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Question-180927

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Question Number 180927 by mr W last updated on 19/Nov/22
Commented by mr W last updated on 19/Nov/22
find the area of the square.
$${find}\:{the}\:{area}\:{of}\:{the}\:{square}. \\ $$
Answered by mr W last updated on 20/Nov/22
Commented by mr W last updated on 20/Nov/22
Method I s=side length of square. its area is s^2 . cos θ=((7^2 +9^2 −((√2)s)^2 )/(2×7×9)) cos θ=((13^2 +11^2 −(2a)^2 )/(2×13×11)) ((13^2 +11^2 −(2a)^2 )/(2×13×11))=((7^2 +9^2 −((√2)s)^2 )/(2×7×9)) ((145)/(143))−((2a^2 )/(143))=((65)/(63))−(s^2 /(63)) ⇒a^2 =((143s^2 −160)/(126)) ...(I) cos α=((a^2 +s^2 −6^2 )/(2as)) cos β=((a^2 +s^2 −2^2 )/(2as)) α+β=90° ⇒cos β=sin α (((a^2 +s^2 −6^2 )/(2as)))^2 +(((a^2 +s^2 −2^2 )/(2as)))^2 =1 (a^2 +s^2 −36)^2 +(a^2 +s^2 −4)^2 =4a^2 s^2 2(a^2 +s^2 )^2 −2(36+4)(a^2 +s^2 )+36^2 +4^2 =4a^2 s^2 ⇒s^4 −40s^2 +(a^2 −20)^2 +256=0 ...(II) put (I) into (II): s^4 −40s^2 +(((143s^2 −160)/(126))−20)^2 +256=0 36 325s^4 −1 401 520s^2 +11 246 656=0 ⇒s^2 =((700 760±(√(700 760^2 −36 325×11 246 656)))/(36 325)) =((700 760±287 280)/(36 325)) =((136)/5) or ((82 696)/(7 265)) (rejected) i.e. the area of the square is ((136)/5).
$$\boldsymbol{{Method}}\:\boldsymbol{{I}} \\ $$$${s}={side}\:{length}\:{of}\:{square}.\:{its}\:{area}\:{is}\:{s}^{\mathrm{2}} . \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{7}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}{s}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{7}×\mathrm{9}} \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{13}^{\mathrm{2}} +\mathrm{11}^{\mathrm{2}} −\left(\mathrm{2}{a}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{13}×\mathrm{11}} \\ $$$$\frac{\mathrm{13}^{\mathrm{2}} +\mathrm{11}^{\mathrm{2}} −\left(\mathrm{2}{a}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{13}×\mathrm{11}}=\frac{\mathrm{7}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}{s}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{7}×\mathrm{9}} \\ $$$$\frac{\mathrm{145}}{\mathrm{143}}−\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{143}}=\frac{\mathrm{65}}{\mathrm{63}}−\frac{{s}^{\mathrm{2}} }{\mathrm{63}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{143}{s}^{\mathrm{2}} −\mathrm{160}}{\mathrm{126}}\:\:\:…\left({I}\right) \\ $$$$\mathrm{cos}\:\alpha=\frac{{a}^{\mathrm{2}} +{s}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} }{\mathrm{2}{as}} \\ $$$$\mathrm{cos}\:\beta=\frac{{a}^{\mathrm{2}} +{s}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }{\mathrm{2}{as}} \\ $$$$\alpha+\beta=\mathrm{90}°\:\Rightarrow\mathrm{cos}\:\beta=\mathrm{sin}\:\alpha \\ $$$$\left(\frac{{a}^{\mathrm{2}} +{s}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} }{\mathrm{2}{as}}\right)^{\mathrm{2}} +\left(\frac{{a}^{\mathrm{2}} +{s}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }{\mathrm{2}{as}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\left({a}^{\mathrm{2}} +{s}^{\mathrm{2}} −\mathrm{36}\right)^{\mathrm{2}} +\left({a}^{\mathrm{2}} +{s}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} {s}^{\mathrm{2}} \\ $$$$\mathrm{2}\left({a}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{36}+\mathrm{4}\right)\left({a}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)+\mathrm{36}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} {s}^{\mathrm{2}} \\ $$$$\Rightarrow{s}^{\mathrm{4}} −\mathrm{40}{s}^{\mathrm{2}} +\left({a}^{\mathrm{2}} −\mathrm{20}\right)^{\mathrm{2}} +\mathrm{256}=\mathrm{0}\:\:\:…\left({II}\right) \\ $$$${put}\:\left({I}\right)\:{into}\:\left({II}\right): \\ $$$${s}^{\mathrm{4}} −\mathrm{40}{s}^{\mathrm{2}} +\left(\frac{\mathrm{143}{s}^{\mathrm{2}} −\mathrm{160}}{\mathrm{126}}−\mathrm{20}\right)^{\mathrm{2}} +\mathrm{256}=\mathrm{0} \\ $$$$\mathrm{36}\:\mathrm{325}{s}^{\mathrm{4}} −\mathrm{1}\:\mathrm{401}\:\mathrm{520}{s}^{\mathrm{2}} +\mathrm{11}\:\mathrm{246}\:\mathrm{656}=\mathrm{0} \\ $$$$\Rightarrow{s}^{\mathrm{2}} =\frac{\mathrm{700}\:\mathrm{760}\pm\sqrt{\mathrm{700}\:\mathrm{760}^{\mathrm{2}} −\mathrm{36}\:\mathrm{325}×\mathrm{11}\:\mathrm{246}\:\mathrm{656}}}{\mathrm{36}\:\mathrm{325}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{700}\:\mathrm{760}\pm\mathrm{287}\:\mathrm{280}}{\mathrm{36}\:\mathrm{325}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{136}}{\mathrm{5}}\:{or}\:\frac{\mathrm{82}\:\mathrm{696}}{\mathrm{7}\:\mathrm{265}}\:\left({rejected}\right) \\ $$$${i}.{e}.\:{the}\:{area}\:{of}\:{the}\:{square}\:{is}\:\frac{\mathrm{136}}{\mathrm{5}}. \\ $$
Answered by mr W last updated on 20/Nov/22
Commented by mr W last updated on 20/Nov/22
Method II ((√2)s)^2 =7^2 +9^2 −2×7×9 cos θ ((√2)s)^2 =2^2 +6^2 −2×2×6 cos (α+β) α+β=180°−θ ⇒cos (α+β)=−cos θ 2s^2 =2^2 +6^2 +2×2×6 cos θ=7^2 +9^2 −2×7×9 cos θ ⇒cos θ=(3/5) ⇒s^2 =((130)/2)−63×(3/5)=((136)/5) i.e. area of square is ((136)/5).
$$\boldsymbol{{Method}}\:\boldsymbol{{II}} \\ $$$$\left(\sqrt{\mathrm{2}}{s}\right)^{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} −\mathrm{2}×\mathrm{7}×\mathrm{9}\:\mathrm{cos}\:\theta \\ $$$$\left(\sqrt{\mathrm{2}}{s}\right)^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} −\mathrm{2}×\mathrm{2}×\mathrm{6}\:\mathrm{cos}\:\left(\alpha+\beta\right) \\ $$$$\alpha+\beta=\mathrm{180}°−\theta\:\Rightarrow\mathrm{cos}\:\left(\alpha+\beta\right)=−\mathrm{cos}\:\theta \\ $$$$\mathrm{2}{s}^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} +\mathrm{2}×\mathrm{2}×\mathrm{6}\:\mathrm{cos}\:\theta=\mathrm{7}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} −\mathrm{2}×\mathrm{7}×\mathrm{9}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\Rightarrow{s}^{\mathrm{2}} =\frac{\mathrm{130}}{\mathrm{2}}−\mathrm{63}×\frac{\mathrm{3}}{\mathrm{5}}=\frac{\mathrm{136}}{\mathrm{5}} \\ $$$${i}.{e}.\:{area}\:{of}\:{square}\:{is}\:\frac{\mathrm{136}}{\mathrm{5}}. \\ $$

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