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

Commented by LPM last updated on 23/Jul/19

KM=?

$$\mathrm{KM}=? \\ $$

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

(6−r)+(8−r)=10 ⇒r=2 KM=((√2)+1)r=2(√2)+2

$$\left(\mathrm{6}−{r}\right)+\left(\mathrm{8}−{r}\right)=\mathrm{10} \\ $$$$\Rightarrow{r}=\mathrm{2} \\ $$$${KM}=\left(\sqrt{\mathrm{2}}+\mathrm{1}\right){r}=\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2} \\ $$

Commented by Tawa1 last updated on 23/Jul/19

I don′t understand sir

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{sir} \\ $$

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

Commented by Tawa1 last updated on 23/Jul/19

Ohh. i understand now. God bless you sir

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

Commented by Tawa1 last updated on 23/Jul/19

Sir, how do you get 6 − r and 8 − r i understand other step sir.

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

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

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

∵ { ((AI=AI)),((∠ACI=∠AMI=90° (RHS congruence))),((CI=MI=r)) :} ∴△ACI≊△AMI ⇒AC=AM ∵CK=r,AK=6 ∴AM=AC=6−r ∵ { ((IB=IB)),((∠IDB=∠IMB=90° (RHS congruence))),((DI=MI)) :} ∴△IDB≊△IMB ⇒DB=MB ∵KD=r,KB=8 ∴MB=DB=8−r ∵AB=AM+MB=10 ∴(6−r)+(8−r)=10 ⇒r=2 ∵CIDK is square ∴KI=(√2)r ⇒KM=KI+IM=(√2)r+r=2(√2)+2

$$\because\begin{cases}{{AI}={AI}}\\{\angle{ACI}=\angle{AMI}=\mathrm{90}°\:\left({RHS}\:{congruence}\right)}\\{{CI}={MI}={r}}\end{cases} \\ $$$$\therefore\bigtriangleup{ACI}\approxeq\bigtriangleup{AMI} \\ $$$$\Rightarrow{AC}={AM} \\ $$$$\because{CK}={r},{AK}=\mathrm{6} \\ $$$$\therefore\boldsymbol{{AM}}=\boldsymbol{{AC}}=\mathrm{6}−\boldsymbol{{r}} \\ $$$$\because\begin{cases}{{IB}={IB}}\\{\angle{IDB}=\angle{IMB}=\mathrm{90}°\:\left({RHS}\:{congruence}\right)}\\{{DI}={MI}}\end{cases} \\ $$$$\therefore\bigtriangleup{IDB}\approxeq\bigtriangleup{IMB} \\ $$$$\Rightarrow{DB}={MB} \\ $$$$\because{KD}={r},{KB}=\mathrm{8} \\ $$$$\therefore\boldsymbol{{MB}}=\boldsymbol{{DB}}=\mathrm{8}−\boldsymbol{{r}} \\ $$$$\because{AB}={AM}+{MB}=\mathrm{10} \\ $$$$\therefore\left(\mathrm{6}−{r}\right)+\left(\mathrm{8}−{r}\right)=\mathrm{10} \\ $$$$\Rightarrow{r}=\mathrm{2} \\ $$$$\because{CIDK}\:{is}\:{square} \\ $$$$\therefore{KI}=\sqrt{\mathrm{2}}{r} \\ $$$$\Rightarrow{KM}={KI}+{IM}=\sqrt{\mathrm{2}}{r}+{r}=\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2} \\ $$

Commented by Tawa1 last updated on 23/Jul/19

Wow, i understand now very well. God bless you more

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

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

it′s wrong sir! since a≠b, KM doesn′t pass through the center of incircle, and KM is not ⊥AB.

$${it}'{s}\:{wrong}\:{sir}! \\ $$$${since}\:{a}\neq{b}, \\ $$$${KM}\:{doesn}'{t}\:{pass}\:{through}\:{the}\:{center} \\ $$$${of}\:{incircle},\:{and}\:{KM}\:{is}\:{not}\:\bot{AB}. \\ $$

Commented by LPM last updated on 23/Jul/19

good

$$\mathrm{good} \\ $$

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

c=(√(6^2 +8^2 ))=10 area of triangle=((6×8)/2) area of triangle=((r(6+8+10))/2) ((6×8)/2)=((r(6+8+10))/2) ⇒r=2 AM=6−r=4 cos A=(6/(10))=(3/5) KM^2 =6^2 +4^2 −2×6×4×cos A =52−48×(3/5)=((116)/5) ⇒KM=(√((116)/5))=2(√((29)/5))=((2(√(145)))/5)=4.82

$${c}=\sqrt{\mathrm{6}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} }=\mathrm{10} \\ $$$${area}\:{of}\:{triangle}=\frac{\mathrm{6}×\mathrm{8}}{\mathrm{2}} \\ $$$${area}\:{of}\:{triangle}=\frac{{r}\left(\mathrm{6}+\mathrm{8}+\mathrm{10}\right)}{\mathrm{2}} \\ $$$$\frac{\mathrm{6}×\mathrm{8}}{\mathrm{2}}=\frac{{r}\left(\mathrm{6}+\mathrm{8}+\mathrm{10}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{r}=\mathrm{2} \\ $$$${AM}=\mathrm{6}−{r}=\mathrm{4} \\ $$$$\mathrm{cos}\:{A}=\frac{\mathrm{6}}{\mathrm{10}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${KM}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} −\mathrm{2}×\mathrm{6}×\mathrm{4}×\mathrm{cos}\:{A} \\ $$$$=\mathrm{52}−\mathrm{48}×\frac{\mathrm{3}}{\mathrm{5}}=\frac{\mathrm{116}}{\mathrm{5}} \\ $$$$\Rightarrow{KM}=\sqrt{\frac{\mathrm{116}}{\mathrm{5}}}=\mathrm{2}\sqrt{\frac{\mathrm{29}}{\mathrm{5}}}=\frac{\mathrm{2}\sqrt{\mathrm{145}}}{\mathrm{5}}=\mathrm{4}.\mathrm{82} \\ $$

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

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

sorry for not noticing this, now i got it

$${sorry}\:{for}\:{not}\:{noticing}\:{this},\:{now}\:{i}\:{got}\:{it} \\ $$

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

yes, tree sir!

$${yes},\:\:{tree}\:{sir}! \\ $$

Commented by Tawa1 last updated on 23/Jul/19

Ohh. Really, and i have understand sir tonny prebvious work. Let me see drawing to understand your solution sir.

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

Commented by Tawa1 last updated on 23/Jul/19

I want to get your second line sir

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

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

i added a comment.

$${i}\:{added}\:{a}\:{comment}. \\ $$

Commented by LPM last updated on 23/Jul/19

good

$$\mathrm{good} \\ $$

Commented by Tawa1 last updated on 23/Jul/19

Ohh. okay. I appreciate sir

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

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