The base radius of two cones are in the ratio 3:4 and their volumes are equal. Find the ratio
of their heights?
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The base radius of two cones are in the ratio 3:4 and their volumes are equal. Find the ratio
of their heights?
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[tex]\large\underline{\sf{Solution-}}[/tex]
Given that, base radius of two cones are in the ratio 3:4.
Let assume that
Let assume that
So, we have to find h : H.
According to statement, it is given that,
[tex]\rm \: Volume_{({1}^{st} \: Cone)} = Volume_{( {2}^{nd} \: Cone)} \\ [/tex]
[tex]\rm \: \dfrac{1}{3}\pi {(3r)}^{2}h \: = \: \dfrac{1}{3}\pi {(4r)}^{2}H \\[/tex]
[tex]\rm \: {9r}^{2} h \: = \: {16r}^{2} H \\[/tex]
[tex]\rm \: 9 h \: = \: 16 H \\[/tex]
[tex]\rm \: \bf\implies \:\dfrac{h}{H} = \dfrac{16}{9} \\ [/tex]
[tex]\bf\implies \:h : H \: = \: 16 : 9 \\ [/tex]
[tex]\rule{190pt}{2pt} \\ [/tex]
Additional Information :-
[tex]\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}[/tex]