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Question-88487




Question Number 88487 by I want to learn more last updated on 11/Apr/20
Commented by I want to learn more last updated on 11/Apr/20
Please help.
$$\mathrm{Please}\:\mathrm{help}. \\ $$
Commented by jagoll last updated on 11/Apr/20
(3+r)^2  = (3−r)^2 +(5−r)^2   (3+r)^2 −(3−r)^2 =(5−r)^2   (6)(2r) = r^2 −10r+25  r^2 −22r+25 = 0  r = ((22 − (√(22^2 −100)))/2) = ((22−19.6)/2)  r ≈ 1.2 cm
$$\left(\mathrm{3}+\mathrm{r}\right)^{\mathrm{2}} \:=\:\left(\mathrm{3}−\mathrm{r}\right)^{\mathrm{2}} +\left(\mathrm{5}−\mathrm{r}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{3}+\mathrm{r}\right)^{\mathrm{2}} −\left(\mathrm{3}−\mathrm{r}\right)^{\mathrm{2}} =\left(\mathrm{5}−\mathrm{r}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{6}\right)\left(\mathrm{2r}\right)\:=\:\mathrm{r}^{\mathrm{2}} −\mathrm{10r}+\mathrm{25} \\ $$$$\mathrm{r}^{\mathrm{2}} −\mathrm{22r}+\mathrm{25}\:=\:\mathrm{0} \\ $$$$\mathrm{r}\:=\:\frac{\mathrm{22}\:−\:\sqrt{\mathrm{22}^{\mathrm{2}} −\mathrm{100}}}{\mathrm{2}}\:=\:\frac{\mathrm{22}−\mathrm{19}.\mathrm{6}}{\mathrm{2}} \\ $$$$\mathrm{r}\:\approx\:\mathrm{1}.\mathrm{2}\:\mathrm{cm} \\ $$
Commented by $@ty@m123 last updated on 11/Apr/20
Nice....
$${Nice}…. \\ $$
Commented by I want to learn more last updated on 11/Apr/20
Oh yes. I get it now. I apprectiate.
$$\mathrm{Oh}\:\mathrm{yes}.\:\mathrm{I}\:\mathrm{get}\:\mathrm{it}\:\mathrm{now}.\:\mathrm{I}\:\mathrm{apprectiate}. \\ $$
Commented by I want to learn more last updated on 11/Apr/20
Wow great.  Sorry to border you sir, am weak in geometry, can you please help me  join how you derive the equation?.
$$\mathrm{Wow}\:\mathrm{great}. \\ $$$$\mathrm{Sorry}\:\mathrm{to}\:\mathrm{border}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{am}\:\mathrm{weak}\:\mathrm{in}\:\mathrm{geometry},\:\mathrm{can}\:\mathrm{you}\:\mathrm{please}\:\mathrm{help}\:\mathrm{me} \\ $$$$\mathrm{join}\:\mathrm{how}\:\mathrm{you}\:\mathrm{derive}\:\mathrm{the}\:\mathrm{equation}?.\: \\ $$
Commented by $@ty@m123 last updated on 11/Apr/20
Commented by $@ty@m123 last updated on 11/Apr/20
Apply Pythagoras theorem in right  angled triangle ABC.
$${Apply}\:{Pythagoras}\:{theorem}\:{in}\:{right} \\ $$$${angled}\:{triangle}\:{ABC}. \\ $$

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