Question-206937

Question Number 206937 by efronzo1 last updated on 01/May/24

Answered by A5T last updated on 01/May/24

Commented by A5T last updated on 01/May/24

8×14=x(x+2r)=x^2 +2rx=112 14^2 =(x+r)^2 +r^2 =x^2 +2rx+2r^2 =112+2r^2 ⇒r^2 =42⇒r=(√(42))⇒x=(√(154))−(√(42)) sinθ=((√(42))/(14))⇒cosθ=((√(154))/(14)) AB^2 =8^2 +((√(154))−(√(42)))^2 −16((√(154))−(√(42)))((√(154))/(14)) ⇒AB=2(√(21−3(√(33)))) BC^2 =4r^2 =AB^2 +AC^2 ⇒AC=2(√(21+3(√(33)))) ⇒((AB×AC)/2)=2(√(21^2 −9×33))=2×12=24

$$\mathrm{8}×\mathrm{14}={x}\left({x}+\mathrm{2}{r}\right)={x}^{\mathrm{2}} +\mathrm{2}{rx}=\mathrm{112} \\ $$$$\mathrm{14}^{\mathrm{2}} =\left({x}+{r}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{2}{rx}+\mathrm{2}{r}^{\mathrm{2}} =\mathrm{112}+\mathrm{2}{r}^{\mathrm{2}} \\ $$$$\Rightarrow{r}^{\mathrm{2}} =\mathrm{42}\Rightarrow{r}=\sqrt{\mathrm{42}}\Rightarrow{x}=\sqrt{\mathrm{154}}−\sqrt{\mathrm{42}} \\ $$$${sin}\theta=\frac{\sqrt{\mathrm{42}}}{\mathrm{14}}\Rightarrow{cos}\theta=\frac{\sqrt{\mathrm{154}}}{\mathrm{14}} \\ $$$${AB}^{\mathrm{2}} =\mathrm{8}^{\mathrm{2}} +\left(\sqrt{\mathrm{154}}−\sqrt{\mathrm{42}}\right)^{\mathrm{2}} −\mathrm{16}\left(\sqrt{\mathrm{154}}−\sqrt{\mathrm{42}}\right)\frac{\sqrt{\mathrm{154}}}{\mathrm{14}} \\ $$$$\Rightarrow{AB}=\mathrm{2}\sqrt{\mathrm{21}−\mathrm{3}\sqrt{\mathrm{33}}} \\ $$$${BC}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} ={AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} \Rightarrow{AC}=\mathrm{2}\sqrt{\mathrm{21}+\mathrm{3}\sqrt{\mathrm{33}}} \\ $$$$\Rightarrow\frac{{AB}×{AC}}{\mathrm{2}}=\mathrm{2}\sqrt{\mathrm{21}^{\mathrm{2}} −\mathrm{9}×\mathrm{33}}=\mathrm{2}×\mathrm{12}=\mathrm{24} \\ $$

Commented by A5T last updated on 01/May/24

After finding r, one can also do this: Let perpendicular from A meet BC at D Then, ((AD)/r)=(8/(14))⇒AD=((4r)/7) [ABC]=((AD×BC)/2)=((4r^2 )/7)=24

$${After}\:{finding}\:{r},\:{one}\:{can}\:{also}\:{do}\:{this}: \\ $$$${Let}\:{perpendicular}\:{from}\:{A}\:{meet}\:{BC}\:{at}\:{D} \\ $$$${Then},\:\frac{{AD}}{{r}}=\frac{\mathrm{8}}{\mathrm{14}}\Rightarrow{AD}=\frac{\mathrm{4}{r}}{\mathrm{7}} \\ $$$$\left[{ABC}\right]=\frac{{AD}×{BC}}{\mathrm{2}}=\frac{\mathrm{4}{r}^{\mathrm{2}} }{\mathrm{7}}=\mathrm{24} \\ $$

Answered by mr W last updated on 01/May/24

Commented by mr W last updated on 01/May/24

((6/2)/r)=(r/(8+6)) ⇒r^2 =42 (h/r)=(8/(8+6))=(4/7) shaded area =((2rh)/2)=((4r^2 )/7)=((4×42)/7)=24

$$\frac{\frac{\mathrm{6}}{\mathrm{2}}}{{r}}=\frac{{r}}{\mathrm{8}+\mathrm{6}}\:\Rightarrow{r}^{\mathrm{2}} =\mathrm{42} \\ $$$$\frac{{h}}{{r}}=\frac{\mathrm{8}}{\mathrm{8}+\mathrm{6}}=\frac{\mathrm{4}}{\mathrm{7}} \\ $$$${shaded}\:{area}\:=\frac{\mathrm{2}{rh}}{\mathrm{2}}=\frac{\mathrm{4}{r}^{\mathrm{2}} }{\mathrm{7}}=\frac{\mathrm{4}×\mathrm{42}}{\mathrm{7}}=\mathrm{24} \\ $$

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