Question-221733

Facebook Tweet Pin

Question Number 221733 by Tawa11 last updated on 09/Jun/25

Commented by Tawa11 last updated on 09/Jun/25

$$\mathrm{Is}\:\:\mathrm{10}.\mathrm{37cm}\:\:\mathrm{correct}? \\ $$

Commented by ajfour last updated on 09/Jun/25

$${nice}\:{geometrical}\:{cartoon} \\ $$

Commented by som(math1967) last updated on 09/Jun/25

$${How}\:{OT}<{AT}\:? \\ $$

Commented by mr W last updated on 09/Jun/25

it′s terrible when in geometry questions unclear and misleading drawings are given, like here or in Q221707. such questions just waste other people′s time.

$${it}'{s}\:{terrible}\:{when}\:{in}\:{geometry} \\ $$$${questions}\:{unclear}\:{and}\:{misleading} \\ $$$${drawings}\:{are}\:{given},\:{like}\:{here}\:{or} \\ $$$${in}\:{Q}\mathrm{221707}.\:{such}\:{questions}\:{just} \\ $$$${waste}\:{other}\:{people}'{s}\:{time}. \\ $$

Commented by mr W last updated on 09/Jun/25

$${i}\:{think}\:{for}\:{this}\:{question}\:{the}\:{correct} \\ $$$${drawing}\:{should}\:{be}\:{like}\:{following}: \\ $$

Commented by mr W last updated on 09/Jun/25

Commented by mr W last updated on 10/Jun/25

then we can find the radius r=((24)/(19))×(√(24^2 −19^2 ))≈18.522 cm

$${then}\:{we}\:{can}\:{find}\:{the}\:{radius} \\ $$$${r}=\frac{\mathrm{24}}{\mathrm{19}}×\sqrt{\mathrm{24}^{\mathrm{2}} −\mathrm{19}^{\mathrm{2}} }\approx\mathrm{18}.\mathrm{522}\:{cm} \\ $$

Commented by Tawa11 last updated on 10/Jun/25

Sir, what formular is this?. And there is a mistake sir: r = ((24)/(19)) × (√(24^2 − 19^2 )) ≈ 18.52

$$\mathrm{Sir},\:\mathrm{what}\:\mathrm{formular}\:\mathrm{is}\:\mathrm{this}?. \\ $$$$\mathrm{And}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{mistake}\:\mathrm{sir}: \\ $$$$\:\:\:\mathrm{r}\:\:=\:\:\frac{\mathrm{24}}{\mathrm{19}}\:×\:\sqrt{\mathrm{24}^{\mathrm{2}} \:−\:\mathrm{19}^{\mathrm{2}} }\:\:\:\approx\:\:\mathrm{18}.\mathrm{52} \\ $$

Commented by Tawa11 last updated on 10/Jun/25

Can the radius just be = (√(24^2 − 19^2 ))

$$\mathrm{Can}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{just}\:\mathrm{be}\:\:\:=\:\:\:\sqrt{\mathrm{24}^{\mathrm{2}} \:\:−\:\:\mathrm{19}^{\mathrm{2}} } \\ $$

Commented by Tawa11 last updated on 10/Jun/25

Can it be: 24^2 = r^2 + 19^2 r = (√(24^2 − 19^2 )) TB = 24,

$$\mathrm{Can}\:\mathrm{it}\:\mathrm{be}:\:\:\:\mathrm{24}^{\mathrm{2}} \:\:=\:\:\:\mathrm{r}^{\mathrm{2}} \:\:+\:\:\mathrm{19}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{r}\:\:\:=\:\:\:\sqrt{\mathrm{24}^{\mathrm{2}} \:\:−\:\:\mathrm{19}^{\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{TB}\:\:=\:\:\mathrm{24},\:\:\:\:\: \\ $$

Commented by Tawa11 last updated on 10/Jun/25

I also saw a formular, r = ((ab)/( (√(a^2 + b^2 ))))

$$\mathrm{I}\:\mathrm{also}\:\mathrm{saw}\:\mathrm{a}\:\mathrm{formular},\:\:\mathrm{r}\:\:=\:\:\frac{\mathrm{ab}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} \:\:+\:\:\mathrm{b}^{\mathrm{2}} }} \\ $$

Commented by Tawa11 last updated on 10/Jun/25

I want to know when to use: r = (a/b)(√(a^2 − b^2 )) r = ((ab)/( (√(a^2 + b^2 )))) r = (√(a^2 − b^2 )) For the same diagram. What condition can change the formula.

$$\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{know}\:\mathrm{when}\:\mathrm{to}\:\mathrm{use}: \\ $$$$\mathrm{r}\:\:=\:\:\frac{\mathrm{a}}{\mathrm{b}}\sqrt{\mathrm{a}^{\mathrm{2}} \:\:−\:\:\mathrm{b}^{\mathrm{2}} } \\ $$$$\mathrm{r}\:\:=\:\:\frac{\mathrm{ab}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} \:\:+\:\:\mathrm{b}^{\mathrm{2}} \:}} \\ $$$$\mathrm{r}\:\:\:=\:\:\:\sqrt{\mathrm{a}^{\mathrm{2}} \:\:−\:\:\mathrm{b}^{\mathrm{2}} } \\ $$$$\mathrm{For}\:\mathrm{the}\:\mathrm{same}\:\mathrm{diagram}. \\ $$$$\mathrm{What}\:\mathrm{condition}\:\mathrm{can}\:\mathrm{change}\:\mathrm{the}\:\mathrm{formula}. \\ $$

Commented by mr W last updated on 10/Jun/25

$${one}\:{can}\:{not}\:{blindly}\:{apply}\:{a}\:{formula} \\ $$$${without}\:{knowing}\:{what}\:{the}\:{symbols} \\ $$$${behind}\:{it}\:{mean}!!!\:{all}\:{the}\:{formulas}\: \\ $$$${you}\:{mentioned}\:{above}\:{are}\:{wrong}\:{or}\: \\ $$$${are}\:{right},\:{depending}\:{on}\:{what}\:{a},\:{b},\:{r}\: \\ $$$${in}\:{the}\:{formulas}\:{mean}. \\ $$

Commented by mr W last updated on 10/Jun/25

Commented by mr W last updated on 10/Jun/25

clearly (√(24^2 −19^2 ))=TC ≠ OB i thought the diagram is clear enough.

$${clearly}\:\sqrt{\mathrm{24}^{\mathrm{2}} −\mathrm{19}^{\mathrm{2}} }={TC}\:\neq\:{OB} \\ $$$${i}\:{thought}\:{the}\:{diagram}\:{is} \\ $$$${clear}\:{enough}. \\ $$

Commented by mr W last updated on 10/Jun/25

my solution (NOT my formula) is ΔTBO∼ΔBCT ⇒((OB)/(TB))=((TC)/(BC)) i.e. (r/(24))=((√(24^2 −19^2 ))/(19)) ⇒r=((24×(√(24^2 −19^2 )))/(19))≈18.522 cm

$${my}\:{solution}\:\left({NOT}\:{my}\:{formula}\right)\:{is} \\ $$$$\Delta{TBO}\sim\Delta{BCT} \\ $$$$\Rightarrow\frac{{OB}}{{TB}}=\frac{{TC}}{{BC}} \\ $$$${i}.{e}.\:\frac{{r}}{\mathrm{24}}=\frac{\sqrt{\mathrm{24}^{\mathrm{2}} −\mathrm{19}^{\mathrm{2}} }}{\mathrm{19}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{24}×\sqrt{\mathrm{24}^{\mathrm{2}} −\mathrm{19}^{\mathrm{2}} }}{\mathrm{19}}\approx\mathrm{18}.\mathrm{522}\:{cm} \\ $$

Commented by mr W last updated on 10/Jun/25

$${depending}\:{on}\:{what}\:{is}\:{given},\:{one} \\ $$$${can}\:{determine}\:{the}\:{radius}\:{in} \\ $$$${different}\:{ways}.\:{but}\:{it}\:{makes}\:{no} \\ $$$${sense}\:{just}\:{to}\:{memorize}\:{the}\: \\ $$$${formulas}. \\ $$

Commented by mr W last updated on 10/Jun/25

Commented by Tawa11 last updated on 10/Jun/25

Wow thanks sir, I really appreciate. God bless you more.

$$\mathrm{Wow}\:\mathrm{thanks}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{more}. \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply