A right circular cone having radius 5 cm and height 12 cm is given. The curved surface area of the cone is ki
Answers
Answer:
We know that,
The slant height of a cone
\sf{ {l}^{2} = r + h}l
2
=r+h
\sf{\implies l = \sqrt{ {r}^{2} + {h}^{2} } }⟹l=
r
2
+h
2
\sf{\implies l = \sqrt{ {5}^{2} + {12}^{2} } }⟹l=
5
2
+12
2
\sf{\implies l = \sqrt{25 + 144} }⟹l=
25+144
\sf{\implies l = \sqrt{169} }⟹l=
169
\sf{\implies l = 13 }⟹l=13
Therefore,the slant height of the cone is 13 cm.
Now, the curved surface area of the cone is (CSA)\sf{= \pi r l}=πrl
\sf{\implies CSA = \frac{22}{7} \times 5 \times 13}⟹CSA=
7
22
×5×13
\sf{\implies CSA = \frac{1430}{7} }⟹CSA=
7
1430
\sf{\implies CSA = 204.28}⟹CSA=204.28
Hence, the curved surface area of the cone = 204.28 cm²
The slant height would be the hypoteneuse of a right triangle whoses are the base radius and height. Pythagorus taught us the rest.
Slant height. = √(5² + 7²). = √(25 + 49)
= √74. Or approx 8.6 cm.
Surface Area = area of base + area of sl0ped arwa.
Aeea of base = πr² = 25π
Area of sloped side: = 1/2 circ of base × slant height
= 1/2 × 2π 5 × √74 =™5√(74)π
Total surface area = (25+5√74)π
.=( 25+5×8.6)π = (25+43)π = 68π or approx 213.63 cm²
Abby R