Question Number 23564 by NECx last updated on 01/Nov/17
$${The}\:{curved}\:{surface}\:{area}\:{of}\:{a}\:{cone} \\ $$$${is}\:\mathrm{21}{cm}^{\mathrm{2}} \:.{Calculate}\:{the}\:{curved} \\ $$$${surface}\:{area}\:{of}\:{a}\:{similar}\:{cone} \\ $$$${whose}\:{height}\:{is}\:\mathrm{4}\:{times}\:{the}\:{other}. \\ $$
Commented by NECx last updated on 01/Nov/17
$${please}\:{show}\:{workings} \\ $$
Commented by ajfour last updated on 01/Nov/17
$${radius}\:{also}\:{becomes}\:\mathrm{4}\:{times}\:{or} \\ $$$${radius}\:{stays}\:{the}\:{same}\:? \\ $$
Commented by NECx last updated on 01/Nov/17
$${thats}\:{how}\:{the}\:{question}\:{is} \\ $$
Commented by NECx last updated on 01/Nov/17
$${i}\:{dont}\:{know}\:{how}\:{possible}\:{it}\:{is}\:{but} \\ $$$${if}\:{theres}\:{any}\:{possibility}\:{of}\:{the} \\ $$$${answer}\:,\:{please}\:{help}\:{out}. \\ $$
Answered by Joel577 last updated on 02/Nov/17
$${A}\:\:=\:\pi{rs}\:\:\:\:\left({s}\:=\:\sqrt{{t}^{\mathrm{2}} \:+\:{r}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:=\:\pi{r}\sqrt{{t}^{\mathrm{2}} \:+\:{r}^{\mathrm{2}} } \\ $$$$\frac{{A}_{\mathrm{1}} }{{A}_{\mathrm{2}} }\:=\:\frac{\pi{r}\sqrt{{t}_{\mathrm{1}} ^{\mathrm{2}} \:+\:{r}^{\mathrm{2}} }}{\pi{r}\sqrt{{t}_{\mathrm{2}} ^{\mathrm{2}} \:+\:{r}^{\mathrm{2}} }}\:=\:\frac{\sqrt{{t}_{\mathrm{1}} ^{\mathrm{2}} \:+\:{r}^{\mathrm{2}} }}{\:\sqrt{\left(\mathrm{4}{t}_{\mathrm{1}} \right)^{\mathrm{2}} \:+\:{r}^{\mathrm{2}} }} \\ $$$${A}_{\mathrm{2}} \:=\:\sqrt{\frac{\mathrm{16}{t}^{\mathrm{2}} \:+\:{r}^{\mathrm{2}} }{{t}^{\mathrm{2}} \:+\:{r}^{\mathrm{2}} }}\:.\:{A}_{\mathrm{1}} \\ $$
Commented by Rasheed.Sindhi last updated on 02/Nov/17
$$\mathrm{From}\:\mathrm{above} \\ $$$$\frac{{A}_{\mathrm{1}} }{{A}_{\mathrm{2}} }\:=\:\frac{\pi{r}_{\mathrm{1}} \sqrt{{t}_{\mathrm{1}} ^{\mathrm{2}} \:+\:{r}_{\mathrm{1}} ^{\mathrm{2}} }}{\pi{r}_{\mathrm{2}} \sqrt{{t}_{\mathrm{2}} ^{\mathrm{2}} \:+\:{r}_{\mathrm{2}} ^{\mathrm{2}} }}\:=\:\frac{{r}_{\mathrm{1}} \sqrt{{t}_{\mathrm{1}} ^{\mathrm{2}} \:+\:{r}_{\mathrm{1}} ^{\mathrm{2}} }}{\left(\mathrm{4}{r}_{\mathrm{1}} \right)\sqrt{\left(\mathrm{4}{t}_{\mathrm{1}} \right)^{\mathrm{2}} \:+\left(\:\mathrm{4r}_{\mathrm{1}} \right)^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\sqrt{{t}_{\mathrm{1}} ^{\mathrm{2}} \:+\:{r}_{\mathrm{1}} ^{\mathrm{2}} }}{\mathrm{4}.\mathrm{4}\sqrt{\left({t}_{\mathrm{1}} \right)^{\mathrm{2}} \:+\left(\:\mathrm{r}_{\mathrm{1}} \right)^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{16}} \\ $$$${A}_{\mathrm{2}} =\mathrm{16A}_{\mathrm{1}} =\mathrm{16}\left(\mathrm{21}\right)=\mathrm{336}\:\mathrm{cm}^{\mathrm{2}} \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 02/Nov/17
$$\mathrm{If}\:\mathrm{the}\:\mathrm{height}\:\mathrm{is}\:\mathrm{4}\:\mathrm{times},\mathrm{the}\:\mathrm{radius} \\ $$$$\mathrm{also}\:\mathrm{should}\:\mathrm{be}\:\mathrm{4}\:\mathrm{times}\:\mathrm{because} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{cones}\:\mathrm{are}\:\boldsymbol{\mathrm{similar}}. \\ $$
Commented by NECx last updated on 02/Nov/17
$${wow}!…\:{thanks} \\ $$
Commented by NECx last updated on 02/Nov/17
$${could}\:{there}\:{be}\:{another}\:{approach} \\ $$$${to}\:{solving}\:{this}\:{question} \\ $$$$ \\ $$