Question Number 48777 by naka3546 last updated on 28/Nov/18 | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Nov/18 | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Nov/18 | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Nov/18 | ||
$${yes}\:{sir}\:{i}\:{found}\:{my}\:{mistake}... \\ $$ | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 28/Nov/18 | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 28/Nov/18 | ||
$${Area}\:{of}\:{big}\:{white}\:{porion}\:{found} \\ $$$${now}\:{from}\:{triangle}\:\bigtriangleup{BOC}\:\:{and}\:{sector}\:{OBXCO} \\ $$$${calculating}\:{small}\:{white}\:{area} \\ $$$${pls}\:{wait}=... \\ $$ | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 28/Nov/18 | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 28/Nov/18 | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Nov/18 | ||
$${who}\:{red}\:{marked}\:{and}\:{why} \\ $$ | ||
Commented by mr W last updated on 28/Nov/18 | ||
$${green}\:{area}={white}\:{area}=\frac{{area}\:{of}\:{circle}}{\mathrm{2}} \\ $$ | ||
Commented by mr W last updated on 28/Nov/18 | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Nov/18 | ||
$${before}\:{pointing}\:{fingure}\:{better}\:{you}\:{judge}\:{yourself} \\ $$$$\boldsymbol{{O}}{K}\:{i}\:{am}\:{leaving}\:{the}\:{forum}... \\ $$ | ||
Commented by mr W last updated on 29/Nov/18 | ||
$${thank}\:{you}\:{sir}! \\ $$$${the}\:{question}\:{is}\:{quite}\:{general},\:{but}\:{in} \\ $$$${your}\:{calculation}\:{you}\:{have}\:{assumed} \\ $$$${that}\:{the}\:{thing}\:{is}\:{symmetric}.\:{in}\:{your} \\ $$$${first}\:{chart}\:{you}\:{assumed}\:{AC}={AD}\:{and} \\ $$$${BC}={ED},\:{so}\:{you}\:{got}\:{BC}={R}\:\mathrm{sin}\:\frac{\mathrm{45}°}{\mathrm{2}}.\:{but} \\ $$$${the}\:{question}\:{doesn}'{t}\:{mean}\:{this}\:{special} \\ $$$${case}.\:{i}.{e}.\:{generally}\:{AC}\neq{AD}\:{and}\: \\ $$$${BC}\neq{ED}\neq{R}\:\mathrm{sin}\:\frac{\mathrm{45}°}{\mathrm{2}}. \\ $$$${that}'{s}\:{to}\:{say},\:{to}\:{be}\:{exact},\:{your}\:{calculation}\: \\ $$$${is}\:{only}\:{a}\:{confirmation}\:{for}\:{the}\:{statement} \\ $$$${that}\:{green}\:{area}\:=\:{circle}\:{area}\:/\:\mathrm{2}\:{in}\:{a} \\ $$$${special}\:{case},\:{but}\:{it}'{s}\:{not}\:{the}\:{general} \\ $$$${proof}\:{for}\:{the}\:{statement}. \\ $$ | ||
Commented by mr W last updated on 29/Nov/18 | ||
$${I}\:{think}\:{someone}\:{red}\:{marked}\:{your}\:{posts} \\ $$$${by}\:{mistake}.\:{in}\:{fact}\:{he}\:{just}\:{wanted}\:{to}\:{give} \\ $$$${your}\:{posts}\:{a}\:{Like}.\:{this}\:{happens}\: \\ $$$${frequentlly}\:{in}\:{the}\:{forum}.\:{so}\:{don}'{t}\:{take} \\ $$$${it}\:{serious}. \\ $$ | ||