Question

The height of the solid cone is 12cm and the area of the base is 81 π sqcm , find the
curved surface area of the cone.[Takeπ =3.14]

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

• Height of cone, h = 12 cm

• Area of base = 81 π sq. cm.

Let assume that

• Radius of cone = r cm

• Slant height of cone = l cm

Now, as it is given that

$\rm \: Area \: of \: base \: of \: cone \: = \: 81\pi$

$\rm \: \pi {r}^{2} \: = \: 81\pi$

$\rm \: {r}^{2} \: = \: 81$

$\rm \: {r}^{2} \: = \: {9}^{2}$

$\rm\implies \:r \: = \: 9 \: cm$

Now, we know slant height of a cone, l is evaluated as

${l}^{2} \: = \: {r}^{2} + {h}^{2}$

${l}^{2} \: = \: {9}^{2} + {12}^{2}$

${l}^{2} \: = \: 81 + 144$

${l}^{2} \: = \: 225$

${l}^{2} \: = \: {15}^{2}$

$\implies \:l \: = \: 15 \: cm$

Now, we know

$\rm \: Curved \: Surface \: Area \: of \: cone \:$

$\rm \: = \: \pi \: r \: l$

$\rm \: = \: 3.14 \times 9 \times 15$

$\rm \: = \: 423.9 \: {cm}^{2}$

Hence,

$\rm\implies \:Curved \: Surface \: Area \: of \: cone=423.9 \:{cm}^{2}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

refer the given attachment