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Question Number 186111 by Rupesh123 last updated on 01/Feb/23

Commented by Rupesh123 last updated on 01/Feb/23

A right angle is rotated and translated as shown in the diagram

Commented by mr W last updated on 01/Feb/23

it′s not a fixed value, it depends on the shape of the triangle.

$${it}'{s}\:{not}\:{a}\:{fixed}\:{value},\:{it}\:{depends}\:{on}\:{the} \\ $$$${shape}\:{of}\:{the}\:{triangle}. \\ $$

Answered by HeferH last updated on 01/Feb/23

Commented by HeferH last updated on 01/Feb/23

I consider the entire diagram as “entire shape”:)

$${I}\:{consider}\:{the}\:{entire}\:{diagram}\:{as} \\ $$$$\left.``{entire}\:{shape}'':\right) \\ $$

Commented by HeferH last updated on 01/Feb/23

Area shaded: 2S Area of the entire shape: 6S ((2S)/(6S)) = (1/3)

$${Area}\:{shaded}:\:\mathrm{2}{S} \\ $$$${Area}\:{of}\:{the}\:{entire}\:{shape}:\:\mathrm{6}{S} \\ $$$$\frac{\mathrm{2}{S}}{\mathrm{6}{S}}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by mr W last updated on 01/Feb/23

((2S)/(4S))=(1/2)

$$\frac{\mathrm{2}{S}}{\mathrm{4}{S}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by mr W last updated on 01/Feb/23

Commented by mr W last updated on 01/Feb/23

θ=the smaller angle total area of triangle A=((c^2 sin θ cos θ)/2) case 0<θ≤30° (x/(c(1−sin θ)))=((sin ((π/2)−θ))/(sin ((π/2)−2θ)))=((cos θ)/(cos 2θ)) A_S =((c^2 (1−sin θ)^2 sin θ cos θ)/(2 cos 2θ)) A_1 =A−A_S (A_1 /A)=1−(A_S /A)=1−(((1−sin θ)^2 )/( cos 2θ))≤(1/2) case 30°≤θ≤45° A_S =(((c cos θ)^2 tan θ)/4)=((c^2 sin θ cos θ)/4)=(A/2) A_1 =(A/2) ⇒(A_1 /A)=(1/2)

$$\theta={the}\:{smaller}\:{angle} \\ $$$${total}\:{area}\:{of}\:{triangle} \\ $$$${A}=\frac{{c}^{\mathrm{2}} \:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\mathrm{2}} \\ $$$${case}\:\mathrm{0}<\theta\leqslant\mathrm{30}° \\ $$$$\frac{{x}}{{c}\left(\mathrm{1}−\mathrm{sin}\:\theta\right)}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right)}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}\theta\right)}=\frac{\mathrm{cos}\:\theta}{\mathrm{cos}\:\mathrm{2}\theta} \\ $$$${A}_{{S}} =\frac{{c}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:\theta\right)^{\mathrm{2}} \:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta} \\ $$$${A}_{\mathrm{1}} ={A}−{A}_{{S}} \\ $$$$\frac{{A}_{\mathrm{1}} }{{A}}=\mathrm{1}−\frac{{A}_{{S}} }{{A}}=\mathrm{1}−\frac{\left(\mathrm{1}−\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{\:\mathrm{cos}\:\mathrm{2}\theta}\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${case}\:\mathrm{30}°\leqslant\theta\leqslant\mathrm{45}° \\ $$$${A}_{{S}} =\frac{\left({c}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} \:\mathrm{tan}\:\theta}{\mathrm{4}}=\frac{{c}^{\mathrm{2}} \:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\mathrm{4}}=\frac{{A}}{\mathrm{2}} \\ $$$${A}_{\mathrm{1}} =\frac{{A}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{A}_{\mathrm{1}} }{{A}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by mr W last updated on 01/Feb/23

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