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प्रश्न
In Fig. 4, from the top of a solid cone of height 12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid. (Use `pi=22/7` and `sqrt5=2.236`)
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उत्तर
The remaining solid is a frustum of the given cone
Total surface area of the frustum = πl(r1+r2)+πr12+πr22
Where
h = Height of the frustum = 12−4 = 8 cm
r1 = Larger radius of the frustum = 6 cm
r2 = Smaller radius of the frustum
l = Slant height of the frustum
In the given figure, ∆ABC ~ ∆ADE by AA similarity criterion.
`:.(BC)/(DC)=(AB)/(AD)`
`=>r_2/6=4/12`
⇒r2=2 cm
We know
`l = sqrt(h^2+(r_1-r_2)^2)`
`=>l=sqrt(8^2+(6-1)^2)`
`=>l = 4sqrt5 `
∴ Total surface area of the frustum = πl(r1+r2)+πr12+πr22
= π×4`sqrt5`(6+2)+π×62+π×22
`=pi(32sqrt5+40)`
`=22/7xx111.552`
= 350.592 cm2
Hence, the total surface area of the remaining solid is 350.592 cm2.
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