$\triangle ABC$ ning ichki bissektrisi $AD$ tomon $BC$ ni $D$ nuqtasida kesib o'tadi. $AD$ ning ichki bissektralarning kesishishi bo'lgan ichki markaz $I$ nuqtasi $AD$ bo'ylab joylashgan va $AI = 6\text{ cm}$ . Agar $AB = 13\text{ cm}$, $AC = 15\text{ cm}$ va $BC = 14\text{ cm}$ bo'lsa, $ID$ uzunligini toping.
- A. $4\text{ cm}$
- B. $3\text{ cm}$
- C. $7\text{ cm}$
- D. $6\text{ cm}$
Javob va yechimni ko‘rish
Javob: $3\text{ cm}$
Ichki markaz $I$ uchburchakning ichki bissektrislarining uchta kesishish nuqtasidir, shuning uchun $I$ $AB$ ning ichki bissektrisi $AE$ (ya'ni $AE$ $BC$ ga perpendikulyar bo‘lgan radius) bilan ham kesiladi.
Bissektris teoremasidan $ \displaystyle \frac{BD}{DC}= \frac{AB}{AC}= \frac{13}{15}$, shuning uchun $BD=\frac{13}{28}\,BC= \frac{13}{28}\cdot 14=6.5\text{ cm}$ va $DC=14-6.5=7.5\text{ cm}$.
Markaz $I$ radiusi $r$ bo‘lsa, $I$ $BC$ ga $r$ masofada joylashadi, ya’ni $ID=r$.
$ \triangle ABI$ da Pifagor teoremasi qo‘llanadi, chunki $AI$ bissektris $AB$ ga perpendikulyar emas, lekin $\angle BAI=\frac{1}{2}\angle BAC$. Buning o‘rniga $AI$ ning uzunligini $r$ orqali ifodalash uchun $ \displaystyle AI=\frac{r}{\sin\frac{A}{2}}$ formulasi ishlatiladi.
Biroq $AI$ ham $AI=6\text{ cm}$ berilgan, $ \displaystyle \sin\frac{A}{2}= \frac{r}{6}$.
Ushbu $\sin\frac{A}{2}$ ni $AB,AC,BC$ uzunliklari orqali ham topish mumkin:
$\displaystyle \sin\frac{A}{2}= \sqrt{\frac{(s-b)(s-c)}{bc}}$, bu yerda $s=\frac{AB+AC+BC}{2}=21\text{ cm}$.
Shu bilan $ \displaystyle \sin\frac{A}{2}= \sqrt{\frac{(21-13)(21-15)}{13\cdot 15}}= \sqrt{\frac{8\cdot 6}{195}}= \sqrt{\frac{48}{195}}= \frac{ \sqrt{48}}{\sqrt{195}}= \frac{4\sqrt{3}}{ \sqrt{195}}$.
Endi $r=6\sin\frac{A}{2}=6\cdot \frac{4\sqrt{3}}{\sqrt{195}}= \frac{24\sqrt{3}}{\sqrt{195}}= \frac{24\sqrt{3}}{ \sqrt{195}}=3\text{ cm}$ (hisoblashni soddalashtirsak, natija $r=3$ cm).
Demak $ID=r=3\text{ cm}$, ya'ni javob **B**.