Question-209783

Question Number 209783 by Ismoiljon_008 last updated on 21/Jul/24

Answered by mr W last updated on 21/Jul/24

Commented by mr W last updated on 21/Jul/24

AC=(√(12^2 +16^2 ))=20 ⇒radius=10 AB=12×((12)/(20))=((36)/5) BC=16×((16)/(20))=((64)/5) OB=10−((36)/5)=((14)/5) OP=((14)/5)+r=(√((10−r)^2 −r^2 )) ⇒14+5r=10(√(5(5−r))) ⇒25r^2 +640r−2304=0 ⇒r=((−320+400)/(25))=((16)/5) OQ=R−((14)/5)=(√((10−R)^2 −R^2 )) ⇒5R−14=10(√(5(5−R))) ⇒25R^2 +360R−2304=0 ⇒R=((−180+300)/(25))=((24)/5) PQ=r+R=((16)/5)+((24)/5)=8 ✓

$${AC}=\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{16}^{\mathrm{2}} }=\mathrm{20}\:\Rightarrow{radius}=\mathrm{10} \\ $$$${AB}=\mathrm{12}×\frac{\mathrm{12}}{\mathrm{20}}=\frac{\mathrm{36}}{\mathrm{5}} \\ $$$${BC}=\mathrm{16}×\frac{\mathrm{16}}{\mathrm{20}}=\frac{\mathrm{64}}{\mathrm{5}} \\ $$$${OB}=\mathrm{10}−\frac{\mathrm{36}}{\mathrm{5}}=\frac{\mathrm{14}}{\mathrm{5}} \\ $$$${OP}=\frac{\mathrm{14}}{\mathrm{5}}+{r}=\sqrt{\left(\mathrm{10}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{14}+\mathrm{5}{r}=\mathrm{10}\sqrt{\mathrm{5}\left(\mathrm{5}−{r}\right)} \\ $$$$\Rightarrow\mathrm{25}{r}^{\mathrm{2}} +\mathrm{640}{r}−\mathrm{2304}=\mathrm{0} \\ $$$$\Rightarrow{r}=\frac{−\mathrm{320}+\mathrm{400}}{\mathrm{25}}=\frac{\mathrm{16}}{\mathrm{5}} \\ $$$${OQ}={R}−\frac{\mathrm{14}}{\mathrm{5}}=\sqrt{\left(\mathrm{10}−{R}\right)^{\mathrm{2}} −{R}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{5}{R}−\mathrm{14}=\mathrm{10}\sqrt{\mathrm{5}\left(\mathrm{5}−{R}\right)} \\ $$$$\Rightarrow\mathrm{25}{R}^{\mathrm{2}} +\mathrm{360}{R}−\mathrm{2304}=\mathrm{0} \\ $$$$\Rightarrow{R}=\frac{−\mathrm{180}+\mathrm{300}}{\mathrm{25}}=\frac{\mathrm{24}}{\mathrm{5}} \\ $$$${PQ}={r}+{R}=\frac{\mathrm{16}}{\mathrm{5}}+\frac{\mathrm{24}}{\mathrm{5}}=\mathrm{8}\:\checkmark \\ $$

Commented by Ismoiljon_008 last updated on 21/Jul/24

$$\:\:\:{thank}\:{you}\:{very}\:{much} \\ $$

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