Question

Find the slant height of a cone whose, Base radius = 14 cm, height 0.4 m Base diameter =70 cm, curved surface area =4070 cm².

- Take care of the units.
- Thank you :)) ​

Answer 1

$\sf \: Given \: \begin{cases} \sf \: base \: radius \: of \: a \: cone = \bold{14 cm} \\ \\ \sf \: height \: of \: the \: cone = 0.4 m = \: \bold{40cm} \\ \\ \sf \: base \: diameter \: of \: the \: cone = \bold{70cm} \\ \\ \sf \: curved \: surface \:area \: of \: the \: cone = \bold{4070cm {}^{2} } \end{cases}$

$\:$

Here we've to find the slant height of the cone.

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As we know that, formula for finding the curved surface area of a cone is :

$\:$

$\boxed{{ \sf \: curved \: surface \: area \: of \: cone = \pi \: rl}}$

• Let's put the values in the given formula.

$\:$

$\sf : \implies \: \pi \: rl = 4070$

$\:$

$\sf : \implies \: l = \dfrac{4070}{ \pi \: \times 35}$

$\:$

$\sf : \implies \: l = \dfrac{4070}{110}$

$\:$

$\sf : \implies \: \boxed{ \sf \: l =37 }$

$\:$

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$\:$

Now, We know that formula for finding slant height is :

$\:$

$\boxed{ \sf \:(slant \: height) {}^{2} + (radius) {}^{2} + (height) {}^{2} }$

• Let's put the values in the given formula.

$\:$

$\sf : \implies \: (l) {}^{2} = (14) {}^{2} + (40) {}^{2}$

$\:$

$\sf : \implies \: l = \sqrt{(14) {}^{2} + (40) {}^{2} }$

$\:$

$\sf : \implies \: l = \sqrt{196 + 1600}$

$\:$

$\sf : \implies \: l = \sqrt{1796}$

$\:$

$\sf : \implies \: \boxed{ \sf \: l = 42.37cm}$

Answer 2

$\color{lime} \bold{Given,} \\ \\ \sf \bull \: Base \: radius \: = \: 14 \: cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \bull \:Base \: diameter \: = \: 70 \: cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \bull \: Height \: = \: 0.4 \: m \: or \: 40 \: cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \bull \: Curved \: surface \: area \: = \: 4070 \: {cm}^{2}$

$\color{red} \bold{To\:find}$

$\sf \bull The\:slant\:height\:(l)$

$\color{skyblue} \bold{Solution,}$

$\sf \: Let,\: \\ \\ \sf \:\bull \: Base \: radius \: = \: r_1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \bull \: Height \: = \: h \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \bull \: Base \: diameter \: = \: \frac{d}{2} \: = \: r_2$

$\bf We\:know\:that,$

$\bf \implies \:CSA\:=\:\pi r l$

$\bf \: Diameter \: = \: 70 \: cm \: \: \: \: \\ \\ \bf Radius \: (r_2) \: = \: \frac{70}{2} \: \: \: \: \\ \\ \bf Radius \: (r_2) \: = \:35 \: cm$

$\color{deeppink}\bold{Substituting \: the \: values,} \: \: \: \: \: \\ \\ \bf \: \implies \: 4070 \: = \: \frac{22}{7} \: \times \: 35 \: \times \: l \\ \\ \bf \implies \: 4070 \: \times \: 110 \: \times \: l \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf \implies \: l \: = \: \frac{4070}{110} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf \implies \: l \: = \: 37 \: cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\color{aqua} \bf \: Hence \: the \: slant \: height \: of \: the \: cone \: is \: 37 \: cm.$

$\color{lime}\rule{200000000 pt}{2pt}$

$\large \bold{Additional\: Information}$

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