A cone has a height of $8 cm$ and its base has a radius of $6 cm$ . If the cone is horizontally cut into two segments $4 cm$ from the base, what would the surface area of the bottom segment be?

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Answer 1 · 2018-06-14T19:36:47

$282.74"cm"^2"to the nearest 2 decimal places"$ enter image source here

$tan theta=8/6=1.333333333=53^@7'48''$

$"top radius"=cot 53^@7'48''=0.75 xx4=3.0cm$

$Lateral area= F=pi(r_1+r_2)sqrt((r_1-r_2)^2+h^2)$

$F=pi(6+3)sqrt((6-3)^2+4^2)$

$F=pi(9)sqrt((9+16)$

$F=9pisqrt25$

$F=28.27433388 xx5=141.3716694"cm"^2$

$S=F+pi(r_1^2+r_2^2)$

$S=141.3716694+pi(6^2+3^2)$

$S=141.3716694+pi(36+9)$

$S=141.3716694+pi(45)$

$S=141.3716694+141.3716694$

$S=282.7433388"cm"^2$

$S="surface area of bottom segment"$

$=282.74"cm"^2"to the nearest 2 decimal places"$

A cone has a height of 8 cm8 cm and its base has a radius of 6 cm6 cm. If the cone is horizontally cut into two segments 4 cm4 cm from the base, what would the surface area of the bottom segment be?