Question

From a solid cylinder whose height is 2.4 cm and diameter 1.4 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

Hope it will help you.

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