Question

A bucket is in the form of a frustum of a cone.If the height of the bucket is 16cm and the radii of the upper and lower ends are 18cm and 6cm respectively, find the height of the cone of which

the bucket is a part




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Answer 1

Given :-

• Height of frustum, h = 16cm

• Radius of frustum of one end, r = 6 cm

• Radius of frustum of other end, R = 18 cm

To Find :-

• Height of cone of which bucket is a part.

Solution :-

Given that,

• Height of frustum, h = 16cm

• Radius of frustum of one end, r = 6 cm

• Radius of frustum of other end, R = 18 cm

Let assume that height of cone of which bucket is a part is 'H' cm.

$\rm :\longmapsto\:In \: \triangle \: AED \: and \: \triangle \: ABC$

$\rm :\longmapsto\: \: \angle \: AED \: = \: \angle \: ABC \: \: \: \: \: \{ \: each \: 90 \degree \: \}$

$\rm :\longmapsto\: \: \angle \: ADE \: = \: \angle \: ACB \: \: \: \: \: \{ \: each \: 90 \degree \: \}$

$\rm :\longmapsto\: \: \triangle \: AED \: \sim \: \triangle \: ABC \: \: \: \: \: \: (AA)$

$\rm :\implies\:\dfrac{AE}{AB} = \dfrac{ED}{BC}$

$\rm :\longmapsto\:\dfrac{H - h}{H} = \dfrac{r}{R}$

$\rm :\longmapsto\:\dfrac{H - 16}{H} = \dfrac{6}{18}$

$\rm :\longmapsto\:\dfrac{H - 16}{H} = \dfrac{1}{3}$

$\rm :\longmapsto\:3H - 48 = H$

$\rm :\longmapsto\:3H - H = 48$

$\rm :\longmapsto\:2H = 48$

$\bf :\longmapsto\:H = 24 \: cm$

Short cut :-

If we have given that

• Height of frustum = h units

• Radius of frustum of one end = r units

• Radius of frustum of other end = R units

• Height of cone of which frustum is part = H units

Then,

Height of cone,

$\bf :\longmapsto\:H = \dfrac{hR}{R - r} \: units$

Additional Information :-

$\rm :\longmapsto\:CSA_{(frustum)} = \pi \: (R + r) \: l$

$\rm :\longmapsto\:TSA_{(frustum)} = \pi \: \bigg( {R}^{2} + {r}^{2} + (R + r)l \bigg)$

$\rm :\longmapsto\:Volume_{(frustum)} = \dfrac{\pi \: h}{3}( {R}^{2} + {r}^{2} + Rr)$