Question-203576

Question Number 203576 by sonukgindia last updated on 22/Jan/24

Answered by mr W last updated on 22/Jan/24

Commented by mr W last updated on 22/Jan/24

both hatched triangles are similar. (a/(0.5)) =(1/b) ⇒ab=0.5 area of rectangle =ab=0.5 ✓

$${both}\:{hatched}\:{triangles}\:{are}\:{similar}. \\ $$$$\frac{{a}}{\mathrm{0}.\mathrm{5}}\:=\frac{\mathrm{1}}{{b}} \\ $$$$\Rightarrow{ab}=\mathrm{0}.\mathrm{5} \\ $$$${area}\:{of}\:{rectangle}\:={ab}=\mathrm{0}.\mathrm{5}\:\checkmark \\ $$

Answered by ajfour last updated on 22/Jan/24

Commented by ajfour last updated on 22/Jan/24

B^( 2) =(p^2 +q^2 ){1+(1−q)^2 } tan θ=t=1−q=(p/q)=(1/(2(1−p))) ⇒ p=1−(1/(2t)) , q=1−t , and 1−(1/(2t))=t(1−t) ⇒ t^3 −t^2 +t−(1/2)=0 ⇒ t^2 (t−1)+(t−1)=−(1/2) or (1−t)(t^2 +1)=(1/2) B^( 2) =q^2 (t^2 +1){1+t^2 ) B =(1−t)(1+t^2 ) B =(1/2)

$${B}^{\:\mathrm{2}} =\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)\left\{\mathrm{1}+\left(\mathrm{1}−{q}\right)^{\mathrm{2}} \right\} \\ $$$$\mathrm{tan}\:\theta={t}=\mathrm{1}−{q}=\frac{{p}}{{q}}=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−{p}\right)} \\ $$$$\Rightarrow\:{p}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{t}}\:\:\:\:,\:\:{q}=\mathrm{1}−{t}\:\:,\:{and} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{t}}={t}\left(\mathrm{1}−{t}\right) \\ $$$$\Rightarrow\:\:{t}^{\mathrm{3}} −{t}^{\mathrm{2}} +{t}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\:\:{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)+\left({t}−\mathrm{1}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${or}\:\:\:\left(\mathrm{1}−{t}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${B}^{\:\mathrm{2}} ={q}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}\right)\left\{\mathrm{1}+{t}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:{B}\:=\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\: \\ $$$$\:\:\:\:\:{B}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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