Question

10. In Fig. 13.12, you see the frame of a lampshade. It is to be
covered with a decorative cloth. The frame has a base
diameter of 20 cm and height of 30 cm. A margin of 2.5 cm
is to be given for folding it over the top and bottom of the
frame. Find how much cloth is required for covering the
lampshade.
Fig. 13.12​

Answer 1

Step-by-step explanation:

Given, Frame-- Diameter= 20 cm

                          Height= 30 cm

∴ Cloth required= πdh

                             = π*20*{30- (2.5+2.5)}

                             = 3.14*20*25              [∵π=3.14]

                             = 3.14*500

                             = 1570 cm

∴ Total cloth required for covering the  lampshade= 1570 cm= 1.57 m

Hope it helps!!  :)

Answer 2

☯ The decorative cloth forms a cylinder

⠀⠀⠀⠀⠀

where,

⠀⠀⠀⠀⠀

• Base diameter = 20 cm
• Height = (30 + 2.5 + 2.5) = 35 cm

⠀⠀⠀⠀⠀

$\setlength{\unitlength}{1.8 cm} \thicklines \begin{picture}(2,0)\qbezier(0,0)(0,0)(0,2.5)\qbezier(2,0)(2,0)(2,2.5)\qbezier(0,0)(1,0.5)(2,0)\qbezier(0,0)( 1, - 0.6)(2,0) \put(2.3,1){\vector(0,1){1.5}}\put(2.3,1){\vector(0, - 1){1.2}}\put(2.3,1){ \bf 30 \: cm }\put(0.6,2.6){ \bf 20 \:cm }\put(0,2.5){\line(1,0){2}}\qbezier(0,2.5)(1,1.9)(2,2.5)\qbezier(0,2.5)(1, 3)(2,2.5)\qbezier(0,0.8)( 1, 0.2)(2,0.8)\qbezier(0,1.8)( 1, 1.1)(2,1.8)\end{picture}$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━

We have,

⠀⠀⠀⠀⠀

• Radius of the decorative cloth cylinder, r = 10 cm
• Height of the decorative cloth cylinder, h = 35 cm

⠀⠀⠀⠀⠀

we know that,

⠀⠀⠀⠀⠀

☯ Area of decorative cloth = LSA of Cylinder

⠀⠀⠀⠀⠀

$\star\;{\boxed{\sf{\purple{LSA_{\;(cylinder)} = 2 \pi rh}}}}$

$:\implies\sf 2 \times \dfrac{22}{ \cancel{7}} \times 10 \times \cancel{35}$

$:\implies\sf 2 \times 22 \times 10 \times 5$

$:\implies{\boxed{\frak{\pink{2200\;cm^2}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;the\;cloth\; required\;for\;covering\;lampshade\;is\; \bf{2200\;cm^2}.}}}$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\boxed{\begin{minipage}{6 cm}\bigstar \:\underline{\textbf{Formulae Related to Cylinder :}}\\\\\sf {\textcircled{\footnotesize\textsf{1}}} \:Area\:of\:Base\:and\:top =\pi r^2 \\\\ \sf {\textcircled{\footnotesize\textsf{2}}} \:\:CSA =2 \pi rh\\\\\sf{\textcircled{\footnotesize\textsf{3}}} \:\:TSA = Area \: of \: Base + CSA + area \: of \: top \\{\quad\:\:\quad=\pi r^2 + 2\pi rh+\pi r^2 = 2 \pi r(h + r)}\\ \\{\textcircled{\footnotesize\textsf{4}}} \: \:Volume=\pi r^2h\end{minipage}}$