Find the volume of a right circular cone, the diameter of whose base is 14cm and the area of curved surface is 550cm squared

Question

Find the volume of a right circular cone, the diameter of whose base is 14cm and the area of curved surface is 550cm squared

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Rose 1 day 2021-11-23T00:12:01+00:00 2 Answers 0 views 0

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    0
    2021-11-23T00:13:23+00:00

    Answer- The above question is from the chapter ‘Surface Areas and Volumes’.

    Concept used: For a right circular cone,

    1) Curved surface area (C.S.A.) = πrl

    2) Volume =  \frac{1}{3} \pi r^{2} h

    3) Diameter = 2 × Radius

    4) l = \sqrt {h^{2} + r^{2}}

     \implies l^{2} = h^{2} + r^{2}

    Here, r = radius of cone

    l = slant height of cone

    h = height of cone

    Given question: Find the volume of a right circular cone, the diameter of whose base is 14 cm and the area of curved surface is 550 cm².

    Solution: For a right circular cone,

    Diameter (d) = 14 cm

    Radius (r) = d ÷ 2 = 14 ÷ 2 = 7 cm

    Curved surface area (C.S.A.) = πrl

    550 = πrl

     \dfrac{22}{7} \times 7 \times l = 550

     l = \dfrac{550}{22}

    l = 25 cm

     l^{2} = h^{2} + r^{2}

     \implies h^{2} = l^{2} - r^{2}

     h^{2} = 25^{2} - 7^{2}

     h^{2} = 625 - 49

     h^{2} = 576

     h = \sqrt{576}

    h = 24 cm

    Volume =  \frac{1}{3} \pi r^{2} h

    Volume =  \frac{1}{3} . \frac{22}{7} . 7^{2} . 24

    Volume =  22 \times 7 \times 8

    Volume = 1232 cm³

    ∴ Required volume of right circular cone = 1232 cm³.

    0
    2021-11-23T00:13:33+00:00

    \bf{\underline{\underline{\bigstar\bigstar\: Figure : }}}\\

    \:\:

    \setlength{\unitlength}{0.99cm}\begin{picture}(6, 4)\linethickness{0.26mm}\qbezier(5.8,2.0)(5.8,2.3728)(4.9799,2.6364)\qbezier(4.9799,2.6364)(4.1598,2.9)(3.0,2.9)\qbezier(3.0,2.9)(1.8402,2.9)(1.0201,2.6364)\qbezier(1.0201,2.6364)(0.2,2.3728)(0.2,2.0)\qbezier(0.2,2.0)(0.2,1.6272)(1.0201,1.3636)\qbezier(1.0201,1.3636)(1.8402,1.1)(3.0,1.1)\qbezier(3.0,1.1)(4.1598,1.1)(4.9799,1.3636)\qbezier(4.9799,1.3636)(5.8,1.6272)(5.8,2.0)\put(0.2,2){\line(1,0){2.8}}\put(3.2,4){\sf{h}}\put(3,2){\line(0,2){4.5}}\put(1.5,1.7){\sf{7cm}}\qbezier(.2,2.05)(.7,3)(3,6.5)\qbezier(5.8,2.05)(5.3,3)(3,6.5)\put(1,4){\sf l}\put(3,2.02){\circle*{0.15}}\put(2.7,2){\dashbox{0.01}(.3,.3)}\end{picture}\\

    \bf{\underline{\underline{\bigstar\bigstar\: Solution : }}}\\

    \:\:

    \footnotesize{ C.S.A \: of \: cone = { 550cm}^{2}}\\

    \footnotesize{\implies  \pi rl = { 550cm}^{2}}\\

    \footnotesize{\implies  \dfrac{22}{7} \times \dfrac{14cm}{2} \times l = { 550cm}^{2}}\\

    \footnotesize{\implies  \dfrac{22}{7} \times 7cm \times l = { 550cm}^{2}}\\

    \footnotesize{\implies  22 \times 1cm \times l = { 550cm}^{2}}\\

    \footnotesize{\implies  22cm \times l = { 550cm}^{2}}\\

    \footnotesize{\implies    l = \dfrac{{ 550cm}^{2}}{22cm} }\\

    \footnotesize{\implies   l = 25cm }\\

    \:\:

    \rule{200}3

    \:\:

    \footnotesize{\implies {l}^{2} = {r}^{2} + {h}^{2} }\\

    \footnotesize{\implies {l}^{2} - {r}^{2} = {h}^{2} }\\

    \footnotesize{\implies {(25cm)}^{2} - {(7cm)}^{2} = {h}^{2} }\\

    \footnotesize{\implies {625cm}^{2} - {49cm}^{2} = {h}^{2} }\\

    \footnotesize{\implies {576cm}^{2} = {h}^{2} }\\

    \footnotesize{\implies \sqrt{{576cm}^{2}} = h}\\

    \footnotesize{\implies 24cm = h}\\

    \:\:

    \rule{200}3

    \:\:

    \footnotesize{Volume = \pi {r}^{2} \dfrac{h}{3}}\\

    \footnotesize{\implies Volume = \dfrac{22}{7} \times {(7cm)}^{2}\times \dfrac{24cm}{3}}\\

    \footnotesize{\implies Volume = \dfrac{22}{7} \times {49cm}^{2} \times 8cm}\\

    \footnotesize{\implies Volume = 22 \times {7cm}^{2} \times 8cm}\\

    \footnotesize{\implies Volume = 22 \times {56cm}^{3} }\\

    \footnotesize{\implies Volume =  {1232cm}^{3} }\\

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