## 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...

## 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 $=(15 \times 12) \mathrm{cm}^{3}$ ...

## 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 ...