Question

The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? (Assume π =22/7)​

Answer 1

Given:-

• r = 3 cm
• h = 10.5 cm

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To Find:-

• Number of cardboard which was required to be bought for the competition.

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Solution:-

★ Now Using Formula

• Surface area of a penholder = CSA of pen holder + Area of base of penholder

$\red{\sf \: 2πrh+πr2}$

$\sf \longrightarrow \: 2×( \frac{22}{7} )×3×10.5+( \frac{22}{7} )×32$

$\sf \longrightarrow \: \frac{1584}{7} \: {cm}^{2}$

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Therefore, Area of cardboard sheet used by one competitor is 1584/7 cm².

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★ For 35 competitors

$\sf \longrightarrow \frac{1584}{7} \times 35$

$\sf \longrightarrow 7920 \: {cm}^{2}$

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Therefore, 7920 cm² cardboard sheet will be needed for the competition.

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Answer 2

$\bf{\dag}\:{\underline{\underline{\sf{Question\::}}}}$

• The students of Vidyalya were asked to participate in a competition for making and decorating pen holders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? (Assume π = 22/7)

$\bf{\dag}\:{\underline{\underline{\sf{Answer\::}}}}$

• 7920 cm² of cardboard was required to be bought for the competition.

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━

$\bf{\dag}\:{\underline{\underline{\sf{Given\::}}}}$

• Radius of pen holder = 3 cm
• Height of pen holder = 10.5 cm
• Vidyalaya was to supply the competitors = cardboard

$\bf{\dag}\:{\underline{\underline{\sf{To\:Find\::}}}}$

• Cardboard required for 35 competitors = ?

$\bf{\dag}\:{\underline{\underline{\sf{Formulae\:used\::}}}}$

$\qquad\bf{\dag}\:{\underline{\boxed{\sf{\red{Area_{(circle)} = \pi r^2}}}}}$

$\qquad\bf{\dag}\:{\underline{\boxed{\sf{\pink{CSA_{(cylinder)} = 2\pi rh}}}}}$

$\bf{\dag}\:{\underline{\underline{\sf{Solution\::}}}}$

☯ Here, we have penholder which is open from top. Firstly let's find it's area. Area of pen holder will be it's CSA + area of it's base. So by using formulas we get;

$\\ :\implies\:\sf Area_{(pen\:holder)} = 2\pi rh + \pi r^2$

$\\ :\implies\:\sf Area_{(pen\:holder)} = \pi r(2h + r)$

$\\ \underline{\sf{\bigstar\:Putting\:all\:values\::}}$

$\\ :\implies\:\sf Area_{(pen\:holder)} = \dfrac{22}{7}\:\times\:3[2(10.5) + 3]$

$\\ :\implies\:\sf Area_{(pen\:holder)} = \dfrac{22}{7}\:\times\:3[(2\:\times\:10.5) + 3]$

$\\ :\implies\:\sf Area_{(pen\:holder)} = \dfrac{22}{7}\:\times\:3[(21) + 3]$

$\\ :\implies\:\sf Area_{(pen\:holder)} = \dfrac{22}{7}\:\times\:3(21 + 3)$

$\\ :\implies\:\sf Area_{(pen\:holder)} = \dfrac{22}{7}\:\times\:3(24)$

$\\ :\implies\:\sf Area_{(pen\:holder)} = \dfrac{22}{7}\:\times\:3\:\times\:24$

$\\ :\implies\:\sf Area_{(pen\:holder)} = \dfrac{22\:\times\:3\:\times\:24}{7}$

$\\ :\implies\:\sf Area_{(pen\:holder)} = \dfrac{66\:\times\:24}{7}$

$\\ :\implies\:{\underline{\underline{\boxed{\bf{Area_{(pen\:holder)} = \dfrac{1584}{7}\:cm^2}}}}}$

• Henceforth, area of one pen holder is 1584/7 cm².

☯ We have area of one pen holder. So, area of 35 pen holders will be;

$\\ :\implies\:\sf Area_{(35\:pen\:holders)} = Area_{(pen\:holder)}\:\times\:35$

$\\ \underline{\sf{\bigstar\:Putting\:all\:values\::}}$

$\\ :\implies\:\sf Area_{(35\:pen\:holders)} = \dfrac{1584}{\cancel{7}}\:\times\:\cancel{35}$

$\\ :\implies\:\sf Area_{(35\:pen\:holders)} = \dfrac{1584}{1}\:\times\:5$

$\\ :\implies\:\sf Area_{(35\:pen\:holders)} = 1584\:\times\:5$

$\\ :\implies\:{\underline{\underline{\boxed{\bf{Area_{(35\:pen\:holders)} = 7920\:cm^2}}}}}$

• Henceforth, 7920 cm² of cardboard was required to be bought for the competition.

$\bf{\dag}\:{\underline{\underline{\sf{Some\:important\:formulae\::}}}}$

↠ TSA of cube = 6a²

↠ CSA of cube = 4a²

↠ Volume of cube = a³

↠ TSA of cuboid = 2(lb + bh + hl)

↠ CSA of cuboid = 2(l + b)h

↠ Volume of cuboid = l × b × h

↠ TSA of cylinder = 2πr(r + h)

↠ CSA of cylinder = 2πrh

↠ Volume of cylinder = πr²h

↠ Volume of hollow cylinder = πh(R²-r²)

↠ TSA of cone = πr(l + r)

↠ CSA of cone = πrl

↠ Volume of cone = 1/3 πr²h

↠ SA of sphere = 4πr²

↠ Volume of sphere = 4/3 πr³

↠ TSA of hemisphere = 3πr²

↠ CSA of hemisphere = 2πr²

↠ Volume of hemisphere = 2/3 πr³

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