Question

A right triangle ABC with sides 5 cm,12 cm and 13 cm is revolved about the side 5 cm. Find volume of the solid so obtained.

If it is now revolved about the side 12 cm, then what would be the ratio of the volumes of the two solids obtained in two cases?

Plz explain with complete steps & justifications. ☺

Answer 1

let hypotenuse =13cm
given triangle revolves around 12cm
radius=5cm
height=12cm
volume=1/3PI×r^2×H
volume=1/3×PI×5×5×12
vol.=100PIcm ^3 (nearly)

second case,
r=12cm
h=5cm
vol.=1/3×PI×12×12×5
vol.=240PI cm^3

the ratio is 100/240
5:12

Answer 2

Given,

A right triangle ABC with sides 5 cm,12 cm and 13 cm is revolved about the side 5 cm.

To find,

(i)Find volume of the solid so obtained.

(ii) If it is now revolved about the side 12 cm, then what would be the ratio of the volumes of the two solids obtained in two cases?

Solution,

(i) let hypotenuse =1 3cm

  Here, we have

  a triangle that revolves around 12cm

   radius = 5cm

   height = 12cm

   volume = 1/3 x π × r² × H

                 = 1/3 × π × 5 × 5 × 12

                 = 100π cm³ = 314 cm³

(ii) r = 12cm

    h = 5cm

volume = 1/3 × π × 12 × 12 × 5

             = 240π cm³ = 753.6 cm³

Ratio =  100π / 240π = 5:12

Hence, (i) volume of the solid so obtained is 314 cm³ and (ii) the ratio of the volumes of the two solids obtained in two cases is 5:12.