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Question Number 194067 by Rupesh123 last updated on 27/Jun/23

Answered by som(math1967) last updated on 27/Jun/23

x^2 +10^2 =2(10−x)^2 ⇒x^2 −40x+100=0 x=10(2−(√3)) side of triangle=(√2)×10(1−2+(√3)) =10(√2)((√3)−1) (a/(10(√2)((√3)−1)−a))=((√3)/2) 2a=30(√2)−10(√6)−(√3)a ∴a=((10(3(√2)−(√6)))/(2+(√3))) a^2 =((100(3(√2)−(√6))^2 )/((2+(√3))^2 ))cm^2

$$\:{x}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{10}−{x}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{40}{x}+\mathrm{100}=\mathrm{0} \\ $$$$\:{x}=\mathrm{10}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$$${side}\:{of}\:{triangle}=\sqrt{\mathrm{2}}×\mathrm{10}\left(\mathrm{1}−\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$$$=\mathrm{10}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right) \\ $$$$\:\frac{{a}}{\mathrm{10}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)−{a}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\: \\ $$$$\mathrm{2}{a}=\mathrm{30}\sqrt{\mathrm{2}}−\mathrm{10}\sqrt{\mathrm{6}}−\sqrt{\mathrm{3}}{a} \\ $$$$\:\therefore{a}=\frac{\mathrm{10}\left(\mathrm{3}\sqrt{\mathrm{2}}−\sqrt{\mathrm{6}}\right)}{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\:{a}^{\mathrm{2}} =\frac{\mathrm{100}\left(\mathrm{3}\sqrt{\mathrm{2}}−\sqrt{\mathrm{6}}\right)^{\mathrm{2}} }{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{cm}^{\mathrm{2}} \\ $$

Commented by som(math1967) last updated on 27/Jun/23

Commented by Frix last updated on 27/Jun/23

I′m afraid there′s a mistake calculating a. I get a=10(9−5(√3))(√2)≈4.80 while your a≈1.24 which seems too small given the side of the triangle is ≈10.35

$$\mathrm{I}'\mathrm{m}\:\mathrm{afraid}\:\mathrm{there}'\mathrm{s}\:\mathrm{a}\:\mathrm{mistake}\:\mathrm{calculating}\:{a}. \\ $$$$\mathrm{I}\:\mathrm{get}\:{a}=\mathrm{10}\left(\mathrm{9}−\mathrm{5}\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{2}}\approx\mathrm{4}.\mathrm{80} \\ $$$$\mathrm{while}\:\mathrm{your}\:{a}\approx\mathrm{1}.\mathrm{24} \\ $$$$\mathrm{which}\:\mathrm{seems}\:\mathrm{too}\:\mathrm{small}\:\mathrm{given}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{triangle}\:\mathrm{is}\:\approx\mathrm{10}.\mathrm{35} \\ $$

Commented by som(math1967) last updated on 27/Jun/23

Thank you sir ,I correct it

$${Thank}\:{you}\:{sir}\:,{I}\:{correct}\:{it} \\ $$

Commented by Rupesh123 last updated on 27/Jun/23

Perfect ��

Commented by Rupesh123 last updated on 27/Jun/23

You're right!

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

s=side length of equilateral s^2 =10^2 +(10−(s/( (√2))))^2 −400+20(√2)s+s^2 =0 ⇒s=10((√6)−(√2)) a=side length of small square a+((2a)/( (√3)))=s ⇒a=(((√3)s)/( (√3)+2))=((10(3(√2)−(√6)))/( (√3)+2))≈4.805

$${s}={side}\:{length}\:{of}\:{equilateral} \\ $$$${s}^{\mathrm{2}} =\mathrm{10}^{\mathrm{2}} +\left(\mathrm{10}−\frac{{s}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$$−\mathrm{400}+\mathrm{20}\sqrt{\mathrm{2}}{s}+{s}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{s}=\mathrm{10}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right) \\ $$$${a}={side}\:{length}\:{of}\:{small}\:{square} \\ $$$${a}+\frac{\mathrm{2}{a}}{\:\sqrt{\mathrm{3}}}={s} \\ $$$$\Rightarrow{a}=\frac{\sqrt{\mathrm{3}}{s}}{\:\sqrt{\mathrm{3}}+\mathrm{2}}=\frac{\mathrm{10}\left(\mathrm{3}\sqrt{\mathrm{2}}−\sqrt{\mathrm{6}}\right)}{\:\sqrt{\mathrm{3}}+\mathrm{2}}\approx\mathrm{4}.\mathrm{805} \\ $$

Commented by Rupesh123 last updated on 27/Jun/23

Perfect ��

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