Question

A joker’s cap is in the form of cone of base radius 7cm and height 24 cm. Find the area of the sheet to make 10 such caps

Answer 1

Answer:

Given  

r=7 cm

h=24 m

Let slant height be   L

L = 25 CM

CSA OF 1 CAP = πRL = 22/7 * 7* 25

CSA OF 10 CAPS = 25 * 22 * 10 = 5500 CM^2

 

 

Answer 2

$\sf Given \begin{cases} & \sf{Radius\;of\;cap,\; = \bf{7\;cm}} \\ & \sf{Height\;of\;cap,\; = \bf{24\;cm}} \end{cases}$

Need to find: Area of sheet to make 10 caps?

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☯ Let's consider slant height of cone be 'l'.

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$\setlength{\unitlength}{1.7mm}\begin{picture}(5,5)\thicklines\put(0,0){\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\put(-0.5,-1){\line(1,2){13}}\put(25.5,-1){\line(-1,2){13}}\multiput(12.5,-1)(2,0){7}{\line(1,0){1}}\multiput(12.5,-1)(0,4){7}{\line(0,1){2}}\put(18,1.6){\sf{7 cm}}\put(9.5,10){\sf{24 cm}}\end{picture}$

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$\dag\;{\underline{\frak{As\;we\;know\;that,}}}$

$\star\;{\boxed{\sf{\pink{(Slant\; height)^2 = (Height)^2 + (Radius)^2}}}}$

$:\implies\sf l^2 = h^2 + r^2\\ \\ :\implies\sf l^2 = (24)^2 + (7)^2\\ \\ :\implies\sf l^2 = 576 + 49 \\ \\ :\implies\sf l^2 = 625\\ \\ :\implies\sf \sqrt{l^2} = \sqrt{625}\\ \\ :\implies{\underline{\boxed{\frak{\purple{l = 25\;cm}}}}}\;\bigstar$

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$\underline{\bigstar\:\boldsymbol{According\:to\:the\:question\::}}$

• Sheet required will be on the curved surface

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$\dag\;{\underline{\frak{As\;we\;know\;that,}}}$

$\star\;{\boxed{\sf{\pink{CSA_{\;(cone)} = \pi rl}}}}$

CSA of 1 joker's cap,

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$:\implies\sf CSA_{\;(1\;cap)} = \dfrac{22}{ \cancel{7}} \times \cancel{7} \times 25\\ \\ :\implies\sf CSA_{\;(1\;cap)} = 22 \times 25\\ \\ :\implies{\underline{\boxed{\frak{\purple{CSA_{\;(1\;cap)} = 550\;cm^2}}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Curved\;surface\;area\;of\;1\;cap\;is\; \bf{550\;cm^2}.}}}$

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CSA of 10 joker's cap,

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$:\implies\sf CSA_{\;(10\;cap)} = 10 \times CSA_{\;(1\;cap)}\\ \\ :\implies\sf CSA_{\;(10\;cap)} = 10 \times 550\\ \\ :\implies{\underline{\boxed{\frak{\purple{CSA_{\;(10\;cap)} = 5500\;cm^2}}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Curved\;surface\;area\;of\;10\;cap\;is\; \bf{5500\;cm^2}.}}}$