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Question Number 64901 by LPM last updated on 23/Jul/19

Commented by LPM last updated on 23/Jul/19

$$\mathrm{KM}=?$$

Commented by Tony Lin last updated on 23/Jul/19

$$(6-r)+(8-r)=10$$$$\Rightarrow r=2$$$$KM=(\sqrt{2}+1)r=2\sqrt{2}+2$$

Commented by Tawa1 last updated on 23/Jul/19

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{understand}\mathrm{sir}$$

Commented by Tony Lin last updated on 23/Jul/19

Commented by Tawa1 last updated on 23/Jul/19

$$\mathrm{Ohh}.\mathrm{i}\mathrm{understand}\mathrm{now}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Tawa1 last updated on 23/Jul/19

$$\mathrm{Sir},\mathrm{how}\mathrm{do}\mathrm{you}\mathrm{get}6-\mathrm{r}\mathrm{and}8-\mathrm{r}$$$$\mathrm{i}\mathrm{understand}\mathrm{other}\mathrm{step}\mathrm{sir}.$$

Commented by Tony Lin last updated on 23/Jul/19

Commented by Tony Lin last updated on 23/Jul/19

$$\because \{\begin{array}{l}AI=AI\\ \mathrm{\angle }ACI=\mathrm{\angle }AMI=90°\left(RHScongruence\right)\\ CI=MI=r\end{array}$$$$\therefore △ACI\approxeq △AMI$$$$\Rightarrow AC=AM$$$$\because CK=r,AK=6$$$$\therefore \mathit{A}\mathit{M}=\mathit{A}\mathit{C}=6-\mathit{r}$$$$\because \{\begin{array}{l}IB=IB\\ \mathrm{\angle }IDB=\mathrm{\angle }IMB=90°\left(RHScongruence\right)\\ DI=MI\end{array}$$$$\therefore △IDB\approxeq △IMB$$$$\Rightarrow DB=MB$$$$\because KD=r,KB=8$$$$\therefore \mathit{M}\mathit{B}=\mathit{D}\mathit{B}=8-\mathit{r}$$$$\because AB=AM+MB=10$$$$\therefore (6-r)+(8-r)=10$$$$\Rightarrow r=2$$$$\because CIDKissquare$$$$\therefore KI=\sqrt{2}r$$$$\Rightarrow KM=KI+IM=\sqrt{2}r+r=2\sqrt{2}+2$$

Commented by Tawa1 last updated on 23/Jul/19

$$\mathrm{Wow},\mathrm{i}\mathrm{understand}\mathrm{now}\mathrm{very}\mathrm{well}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{more}$$

Commented by mr W last updated on 23/Jul/19

$${it}^{\prime }swrongsir!$$$$sincea\ne b,$$$$KM{doesn}^{\prime }tpassthroughthecenter$$$$ofincircle,andKMisnot\mathrm{\perp }AB.$$

Commented by LPM last updated on 23/Jul/19

$$\mathrm{good}$$

Answered by mr W last updated on 23/Jul/19

$$c=\sqrt{6^2+8^2}=10$$$$areaoftriangle=\frac{6\times 8}{2}$$$$areaoftriangle=\frac{r(6+8+10)}{2}$$$$\frac{6\times 8}{2}=\frac{r(6+8+10)}{2}$$$$\Rightarrow r=2$$$$AM=6-r=4$$$$\cos A=\frac{6}{10}=\frac{3}{5}$$$${KM}^2=6^2+4^2-2\times 6\times 4\times \cos A$$$$=52-48\times \frac{3}{5}=\frac{116}{5}$$$$\Rightarrow KM=\sqrt{\frac{116}{5}}=2\sqrt{\frac{29}{5}}=\frac{2\sqrt{145}}{5}=4.82$$

Commented by Tony Lin last updated on 23/Jul/19

Commented by Tony Lin last updated on 23/Jul/19

$$sorryfornotnoticingthis,nowigotit$$

Commented by mr W last updated on 23/Jul/19

$$yes,treesir!$$

Commented by Tawa1 last updated on 23/Jul/19

$$\mathrm{Ohh}.\mathrm{Really},\mathrm{and}\mathrm{i}\mathrm{have}\mathrm{understand}\mathrm{sir}\mathrm{tonny}\mathrm{prebvious}\mathrm{work}.$$$$\mathrm{Let}\mathrm{me}\mathrm{see}\mathrm{drawing}\mathrm{to}\mathrm{understand}\mathrm{your}\mathrm{solution}\mathrm{sir}.$$

Commented by Tawa1 last updated on 23/Jul/19

$$\mathrm{I}\mathrm{want}\mathrm{to}\mathrm{get}\mathrm{your}\mathrm{second}\mathrm{line}\mathrm{sir}$$

Commented by mr W last updated on 23/Jul/19

$$iaddedacomment.$$

Commented by LPM last updated on 23/Jul/19

$$\mathrm{good}$$

Commented by Tawa1 last updated on 23/Jul/19

$$\mathrm{Ohh}.\mathrm{okay}.\mathrm{I}\mathrm{appreciate}\mathrm{sir}$$

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