Question-216180

Question Number 216180 by mr W last updated on 29/Jan/25

Answered by AntonCWX last updated on 29/Jan/25

Commented by AntonCWX last updated on 29/Jan/25

{F G}^{2} = F C \times F B

6^{2} = 4 (C B + 4) \Rightarrow C B = 5

C E = C B = 5

G E = a

E H = b

E D = 2 r - a

G E \times E D = C E \times E H

a (2 r - a) = 5 b

2 r - a = \frac{5 b}{a}

2 r = \frac{5 b + a^{2}}{a}

r = \frac{5 b + a^{2}}{2 a}

F r o m Δ F C E,

{F E}^{2} = {F C}^{2} + {C E}^{2}

{F E}^{2} = 4^{2} + 5^{2} \Rightarrow {F E}^{2} = 41

F r o m Δ F G E,

{F E}^{2} = {F G}^{2} + {G E}^{2}

41 = 6^{2} + a^{2}

a^{2} = 5 \Rightarrow a = \sqrt{5}

r = \frac{5 b + 5}{2 \sqrt{5}}

F r o m Δ C B H,

{B H}^{2} = {B C}^{2} + {C H}^{2}

{(2 r)}^{2} = 5^{2} + {(5 + b)}^{2}

4 {(\frac{5 b + 5}{2 \sqrt{5}})}^{2} = 25 + 25 + 10 b + b^{2}

\frac{25 b^{2} + 50 b + 25}{5} = 50 + 10 b + b^{2}

25 b^{2} + 50 b + 25 = 250 + 50 b + 5 b^{2}

20 b^{2} = 225 \Rightarrow b = \sqrt{\frac{45}{4}} = \frac{3 \sqrt{5}}{2}

r = \frac{5 (\frac{3 \sqrt{5}}{2}) + 5}{2 \sqrt{5}} = \frac{2 \sqrt{5} + 15}{4}

Commented by mr W last updated on 29/Jan/25

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Answered by A5T last updated on 29/Jan/25

Commented by AntonCWX last updated on 30/Jan/25

I t h i n k y o u m i s t a k e n l y t y p e d i n t h e 1 t h e r e \dots

\frac{2 \sqrt{5} + 15}{41}

R e m o v e t h e 1 a n d i t s d o n e .

Commented by A5T last updated on 30/Jan/25

y = ED = EF \Rightarrow 6^{2} = 4 (4 + y) \Rightarrow y = 5

6^{2} + {AD}^{2} = 4^{2} + 5^{2} \Rightarrow AD = \sqrt{5} \Rightarrow CD = \sqrt{41}

(AD \times CE) + (AC \times DE) = CD \times EA

\Rightarrow EA = \frac{4 \sqrt{5} + 30}{\sqrt{41}} = z

2 r (\sqrt{5}) (2 r - \sqrt{5}) + 5^{2} (2 r) = ({4 r}^{2} - {EA}^{2}) \sqrt{5} + {EA}^{2} (2 r - \sqrt{5})

\Rightarrow r ({EA}^{2} - 20) = {EA}^{2} \sqrt{5} \Rightarrow r = \frac{{EA}^{2} \sqrt{5}}{{EA}^{2} - 20}

\Rightarrow r = \frac{49 \sqrt{5} + 60}{8 + 12 \sqrt{5}} = \frac{2 \sqrt{5} + 15}{4} \approx 4.868

Commented by A5T last updated on 30/Jan/25

It was a typo, thanks .

Commented by mr W last updated on 29/Jan/25

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Answered by mr W last updated on 29/Jan/25

Commented by mr W last updated on 29/Jan/25

4 (4 + x) = 6^{2}

\Rightarrow x = 5

6^{2} + y^{2} = 4^{2} + 5^{2}

\Rightarrow y = \sqrt{5}

{(R - \sqrt{5})}^{2} = {(\frac{x}{2})}^{2} + {(x - \sqrt{R^{2} - {(\frac{x}{2})}^{2}})}^{2}

10 + R \sqrt{5} = 5 \sqrt{R^{2} - 2 . 5^{2}}

R^{2} - \sqrt{5} R - \frac{205}{16} = 0

\Rightarrow R = \frac{\sqrt{5} + \sqrt{5 + \frac{205}{4}}}{2} = \frac{2 \sqrt{5} + 15}{4} \approx 4.868

Answered by A5T last updated on 29/Jan/25

Commented by A5T last updated on 29/Jan/25

4 \times (4 + EF) = 6^{2} \Rightarrow EF = ED = 5

6^{2} + {AD}^{2} = {CE}^{2} + {ED}^{2} \Rightarrow AD = \sqrt{5}

AD \times DH = ED \times DG \Rightarrow \sqrt{5} (2 r - \sqrt{5}) = 5 DG \dots (i)

In △ EFG; {GF}^{2} = {EF}^{2} + {EG}^{2}

\Rightarrow {4 r}^{2} = 25 + {(5 + DG)}^{2} \Rightarrow {4 r}^{2} = {DG}^{2} + 10 DG \dots (ii)

(i) in (ii)

\Rightarrow {4 r}^{2} = 25 + {(5 + \frac{2 r - \sqrt{5}}{\sqrt{5}})}^{2} \Rightarrow {4 r}^{2} = \frac{125 + {(2 r + 4 \sqrt{5})}^{2}}{5}

\Rightarrow {16 r}^{2} - 16 \sqrt{5} r - 205 = 0

\Rightarrow r = \frac{16 \sqrt{5} \underline{+} \sqrt{16^{2} (5) - 4 (16) (- 205)}}{32} = \frac{16 \sqrt{5} \underline{+} 120}{32} > 0

\Rightarrow r = \frac{2 \sqrt{5} + 15}{4}

Commented by AntonCWX last updated on 31/Jan/25

S a m e i d e a a s m i n e \underset{―}{\underset{⏟}{⋚}}