A cone has a height of $15 cm$ and its base has a radius of $9 cm$ . If the cone is horizontally cut into two segments $7 cm$ from the base, what would the surface area of the bottom segment be?

Question

1433 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-20T00:27:31

Total Surface Area = 38.6979 $cm^2$

enter image source here
$OA = h = 15 cm, OB r = 9 cm, , AF = h_1 = 15-7 = 8 cm$

$FD = r_2 =( h_1 /h)*r = (8/15)*9 = 24/5 cm$

$AC = l = sqrt(h^2 + r^2) = sqrt(15^2 + 9^2) = 17.4929$

$AE = l_1 = sqrt(h_1^2 + r_2^2) = sqrt(8^2 + (24/5)^2) = 9.33 cm$

$pir^2 = pi*9^2 = 254.469 cm^2$

$pir_2^2 = pi*(24/5)^2 = 72.3823 cm^2$

#pirl= pi917.4929 = 494.6 cm^2

$pir_2l_1 = pi* (25/4)* 9.33 = 183.1941$

Total surface area = $(pir^2 + pir_1^2 + pi.r.l - pi.r_2.l_1)$

$T S A = 254.469 + 72.823 + 494.6 - 183.1941 =**638.6979cm^2**$

A cone has a height of 15 cm15 cm and its base has a radius of 9 cm9 cm. If the cone is horizontally cut into two segments 7 cm7 cm from the base, what would the surface area of the bottom segment be?