Question-185774

Question Number 185774 by cherokeesay last updated on 27/Jan/23

Answered by HeferH last updated on 27/Jan/23

Commented by HeferH last updated on 27/Jan/23

R = 37 s = ((37+37+24)/2) = 49 Area(MGI) = (√(49(49−37)^2 (49 − 24))) = 420 GC = (√(37^2 −((23^2 ∙2)/4))) − ((23(√2))/2) ≈ 17 2Area(GCH) = 2(((17∙((23(√2))/2))/2))= ((391(√2))/2) 420 + ((391(√2))/2) ≈ 696 u^2

$${R}\:=\:\mathrm{37}\: \\ $$$$\:{s}\:=\:\frac{\mathrm{37}+\mathrm{37}+\mathrm{24}}{\mathrm{2}}\:=\:\mathrm{49} \\ $$$${Area}\left({MGI}\right)\:=\:\sqrt{\mathrm{49}\left(\mathrm{49}−\mathrm{37}\right)^{\mathrm{2}} \left(\mathrm{49}\:−\:\mathrm{24}\right)}\:=\:\mathrm{420} \\ $$$${GC}\:=\:\sqrt{\mathrm{37}^{\mathrm{2}} −\frac{\mathrm{23}^{\mathrm{2}} \centerdot\mathrm{2}}{\mathrm{4}}}\:\:−\:\frac{\mathrm{23}\sqrt{\mathrm{2}}}{\mathrm{2}}\:\approx\:\mathrm{17}\: \\ $$$$\:\mathrm{2}{Area}\left({GCH}\right)\:=\:\mathrm{2}\left(\frac{\mathrm{17}\centerdot\frac{\mathrm{23}\sqrt{\mathrm{2}}}{\mathrm{2}}}{\mathrm{2}}\right)=\:\frac{\mathrm{391}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\:\mathrm{420}\:+\:\frac{\mathrm{391}\sqrt{\mathrm{2}}}{\mathrm{2}}\:\approx\:\mathrm{696}\:{u}^{\mathrm{2}} \\ $$

Commented by HeferH last updated on 27/Jan/23

Commented by mr W last updated on 27/Jan/23

$${it}\:{is}\:{to}\:{prove}\:{that}\:{HC}={IC}. \\ $$

Commented by HeferH last updated on 27/Jan/23

((CA)/(24)) = ((23)/(70)) ⇒ CA = ((23∙12)/(35)) = ((276)/(35)) ((QQ′)/(24)) = ((47)/(70)) ⇒ QQ′ = ((47∙12)/(35))=((564)/(35)) (((276+564)/(35)))∙(1/2)∙24∙(1/2)+ 24∙23 = 696 u^2

$$\:\frac{{CA}}{\mathrm{24}}\:=\:\frac{\mathrm{23}}{\mathrm{70}}\:\Rightarrow\:{CA}\:=\:\frac{\mathrm{23}\centerdot\mathrm{12}}{\mathrm{35}}\:=\:\frac{\mathrm{276}}{\mathrm{35}} \\ $$$$\:\frac{{QQ}'}{\mathrm{24}}\:=\:\frac{\mathrm{47}}{\mathrm{70}}\:\Rightarrow\:{QQ}'\:=\:\frac{\mathrm{47}\centerdot\mathrm{12}}{\mathrm{35}}=\frac{\mathrm{564}}{\mathrm{35}} \\ $$$$\:\:\left(\frac{\mathrm{276}+\mathrm{564}}{\mathrm{35}}\right)\centerdot\frac{\mathrm{1}}{\mathrm{2}}\centerdot\mathrm{24}\centerdot\frac{\mathrm{1}}{\mathrm{2}}+\:\mathrm{24}\centerdot\mathrm{23}\:\:=\:\mathrm{696}\:{u}^{\mathrm{2}} \\ $$

Commented by HeferH last updated on 27/Jan/23

$${yes} \\ $$

Answered by mr W last updated on 27/Jan/23

Commented by mr W last updated on 27/Jan/23

a=24 b=23 c=? sin α=(b/( (√2)R))=((√(b^2 +c^2 ))/(2R)) (√2)b=(√(b^2 +c^2 )) ⇒b^2 =c^2 ⇒c=b ✓ ((√2)R)^2 =(a+c)^2 +b^2 =a^2 +2b(a+b) ⇒R=(√((a^2 +2b(a+b))/2)) =(√((24^2 +2×23×(24+23))/2))=37 green=(a/2)(√(R^2 −(a^2 /4)))+(((√2)b)/2)(√(R^2 −(b^2 /2)))−(b^2 /2) =12×58=696

$${a}=\mathrm{24} \\ $$$${b}=\mathrm{23} \\ $$$${c}=? \\ $$$$\mathrm{sin}\:\alpha=\frac{{b}}{\:\sqrt{\mathrm{2}}{R}}=\frac{\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }}{\mathrm{2}{R}} \\ $$$$\sqrt{\mathrm{2}}{b}=\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$\Rightarrow{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \:\Rightarrow{c}={b}\:\checkmark \\ $$$$\left(\sqrt{\mathrm{2}}{R}\right)^{\mathrm{2}} =\left({a}+{c}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{2}{b}\left({a}+{b}\right) \\ $$$$\Rightarrow{R}=\sqrt{\frac{{a}^{\mathrm{2}} +\mathrm{2}{b}\left({a}+{b}\right)}{\mathrm{2}}} \\ $$$$\:\:\:\:=\sqrt{\frac{\mathrm{24}^{\mathrm{2}} +\mathrm{2}×\mathrm{23}×\left(\mathrm{24}+\mathrm{23}\right)}{\mathrm{2}}}=\mathrm{37} \\ $$$${green}=\frac{{a}}{\mathrm{2}}\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}+\frac{\sqrt{\mathrm{2}}{b}}{\mathrm{2}}\sqrt{{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{2}}}−\frac{{b}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:=\mathrm{12}×\mathrm{58}=\mathrm{696} \\ $$

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