The slant height of a frustum of cone is 4 cm and the circumference of its circular ends are 22cm and 10 cm. Find the curved surface area of

Question

The slant height of a frustum of cone is 4 cm and the circumference of its circular ends are 22cm and 10 cm. Find the curved surface area of the frustum .
4 points
64 sq cm
128 sq cm
880 sq cm

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Julia 3 weeks 2021-10-05T06:55:38+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-05T06:57:29+00:00

    Given :

    • Slant height of frustum = 4 cm
    • Circumference of 1st end = 22 cm
    • Circumference of 2nd end = 10 cm

    To Find :

    • Curved surface area of frustum

    Solution :

    \large \implies \boxed{ \sf Circumference = 2\pi r}

    Case 1 :-

     \sf \implies22 = 2 \times  \frac{22}{7} \times r_1 \\  \\ \sf \implies22 \times  \frac{7}{44}  =r_1 \\  \\ \implies  \boxed{\sf  r_1 = \frac{77}{44} \: cm}

    Case 2 :-

     \sf \implies10 = 2 \times  \frac{22}{7} \times r_2 \\  \\ \sf \implies10 \times  \frac{7}{44}  =r_2 \\  \\ \implies  \boxed{\sf  r_2 = \frac{70}{44} \: cm}

    Now Curved surface area of frustum is

     \large\implies \boxed{ \boxed{ \sf CSA_{frustum} = \pi(r_1 + r_2)l}} \\\\\sf \implies \frac{22}{7} \times \bigg( \frac{77}{44}+ \frac{70}{44}\bigg) \times 4 \\  \\\sf \implies \frac{22}{7} \times\frac{77 + 70}{44}  \times 4\\\\\sf \implies \frac{22}{7} \times\frac{147}{44}  \times 4\\\\\sf \implies \frac{1}{7} \times147\times 2 \\ \\ \sf \implies 21 \times 2 \\\\\large\implies \boxed{ \boxed{ \sf CSA_{frustum} = 42 \:  {cm}^{2}}}

    0
    2021-10-05T06:57:31+00:00

    \rule{200}3

    \Huge{\orange{\underline{\textsf{Answer}}}}

    \sf\ We \: know \: that

    \sf\ Circumference \: surface \: area \: of \:frustum = \pi (r_1 + r_2)l

    \sf\ Here, l = 4 cm

    \sf\ We \: need \: to \: find \: r_1 \: and \: r_2

    \rule{200}3

    \large{\blue{\underline{\tt{For \: r_1}}}}

    \sf Circumference\:of\:first\:end = 22cm

    \leadsto \sf\ 2\pi r_1 = 22

    \leadsto \sf r_1 = \dfrac{22}{2}\times\dfrac{1}{\pi}

    \leadsto \sf r_1 = \dfrac{11}{\pi}

    \rule{200}3

    \large{\red{\underline{\tt{For \: r_2}}}}

    \sf Circumference\:of\:second\:end = 10cm

    \leadsto \sf\ 2\pi r_2 = 10

    \leadsto \sf r_2 = \dfrac{10}{2}\times\dfrac{1}{\pi}

    \leadsto \sf r_2 = \dfrac{5}{\pi}

    \rule{200}3

    \sf\ Now,

    Curved surface area of frustum =  \pi (r_1 + r_2)l

    \leadsto[tex]\sf\pi\times\dfrac{11+5}{\pi}\times 4[/tex]

    \leadsto[tex]\sf\pi\times\dfrac{16}{\pi}\times 4[/tex]

    \leadsto[tex]\sf\ 64cm^{2}[/tex]

    \rule{200}3

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