In the following figure
$AB=5$ and $BC=4$. I have to find radius of the sector. Somebody help me with this question
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3Use the power of the point $A$. It gives you a relation between $CD$ and $DA$: $DA^2-CD^2=5\cdot (5+4)$. Pythagoras gives you another: $DA^2+CD^2=(5+4)^2$. – MoonLightSyzygy Dec 25 '19 at 19:44
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I don't know how much geometry you have learned. Actually there are a variety of ways to solve this problem. The methods provided by MoonLightSyzygy both works. Here I just want to present a very fundamental solution.
We can draw a line orthogonal to $AC$ from $D$ intersecting $AC$ by $F$. In addition, we can just connect $B$ and $D$. Clearly, $CD=BD=r$, where $r$ is the radius of the sector. Thus, $\triangle BDC$ is an isosceles triangle. Thus, $F$ is also the midpoint of $BC$, implying that $CF=BF=BC/2=2$.
Also notice that $\triangle CDF$ is a right triangle, which is similar to $\triangle ACD$. Thus,
$$\frac{CF}{CD}=\frac{CD}{AC}\implies\frac{2}{r}=\frac{r}{4+5}\implies r^2=18\implies r=3\sqrt{2}.$$
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By Pythagoras we have $$(AE+r)^2 = 9^2-r^2\implies \boxed{AE^2+2rAE = 81-2r^2}$$
By the power of the point $A$ we have $$AB\cdot AC = AE\cdot (AE+2r)$$So
$$ 5\cdot 9 = 81-2r^2 \implies r=3\sqrt{2}$$
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