Question Number 180299 by mnjuly1970 last updated on 10/Nov/22
Answered by som(math1967) last updated on 10/Nov/22
Commented by som(math1967) last updated on 10/Nov/22
![let O is centre of semicircle AO= (((a+b)/2) −r) [semicircle and circle touch internally] BO=b−((a+b)/2)=((b−a)/2) (((a+b)/2)−r)^2 =r^2 +(((b−a)/2))^2 (((a+b)^2 )/4) −r(a+b)+r^2 =r^2 +(((b−a)^2 )/4) ⇒(((a+b)^2 −(b−a)^2 )/4)=r(a+b) ⇒((4ab)/4)=r(a+b) ⇒ (1/r)=((a+b)/(ab)) ∴ (1/r)=(1/a)+(1/b)](Q180303.png)
$${let}\:{O}\:{is}\:{centre}\:{of}\:{semicircle} \\ $$$${AO}=\:\left(\frac{{a}+{b}}{\mathrm{2}}\:−{r}\right)\:\left[{semicircle}\:{and}\:{circle}\right. \\ $$$$\left.{touch}\:{internally}\right] \\ $$$${BO}={b}−\frac{{a}+{b}}{\mathrm{2}}=\frac{{b}−{a}}{\mathrm{2}} \\ $$$$\left(\frac{{a}+{b}}{\mathrm{2}}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} +\left(\frac{{b}−{a}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\frac{\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{4}}\:−{r}\left({a}+{b}\right)+{r}^{\mathrm{2}} ={r}^{\mathrm{2}} +\frac{\left({b}−{a}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\Rightarrow\frac{\left({a}+{b}\right)^{\mathrm{2}} −\left({b}−{a}\right)^{\mathrm{2}} }{\mathrm{4}}={r}\left({a}+{b}\right) \\ $$$$\Rightarrow\frac{\mathrm{4}{ab}}{\mathrm{4}}={r}\left({a}+{b}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{r}}=\frac{{a}+{b}}{{ab}} \\ $$$$\therefore\:\frac{\mathrm{1}}{{r}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}} \\ $$
Commented by mnjuly1970 last updated on 10/Nov/22
$$\:\:\:{thanks}\:{alot}\:{sir}\:..{excellent} \\ $$$${solution}... \\ $$