Question

38. A heap of wheat is in the form of a cone whose diameter is 48 m and height is 7m. Find its
volume. If the heap is to be covered by canvas to protect it from rain, find the area of the canvasrequired.
​

Answer 1

Answer:

Area of canvas required=1885.71$m^{2}$

Volume of heap=4225$m^{3}$

Step-by-step explanation:

r=d/2

r=48/2

r=24m

Volume of cone=1/3*π*$r^{2}$*h

1/3*22/7*$24^{2}$*7=Volume of cone

576*22*7/7*3=Volume of cone\

192*22=Volume of cone

4224$m^{3}$=Volume of cone

$l^{2}$=$r^{2}$+$h^{2}$

$l^{2}$=576+49

$l^{2}$=625

Taking root,

l=±25m

l=25m(Measurements cannot be negative)

CSA=π×r×l

CSA-22×24×25/7

CSA=13200/7

CSA=1885.71$m^{2}$

Area of canvas required=1885.71$m^{2}$

Volume of heap=4225$m^{3}$

Hope it helps..

Answer 2

$\LARGE\underline\mathfrak{AnSwEr:}$

•ᴠᴏʟᴜᴍᴇ ᴏғ ᴄᴏɴᴇ = $\bf\ 4424cm^3$

• ᴀʀᴇᴀ ᴏғ ᴄᴀɴᴠᴀs ʀᴇǫᴜɪʀᴇᴅ = $\bf\ 1885.714cm^2$

$\LARGE\underline\mathfrak{Step \: by \: step \: explanation:}$

$\normalsize{\sf{\underline{\underline \orange{Given}}}}$

• ᴅɪᴀᴍᴇᴛᴇʀ ᴏғ ᴄᴏɴᴇ = 48ᴄᴍ

•ʀᴀᴅɪᴜs ᴏғ ᴄᴏɴᴇ = $\bf\frac{D}{2} \: = \: \frac{48}{2} \: =$ 24ᴄᴍ

• ʜᴇɪɢʜᴛ ᴏғ ᴄᴏɴᴇ = 7ᴄᴍ

•sʟᴀɴᴛ ʜᴇɪɢʜᴛ ᴏғ ᴄᴏɴᴇ(ʟ)=

$\bf\displaystyle\sqrt{ h^2 \: + r^2} \: = \: \sqrt{49 \: + \: 576} \: =$ 25ᴄᴍ

$\normalsize{\sf{\underline{\underline \orange{To \: find}}}}$

•ᴠᴏʟᴜᴍᴇ ᴏғ ᴄᴏɴᴇ

• ᴀʀᴇᴀ ᴏғ ᴄᴀɴᴠᴀs ʀᴇǫᴜɪʀᴇᴅ

$\normalsize{\sf{\underline{\underline \orange{Volume \: of \: cone}}}}$

${\boxed{\bf\red{Volume \: of \: cone \: = \: \frac{1}{3} \pi r^2h}}}$

ʙʟᴏᴄᴋ ᴛʜᴇ ᴠᴀʟᴜᴇs ɪɴ ᴀᴠᴀɪʟᴀʙʟᴇ ᴅᴀᴛᴀ ;

➠$\large\tt\frac{1}{3} \times\frac{22}{7} \times\ (24)^2 \times\ 7$

➠$\large\tt\frac{ 1 \times\ 22 \times\ 24 \times\ 24 \times\ 7}{ 3 \times\ 7}$

➠$\large\tt\ 1 \times\ 22 \times\ 8 \times\ 24$

ᴄᴀɴᴄᴇʟ ᴛʜᴇ ᴠᴀʟᴜᴇs;( 24 ʙʏ 3 ᴀɴᴅ 7 ʙʏ 7)

➠$\large\tt\ 4424cm^3$

➠${\boxed{\sf{Volume \: of \: cone \: = \: 4424cm^{3} }}}$

$\normalsize{\sf{\underline{\underline \orange{Area \: of \: canvas \: required}}}}$

•ᴀʀᴇᴀ ᴏғ ᴄᴀɴᴠᴀs ʀᴇǫᴜɪʀᴇᴅ ᴍᴇᴀɴs ᴛʜᴇ ᴏᴜᴛᴇʀ ᴄᴏᴠᴇʀɪɴɢ ʀᴇǫᴜɪʀᴇᴅ,sᴏ ᴡᴇ ᴜsᴇ ᴛʜᴇ ᴄᴜʀᴠᴇᴅ sᴜʀғᴀᴄᴇ ᴀʀᴇᴀ ғᴏʀᴍᴜʟᴀ

${\boxed{\bf\red{Curved \: surface \: area \: of \: cone \: = \: \pi rl}}}$

ʙʟᴏᴄᴋ ᴛʜᴇ ᴠᴀʟᴜᴇs ɪᴍ ᴀᴠᴀɪʟᴀʙʟᴇ ᴅᴀᴛᴀ;

➠$\large\tt\frac{22}{7} \times\ 24 \times\ 25$

➠$\large\tt\frac{ 22 \times\ 24 \times\ 25}{7}$

➠$\large\tt\ 1885.714cm^2$

➠${\boxed{\sf{Area \: of \: canvas\: = \: 1885.714cm^2}}}$