Question

8. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find
the volume of the solid so obtained. Find also the ratio of the volumes of the two
solids obtained in Questions 7 and 8.​

Answer 1

$\large\underline\bold{\mathcal{\orange{ANSWER}} }$

$\rm{\boxed{\sf{ V_1= 240\pi cm^3 \: \: V_2= 100\pi cm^3\:\: ratio= 5:12 }}}$

$\rm\underline{\underline{\bold{\pink{\mathfrak{explanation\:in\:detail:-}}}}}$

$\large\underline\bold{GIVEN:-}$

$\dashrightarrow \triangle ABC \: With \: \\ \dashrightarrow radius(r)= 12cm \\ \dashrightarrow height(H) = 5cm \\ \dashrightarrow slant \:height(L)=13cm$

$\large\underline\bold{\pink{TO\:FIND:-}}$

$\rm{\green{\star}}\: \red{ volume\:of\:cone.} \\ \\ \rm{\green{\star}} \: \red{volume \:ratios\:of\:two\:solid\:obtained.}$

$\large\underline{\underline\bold{ANSWER:}}$

$\therefore \: Now\: finding\:volume \: of \: solid \:obtained.(V_1)$

$\rm{\boxed{\underline{ \pink{ volume \:of \:solid(cone) = \dfrac{1}{3} \: \pi\: r^{2} \:h }}}}$

$\therefore Now \:finding\: Volumes,$

$\\ \\ \therefore finding \:V_1 \\ :\implies \dfrac{1}{3} \times \pi \times (12) \times (12) \times 5 \\ \\ :\implies \dfrac{1}{\cancel{\:3\:}} \times \pi \times \cancel{(12)} \times (12) \times 5 \\ :\implies 4\times 12\times 5 \times \pi \\ :\implies 240\pi cm^2 \\ \\ \rm{\underline{\boxed{\mathtt{ \green{ V_1= 240\pi cm^3.}}}}}$

$\red{\underline { in\:previous\: question\: V_2\: obtained\:by\: revoving\: \triangle ABC \: is \:}} \\ \\ \pink{\underline{ V_2= 100 \pi cm^3}}$

$\red{\underline{ Now\:finding\:ratios\:of\:the\:volumes, }} \\ \\ \\ :\dashrightarrow ratio = \dfrac{V_2}{V_1 } \\ \\ :\implies \dfrac{100\pi }{ 240\pi } \\ \\ :\implies \dfrac{10 \: \cancel{0\:} \:\: \cancel{\pi\:}}{24 \: \cancel{0\:} \: \cancel{\pi \:}} \\ :\implies \dfrac{10}{24} \\ \\ :\implies \dfrac{\cancel{10}}{\cancel{24}} \\ \\ :\implies \dfrac{5}{12} \\ \\ :\implies ratios= 5:12 \\ \\ \rm{\underline{\boxed{\mathtt{ \green{ ratio= 5:12}}}}}$

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