Question

The slant hieght of the frustum of a cone is 4cm and perimeter os its circular bases are 18 cm and 6cm .find its curved surface area?

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{C.S.A\:of\:frustom=48\:cm}}^{2}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }} \\ : \implies \text{Slant \: height(l) = 4 \: cm} \\ \\ : \implies \text{Perimeter \: of \: 1st \: circular \: base = 18 \: cm} \\ \\ : \implies \text{Perimeter \: of \: 2nd\: circular \: base = 6 \: cm} \\ \\ \red{ \underline \bold{To \:Find : }} \\ : \implies \text{C.S.A \: of \: frustom = ?}$

• According to given question :

$\bold{For\:r_{1}} \\: \implies \text{Circumference \: of \: 1st \: circular \: base = 18 \: cm} \\ \\ : \implies 2\pi r_{1}= 18 \\ \\ : \implies r_{1}= \frac{18}{2\pi} \\ \\ \green{: \implies r_{1} = \frac{9}{\pi} } \\ \\ \bold{For \: r_{2 }} \\ : \implies \text{Circumference \: of \: 2nd \: circular \: base =6 \: cm} \\ \\ : \implies 2\pi r_{2} = 6 \\ \\ : \implies r_{2} = \frac{6}{2\pi} \\ \\ \green{: \implies r_{2} = \frac{3}{\pi} } \\ \\ \bold{For \: C.S.A \: of \: frustom} \\ : \implies C.S.A \: of \: frustom = \pi \times l( r_{1} + r_{2} \\ \\ : \implies C.S.A \: of \: frustom = \pi \times 4( \frac{9}{\pi} + \frac{3}{\pi} ) \\ \\ : \implies C.S.A \: of \: frustom =\pi \times 4 \times \frac{9 + 3}{\pi} \\ \\ : \implies C.S.A \: of \: frustom =\pi \times 4 \times \frac{12}{\pi} \\ \\ \green{: \implies \text{C.S.A \: of \: frustom =48 \: cm}^{2} }$

Answer 2

Answer:

${\boxed{\bold{C.S.A. = 48\:{cm}^{2} }}}$

Step-by-step explanation:

Given:-

• Slant height = 4 cm
• Perimeter of 1st circular base =18 cm
• Perimeter of 2nd circular base =6 cm
• C.S.A. = ?

Formula used:-

${\boxed{\bold{C.S.A. = \pi \times l (r_1 + r_2) }}}$

$\tt r_1 = 2\pi r = 18 \\\rightarrow\tt r_1 = \dfrac{18}{2\pi} \\\rightarrow{\boxed{\tt {r_1 = \dfrac{9}{\pi} }}}$

$\tt r_2 = 2\pi r = 6 \\\rightarrow\tt r_2 =\dfrac{6}{2\pi} \\\rightarrow{\boxed{\tt{r_2 = \dfrac{3}{\pi} }}}$

$\rightarrow\tt CSA = \pi \times l (r_1+r_2) \\\\\rightarrow\tt CSA = \pi \times 4(\frac{9}{\pi} + \frac{3}{\pi} ) \\\\\rightarrow\tt CSA = 4\pi (\frac{12}{\pi}) \\\\\rightarrow\tt CSA = 4 \cancel{\pi} \times \frac{12}{\cancel{\pi} }\\\\\rightarrow{\boxed{\bold\green{CSA = 48\:{cm}^{2} }}}$