Question-208021

Question Number 208021 by efronzo1 last updated on 02/Jun/24

Answered by A5T last updated on 02/Jun/24

\frac{C D}{C D + r} = \frac{r}{r + A B} \Rightarrow \frac{r + A B}{r + C D} = \frac{r}{C D}

\frac{r + A B}{8 + r} = \frac{r + C D}{3 + r} \Rightarrow \frac{r + A B}{r + C D} = \frac{8 + r}{3 + r} = \frac{r}{C D}

\Rightarrow C D = \frac{r (3 + r)}{8 + r}

\frac{A B}{A B + r} = \frac{r}{C D + r} = A B (C D + r) = r (A B) + r^{2}

A B = \frac{r^{2}}{C D} = \frac{r (8 + r)}{3 + r}

{(r + A B)}^{2} + {(r + C D)}^{2} = {(11 + 2 r)}^{2}

\Rightarrow {(r + A B)}^{2} = {(11 + 2 r)}^{2} - r^{2} {(1 + \frac{3 + r}{8 + r})}^{2}

A B = \sqrt{{(11 + 2 r)}^{2} - r^{2} {(1 + \frac{3 + r}{8 + r})}^{2}} - r = \frac{r (8 + r)}{3 + r}

\Rightarrow r = 12 \Rightarrow D i a m e t e r = 24

Answered by A5T last updated on 02/Jun/24

A B = \sqrt{8 (8 + 2 r)} = \frac{r (8 + r)}{3 + r} \Rightarrow r = 12

Answered by A5T last updated on 02/Jun/24

\frac{r}{r + A B} = \frac{3 + r}{11 + 2 r} \Rightarrow 11 r + 2 r^{2} = r (3 + r) + 3 A B + r A B

A B = \frac{r (8 + r)}{3 + r}

{(8 + r)}^{2} = r^{2} + \frac{r^{2} {(8 + r)}^{2}}{{(3 + r)}^{2}} \Rightarrow r = 12

Answered by efronzo1 last updated on 02/Jun/24

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