Sheeza has a piece of canvas whose area is 553 m² She uses it to make a corucal tent with a base diameter of 14 m Assuming all stitching margins and wastage incurred while cutting amounts 3 m Find the volume of the air happed in the conical tent.
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Answers
Let slant height=l m
Let height=h m
Area of canvas=Curved surface area of conical tent+wastage
553m²=CSA of conical tent+3m²
CSA of conical tent=553m²-3m²
CSA of conical tent=550m²
550=πrl
550=22l
l=25m
l²=h²+r²
25²=h²+7²
h²=25²-7²
h²=(25-7)(25+7)
h²=(18)(32)
h²=576
h=24m
Volume of air in the tent=22×7×8 m³
Step-by-step explanation:
Solution!!
radius=\frac{14}{2}=7\:mradius=
2
14
=7m
Let slant height=l m
Let height=h m
Area of canvas=Curved surface area of conical tent+wastage
553m²=CSA of conical tent+3m²
CSA of conical tent=553m²-3m²
CSA of conical tent=550m²
550=πrl
550=\frac{22}{\cancel 7}\times \cancel 7\times l550=
7
22
×
7
×l
550=22l
l=\frac{550}{22}l=
22
550
l=25m
l²=h²+r²
25²=h²+7²
h²=25²-7²
h²=(25-7)(25+7)
h²=(18)(32)
h²=576
h=\sqrt{576}h=
576
h=24m
Volume\:of\:air\:in\:the\:tent=\frac{1}{3}\pi r^{2}hVolumeofairinthetent=
3
1
πr
2
h
Volume\:of\:air\:in\:the\:tent=\frac{1}{3}\times \frac{22}{7}\times 7^{2}\times 24 m³Volumeofairinthetent=
3
1
×
7
22
×7
2
×24m³
Volume\:of\:air\:in\:the\:tent=\frac{1}{3}\times \frac{22}{7}\times 49\times 24 m³Volumeofairinthetent=
3
1
×
7
22
×49×24m³
Volume of air in the tent=22×7×8 m³
\boxed{\bigstar Volume\:of\:air\:in\:the\:tent=1232m^{3}}
★Volumeofairinthetent=1232m
3. please help new here