Question-208639

Question Number 208639 by Tawa11 last updated on 20/Jun/24

Commented by mr W last updated on 20/Jun/24

$${is}\:{it}\:{a}\:{red}\:{square}? \\ $$$${does}\:{one}\:{corner}\:{of}\:{the}\:{square}\:{lie} \\ $$$${on}\:{the}\:{center}\:{of}\:{semicircle}? \\ $$

Commented by Tawa11 last updated on 20/Jun/24

$$\mathrm{Yes}\:\mathrm{sir}. \\ $$

Commented by som(math1967) last updated on 20/Jun/24

$${i}\:{got}\:\mathrm{4}×\sqrt{\frac{\mathrm{5}}{\mathrm{3}}} \\ $$

Answered by AlagaIbile last updated on 20/Jun/24

$$\:\boldsymbol{{x}}\:=\:\sqrt{\frac{\mathrm{80}}{\mathrm{3}}}\:\boldsymbol{{cm}}\:=\:\mathrm{4}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\:\boldsymbol{{cm}} \\ $$

Commented by Tawa11 last updated on 20/Jun/24

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 20/Jun/24

side length of square a=(√5) radius of semicircle r a^2 =(r−a)(r+a) ⇒r^2 =2a^2 (x/(2r))=(r/( (√(r^2 +a^2 )))) ⇒x=((2r^2 )/( (√(r^2 +a^2 ))))=((4a)/( (√3)))=((4(√5))/( (√3)))=((4(√(15)))/3) ✓

$${side}\:{length}\:{of}\:{square}\:{a}=\sqrt{\mathrm{5}} \\ $$$${radius}\:{of}\:{semicircle}\:{r} \\ $$$${a}^{\mathrm{2}} =\left({r}−{a}\right)\left({r}+{a}\right) \\ $$$$\Rightarrow{r}^{\mathrm{2}} =\mathrm{2}{a}^{\mathrm{2}} \\ $$$$\frac{{x}}{\mathrm{2}{r}}=\frac{{r}}{\:\sqrt{{r}^{\mathrm{2}} +{a}^{\mathrm{2}} }} \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}{r}^{\mathrm{2}} }{\:\sqrt{{r}^{\mathrm{2}} +{a}^{\mathrm{2}} }}=\frac{\mathrm{4}{a}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{4}\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{4}\sqrt{\mathrm{15}}}{\mathrm{3}}\:\checkmark \\ $$

Commented by Tawa11 last updated on 20/Jun/24

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 20/Jun/24

Commented by Tawa11 last updated on 20/Jun/24

$$\mathrm{Wow},\:\mathrm{I}\:\mathrm{understand}\:\mathrm{now}\:\mathrm{sir}. \\ $$

Answered by A5T last updated on 20/Jun/24

Commented by A5T last updated on 20/Jun/24

s×s=5⇒s=(√5); s^2 +s^2 =r^2 ⇒r=(√(10)) ((x−k)/(2r))=(r/x)⇒2r^2 =x^2 −kx⇒20=x(x−k)...(i) (x−k)^2 =s^2 +r^2 =15⇒x−k=(√(15))...(ii) (i)&(ii)⇒x=((20)/( (√(15))))=((4(√(15)))/3)

$${s}×{s}=\mathrm{5}\Rightarrow{s}=\sqrt{\mathrm{5}};\:{s}^{\mathrm{2}} +{s}^{\mathrm{2}} ={r}^{\mathrm{2}} \Rightarrow{r}=\sqrt{\mathrm{10}} \\ $$$$\frac{{x}−{k}}{\mathrm{2}{r}}=\frac{{r}}{{x}}\Rightarrow\mathrm{2}{r}^{\mathrm{2}} ={x}^{\mathrm{2}} −{kx}\Rightarrow\mathrm{20}={x}\left({x}−{k}\right)…\left({i}\right) \\ $$$$\left({x}−{k}\right)^{\mathrm{2}} ={s}^{\mathrm{2}} +{r}^{\mathrm{2}} =\mathrm{15}\Rightarrow{x}−{k}=\sqrt{\mathrm{15}}…\left({ii}\right) \\ $$$$\left({i}\right)\&\left({ii}\right)\Rightarrow{x}=\frac{\mathrm{20}}{\:\sqrt{\mathrm{15}}}=\frac{\mathrm{4}\sqrt{\mathrm{15}}}{\mathrm{3}} \\ $$

Commented by Tawa11 last updated on 20/Jun/24

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

Commented by Tawa11 last updated on 20/Jun/24

Sir, am confused on. ((x − k)/(2r)) = (r/x) Please explain the rule that gives us that sir.

$$\mathrm{Sir},\:\mathrm{am}\:\mathrm{confused}\:\mathrm{on}. \\ $$$$\:\:\frac{\mathrm{x}\:\:−\:\:\mathrm{k}}{\mathrm{2r}}\:\:\:=\:\:\:\frac{\mathrm{r}}{\mathrm{x}} \\ $$$$\mathrm{Please}\:\mathrm{explain}\:\mathrm{the}\:\mathrm{rule}\:\mathrm{that}\:\mathrm{gives}\:\mathrm{us}\:\mathrm{that}\:\mathrm{sir}. \\ $$

Commented by A5T last updated on 20/Jun/24

Commented by A5T last updated on 20/Jun/24

△DFC is similar to ABC ⇒((DC=x−k)/(AC=2r))=((FC=r)/(BC=x))⇒2r^2 =x(x−k)

$$\bigtriangleup{DFC}\:{is}\:{similar}\:{to}\:{ABC} \\ $$$$\Rightarrow\frac{{DC}={x}−{k}}{{AC}=\mathrm{2}{r}}=\frac{{FC}={r}}{{BC}={x}}\Rightarrow\mathrm{2}{r}^{\mathrm{2}} ={x}\left({x}−{k}\right) \\ $$

Commented by Tawa11 last updated on 20/Jun/24

$$\mathrm{Wow},\:\mathrm{I}\:\mathrm{understand}\:\mathrm{now}\:\mathrm{sir} \\ $$

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