A cone has a height of #24 cm# and its base has a radius of #15 cm#. If the cone is horizontally cut into two segments #12 cm# from the base, what would the surface area of the bottom segment be?

2 Answers
sankarankalyanam · saumyanand
Feb 23, 2018

22.1 SQ CM

Explanation:

Requires reworking based on the details given below

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#A_T= A_B + A_b + L_R - L_r#

Where #A_T# is Total Surface Area of the bottom truncated cone, #A_B# is the area of the bottom circular base, #A_b# is the area of the circular base of top cut cone, #L_R# is the Lateral Surface Area of the whole cone and #l_r# is the Lateral Surface Area of the top half cone.

sankarankalyanam
Feb 23, 2018

Total Surface Area of the bottom segment of the cone

#A_(R-r) = color(purple)(1882.6)cm^3#

Explanation:

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#r / R = h / H#

#r = (15 * (24-12))/24 = 7.5 cm#

Lateral surface are of a cone #L_R= pi R L#

Now we have to find L

#L = sqrt(R^2 + H^2) = sqrt(15^2 + 24^2) ~~ 28.3 cm#

#L_R = pi * 15 * 28.3 ~~ 1333.6 cm^2#

Similarly,#l = sqrt(12^2 + 7.5^2) ~~ 14.2 cm#

#l_r = pi r l = pi * 7.5 * 14.2 ~~ 334.6#

Area of base of a cylinder #A_R = pi R^2 = pi * 15^2 ~~ 706.9#

Similarly, #A_r = pi r^2 = pi * 7.5^2 ~~ 176.7#

Total surface area of the truncated cone

#A_(R-r) = L_R - l_r + A_R + A_r = 1333.6 - 334.6 + 706.9 + 176.7 = color(purple)(1882.6)cm^3#

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