Two cones of the height of 6cm and base radius 7 cm is cut from a right circular cylinder of base radius 14 cm and height 10 cm, such that bases of each cone and cylinder lie on the same plane, then find the total surface area of the remaining cylinder.
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(574+14√85)π
Plzz confirm sir
574pi+14pi square root of 85
Cylinder is hollow or solid?? Please clear this
2058.194
Will solve later.
1804+44(85)^1/2
44(41+√85)
isme cylinder ka C. S. A. me koi change nhi hoga& 2 cone ka CSA+2 rings ka area kyoki cylinder ke top & bottom par 2 circle 7 & 14 radius k banege
=> TSA= (2pi ×14×10) + [2pi ×7× (85)^1/2] + 2pi[ (14)^1/2 – (7)^1/2]
574π+( (245√85)*π/36)
574π+(245*√85*π/18) final
2198.26
1804+44√85
44(161+rt85)
1804+44√85
44(161+√85)
(574 plus 16under root 143)pie
478-44√85
1804+44√85
14π[41+√85]
Total surface area of the remaining cylinder is nothing but the (Curve surface area of the cylinder + area of the one base +curve surface area of the two cone +area of the one base after removing of two circles of radius 7cm). =1804 + 44✓85
pi( 574 + 245root85/18)
1708 + 44sqroot85
sorry its 1808
1804+44√85
2024 + 44( sqrt(85))
(574+14√85)π
Total surface area of remaining part is Nothing bt CSA of cylinder + area of top+2*CSA of cone +(area of base of cylinder -2* base area of cone) =
(2*Pi*R*H)+(Pi*R^2)+(2*Pi*r*l) +(Pi*R^2-2*Pi*r^2)
=
(574+14*sq rt of85)*pi
=44(41+sq rt of85)
572+44√85
1804+44 root85
44(41-√85)
1293
Sol https://youtu.be/VjogGgthDNY