Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 77390 by Maclaurin Stickker last updated on 05/Jan/20

Commented by Maclaurin Stickker last updated on 05/Jan/20

find the radius of the shaded   circumference as a function of  side a of the square.

$${find}\:{the}\:{radius}\:{of}\:{the}\:{shaded}\: \\ $$$${circumference}\:{as}\:{a}\:{function}\:{of} \\ $$$${side}\:{a}\:{of}\:{the}\:{square}. \\ $$

Answered by jagoll last updated on 06/Jan/20

r(1+(√2)) = a((√2)−1)  r = a(3−2(√2))

$$\mathrm{r}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:=\:\mathrm{a}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\mathrm{r}\:=\:\mathrm{a}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$

Commented by mr W last updated on 06/Jan/20

it′s not correct sir!  the center of the small circle doesn′t  lie on the diagonal of the square!

$${it}'{s}\:{not}\:{correct}\:{sir}! \\ $$$${the}\:{center}\:{of}\:{the}\:{small}\:{circle}\:{doesn}'{t} \\ $$$${lie}\:{on}\:{the}\:{diagonal}\:{of}\:{the}\:{square}! \\ $$

Commented by jagoll last updated on 06/Jan/20

  o yes sir i agree

$$\:\:{o}\:{yes}\:{sir}\:{i}\:{agree} \\ $$

Answered by mr W last updated on 06/Jan/20

Commented by mr W last updated on 06/Jan/20

AC=a−r  BC=a+r  sin β=((DC)/(AC))=(r/(a−r))=cos α  BC^2 =AC^2 +AB^2 −2×AC×AB×cos α  (a+r)^2 =(a−r)^2 +a^2 −2(a−r)a×(r/(a−r))  a^2 +r^2 +2ar=a^2 −2ar+r^2 +a^2 −2ar  0=a^2 −6ar  ⇒r=(a/6)

$${AC}={a}−{r} \\ $$$${BC}={a}+{r} \\ $$$$\mathrm{sin}\:\beta=\frac{{DC}}{{AC}}=\frac{{r}}{{a}−{r}}=\mathrm{cos}\:\alpha \\ $$$${BC}^{\mathrm{2}} ={AC}^{\mathrm{2}} +{AB}^{\mathrm{2}} −\mathrm{2}×{AC}×{AB}×\mathrm{cos}\:\alpha \\ $$$$\left({a}+{r}\right)^{\mathrm{2}} =\left({a}−{r}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}\left({a}−{r}\right){a}×\frac{{r}}{{a}−{r}} \\ $$$${a}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{ar}={a}^{\mathrm{2}} −\mathrm{2}{ar}+{r}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{ar} \\ $$$$\mathrm{0}={a}^{\mathrm{2}} −\mathrm{6}{ar} \\ $$$$\Rightarrow{r}=\frac{{a}}{\mathrm{6}} \\ $$

Commented by mr W last updated on 06/Jan/20

or:  AD^2 =AC^2 −DC^2 =(a−r)^2 −r^2 =a^2 −2ar  BC^2 =AD^2 +(AB−DC)^2   (a+r)^2 =a^2 −2ar+(a−r)^2   ⇒r=(a/6)

$${or}: \\ $$$${AD}^{\mathrm{2}} ={AC}^{\mathrm{2}} −{DC}^{\mathrm{2}} =\left({a}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} ={a}^{\mathrm{2}} −\mathrm{2}{ar} \\ $$$${BC}^{\mathrm{2}} ={AD}^{\mathrm{2}} +\left({AB}−{DC}\right)^{\mathrm{2}} \\ $$$$\left({a}+{r}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} −\mathrm{2}{ar}+\left({a}−{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{{a}}{\mathrm{6}} \\ $$

Commented by jagoll last updated on 06/Jan/20

how AC = a − r sir?

$${how}\:{AC}\:=\:{a}\:−\:{r}\:{sir}? \\ $$

Commented by mr W last updated on 06/Jan/20

Commented by mr W last updated on 06/Jan/20

AC=AE−CE=a−r

$${AC}={AE}−{CE}={a}−{r} \\ $$

Commented by Maclaurin Stickker last updated on 06/Jan/20

Yes, Mr W, you are right!

$${Yes},\:{Mr}\:{W},\:{you}\:{are}\:{right}! \\ $$

Commented by mr W last updated on 06/Jan/20

thanks for comfirming sir!

$${thanks}\:{for}\:{comfirming}\:{sir}! \\ $$

Commented by Tawa11 last updated on 29/Dec/21

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com