Question-193975

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Question Number 193975 by mr W last updated on 24/Jun/23

Commented by mr W last updated on 25/Jun/23

$${find}\:{the}\:{area}\:{of}\:{the}\:{hatched}\:{square}. \\ $$

Answered by mr W last updated on 25/Jun/23

Commented by mr W last updated on 25/Jun/23

cos α=((7^2 +9^2 −((√2)s)^2 )/(2×7×9)) cos (β+γ)=((2^2 +6^2 −((√2)s)^2 )/(2×2×6))=−cos α ⇒((2^2 +6^2 −((√2)s)^2 )/(2×2×6))=−((7^2 +9^2 −((√2)s)^2 )/(2×7×9)) ⇒s^2 =((136)/5)=area of square

$$\mathrm{cos}\:\alpha=\frac{\mathrm{7}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}{s}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{7}×\mathrm{9}} \\ $$$$\mathrm{cos}\:\left(\beta+\gamma\right)=\frac{\mathrm{2}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}{s}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}×\mathrm{6}}=−\mathrm{cos}\:\alpha \\ $$$$\Rightarrow\frac{\mathrm{2}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}{s}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}×\mathrm{6}}=−\frac{\mathrm{7}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}{s}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{7}×\mathrm{9}} \\ $$$$\Rightarrow{s}^{\mathrm{2}} =\frac{\mathrm{136}}{\mathrm{5}}={area}\:{of}\:{square} \\ $$

Answered by Subhi last updated on 25/Jun/23

put ef^� b = α , fb^� e = β , each side of the square = l fd = (√(l^2 +l^2 ))=(√2)l Δfbd: BD^2 = 2l^2 +6^2 −12(√2)lcos(45+α) 2+13^2 +36Q^2 =11(18+36+2l^2 −12(√2)l.cos(45+α) (stewart′s theorem) 36Q^2 −22l^2 +132(√2)l.cos(45+α)=256 →(i) ΔAEB: h^2 = Q^2 +13^2 −26Qcos(β) 11^2 +13^2 =2Q^2 +2h^2 (Apollonius theorem) 11^2 +13^2 =4Q^2 +2×13^2 −52Qcos(β) (1) Δefb: cos(β) = ((Q^2 +6^2 −l^2 )/(12Q)) (2) compute 1,2 108 = ((13)/3)l^2 −(1/3)Q^2 (ii) compute (i),(ii) 468l^2 −11664−22l^2 +132(√2)l.cos(45+α) = 256 446l^2 +132(√2)l.cos(45+α) = 11920 → (iii) ΔADF: cos(135−α) = ((7^2 +2l^2 −9^2 )/(14(√2) l))=−cos(45+α) (3) compute (iii), (3) 446l^2 −132(√2).(((2l^2 −32)/(14(√2))))=11920 l^2 =((11920−((132(√2).32)/(14(√2))))/(446−((264)/(14))))=((136)/5)=area of the square

$${put}\:{e}\hat {{f}b}\:=\:\alpha\:,\:\:{f}\hat {{b}e}\:=\:\beta\:\:,\:{each}\:{side}\:{of}\:{the}\:{square}\:=\:{l} \\ $$$${fd}\:=\:\sqrt{{l}^{\mathrm{2}} +{l}^{\mathrm{2}} }=\sqrt{\mathrm{2}}{l} \\ $$$$\Delta{fbd}:\:{BD}^{\mathrm{2}} \:=\:\mathrm{2}{l}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} −\mathrm{12}\sqrt{\mathrm{2}}{lcos}\left(\mathrm{45}+\alpha\right) \\ $$$$\mathrm{2}+\mathrm{13}^{\mathrm{2}} +\mathrm{36}{Q}^{\mathrm{2}} =\mathrm{11}\left(\mathrm{18}+\mathrm{36}+\mathrm{2}{l}^{\mathrm{2}} −\mathrm{12}\sqrt{\mathrm{2}}{l}.{cos}\left(\mathrm{45}+\alpha\right)\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({stewart}'{s}\:{theorem}\right) \\ $$$$\mathrm{36}{Q}^{\mathrm{2}} −\mathrm{22}{l}^{\mathrm{2}} +\mathrm{132}\sqrt{\mathrm{2}}{l}.{cos}\left(\mathrm{45}+\alpha\right)=\mathrm{256}\:\rightarrow\left({i}\right) \\ $$$$\Delta{AEB}:\:{h}^{\mathrm{2}} \:=\:{Q}^{\mathrm{2}} +\mathrm{13}^{\mathrm{2}} −\mathrm{26}{Qcos}\left(\beta\right) \\ $$$$\mathrm{11}^{\mathrm{2}} +\mathrm{13}^{\mathrm{2}} =\mathrm{2}{Q}^{\mathrm{2}} +\mathrm{2}{h}^{\mathrm{2}} \:\:\:\:\:\:\left({Apollonius}\:{theorem}\right) \\ $$$$\mathrm{11}^{\mathrm{2}} +\mathrm{13}^{\mathrm{2}} =\mathrm{4}{Q}^{\mathrm{2}} +\mathrm{2}×\mathrm{13}^{\mathrm{2}} −\mathrm{52}{Qcos}\left(\beta\right)\:\left(\mathrm{1}\right) \\ $$$$\Delta{efb}:\:{cos}\left(\beta\right)\:=\:\frac{{Q}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} −{l}^{\mathrm{2}} }{\mathrm{12}{Q}}\:\:\left(\mathrm{2}\right) \\ $$$${compute}\:\mathrm{1},\mathrm{2} \\ $$$$\mathrm{108}\:=\:\frac{\mathrm{13}}{\mathrm{3}}{l}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}{Q}^{\mathrm{2}} \:\:\left({ii}\right) \\ $$$${compute}\:\left({i}\right),\left({ii}\right) \\ $$$$\mathrm{468}{l}^{\mathrm{2}} −\mathrm{11664}−\mathrm{22}{l}^{\mathrm{2}} +\mathrm{132}\sqrt{\mathrm{2}}{l}.{cos}\left(\mathrm{45}+\alpha\right)\:=\:\mathrm{256} \\ $$$$\mathrm{446}{l}^{\mathrm{2}} +\mathrm{132}\sqrt{\mathrm{2}}{l}.{cos}\left(\mathrm{45}+\alpha\right)\:=\:\mathrm{11920}\:\rightarrow\:\left({iii}\right) \\ $$$$\Delta{ADF}:\:{cos}\left(\mathrm{135}−\alpha\right)\:=\:\frac{\mathrm{7}^{\mathrm{2}} +\mathrm{2}{l}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} }{\mathrm{14}\sqrt{\mathrm{2}}\:{l}}=−{cos}\left(\mathrm{45}+\alpha\right)\:\left(\mathrm{3}\right) \\ $$$${compute}\:\left({iii}\right),\:\left(\mathrm{3}\right) \\ $$$$\mathrm{446}{l}^{\mathrm{2}} −\mathrm{132}\sqrt{\mathrm{2}}.\left(\frac{\mathrm{2}{l}^{\mathrm{2}} −\mathrm{32}}{\mathrm{14}\sqrt{\mathrm{2}}}\right)=\mathrm{11920} \\ $$$${l}^{\mathrm{2}} =\frac{\mathrm{11920}−\frac{\mathrm{132}\sqrt{\mathrm{2}}.\mathrm{32}}{\mathrm{14}\sqrt{\mathrm{2}}}}{\mathrm{446}−\frac{\mathrm{264}}{\mathrm{14}}}=\frac{\mathrm{136}}{\mathrm{5}}={area}\:{of}\:{the}\:{square} \\ $$

Commented by Subhi last updated on 25/Jun/23