore / \times b \times h=380$ $[(10)(8) h] \mathrm{m}^{2}=380 \mathrm{~m}^{3}$ $h=4.75 \mathrm{~m}$ Therefore, the height of the vessel should be 4.75 m. [/solution]...

## A cuboidal water tank is 6 m long,

[question] Question. A cuboidal water tank is $6 \mathrm{~m}$ long, $5 \mathrm{~m}$ wide and $4.5 \mathrm{~m}$ deep. How many litres of water can it hold? $\left(1 \mathrm{~m}^{3}=1000 /\right)$ [/question] [solution] Solution: The given cuboidal water tank has its length $(l)$ as $6 \mathrm{~m}$, breadth $(b)$ as $5 \mathrm{~m}$, and height $(h)$ as $4.5 \mathrm{~m}$. Volume of tank $=l \times b \times h$ $=(6 \times 5 \times 4.5) \mathrm{m}^{3}=135 \mathrm{~m}^{3}$ Amount of water that 1 m3 vo...

## A matchbox measures $4 \mathrm{~cm} \times 2.5 \mathrm{~cm} \times 1.5 \mathrm{~cm}$.

[question] Question. A matchbox measures $4 \mathrm{~cm} \times 2.5 \mathrm{~cm} \times 1.5 \mathrm{~cm}$. What will be the volume of a packet containing 12 such boxes? [solution] Solution: Matchbox is a cuboid having its length $(l)$, breadth $(b)$, height $(h)$ as $4 \mathrm{~cm}, 2.5 \mathrm{~cm}$, and $1.5 \mathrm{~cm}$. Volume of 1 match box $=1 \times b \times h$ $=(4 \times 2.5 \times 1.5) \mathrm{cm}^{3}=15 \mathrm{~cm}^{3}$ Volume of 12 such matchboxes $=(15 \times 12) \mathrm{cm}^{3}$ ...

## A right circular cylinder just encloses a sphere of radius r (see figure).

[question] Question. A right circular cylinder just encloses a sphere of radius r (see figure). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in (i) and (ii). [/question] [solution] Solution: (i) Surface area of sphere $=4 \pi r^{2}$ (ii) Height of cylinder $=r+r=2 r$ Radius of cylinder $=r$ CSA of cylinder $=2 \pi r h$ $=2 \pi r(2 r)$ $=4 \pi r^{2}$ (iii) Required ratio $=\frac{\text { Surface area of sphere }}{\text { CSA of cy...

## A hemispherical bowl is made of steel, 0.25 cm thick.

[question] Question. A hemispherical bowl is made of steel, $0.25 \mathrm{~cm}$ thick. The inner radius of the bowl is $5 \mathrm{~cm}$. Find the outer curved surface area of the bowl. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Inner radius of hemispherical bowl = 5 cm Thickness of the bowl = 0.25 cm $\therefore$ Outer radius $(r)$ of hemispherical bowl $=(5+0.25) \mathrm{cm}$ $=5.25 \mathrm{~cm}$ Outer CSA of hemispherical bowl $=2 \pi r^{2}$ $=2 \ti...

## The diameter of the moon is approximately one-fourth of the diameter of the earth.

[question] Question. The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface area. [/question] [solution] Solution: Let the diameter of earth be $d$. Therefore, the diameter of moon will be $\frac{d}{4}$. Radius of earth $=\frac{d}{2}$ Radius of moon $=\frac{1}{2} \times \frac{d}{4}=\frac{d}{8}$ Surface area of moon $=4 \pi\left(\frac{d}{8}\right)^{2}$ Surface area of earth $=4 \pi\left(\frac{d}{2}\right)^{2}$ Required ratio $=\frac{4 \p...

## Find the radius of a sphere whose surface area is $154 \mathrm{~cm}^{2}$

[question] Question. Find the radius of a sphere whose surface area is $154 \mathrm{~cm}^{2} .\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Let the radius of the sphere be r. Surface area of sphere = 154 $\therefore 4 m r^{2}=154 \mathrm{~cm}^{2}$ $r^{2}=\left(\frac{154 \times 7}{4 \times 22}\right) \mathrm{cm}^{2}=\left(\frac{7 \times 7}{2 \times 2}\right) \mathrm{cm}^{2}$ $r=\left(\frac{7}{2}\right) \mathrm{cm}=3.5 \mathrm{~cm}$ Therefore, the radius of...

## A hemispherical bowl made of brass has inner diameter 10.5 cm.

[question] Question. A hemispherical bowl made of brass has inner diameter $10.5 \mathrm{~cm}$. Find the cost of tin-plating it on the inside at the rate of Rs 16 per $100 \mathrm{~cm}^{2} .\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Inner radius $(r)$ of hemispherical bowl $=\left(\frac{10.5}{2}\right) \mathrm{cm}=5.25 \mathrm{~cm}$ Surface area of hemispherical bowl $=2 \pi r^{2}$ $=\left[2 \times \frac{22}{7} \times(5.25)^{2}\right] \mathrm{cm}^{2}$ ...

## The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it.

[question] Question. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. [/question] [solution] Solution: Radius (r1) of spherical balloon = 7 cm Radius (r2) of spherical balloon, when air is pumped into it = 14 cm Required ratio $=\frac{\text { Initial surface area }}{\text { Surface area after pumping air into balloon }}$ $=\frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}=\left(\frac{r_{1}}{r_{2}...

## Find the total surface area of a hemisphere of radius 10 cm.

[question] Question. Find the total surface area of a hemisphere of radius $10 \mathrm{~cm}$. [Use $\pi=3.14$ ] [/question] [solution] Solution: Radius (r) of hemisphere = 10 cm Total surface area of hemisphere = CSA of hemisphere + Area of circular end of hemisphere $=2 \pi r^{2}+\pi r^{2}$ $=3 \pi r^{2}$ $=\left[3 \times 3.14 \times(10)^{2}\right] \mathrm{cm}^{2}$ $=942 \mathrm{~cm}^{2}$ Therefore, the total surface area of such a hemisphere is $942 \mathrm{~cm}^{2}$. [/solution]...

## Find the surface area of a sphere of diameter:

[question] Question. Find the surface area of a sphere of diameter: (i) $14 \mathrm{~cm}$ (ii) $21 \mathrm{~cm}$ (iii) $3.5 \mathrm{~m}$ [Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: (i) Radius $(r)$ of sphere $=\frac{\text { Diameter }}{2}=\left(\frac{14}{2}\right) \mathrm{cm}=7 \mathrm{~cm}$ Surface area of sphere $=4 \pi r^{2}$ $=\left(4 \times \frac{22}{7} \times(7)^{2}\right) \mathrm{cm}^{2}$ $=(88 \times 7) \mathrm{cm}^{2}$ $=616 \mathrm{~cm}^{2}$ Therefore, the ...

## A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24 cm.

[question] Question. A joker's cap is in the form of right circular cone of base radius $7 \mathrm{~cm}$ and height $24 \mathrm{~cm}$. Find the area of the sheet required to make 10 such caps. [ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Radius $(r)$ of conical cap $=7 \mathrm{~cm}$ Height $(h)$ of conical cap $=24 \mathrm{~cm}$ Slant height ( $l$ ) of conical cap $=\sqrt{r^{2}+h^{2}}$ $=\left[\sqrt{(7)^{2}+(24)^{2}}\right] \mathrm{cm}=(\sqrt{625}) \mathrm{cm}=25 \ma...

## A conical tent is 10 m high and the radius of its base is 24 m. Find

[question] Question. A conical tent is 10 m high and the radius of its base is 24 m. Find (i) slant height of the tent (ii) cost of the canvas required to make the tent, if the cost of $1 \mathrm{~m}^{2}$ canvas is Rs 70 . $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: (i) Let ABC be a conical tent. Height (h) of conical tent = 10 m Radius (r) of conical tent = 24 m Let the slant height of the tent be l. In $\triangle \mathrm{ABO}$ $\mathrm{AB}^{2}=\mathr...

## Curved surface area of a cone is $308 \mathrm{~cm}^{2}$ and its slant height is $14 \mathrm{~cm}$.

[question] Question. Curved surface area of a cone is $308 \mathrm{~cm}^{2}$ and its slant height is $14 \mathrm{~cm}$. Find (i) radius of the base and (ii) total surface area of the cone. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: (i) Slant height (l) of cone $=14 \mathrm{~cm}$ Let the radius of the circular end of the cone be $r$. We know, CSA of cone $=\pi r$ $(308) \mathrm{cm}^{2}=\left(\frac{22}{7} \times r \times 14\right) \mathrm{cm}$ $\Rightar...

## Find the total surface area of a cone,

[question] Question. Find the total surface area of a cone, if its slant height is $21 \mathrm{~m}$ and diameter of its base is $24 \mathrm{~m}$. [ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Radius $(r)$ of the base of cone $=\left(\frac{24}{2}\right) \mathrm{m}=12 \mathrm{~m}$ Slant height (l) of cone $=21 \mathrm{~m}$ Total surface area of cone $=\pi r(r+l)$ $=\left[\frac{22}{7} \times 12 \times(12+21)\right] \mathrm{m}^{2}$ $=\left(\frac{22}{7} \times 12 \times 33...

## Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm.

[question] Question. Diameter of the base of a cone is $10.5 \mathrm{~cm}$ and its slant height is $10 \mathrm{~cm}$. Find its curved surface area. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Radius $(r)$ of the base of cone $=\left(\frac{10.5}{2}\right) \mathrm{cm}=5.25 \mathrm{~cm}$ Slant height (l) of cone $=10 \mathrm{~cm}$ CSA of cone $=\pi r l$ $=\left(\frac{22}{7} \times 5.25 \times 10\right) \mathrm{cm}^{2}=(22 \times 0.75 \times 10) \mathrm{cm...

## Find

[question] Question. Find (i) The lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high. (ii) How much steel was actually used, if $\frac{1}{12}$ of the steel actually used was wasted in making the tank. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Height (h) of cylindrical tank = 4.5 m Radius $(r)$ of the circular end of cylindrical tank $=\left(\frac{4.2}{2}\right) \mathrm{m}=2.1 \mathrm{~m...

## In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm.

[question] Question. In a hot water heating system, there is a cylindrical pipe of length $28 \mathrm{~m}$ and diameter $5 \mathrm{~cm}$. Find the total radiating surface in the system. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Height (h) of cylindrical pipe = Length of cylindrical pipe = 28 m Radius (r) of circular end of pipe = = 2.5 cm = 0.025 m CSA of cylindrical pipe $=2 \pi r h$ $=\left(2 \times \frac{22}{7} \times 0.025 \times 28\right) \mathr...

## The inner diameter of a circular well is 3.5 m.

[question] Question. The inner diameter of a circular well is $3.5 \mathrm{~m}$. It is $10 \mathrm{~m}$ deep. Find (i) Its inner curved surface area, (ii) The cost of plastering this curved surface at the rate of Rs 40 per $\mathrm{m}^{2} .\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Inner radius $(r)$ of circular well $=\left(\frac{3.5}{2}\right) \mathrm{m}=1.75 \mathrm{~m}$ Depth $(h)$ of circular well $=10 \mathrm{~m}$ Inner curved surface area $=2 \p...

## Curved surface area of a right circular cylinder is $4.4 \mathrm{~m}^{2}$

[question] Question. Curved surface area of a right circular cylinder is $4.4 \mathrm{~m}^{2}$. If the radius of the base of the cylinder is $0.7 \mathrm{~m}$, find its height. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Let the height of the circular cylinder be h. Radius $(r)$ of the base of cylinder $=0.7 \mathrm{~m}$ CSA of cylinder $=4.4 \mathrm{~m}^{2}$ $2 \pi r h=4.4 \mathrm{~m}^{2}$ $\left(2 \times \frac{22}{7} \times 0.7 \times h\right) \mathr...

## A cylindrical pillar is $50 \mathrm{~cm}$ in diameter and $3.5 \mathrm{~m}$ in height.

[question] Question. A cylindrical pillar is $50 \mathrm{~cm}$ in diameter and $3.5 \mathrm{~m}$ in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. $12.50$ per $\mathrm{m}^{2} .\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Height $(h)$ cylindrical pillar $=3.5 \mathrm{~m}$ Radius $(r)$ of the circular end of pillar $=\frac{50}{2}=25 \mathrm{~cm}$ $=0.25 \mathrm{~m}$ CSA of pillar $=2 \pi r h$ $=\left(2 \times \frac{22...

## A metal pipe is 77 cm long.

[question] Question. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. (i) Inner curved surface area, (ii) Outer curved surface area (iii) Total surface area. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Inner radius $\left(r_{1}\right)$ of cylindrical pipe $=\left(\frac{4}{2}\right) \mathrm{cm}=2 \mathrm{~cm}$ Outer radius $\left(r_{2}\right)$ of cylindrical pipe $=\left(\frac{4.4}{2}\right) \ma...

## It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet.

[question] Question. It is required to make a closed cylindrical tank of height $1 \mathrm{~m}$ and base diameter $140 \mathrm{~cm}$ from a metal sheet. How many square meters of the sheet are required for the same? $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: Height (h) of cylindrical tank = 1 m Base radius $(r)$ of cylindrical $\operatorname{tank}=\left(\frac{140}{2}\right) \mathrm{cm}=70 \mathrm{~cm}=0.7 \mathrm{~m}$ Area of sheet required $=$ Total ...

## A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape.

[question] Question. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high. (i) What is the area of the glass? (ii) How much of tape is needed for all the 12 edges? [/question] [solution] Solution: (i) Length $(l)$ of green house $=30 \mathrm{~cm}$ Breadth $(b)$ of green house $=25 \mathrm{~cm}$ Height $(h)$ of green house $=25 \mathrm{~cm}$ Total surface area of green house $=2[1 b+1 h+b h]$ $=...

## A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long,

[question] Question. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high. (i) Which box has the greater lateral surface area and by how much? (ii) Which box has the smaller total surface area and by how much? [/question] [solution] Solution: (i) Edge of cube $=10 \mathrm{~cm}$ Length $(l)$ of box $=12.5 \mathrm{~cm}$ Breadth $(b)$ of box $=10 \mathrm{~cm}$ Height $(h)$ of box $=8 \mathrm{~cm}$ Lateral surface area of cubical box $=4(\text { edge }...

## The paint in a certain container is sufficient to paint an area equal to 9.375 m2.

[question] Question. The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container? [/question] [solution] Solution: Total surface area of one brick = 2(lb + bh + lh) $=[2(22.5 \times 10+10 \times 7.5+22.5 \times 7.5)] \mathrm{cm}^{2}$ $=2(225+75+168.75) \mathrm{cm}^{2}$ $=(2 \times 468.75) \mathrm{cm}^{2}$ $=9375 \mathrm{~cm}^{2}$ Let n bricks can be painted out by the paint of the c...

## The floor of a rectangular hall has a perimeter 250 m.

[question] Question. The floor of a rectangular hall has a perimeter 250 m. If the cost of panting the four walls at the rate of Rs.10 per m2 is Rs.15000, find the height of the hall. [/question] [solution] Solution: Let length, breadth, and height of the rectangular hall be l m, b m, and h m respectively. Area of four walls $=2 / h+2 b h$ $=2(I+b) h$ Perimeter of the floor of hall $=2(I+b)$ $=250 \mathrm{~m}$ $\therefore$ Area of four walls $=2(I+b) h=250 h \mathrm{~m}^{2}$ Cost of painting per...

## The length, breadth and height of a room are 5 m, 4 m and 3 m respectively.

[question] Question. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per m2. [/question] [solution] Solution: It is given that Length (l) of room = 5 m Breadth (b) of room = 4 m Height (h) of room = 3 m It can be observed that four walls and the ceiling of the room are to be white-washed. The floor of the room is not to be white-washed. Area to be white-washed = Area of walls +...

## A plastic box 1.5 m long, 1.25 m wide and 65 cm deep, i

[question] Question. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep, is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine: (i) The area of the sheet required for making the box. (ii) The cost of sheet for it, if a sheet measuring 1 m2 costs Rs 20. [/question] [solution] Solution: It is given that, length (l) of box = 1.5 m Breadth (b) of box = 1.25 m Depth (h) of box = 0.65 m (i) Box is to be open at top. Area of sheet required $=2 h+2 b h+l b$...