The-curved-surface-area-of-a-cone-is-21cm-2-Calculate-the-curved-surface-area-of-a-similar-cone-whose-height-is-4-times-the-other-

Question Number 23564 by NECx last updated on 01/Nov/17

T h e c u r v e d s u r f a c e a r e a o f a c o n e

i s 21 {c m}^{2} . C a l c u l a t e t h e c u r v e d

s u r f a c e a r e a o f a s i m i l a r c o n e

w h o s e h e i g h t i s 4 t i m e s t h e o t h e r .

Commented by NECx last updated on 01/Nov/17

p l e a s e s h o w w o r k i n g s

Commented by ajfour last updated on 01/Nov/17

r a d i u s a l s o b e c o m e s 4 t i m e s o r

r a d i u s s t a y s t h e s a m e ?

Commented by NECx last updated on 01/Nov/17

t h a t s h o w t h e q u e s t i o n i s

Commented by NECx last updated on 01/Nov/17

i d o n t k n o w h o w p o s s i b l e i t i s b u t

i f t h e r e s a n y p o s s i b i l i t y o f t h e

a n s w e r, p l e a s e h e l p o u t .

Answered by Joel577 last updated on 02/Nov/17

A = π r s (s = \sqrt{t^{2} + r^{2}})

= π r \sqrt{t^{2} + r^{2}}

\frac{A_{1}}{A_{2}} = \frac{π r \sqrt{t_{1}^{2} + r^{2}}}{π r \sqrt{t_{2}^{2} + r^{2}}} = \frac{\sqrt{t_{1}^{2} + r^{2}}}{\sqrt{{(4 t_{1})}^{2} + r^{2}}}

A_{2} = \sqrt{\frac{16 t^{2} + r^{2}}{t^{2} + r^{2}}} . A_{1}

Commented by Rasheed.Sindhi last updated on 02/Nov/17

From above

\frac{A_{1}}{A_{2}} = \frac{π r_{1} \sqrt{t_{1}^{2} + r_{1}^{2}}}{π r_{2} \sqrt{t_{2}^{2} + r_{2}^{2}}} = \frac{r_{1} \sqrt{t_{1}^{2} + r_{1}^{2}}}{(4 r_{1}) \sqrt{{(4 t_{1})}^{2} + {({4 r}_{1})}^{2}}}

= \frac{\sqrt{t_{1}^{2} + r_{1}^{2}}}{4.4 \sqrt{{(t_{1})}^{2} + {(r_{1})}^{2}}} = \frac{1}{16}

A_{2} = {16 A}_{1} = 16 (21) = 336 {cm}^{2}

Commented by Rasheed.Sindhi last updated on 02/Nov/17

If the height is 4 times, the radius

also should be 4 times because

the two cones are similar .

Commented by NECx last updated on 02/Nov/17

w o w! \dots t h a n k s

Commented by NECx last updated on 02/Nov/17

c o u l d t h e r e b e a n o t h e r a p p r o a c h

t o s o l v i n g t h i s q u e s t i o n

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