Question-64901

Tinku Tara June 4, 2023 Geometry 0 Comments

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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=?

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

(6 - r) + (8 - r) = 10

\Rightarrow r = 2

K M = (\sqrt{2} + 1) r = 2 \sqrt{2} + 2

Commented by Tawa1 last updated on 23/Jul/19

I don′t understand sir

I {don}^{'} t understand 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

Ohh . i understand now . God bless you 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.

Sir, how do you get 6 - r and 8 - r

i understand other step 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

∵ {\begin{cases} A I = A I \\ ∠ A C I = ∠ A M I = 90 ° (R H S c o n g r u e n c e) \\ C I = M I = r \end{cases}

∴ △ A C I ≊ △ A M I

\Rightarrow A C = A M

∵ C K = r, A K = 6

∴ A M = A C = 6 - r

∵ {\begin{cases} I B = I B \\ ∠ I D B = ∠ I M B = 90 ° (R H S c o n g r u e n c e) \\ D I = M I \end{cases}

∴ △ I D B ≊ △ I M B

\Rightarrow D B = M B

∵ K D = r, K B = 8

∴ M B = D B = 8 - r

∵ A B = A M + M B = 10

∴ (6 - r) + (8 - r) = 10

\Rightarrow r = 2

∵ C I D K i s s q u a r e

∴ K I = \sqrt{2} r

\Rightarrow K M = K I + I M = \sqrt{2} r + r = 2 \sqrt{2} + 2

Commented by Tawa1 last updated on 23/Jul/19

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

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

{i t}^{'} s w r o n g s i r!

s i n c e a \neq b,

K M {d o e s n}^{'} t p a s s t h r o u g h t h e c e n t e r

o f i n c i r c l e, a n d K M i s n o t ⊥ A B .

Commented by LPM last updated on 23/Jul/19

good

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{6^{2} + 8^{2}} = 10

a r e a o f t r i a n g l e = \frac{6 \times 8}{2}

a r e a o f t r i a n g l e = \frac{r (6 + 8 + 10)}{2}

\frac{6 \times 8}{2} = \frac{r (6 + 8 + 10)}{2}

\Rightarrow r = 2

A M = 6 - r = 4

\cos A = \frac{6}{10} = \frac{3}{5}

{K M}^{2} = 6^{2} + 4^{2} - 2 \times 6 \times 4 \times \cos A

= 52 - 48 \times \frac{3}{5} = \frac{116}{5}

\Rightarrow K M = \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

sorry for not noticing this, now i got it

s o r r y f o r n o t n o t i c i n g t h i s, n o w i g o t i t

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

yes, tree sir!

y e s, t r e e s i r!

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.

Ohh . Really, and i have understand sir tonny prebvious work .

Let me see drawing to understand your solution sir .

Commented by Tawa1 last updated on 23/Jul/19

I want to get your second line sir

I want to get your second line sir

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

i added a comment.

i a d d e d a c o m m e n t .

Commented by LPM last updated on 23/Jul/19

good

good

Commented by Tawa1 last updated on 23/Jul/19

Ohh. okay. I appreciate sir

Ohh . okay . I appreciate sir

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