Question

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

the area of the sheet required to make 10 such caps. ( pi = 22/ 7 )​

Answer 1

Answer:

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Answer 2

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

Exigency To Find : The area of the sheet required to make 10 such caps .

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❍ Let's Consider Radius and Height of the cone be r & h , respectively.

⠀⠀⠀⠀⠀As , We have to find Area Sheet required to make joker caps , therefore the area of sheet will be the C.S.A ( Curved surface area ) of cone .

$\dag\:\:\frak{ As,\:We\:know\:that\::}\\\\ \qquad\maltese\bf\:\:Curved\:Surface\:Area\:of\:cone\::$

$\qquad \dag\:\:\bigg\lgroup \sf{ C.S.A \:_{(Cone)}\:: \:\pi rl }\bigg\rgroup$

⠀⠀⠀⠀⠀Here , r is the Base Radius of cone , l is the slant Height of cone & $\sf \pi \:\:=\:\:\dfrac{22}{7}$

⠀⠀⠀⠀⠀Now , For C.S.A First we have to find slant height of cone & it's Given by :

$\qquad :\implies \sf Slant \: Height \: (l)\:\:= \: \sqrt { (r)^2 \: + \: (h)^2 \: }$

$\qquad :\implies\sf Slant \: Height \: (l)\:\:= \: \sqrt { (7)^2 \: + \: (24)^2 \: }$

$\qquad :\implies \sf Slant \: Height \: (l)\:\:= \: \sqrt { 49 \: + \:576 \: }$

$\qquad :\implies \sf Slant \: Height \: (l)\:\:= \: \sqrt { 625\: }$

$\qquad :\implies \sf Slant \: Height \: (l)\:\:= \: 25$

$\qquad \therefore \pmb{\underline{\purple{\: Slant \: Height \: (l)\:\:=\:25 cm }} }\:\;\bigstar$

⠀⠀⠀⠀⠀Now, Finding Area of Sheet Required in making of one joker caps :

$\qquad \dashrightarrow \sf C.S.A \:_{(Cone)}\:= \:\pi rl$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf C.S.A \:_{(Cone)}\:= \:\pi rl$

$\qquad \dashrightarrow \sf C.S.A \:_{(Cone)}\:= \:\dfrac{22}{7} \times 7 \times 25$

$\qquad \dashrightarrow \sf C.S.A \:_{(Cone)}\:= \:\dfrac{22}{\cancel {7}} \times \cancel {7} \times 25$

$\qquad \dashrightarrow \sf C.S.A \:_{(Cone)}\:= \:22 \times 25$

$\qquad \dashrightarrow \sf C.S.A \:_{(Cone)}\:= \: 550 \:$

$\qquad \therefore \pmb{\underline{\purple{\:C.S.A \:_{(Cone)}\:= \: 550 \:cm^2 }} }\:\;\bigstar$

⠀⠀⠀⠀⠀Therefore, Area of Sheet required to make one joker cap is 550 cm ²

⠀⠀⠀⠀⠀Now , Finding Area of Sheet required to make 10 Joker caps :

$\qquad\dashrightarrow \sf Area\:of\:sheet \:_{(10\:Joker \:cap \: )} = \: 10 \times Area\:of\:sheet \:_{(1\:Joker \:cap \: )}$

$\qquad\dashrightarrow \sf Area\:of\:sheet \:_{(10\:Joker \:cap \: )} = \: 10 \times 550$

$\qquad\dashrightarrow \sf Area\:of\:sheet \:_{(10\:Joker \:cap \: )} = \: 5500$

$\qquad \therefore \pmb{\underline{\purple{\:Area\:of\:sheet \:_{(10\:Joker \:cap \: )} = \: 5500 \:cm^2 }} }\:\;\bigstar$

$\therefore \underline {\:\:\sf Hence, \:Area \:of\:sheet \:required \:to \:make \:10 \:joker\:caps \: is \:\bf 5,500 \: cm^2 }$