Question

From a solid cylinder whose height is 24 cm and diameter 14 cm, a conical cavity of the
same height and same diameter is hollowed out. Find the total surface area of the
remaining solid to the nearest cm?
​

Answer 1

Answers

radius of cy=7cm =r of cone

h of my = h of cone = 24 cm

l^2= h^2+ r^2

l=√(24)^2 + (7)^2

l= √576+49

l=√625

l= 25cm

total surface of the remaining solid

= C.S.A of cylinder - C.S.A of cone + Ar of circular base of cylinder

= 2 pie r h - pie r l + pie r^2

=2×22/7×7×24 - 22/7×7×25 + 22/7×7×7

= 22/7×7( 2×24-25+7)

=22(48-18)

= 22×30

= 660cm^2

Answer 2

Answer:

Given;

Height of cylinder(h)=24cm(h)=24cm and diameter(d)=14cm(d)=14cm

Radius(r)=\frac{d}{2}=7cm(r)=

2

d

=7cm

From figure,

∴cb(l)=\sqrt{h^{2} +r^{2} }cb(l)=

h

2

+r

2

⇒l=\sqrt{24^{2}+7^{2} }l=

24

2

+7

2

∴l=25l=25 cmcm

∴Total surface area of remaining solid== Area of cylinder++ Area of cone++ Area of base

⇒Total surface area of remaining solid=2\pi r h+\pi rl+\pi r^{2}=2πrh+πrl+πr

2

⇒Total surface area of remaining solid=(2\pi \times7\times 24)+(\pi\times 7\times25)+(\pi7^{2})=(2π×7×24)+(π×7×25)+(π7

2

)

⇒Total surface area of remaining solid=\frac{22\times7}{7}( (2 \times 24)+25+7)=

7

22×7

((2×24)+25+7)

⇒Total surface area of remaining solid=22(48+32)=22(48+32)

∴Total surface area of remaining solid=1760cm^{2}=1760cm

2

∴Total surface area of remaining solid is 1760cm^{2}1760cm

2