Question

The side of a right-angled triangle are 3cm, 4cm and 5cm respectively. It revalues with respect to its longest side. What is the volume of so formed new figure. (a) 37 7 cm 3 (b) 48 -7 cm 5 (c) 211637 cm (d) None of these​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that, side of a right-angled triangle are 3cm, 4cm and 5cm respectively. It revolves with respect to its longest side.

Let assume that triangle be ABC right-angled at A, such that AB= 3 cm, AC = 4 cm and BC = 5 cm

Now, when triangle ABC is revolved along its longest side BC, two cones are generated, ABA' and ACA'

Now,

$\rm \: Area_{(\triangle\:ABC)} = \dfrac{1}{2} \times AB \times AC = \dfrac{1}{2} \times OA \times BC$

$\rm \: \dfrac{1}{2} \times 3 \times 4 = \dfrac{1}{2} \times OA \times 5$

$\rm \: 12 = 5 \: OA$

$\rm\implies \:OA = \dfrac{12}{5} \: cm$

Now,

$\rm \: Volume_{(Solid\:thus\:formed)}$

$\rm \: = \: Volume_{(Cone \: ABA')} + Volume_{(Cone \: ACA')}$

$\rm \: = \: \dfrac{1}{3}\pi {(OA)}^{2} \times (BO) + \dfrac{1}{3}\pi {(OA)}^{2} \times (OC)$

$\rm \: = \: \dfrac{1}{3}\pi {(OA)}^{2} \times( BO + OC)$

$\rm \: = \: \dfrac{1}{3}\pi {(OA)}^{2} \times( BC)$

$\rm \: = \: \dfrac{1}{3}\pi \times {\bigg(\dfrac{12}{5} \bigg) }^{2} \times \: 5$

$\rm \: = \: \dfrac{1}{3}\pi \times \dfrac{144}{25} \times \: 5$

$\rm \: = \: \dfrac{48}{5}\pi \: {cm}^{3}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:Volume_{(Solid\:thus\:formed)} = \: \dfrac{48}{5}\pi \: {cm}^{3} \: }}$

So, option (b) is correct.

$\rule{190pt}{2pt}$

Additional Information :-

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