Question-221012

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Question Number 221012 by mr W last updated on 22/May/25

Commented by mr W last updated on 24/May/25

$$\cancel{{yes}}\:{it}'{s}\:{not}\:{given}\:{that}\:{it}\:{is}\:{a}\:{square}. \\ $$

Commented by fantastic last updated on 22/May/25

$${is}\:{that}\:{a}\:{squar} \\ $$

Commented by universe last updated on 22/May/25

$$\mathrm{21}+\mathrm{6}\sqrt{\mathrm{6}\:}\:?? \\ $$

Commented by A5T last updated on 22/May/25

$$\mathrm{Q213550} \\ $$

Answered by universe last updated on 22/May/25

Commented by universe last updated on 22/May/25

AB^2 +OB^2 =OA^2 (((a−3)/2))^2 +(r−6)^2 =r^2 (a−3)^2 +4(r−6)^2 =4r^2 ...(1) now ΔACD (a−3)^2 +(a−6)^2 =4r^2 ...(2) by eq(1) and eq(2) 4(r−6)^2 =(a−6)^2 2r−12 = a−6 ⇒a = 2r−6 by equation 2 (2r−9)^2 +(2r−12)^2 = 4r^2 4r^2 +81−36r+4r^2 +144−48r=4r^2 4r^2 −84r+225=0 r = ((84±(√(84^2 −4.4.225)))/8) r = ((84±(√(7056−3600)))/8) r= ((84±(√(3456)))/8) = ((84±24(√6))/8) 2r = 21±6(√6) 2r = 21+6(√6)

$$ \\ $$$${AB}^{\mathrm{2}} +{OB}^{\mathrm{2}} ={OA}^{\mathrm{2}} \\ $$$$\left(\frac{{a}−\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({r}−\mathrm{6}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left({a}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{4}\left({r}−\mathrm{6}\right)^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} \:\:\:\:\:…\left(\mathrm{1}\right) \\ $$$${now}\:\Delta{ACD} \\ $$$$\left({a}−\mathrm{3}\right)^{\mathrm{2}} +\left({a}−\mathrm{6}\right)^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} \:\:\:\:…\left(\mathrm{2}\right) \\ $$$${by}\:\:{eq}\left(\mathrm{1}\right)\:{and}\:{eq}\left(\mathrm{2}\right) \\ $$$$\:\mathrm{4}\left({r}−\mathrm{6}\right)^{\mathrm{2}} =\left({a}−\mathrm{6}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{r}−\mathrm{12}\:=\:{a}−\mathrm{6}\:\Rightarrow{a}\:=\:\mathrm{2}{r}−\mathrm{6} \\ $$$${by}\:{equation}\:\mathrm{2} \\ $$$$\left(\mathrm{2}{r}−\mathrm{9}\right)^{\mathrm{2}} +\left(\mathrm{2}{r}−\mathrm{12}\right)^{\mathrm{2}} \:=\:\mathrm{4}{r}^{\mathrm{2}} \\ $$$$\mathrm{4}{r}^{\mathrm{2}} +\mathrm{81}−\mathrm{36}{r}+\mathrm{4}{r}^{\mathrm{2}} +\mathrm{144}−\mathrm{48}{r}=\mathrm{4}{r}^{\mathrm{2}} \\ $$$$\mathrm{4}{r}^{\mathrm{2}} −\mathrm{84}{r}+\mathrm{225}=\mathrm{0} \\ $$$$\:{r}\:=\:\frac{\mathrm{84}\pm\sqrt{\mathrm{84}^{\mathrm{2}} −\mathrm{4}.\mathrm{4}.\mathrm{225}}}{\mathrm{8}} \\ $$$$\:{r}\:=\:\frac{\mathrm{84}\pm\sqrt{\mathrm{7056}−\mathrm{3600}}}{\mathrm{8}} \\ $$$${r}=\:\frac{\mathrm{84}\pm\sqrt{\mathrm{3456}}}{\mathrm{8}}\:=\:\frac{\mathrm{84}\pm\mathrm{24}\sqrt{\mathrm{6}}}{\mathrm{8}} \\ $$$$\mathrm{2}{r}\:=\:\mathrm{21}\pm\mathrm{6}\sqrt{\mathrm{6}} \\ $$$$\mathrm{2}{r}\:=\:\mathrm{21}+\mathrm{6}\sqrt{\mathrm{6}} \\ $$

Commented by universe last updated on 22/May/25

sir A5T can u check my solution where i am wrong

$${sir}\:\mathrm{A5T}\:{can}\:{u}\:{check}\:{my}\:{solution}\:{where} \\ $$$${i}\:{am}\:{wrong}\: \\ $$

Commented by A5T last updated on 22/May/25

Commented by A5T last updated on 22/May/25

The solution is logically correct if the figure were a square. But the figure can′t actually be correct for a square, as we get a contradiction: a=2r−3 and a=2r−6. If you check my solution in Q213550, I used two different variables, s and t, for the perpendicular sides. I also assumed the figure to be a square initially until I constructed it and got the figure below, hinting that something went wrong.

$$\mathrm{The}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{logically}\:\mathrm{correct}\:\mathrm{if}\:\mathrm{the}\:\mathrm{figure}\: \\ $$$$\mathrm{were}\:\mathrm{a}\:\mathrm{square}.\:\mathrm{But}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{can}'\mathrm{t}\:\mathrm{actually}\:\mathrm{be} \\ $$$$\mathrm{correct}\:\mathrm{for}\:\mathrm{a}\:\mathrm{square},\:\mathrm{as}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{contradiction}: \\ $$$$\mathrm{a}=\mathrm{2r}−\mathrm{3}\:\mathrm{and}\:\mathrm{a}=\mathrm{2r}−\mathrm{6}. \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{check}\:\mathrm{my}\:\mathrm{solution}\:\mathrm{in}\:\mathrm{Q213550},\:\mathrm{I}\:\mathrm{used}\:\mathrm{two}\: \\ $$$$\mathrm{different}\:\mathrm{variables},\:\mathrm{s}\:\mathrm{and}\:\mathrm{t},\:\mathrm{for}\:\mathrm{the}\:\mathrm{perpendicular} \\ $$$$\mathrm{sides}. \\ $$$$\:\mathrm{I}\:\mathrm{also}\:\mathrm{assumed}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{to}\:\mathrm{be}\:\mathrm{a}\:\mathrm{square} \\ $$$$\mathrm{initially}\:\mathrm{until}\:\mathrm{I}\:\mathrm{constructed}\:\mathrm{it}\:\mathrm{and}\:\mathrm{got}\:\mathrm{the}\:\mathrm{figure} \\ $$$$\mathrm{below},\:\mathrm{hinting}\:\mathrm{that}\:\mathrm{something}\:\mathrm{went}\:\mathrm{wrong}. \\ $$

Commented by universe last updated on 22/May/25

$${got}\:{it}\: \\ $$$${thank}\:{you}\:{so}\:{much}\:{sir}\: \\ $$

Commented by fantastic last updated on 23/May/25

$${what}\:{application}\:{are}\:{you}\:{using}????? \\ $$

Commented by A5T last updated on 24/May/25

$$\mathrm{Geometry}/\mathrm{Geogebra} \\ $$

Answered by mr W last updated on 22/May/25

Commented by mr W last updated on 22/May/25

(R−6)^2 +(R−3)^2 =R^2 R^2 −18R+45=0 (R−15)(R+3)=0 ⇒R=15 ⇒?=2R=30 ✓

$$\left({R}−\mathrm{6}\right)^{\mathrm{2}} +\left({R}−\mathrm{3}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} −\mathrm{18}{R}+\mathrm{45}=\mathrm{0} \\ $$$$\left({R}−\mathrm{15}\right)\left({R}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow{R}=\mathrm{15}\:\Rightarrow?=\mathrm{2}{R}=\mathrm{30}\:\checkmark \\ $$

Commented by mehdee7396 last updated on 23/May/25