Question

diameter of a base of a cone is 10.5 cm and its slant height is 10 cm. find its csa and tsa
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Answer 1

We have :-

• Diameter of the base of Cone = 10.5 cm . ∴ Radius = 10.5/2 cm .
• Slant height of Cone = 10 cm .

To find :-

• find T.S.A (Total Surface Area) and C.S.A (Curved Surface Area ) of the Cone .

Formula used :-

• C.S.A of Cone = π.r.l
• T.S.A of Cone = πr ( l + 2r )

Solution :-

C.S.A of Cone :-

C.S.A of Cone :- π.r.l

$\sf \: C.S.A \: of \: Cone \: = \dfrac{22}{7} \times \dfrac{10.5}{2} \times 10 \\ \\ \sf \: C.S.A \: of \: Cone \: = \dfrac{ \cancel{22}}{ \cancel7} \times \dfrac{ \cancel{10.5}}{ \cancel2} \times 10 \\ \\ \sf \: C.S.A \: of \: Cone \: = 11 \times 1.5 \times 10 \\ \\ \sf \: C.S.A \: of \: Cone \: =1 65 \: {cm}^{2}$

hence, C.S.A of the Cone = 165 cm² .

Now, T.S.A of Cone :- π.r ( l + r )

$\sf \: T.S.A \: of \: Cone = \dfrac{22}{7} \times \dfrac{10.5}{2} \bigg (10 + \dfrac{10.5}{2} \bigg) \\ \\ \sf \: T.S.A \: of \: Cone = \dfrac{\cancel{22}}{\cancel7} \times \dfrac{ \cancel{10.5}}{\cancel2} \bigg (10 + \dfrac{10.5}{2} \bigg) \\ \\\sf \: T.S.A \: of \: Cone = 11 \times 1.5 \times \dfrac{30.5}{2}\\ \\ \sf \: T.S.A \: of \: Cone = 11 \times 1.5 \times 15.25 \\ \\ \sf \: T.S.A \: of \: Cone = 251.625 \: {cm}^{2}$

Hence, T.S.A of Cone = 251.625 cm² .

the required C.S.A and T.S.A of the Cone is 165 cm² and 251.625 cm² respectively .