From a solid cylinder whose height 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 cm².
Answers
Answer:
State true or false: If the inner surface of the spherical mirror is reflecting surface, then it is called a concave mirror.
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